Lin Lin

Oct 10th, 2012, 12:58 AM

Calculate carefully;):wavey::wavey::wavey:

View Full Version : 6÷2（2+1）=？

Lin Lin

Oct 10th, 2012, 12:58 AM

Calculate carefully;):wavey::wavey::wavey:

RenaSlam.

Oct 10th, 2012, 01:16 AM

1.

kiwifan

Oct 10th, 2012, 01:24 AM

I wasn't a great Math guy but I remember "orders of operation" from way back when...9

Nicolás89

Oct 10th, 2012, 01:26 AM

Hard to say but the answer is 9.

New_balls_please

Oct 10th, 2012, 01:27 AM

What the hell is this about? :help: :spit:

6÷2（2+1)

6÷2 x（2+1)

3 x 3

9

:unsure:

6÷2（2+1)

6÷2 x（2+1)

3 x 3

9

:unsure:

Nicolás89

Oct 10th, 2012, 01:31 AM

What the hell is this about? :help: :spit:

It is a math problem that appeared on facebook some time ago, some people argue to death and defriend each other arguing which is the correct answer for it, 9 or 1. Scandalous.

It is a math problem that appeared on facebook some time ago, some people argue to death and defriend each other arguing which is the correct answer for it, 9 or 1. Scandalous.

New_balls_please

Oct 10th, 2012, 01:35 AM

It is a math problem that appeared on facebook some time ago, some people argue to death and defriend each other arguing which is the correct answer for it, 9 or 1. Scandalous.

Oh really? I've never heard of it :lol:

*subscribes*

http://www.alien109.com/junk/images/buffy_popcorn.gif

Oh really? I've never heard of it :lol:

*subscribes*

http://www.alien109.com/junk/images/buffy_popcorn.gif

Dani12

Oct 10th, 2012, 01:38 AM

I got 9 as well, but I'm not very good a maths :lol:

slamchamp

Oct 10th, 2012, 01:44 AM

9:shrug:

Dav.

Oct 10th, 2012, 01:50 AM

It's 1 following the parentheses rule, no?

Pump-it-UP

Oct 10th, 2012, 01:52 AM

1 :shrug:

Inger67

Oct 10th, 2012, 01:56 AM

You do the parenthesis first. Multiplication AND Division, you do whatever one comes first in the equation. So in this case, you divide first and then multiply.

(2+1) = (3) and then 6 ÷ 2 = 3, and 3(3) = 9

:cheer:

(2+1) = (3) and then 6 ÷ 2 = 3, and 3(3) = 9

:cheer:

Rui.

Oct 10th, 2012, 01:57 AM

Technically can't that be written as 6÷2（2+1）= 6 ÷(2x2 + 2x1) = 1?

Then again I'm not sure. :lol:

Then again I'm not sure. :lol:

Inger67

Oct 10th, 2012, 02:00 AM

It's 1 following the parentheses rule, no?

Chris :sobbing:

Chris :sobbing:

Inger67

Oct 10th, 2012, 02:02 AM

Technically can't that be written as 6÷2（2+1）= 6 ÷(2x2 + 2x1) = 1?

Then again I'm not sure. :lol:

No, because you do whatever is in the parenthesis first. So you Add 2 and 1 to get 3.

Then it's 6÷2（3）

Then again I'm not sure. :lol:

No, because you do whatever is in the parenthesis first. So you Add 2 and 1 to get 3.

Then it's 6÷2（3）

Rui.

Oct 10th, 2012, 02:04 AM

No, because you do whatever is in the parenthesis first. So you Add 2 and 1 to get 3.

Then it's 6÷2（3）

so,what would you do if you have this 6 (Y+1) = ?

Then it's 6÷2（3）

so,what would you do if you have this 6 (Y+1) = ?

Inger67

Oct 10th, 2012, 02:10 AM

so,what would you do if you have this 6 (Y+1) = ?

That's different. You skipped out of the division aspect of the problem, that's what makes this unique.

6 / 2 (1 + 2) is really

6 / 2 x (1 + 2)

If the problem you said was 6/2 (Y + 1) = ...

You would get 3Y + 3 = ...

That's different. You skipped out of the division aspect of the problem, that's what makes this unique.

6 / 2 (1 + 2) is really

6 / 2 x (1 + 2)

If the problem you said was 6/2 (Y + 1) = ...

You would get 3Y + 3 = ...

New_balls_please

Oct 10th, 2012, 02:11 AM

I don't want to sound like a smartass, but if you can't solve this.... there's probably something REALLY wrong with you :unsure: :tape:

The 2nd Law

Oct 10th, 2012, 02:11 AM

so,what would you do if you have this 6 (Y+1) = ?

= 6Y + 6 :shrug:

= 6Y + 6 :shrug:

The 2nd Law

Oct 10th, 2012, 02:12 AM

That's different. You skipped out of the division aspect of the problem, that's what makes this unique.

6 / 2 (1 + 2) is really

6 / 2 x (1 + 2)

If the problem you said was 6/2 (Y + 1) = ...

You would get 3Y + 3 = ...

:yeah:

6 / 2 (1 + 2) is really

6 / 2 x (1 + 2)

If the problem you said was 6/2 (Y + 1) = ...

You would get 3Y + 3 = ...

:yeah:

Inger67

Oct 10th, 2012, 02:13 AM

I can't believe 1 is winning this right now :eek:

Rui.

Oct 10th, 2012, 02:19 AM

That's different. You skipped out of the division aspect of the problem, that's what makes this unique.

6 / 2 (1 + 2) is really

6 / 2 x (1 + 2)

If the problem you said was 6/2 (Y + 1) = ...

You would get 3Y + 3 = ...

I guess you're correct :lol: Then again, I'm still not sure. :lol: I just meant to say that you could play with what's inside the parentheses and then division and multiplication should be interchangeable but i'm not sure anymore. :lol: :o

6 / 2 (1 + 2) is really

6 / 2 x (1 + 2)

If the problem you said was 6/2 (Y + 1) = ...

You would get 3Y + 3 = ...

I guess you're correct :lol: Then again, I'm still not sure. :lol: I just meant to say that you could play with what's inside the parentheses and then division and multiplication should be interchangeable but i'm not sure anymore. :lol: :o

delicatecutter

Oct 10th, 2012, 02:34 AM

I don't want to sound like a smartass, but if you can't solve this.... there's probably something REALLY wrong with you :unsure: :tape:

I know, right? Even I solved it. :hysteric:

I know, right? Even I solved it. :hysteric:

Dav.

Oct 10th, 2012, 02:41 AM

That's different. You skipped out of the division aspect of the problem, that's what makes this unique.

6 / 2 (1 + 2) is really

6 / 2 x (1 + 2)

If the problem you said was 6/2 (Y + 1) = ...

You would get 3Y + 3 = ...

You have to distribute the 3 into the parenthesis first before you divide, no?

6 / 2 (1 + 2) is really

6 / 2 x (1 + 2)

If the problem you said was 6/2 (Y + 1) = ...

You would get 3Y + 3 = ...

You have to distribute the 3 into the parenthesis first before you divide, no?

In The Zone

Oct 10th, 2012, 02:51 AM

Right. Just because it would be 2(3) doesn't mean make it a parentheses rule.

Inger67

Oct 10th, 2012, 02:51 AM

You have to distribute the 3 into the parenthesis first before you divide, no?

If the problem was like this what would you do?

6/2 x (3)

By using Please Excuse My Dear Aunt Sally, which states Multiplication AND Division, you do whatever comes first from left to right. You divide first and then multiply.

If the problem was like this what would you do?

6/2 x (3)

By using Please Excuse My Dear Aunt Sally, which states Multiplication AND Division, you do whatever comes first from left to right. You divide first and then multiply.

Inger67

Oct 10th, 2012, 02:52 AM

Right. Just because it would be 2(3) doesn't mean make it a parentheses rule.

Exactly, you distribute INSIDE the parenthesis not if it's outside like that. That's multiplication.

Exactly, you distribute INSIDE the parenthesis not if it's outside like that. That's multiplication.

slamchamp

Oct 10th, 2012, 02:54 AM

this is like doing 6*(1/2)*(2+1) It's 3 multiplications..

New_balls_please

Oct 10th, 2012, 02:54 AM

If the problem was like this what would you do?

6/2 x (3)

By using Please Excuse My Dear Aunt Sally, which states Multiplication AND Division, you do whatever comes first from left to right. You divide first and then multiply.

Exactly.

6/2 x (3)

By using Please Excuse My Dear Aunt Sally, which states Multiplication AND Division, you do whatever comes first from left to right. You divide first and then multiply.

Exactly.

New_balls_please

Oct 10th, 2012, 02:55 AM

I know, right? Even I solved it. :hysteric:

Wally :rolls: :rolls:

:kiss:

Wally :rolls: :rolls:

:kiss:

égalité

Oct 10th, 2012, 03:04 AM

It's 9.

If you want it to be 1 then write 6÷(2 (2+1)).

If you want it to be 1 then write 6÷(2 (2+1)).

égalité

Oct 10th, 2012, 03:07 AM

You have to distribute the 3 into the parenthesis first before you divide, no?

nonono :tears:

This problem is the same as 6 x (1/2) x (2+1).

nonono :tears:

This problem is the same as 6 x (1/2) x (2+1).

Dav.

Oct 10th, 2012, 03:16 AM

I'm a psychology student. Math isn't my thing. :sobbing:

delicatecutter

Oct 10th, 2012, 03:16 AM

It's 9.

If you want it to be 1 then write 6÷(2 (2+1)).

Mathpert. :hearts:

If you want it to be 1 then write 6÷(2 (2+1)).

Mathpert. :hearts:

Dav.

Oct 10th, 2012, 03:16 AM

nonono :tears:

This problem is the same as 6 x (1/2) x (2+1).

I meant the two, but still :hysteric:

This problem is the same as 6 x (1/2) x (2+1).

I meant the two, but still :hysteric:

Inger67

Oct 10th, 2012, 03:23 AM

We love you, Chris :cheer:

égalité

Oct 10th, 2012, 03:31 AM

I meant the two, but still :hysteric:

PEMDAS is a lie. Multiplication and division happen at the same time. Addition and subtraction happen at the same time. :oh:

PEMDAS is a lie. Multiplication and division happen at the same time. Addition and subtraction happen at the same time. :oh:

Yoncé

Oct 10th, 2012, 03:36 AM

It's 9. How can 1 have so many votes :confused:

Hurley

Oct 10th, 2012, 04:10 AM

P, E, M/D, A/S

The answer is fucking 9.

The answer is fucking 9.

young_gunner913

Oct 10th, 2012, 04:47 AM

I thought it was 9 then felt like a dipshit when I saw all the 1's on the first page. :tears: Glad to see that TF's Mathematical Legend, Egalite is here to grace us with the correct answer and a math lesson. :worship:

Pump-it-UP

Oct 10th, 2012, 05:03 AM

Setting the equation equal to 1 and solving for k:

6/k(2+1) = 1

k(2+1) = 6*1

k(3) = 6

k = 6/(3)

k = 2

Setting the equation equal to 9 and solving for k:

6/k(2+1) = 9

k(2+1) = 6*9

k(3) = 54

k = 54/(3)

k = 18

k = 2 in the OP, ∴ the solution is 1. :shrug::angel:

6/k(2+1) = 1

k(2+1) = 6*1

k(3) = 6

k = 6/(3)

k = 2

Setting the equation equal to 9 and solving for k:

6/k(2+1) = 9

k(2+1) = 6*9

k(3) = 54

k = 54/(3)

k = 18

k = 2 in the OP, ∴ the solution is 1. :shrug::angel:

Moveyourfeet

Oct 10th, 2012, 05:13 AM

This stupid question again. I want to find the person that created it and slap some sense into that fool. I hate dumb questions like this because it's not designed to test actual knowledge. If a developer wrote code as poorly formatted as this problem he would be working for some low level city office somewhere instead of making big bucks in the private sector.

But don't let me stop those of you from feeling smarter than others who thought the answer was 1. Carry on.

But don't let me stop those of you from feeling smarter than others who thought the answer was 1. Carry on.

dybbuk

Oct 10th, 2012, 05:21 AM

Setting the equation equal to 1 and solving for k:

Setting the equation equal to 9 and solving for k:

k = 2 in the OP, ∴ the solution is 1. :shrug::angel:

You do realize you're disagreeing with a professional mathematician here (Egalite) right?

Setting the equation equal to 9 and solving for k:

k = 2 in the OP, ∴ the solution is 1. :shrug::angel:

You do realize you're disagreeing with a professional mathematician here (Egalite) right?

young_gunner913

Oct 10th, 2012, 05:56 AM

You do realize you're disagreeing with a professional mathematician here (Egalite) right?

He's from Boston, you'll have to excuse him. :oh:

He's from Boston, you'll have to excuse him. :oh:

Pump-it-UP

Oct 10th, 2012, 06:18 AM

You do realize you're disagreeing with a professional mathematician here (Egalite) right?

Yes? I've been unfortunate enough to sit through debates of this type of problem in multiple math classes, so I'm cognizant of each argument. Both are logically sound IMO due to the horrific manner in which the equation is structured. I just happen to think that answering it with 1 is more appropriate. :shrug:

Yes? I've been unfortunate enough to sit through debates of this type of problem in multiple math classes, so I'm cognizant of each argument. Both are logically sound IMO due to the horrific manner in which the equation is structured. I just happen to think that answering it with 1 is more appropriate. :shrug:

young_gunner913

Oct 10th, 2012, 06:22 AM

Throw down!! Egalite, come here and defend your honor! Pump is saying you are old and slow and have a lack of a formal education. :oh:

Pump-it-UP

Oct 10th, 2012, 06:30 AM

Throw down!! Egalite, come here and defend your honor! Pump is saying you are old and slow and have a lack of a formal education. :oh:

Ch! Stop stirring shit. That's someone else's job. :lol:

Ch! Stop stirring shit. That's someone else's job. :lol:

Sam L

Oct 10th, 2012, 08:07 AM

It's 9, isn't it?

eck

Oct 10th, 2012, 08:45 AM

At first I thought it was 1, but now I think it's 9 :hysteric:

RYNJ

Oct 10th, 2012, 09:03 AM

PEMDAS

answer is 9

answer is 9

Ashi

Oct 10th, 2012, 09:05 AM

Following BODMAS

6÷2（2+1）=？

(2+1) = 3

6 ÷ 2 = 3

3 x 3 = 9

6÷2（2+1）=？

(2+1) = 3

6 ÷ 2 = 3

3 x 3 = 9

Just Do It

Oct 10th, 2012, 09:13 AM

6÷2 ?（2+1）=？

It depends what do you put instead first ? If it's x then it's 9, if it's % then it's 1 :shrug:

It depends what do you put instead first ? If it's x then it's 9, if it's % then it's 1 :shrug:

Moveyourfeet

Oct 10th, 2012, 09:18 AM

Yes? I've been unfortunate enough to sit through debates of this type of problem in multiple math classes, so I'm cognizant of each argument. Both are logically sound IMO due to the horrific manner in which the equation is structured. I just happen to think that answering it with 1 is more appropriate. :shrug:

If you write a simple SQL expression with this formula, it doesn't even work. You get an error. It's just a 'gotcha' question.

That being said, 9 is the correct answer.

If you write a simple SQL expression with this formula, it doesn't even work. You get an error. It's just a 'gotcha' question.

That being said, 9 is the correct answer.

FORZA SARITA

Oct 10th, 2012, 10:41 AM

seriously this is so elementary :lol:

HippityHop

Oct 10th, 2012, 11:43 AM

Order of operations: follow them.

Mary Cherry.

Oct 10th, 2012, 12:07 PM

EDIT: What the fuck is "parenthesis"?

hectopascal

Oct 10th, 2012, 12:24 PM

EDIT: What the fuck is "parenthesis"?

Parenthesis = the brackets

The way I see it, the 6/2 is basically a fraction in front of the parenthesis.

Therefore it is 3(2+1) = 9

Parenthesis = the brackets

The way I see it, the 6/2 is basically a fraction in front of the parenthesis.

Therefore it is 3(2+1) = 9

$uricate

Oct 10th, 2012, 12:27 PM

122.6

Mary Cherry.

Oct 10th, 2012, 12:30 PM

Ohh...people should just called them brackets :oh:

Some serious math snobs in here, it's pretty much a trick question otherwise there wouldn't be a thread about it.

Some serious math snobs in here, it's pretty much a trick question otherwise there wouldn't be a thread about it.

Lin Lin

Oct 10th, 2012, 12:32 PM

:eek:

ElusiveChanteuse

Oct 10th, 2012, 01:19 PM

I'm a Chinese and I say it's 9.:oh:

SilverSlam

Oct 10th, 2012, 01:37 PM

BIMDAS PEOPLE.

Brackets

Indicies

Multiplication

Division

Addition

Subtraction!

Brackets

Indicies

Multiplication

Division

Addition

Subtraction!

Super Dave

Oct 10th, 2012, 01:49 PM

http://boards.weddingbee.com/?bb_attachments=141818&bbat=6516&inline

Mary Cherry.

Oct 10th, 2012, 02:33 PM

BIMDAS PEOPLE.

Brackets

Indicies

Multiplication

Division

Addition

Subtraction!

Even in that order I'm still confused :sobbing: From what I remember (bear in mind it's been at least 7 years since I did this stuff and in England the exams are "dumbed down" anyway) dealing brackets first means you should get rid of them, so you'd multiply the (2+1) with the 2 rather than leaving it as 6/2(3) and thus end up with 6/6 = 1.

Another possibility is that I may have remembered this all wrong. Apparently that's the case :sobbing:

Brackets

Indicies

Multiplication

Division

Addition

Subtraction!

Even in that order I'm still confused :sobbing: From what I remember (bear in mind it's been at least 7 years since I did this stuff and in England the exams are "dumbed down" anyway) dealing brackets first means you should get rid of them, so you'd multiply the (2+1) with the 2 rather than leaving it as 6/2(3) and thus end up with 6/6 = 1.

Another possibility is that I may have remembered this all wrong. Apparently that's the case :sobbing:

ElusiveChanteuse

Oct 10th, 2012, 02:41 PM

BIMDAS PEOPLE.

Brackets

Indicies

Multiplication

Division

Addition

Subtraction!

To me both multiplication and division or addition and substraction are equal. So I will start from left to right in the equation:

6÷2（2+1）= 6/2 (3) = 3 (3) = 9! :cheer:

Brackets

Indicies

Multiplication

Division

Addition

Subtraction!

To me both multiplication and division or addition and substraction are equal. So I will start from left to right in the equation:

6÷2（2+1）= 6/2 (3) = 3 (3) = 9! :cheer:

Sammo

Oct 10th, 2012, 02:50 PM

It's 9

6/2=3

2+1=3

First you have to opperate the parenthesis, then you opperate from let to right. 3 x 3 = 9

It is a math problem that appeared on facebook some time ago, some people argue to death and defriend each other arguing which is the correct answer for it, 9 or 1. Scandalous.

:spit:

Well I would definitely defriend someone if he started insulting me because of saying the correct answer :unsure::lol:

6/2=3

2+1=3

First you have to opperate the parenthesis, then you opperate from let to right. 3 x 3 = 9

It is a math problem that appeared on facebook some time ago, some people argue to death and defriend each other arguing which is the correct answer for it, 9 or 1. Scandalous.

:spit:

Well I would definitely defriend someone if he started insulting me because of saying the correct answer :unsure::lol:

Shadowcat

Oct 10th, 2012, 03:42 PM

It's 9. I don't get the point of it actually.. :unsure:

(6÷2)（2+1）

(3)(3)

9

(6÷2)（2+1）

(3)(3)

9

Morning Morgan

Oct 10th, 2012, 04:10 PM

Seriously the aim of these kind of questions is just to trick people, because with sufficient use of brackets all ambiguity can be avoided.

SilverSlam

Oct 10th, 2012, 04:26 PM

Even in that order I'm still confused :sobbing: From what I remember (bear in mind it's been at least 7 years since I did this stuff and in England the exams are "dumbed down" anyway) dealing brackets first means you should get rid of them, so you'd multiply the (2+1) with the 2 rather than leaving it as 6/2(3) and thus end up with 6/6 = 1.

Another possibility is that I may have remembered this all wrong. Apparently that's the case :sobbing:

AAAAAAAAND now I know your age :oh:

Another possibility is that I may have remembered this all wrong. Apparently that's the case :sobbing:

AAAAAAAAND now I know your age :oh:

Mary Cherry.

Oct 10th, 2012, 04:37 PM

AAAAAAAAND now I know your age :oh:

Depends, do you know the GCSE Maths/Key Stage 3 curricula for 2005? :oh:

Depends, do you know the GCSE Maths/Key Stage 3 curricula for 2005? :oh:

égalité

Oct 10th, 2012, 04:37 PM

Yes? I've been unfortunate enough to sit through debates of this type of problem in multiple math classes, so I'm cognizant of each argument. Both are logically sound IMO due to the horrific manner in which the equation is structured. I just happen to think that answering it with 1 is more appropriate. :shrug:

Well mathematics is a language with very fussy, standardized grammar, and the way the expression is written makes it equal to 9.

1 is neither a logically sound nor appropriate answer because the expression 6÷2(2+1) means "take 6, divide by two, multiply by 3." It does not mean "multiply 2 by 3 and divide 6 by that whole thing." There is one way of interpreting what is written here.

Well mathematics is a language with very fussy, standardized grammar, and the way the expression is written makes it equal to 9.

1 is neither a logically sound nor appropriate answer because the expression 6÷2(2+1) means "take 6, divide by two, multiply by 3." It does not mean "multiply 2 by 3 and divide 6 by that whole thing." There is one way of interpreting what is written here.

Mary Cherry.

Oct 10th, 2012, 04:40 PM

Well mathematics is a language with very fussy, standardized grammar, and the way the expression is written makes it equal to 9.

1 is neither a logically sound nor appropriate answer because the expression 6÷2(2+1) means "take 6, divide by two, multiply by 3." It does not mean "multiply 2 by 3 and divide 6 by that whole thing." There is one and only one way of interpreting what is written here.

So basically we should dismiss what they taught us in school?

I'm not even surprised :lol:

1 is neither a logically sound nor appropriate answer because the expression 6÷2(2+1) means "take 6, divide by two, multiply by 3." It does not mean "multiply 2 by 3 and divide 6 by that whole thing." There is one and only one way of interpreting what is written here.

So basically we should dismiss what they taught us in school?

I'm not even surprised :lol:

égalité

Oct 10th, 2012, 04:48 PM

So basically we should dismiss what they taught us in school?

I'm not even surprised :lol:

No, this is what they taught you in school :oh:

You do what's inside the brackets first, i.e. change 2+1 to 3...then the expression is 6÷2 x 3.

I'm not even surprised :lol:

No, this is what they taught you in school :oh:

You do what's inside the brackets first, i.e. change 2+1 to 3...then the expression is 6÷2 x 3.

young_gunner913

Oct 10th, 2012, 04:49 PM

Oh Mary, such a hot, mathematical, lesbian mess. :oh:

Mary Cherry.

Oct 10th, 2012, 05:06 PM

EDIT: I can't even be arsed.

pov

Oct 10th, 2012, 06:14 PM

Order as I know it is:

parentheses 6/2(3)

multiplication 6/6

division 1

I'm surprised so many did division before multiplication giving 9

parentheses 6/2(3)

multiplication 6/6

division 1

I'm surprised so many did division before multiplication giving 9

$uricate

Oct 10th, 2012, 06:26 PM

1.

I said 9 before but I consulted my old textbook on parenthesis and I change my answer.

BOMDAS

Brackets off Multiply Divide Add Subtract.

I said 9 before but I consulted my old textbook on parenthesis and I change my answer.

BOMDAS

Brackets off Multiply Divide Add Subtract.

silverwhite

Oct 10th, 2012, 07:47 PM

I thought it was 1 too but I was complacent and didn't think carefully :facepalm:

6 ÷ 2 (2+1) = 6 ÷ 2 x (2+1) = 9

The first and second expressions are equivalent :shrug:

Seriously the aim of these kind of questions is just to trick people, because with sufficient use of brackets all ambiguity can be avoided.

Agreed. Personally, I'm used to seeing and I'm used to using brackets for multiplication when that particular term is involved in addition/subtraction (eg. 6 + 2 (2+1) ). That's why I instinctively did "2 (2+1) = 6" in my head first. It's a reflex :lol:

If I had to write this particular expression, I would always use the multiplication sign to avoid any ambiguity (i.e. 6 ÷ 2 x (2+1) ) :lol:

6 ÷ 2 (2+1) = 6 ÷ 2 x (2+1) = 9

The first and second expressions are equivalent :shrug:

Seriously the aim of these kind of questions is just to trick people, because with sufficient use of brackets all ambiguity can be avoided.

Agreed. Personally, I'm used to seeing and I'm used to using brackets for multiplication when that particular term is involved in addition/subtraction (eg. 6 + 2 (2+1) ). That's why I instinctively did "2 (2+1) = 6" in my head first. It's a reflex :lol:

If I had to write this particular expression, I would always use the multiplication sign to avoid any ambiguity (i.e. 6 ÷ 2 x (2+1) ) :lol:

PhilePhile

Oct 10th, 2012, 08:04 PM

I thought it was 1 too but I was complacent and didn't think carefully :facepalm:

6 ÷ 2 (2+1) = 6 ÷ 2 x (2+1) = 9

The first and second expressions are equivalent :shrug:

Agreed. Personally, I'm used to seeing and I'm used to using brackets for multiplication when that particular term is involved in addition/subtraction (eg. 6 + 2 (2+1) ). That's why I instinctively did "2 (2+1) = 6" in my head first. It's a reflex :lol:

If I had to write this particular expression, I would always use the multiplication sign to avoid any ambiguity (i.e. 6 ÷ 2 x (2+1) ) :lol:

No. To avoid ambiguity, (6÷2)x(2+1) if the desired answer is 9.

6 ÷ 2 (2+1) = 6 ÷ 2 x (2+1) = 9

The first and second expressions are equivalent :shrug:

Agreed. Personally, I'm used to seeing and I'm used to using brackets for multiplication when that particular term is involved in addition/subtraction (eg. 6 + 2 (2+1) ). That's why I instinctively did "2 (2+1) = 6" in my head first. It's a reflex :lol:

If I had to write this particular expression, I would always use the multiplication sign to avoid any ambiguity (i.e. 6 ÷ 2 x (2+1) ) :lol:

No. To avoid ambiguity, (6÷2)x(2+1) if the desired answer is 9.

silverwhite

Oct 10th, 2012, 08:08 PM

No. To avoid ambiguity, (6÷2)x(2+1) if the desired answer is 9.

It depends on how you see it. I personally prefer 6 ÷ 2 x (2+1) :shrug:

It depends on how you see it. I personally prefer 6 ÷ 2 x (2+1) :shrug:

Kon.

Oct 10th, 2012, 08:12 PM

Order as I know it is:

parentheses 6/2(3)

multiplication 6/6

division 1

I'm surprised so many did division before multiplication giving 9

This isn't the correct order of operations though.

It is:

Parentheses

Exponents

Multiplication-Division

Addition- Subtraction

Multiplication doesn't go before division. Neither has precedence over the other. Same thing with addition and subtraction. They happen at the same time.

Then you just operate from left to right.

In this case we have 6 ÷ 2 (3) so the reason it's 9 is than you first have to do the division because it's on the left, getting 3 (3) =9.

parentheses 6/2(3)

multiplication 6/6

division 1

I'm surprised so many did division before multiplication giving 9

This isn't the correct order of operations though.

It is:

Parentheses

Exponents

Multiplication-Division

Addition- Subtraction

Multiplication doesn't go before division. Neither has precedence over the other. Same thing with addition and subtraction. They happen at the same time.

Then you just operate from left to right.

In this case we have 6 ÷ 2 (3) so the reason it's 9 is than you first have to do the division because it's on the left, getting 3 (3) =9.

PhilePhile

Oct 10th, 2012, 08:14 PM

It depends on how you see it. I personally prefer 6 ÷ 2 x (2+1) :shrug:

There is no "personally prefer" in arithmetics, either you follow the rules or be as clear as possible.

There is no "personally prefer" in arithmetics, either you follow the rules or be as clear as possible.

silverwhite

Oct 10th, 2012, 08:15 PM

There is no "personally prefer" in arithmetics, either you follow the rules or be as clear as possible.

Explain how (6÷2)x(2+1) is objectively clearer :scratch:

Explain how (6÷2)x(2+1) is objectively clearer :scratch:

Mary Cherry.

Oct 10th, 2012, 08:23 PM

I still don't get how you can put 2(2+1) as 2 x 3. I know the number outside the bracket means you should times it by what's inside the bracket and since brackets come first, shouldn't you do 2(3) before the division?

I don't remember them teaching us that you get marks for re-writing the question.

I don't remember them teaching us that you get marks for re-writing the question.

silverwhite

Oct 10th, 2012, 08:24 PM

I still don't get how you can put 2(2+1) as 2 x 3. I know the number outside the bracket means you should times it by what's inside the bracket and since brackets come first, shouldn't you do 2(3) before the division?

I don't remember them teaching us that you get marks for re-writing the question.

What's inside the brackets comes first ;)

I don't remember them teaching us that you get marks for re-writing the question.

What's inside the brackets comes first ;)

LoveFifteen

Oct 10th, 2012, 08:24 PM

The answer is 1, but if anything, this problem just demonstrates how stupid random math problems are.

Mary Cherry.

Oct 10th, 2012, 08:28 PM

What's inside the brackets comes first ;)

Well yes, that's how you know to times the 2 by 3. I still don't get how you can rewrite the question and add the x in yourself though. The 2 is grouped with what's in the bracket, not with the 6.

Well yes, that's how you know to times the 2 by 3. I still don't get how you can rewrite the question and add the x in yourself though. The 2 is grouped with what's in the bracket, not with the 6.

Mary Cherry.

Oct 10th, 2012, 08:28 PM

The answer is 1, but if anything, this problem just demonstrates how stupid random math problems are.

Pretty much.

Let's end the thread here :lol:

Pretty much.

Let's end the thread here :lol:

young_gunner913

Oct 10th, 2012, 08:30 PM

Pretty much.

Let's end the thread here :lol:

We can end the thread with that statement and the correct answer: 9. :oh:

Let's end the thread here :lol:

We can end the thread with that statement and the correct answer: 9. :oh:

Inger67

Oct 10th, 2012, 08:30 PM

I still don't get how you can put 2(2+1) as 2 x 3. I know the number outside the bracket means you should times it by what's inside the bracket and since brackets come first, shouldn't you do 2(3) before the division?

I don't remember them teaching us that you get marks for re-writing the question.

That is multiplication then, not considered brackets anymore. You just do what is inside the brackets first.

The answer is 1, but if anything, this problem just demonstrates how stupid random math problems are.

Can you please explain how you get 1?

The answer is 9.

As many people have said before (as myself) you do what's INSIDE the parenthesis (brackets) first and then divide and multiply. Since those steps come in the same order of PEMDAS (P, E, M&D, A&S to make it clearer) you do what comes first, i.e. LEFT to RIGHT.

6 divided by 2 and then 3(3) = 9

I don't remember them teaching us that you get marks for re-writing the question.

That is multiplication then, not considered brackets anymore. You just do what is inside the brackets first.

The answer is 1, but if anything, this problem just demonstrates how stupid random math problems are.

Can you please explain how you get 1?

The answer is 9.

As many people have said before (as myself) you do what's INSIDE the parenthesis (brackets) first and then divide and multiply. Since those steps come in the same order of PEMDAS (P, E, M&D, A&S to make it clearer) you do what comes first, i.e. LEFT to RIGHT.

6 divided by 2 and then 3(3) = 9

$uricate

Oct 10th, 2012, 08:32 PM

It's 1

Mary Cherry.

Oct 10th, 2012, 08:32 PM

We can end the thread with that statement and the correct answer: 9. :oh:

But the thread has already ended so your answer is irrelevant :oh:

But the thread has already ended so your answer is irrelevant :oh:

silverwhite

Oct 10th, 2012, 08:32 PM

Well yes, that's how you know to times the 2 by 3. I still don't get how you can rewrite the question and add the x in yourself though. The 2 is grouped with what's in the bracket, not with the 6.

We can because A(B) = A x B basically :lol:

We can because A(B) = A x B basically :lol:

PhilePhile

Oct 10th, 2012, 08:33 PM

I still don't get how you can put 2(2+1) as 2 x 3. I know the number outside the bracket means you should times it by what's inside the bracket and since brackets come first, shouldn't you do 2(3) before the division?

I don't remember them teaching us that you get marks for re-writing the question.

2x(2+1)= 2x2 + 2 = 4 + 2 = 6 as well.

What's inside the brackets comes first ;)

Is that by following rules or common sense?

I don't remember them teaching us that you get marks for re-writing the question.

2x(2+1)= 2x2 + 2 = 4 + 2 = 6 as well.

What's inside the brackets comes first ;)

Is that by following rules or common sense?

$uricate

Oct 10th, 2012, 08:33 PM

Can this thread end please?

http://i1208.photobucket.com/albums/cc371/zoraluv/gifs/tellmemorecat_zps8fd5f6f4.png

http://i1208.photobucket.com/albums/cc371/zoraluv/gifs/tellmemorecat_zps8fd5f6f4.png

Inger67

Oct 10th, 2012, 08:34 PM

Well yes, that's how you know to times the 2 by 3. I still don't get how you can rewrite the question and add the x in yourself though. The 2 is grouped with what's in the bracket, not with the 6.

No it's not.

6÷2(2+1) is the same as writing 6÷2 x (3).

The first two are grouped and then the second. You go in order of operations at that point and you go from left to right so you divide first and then multiply.

This problem is pretty clear if you know the rules properly :shrug:

No it's not.

6÷2(2+1) is the same as writing 6÷2 x (3).

The first two are grouped and then the second. You go in order of operations at that point and you go from left to right so you divide first and then multiply.

This problem is pretty clear if you know the rules properly :shrug:

silverwhite

Oct 10th, 2012, 08:36 PM

Is that by following rules or common sense?

You haven't answered my question :bigwave:

You haven't answered my question :bigwave:

Super Dave

Oct 10th, 2012, 08:38 PM

I think this thread should just burn to death in a pile of random gifs.

http://i.imgur.com/hcxTO.gif

http://i.imgur.com/hcxTO.gif

Mary Cherry.

Oct 10th, 2012, 08:44 PM

We can because A(B) = A x B basically :lol:

Like I said, I know that but removing the bracket and writing it as 2 x 3 separates them both so the 6 gets involved with the 2 and 3 is like "wtf bitch back off" and 7 ate 9 and they all got into a fight.

Like I said, I know that but removing the bracket and writing it as 2 x 3 separates them both so the 6 gets involved with the 2 and 3 is like "wtf bitch back off" and 7 ate 9 and they all got into a fight.

young_gunner913

Oct 10th, 2012, 08:47 PM

But the thread has already ended so your answer is irrelevant :oh:

Then the thread is reopen for discussion due to the answer being incorrect. :oh:

Then the thread is reopen for discussion due to the answer being incorrect. :oh:

silverwhite

Oct 10th, 2012, 08:49 PM

Like I said, I know that but removing the bracket and writing it as 2 x 3 separates them both so the 6 gets involved with the 2 and 3 is like "wtf bitch back off" and 7 ate 9 and they all got into a fight.

The brackets don't give whatever is adjacent to them any special priority when it comes to multiplication. Read Inger67's post for a more detailed explanation ;)

The brackets don't give whatever is adjacent to them any special priority when it comes to multiplication. Read Inger67's post for a more detailed explanation ;)

Mary Cherry.

Oct 10th, 2012, 08:52 PM

Then the thread is reopen for discussion due to the answer being incorrect. :oh:

Life is incorrect :sobbing:

Life is incorrect :sobbing:

young_gunner913

Oct 10th, 2012, 08:56 PM

Life is incorrect :sobbing:

The answer 1 is incorrect. :p

The answer 1 is incorrect. :p

Cajka

Oct 10th, 2012, 09:18 PM

It is a math problem that appeared on facebook some time ago, some people argue to death and defriend each other arguing which is the correct answer for it, 9 or 1. Scandalous.

:rolls:

:rolls:

$uricate

Oct 10th, 2012, 11:32 PM

http://i1208.photobucket.com/albums/cc371/zoraluv/gifs/tellmemorecat_zps8fd5f6f4.png

Novichok

Oct 11th, 2012, 12:03 AM

Everyone who posts here can solve this problem. It's just a poorly written (too ambiguous) problem.

égalité

Oct 11th, 2012, 12:32 AM

Order as I know it is:

parentheses 6/2(3)

multiplication 6/6

division 1

I'm surprised so many did division before multiplication giving 9

Multiplication does not take precedence over division. Any division can be rephrased as multiplication, so how can one take precedence over the other? e.g. 4/2 = 4x(1/2). The MD in "PEMDAS" is ONE step, not two.

parentheses 6/2(3)

multiplication 6/6

division 1

I'm surprised so many did division before multiplication giving 9

Multiplication does not take precedence over division. Any division can be rephrased as multiplication, so how can one take precedence over the other? e.g. 4/2 = 4x(1/2). The MD in "PEMDAS" is ONE step, not two.

égalité

Oct 11th, 2012, 12:42 AM

Well yes, that's how you know to times the 2 by 3. I still don't get how you can rewrite the question and add the x in yourself though. The 2 is grouped with what's in the bracket, not with the 6.

The 2 is not grouped with anything. The 6, the 2, and the (2+1) are doing their own things gurl. Until someone comes along and takes the 6 divides by the 2 and then multiplies by the (2+1). The only operation that gets done before all this is the "2+1" inside the brackets. There's no rule that says what's next to the brackets takes precedence over everything else. Outsiders are not invited into the brackets. :oh:

The 2 is not grouped with anything. The 6, the 2, and the (2+1) are doing their own things gurl. Until someone comes along and takes the 6 divides by the 2 and then multiplies by the (2+1). The only operation that gets done before all this is the "2+1" inside the brackets. There's no rule that says what's next to the brackets takes precedence over everything else. Outsiders are not invited into the brackets. :oh:

Wigglytuff

Oct 11th, 2012, 01:10 AM

Order as I know it is:

parentheses 6/2(3)

multiplication 6/6

division 1

I'm surprised so many did division before multiplication giving 9

That's because you are wrong!

parentheses 6/2(3)

multiplication 6/6

division 1

I'm surprised so many did division before multiplication giving 9

That's because you are wrong!

Wigglytuff

Oct 11th, 2012, 01:14 AM

To me both multiplication and division or addition and substraction are equal. So I will start from left to right in the equation:

6÷2（2+1）= 6/2 (3) = 3 (3) = 9! :cheer:

This is correct. I don't understand how anyone gets this wrong?

6÷2（2+1）= 6/2 (3) = 3 (3) = 9! :cheer:

This is correct. I don't understand how anyone gets this wrong?

Wigglytuff

Oct 11th, 2012, 01:17 AM

http://en.wikipedia.org/wiki/Order_of_operations

Novichok

Oct 11th, 2012, 01:38 AM

This is correct. I don't understand how anyone gets this wrong?

Because most people don't remember order of operations from elementary school. :lol: Which is understandable, since one would almost never have to calculate a problem written like this in real life.

Because most people don't remember order of operations from elementary school. :lol: Which is understandable, since one would almost never have to calculate a problem written like this in real life.

Nicolás89

Oct 11th, 2012, 01:47 AM

Lol at some people, the "x" doesn't need to be written to set a multiplication.

Philip

Oct 11th, 2012, 01:50 AM

I got 9 from what I remember from school :tape:

mc8114

Oct 11th, 2012, 01:53 AM

I would say 9

Philip

Oct 11th, 2012, 01:54 AM

And what I remember from school:

6/2 (2+1)

- Do your bracket calculations first - 6/2 (3)

- Then do your calculations outside of the bracket - 3 (3)

- Then muliply whats outside the bracket with whats inside - 9

:tape: :lol:

6/2 (2+1)

- Do your bracket calculations first - 6/2 (3)

- Then do your calculations outside of the bracket - 3 (3)

- Then muliply whats outside the bracket with whats inside - 9

:tape: :lol:

Morning Morgan

Oct 11th, 2012, 02:40 AM

Do I really care when I'm trying to quotient out my torsional modules from my multilinear map to get my tensor product?

Shadowcat

Oct 11th, 2012, 03:34 AM

No. To avoid ambiguity, (6÷2)x(2+1) if the desired answer is 9.

Somebody who finally gets it.

Somebody who finally gets it.

Shadowcat

Oct 11th, 2012, 03:37 AM

Explain how (6÷2)x(2+1) is objectively clearer :scratch:

Well for me, anything I see which has multiplication or division in it I put them in brackets so it is easier.

(6÷2)(2+1)

(3)(3)

9

The brackets organise things for you so you don't mix up stuff.

Just like example (2a + 2b)(2b + 2a)

Well for me, anything I see which has multiplication or division in it I put them in brackets so it is easier.

(6÷2)(2+1)

(3)(3)

9

The brackets organise things for you so you don't mix up stuff.

Just like example (2a + 2b)(2b + 2a)

Shadowcat

Oct 11th, 2012, 03:39 AM

I think this thread should just burn to death in a pile of random gifs.

http://i.imgur.com/hcxTO.gif

Agreed. :lol:

http://i.imgur.com/hcxTO.gif

Agreed. :lol:

Hurley

Oct 11th, 2012, 05:35 AM

All I know is if you want to get into a decent graduate school via the GRE, you better answer this question as fucking 9, otherwise you're going to Arizona State.

P: do any fucking thing inside the parentheses fucking first. 2+1 is fucking 3.

E: there are no exponents.

M/D: All multiplication and division is done left to right. 6 divided by 2 is 3. That 3 times 3 is fucking 9.

A/S: there is no addition or subtraction which has not already been completed (cf. P). The answer is fucking 9.

Don't like it? Take it up with ETS. :shrug:

P: do any fucking thing inside the parentheses fucking first. 2+1 is fucking 3.

E: there are no exponents.

M/D: All multiplication and division is done left to right. 6 divided by 2 is 3. That 3 times 3 is fucking 9.

A/S: there is no addition or subtraction which has not already been completed (cf. P). The answer is fucking 9.

Don't like it? Take it up with ETS. :shrug:

Hurley

Oct 11th, 2012, 05:42 AM

HOWEVER - I will admit that the formatting of the question in the thread title is ridiculous and could lead to issues. I can forgive people wanting to distribute the 2 into the parentheses, which would give you 6 divided by 6, which is non-fucking 1.

But this question is "6÷2 x (2+1)" and that has only one answer per order of operations, which is fucking 9.

But this question is "6÷2 x (2+1)" and that has only one answer per order of operations, which is fucking 9.

Shadowcat

Oct 11th, 2012, 05:53 AM

Dude chill on the f-word. :weirdo:

Hurley

Oct 11th, 2012, 06:05 AM

Dude chill on the f-word. :weirdo:

Hi. Never. Bye.

Hi. Never. Bye.

silverwhite

Oct 11th, 2012, 06:36 AM

Well for me, anything I see which has multiplication or division in it I put them in brackets so it is easier.

(6÷2)(2+1)

(3)(3)

9

The brackets organise things for you so you don't mix up stuff.

Just like example (2a + 2b)(2b + 2a)

It's not clearer. It's just slightly different.

(6÷2) x (2+1) = 3 x 3 = 9

6 ÷ 2 x (2+1) = 6 ÷ 2 x 3 = 9

(6÷2)(2+1)

(3)(3)

9

The brackets organise things for you so you don't mix up stuff.

Just like example (2a + 2b)(2b + 2a)

It's not clearer. It's just slightly different.

(6÷2) x (2+1) = 3 x 3 = 9

6 ÷ 2 x (2+1) = 6 ÷ 2 x 3 = 9

SilverPersian

Oct 11th, 2012, 06:40 AM

x/y(y+z)=x/(y2+yz)

=6/4+2

=1

But I don't know :shrug:

=6/4+2

=1

But I don't know :shrug:

taloki

Oct 11th, 2012, 07:29 AM

My first attempt at the question was 1...

Then I looked at the responses and realised where I went wrong. :sobbing:

Then I looked at the responses and realised where I went wrong. :sobbing:

Remix13

Oct 11th, 2012, 08:33 AM

Windows calculator says 9

Vartan

Oct 11th, 2012, 08:46 AM

Setting the equation equal to 1 and solving for k:

Setting the equation equal to 9 and solving for k:

k = 2 in the OP, ∴ the solution is 1. :shrug::angel:

Please tell me you are trolling. You solved both equations you set up wrong.

Half of the posts in this thread are the reasons why Romney cannot be allowed to win the election.

Setting the equation equal to 9 and solving for k:

k = 2 in the OP, ∴ the solution is 1. :shrug::angel:

Please tell me you are trolling. You solved both equations you set up wrong.

Half of the posts in this thread are the reasons why Romney cannot be allowed to win the election.

Singleniacki

Oct 11th, 2012, 09:38 AM

9. Bedmas :p

LOL jk it's:

6/2 (2 + 1) = x

x = (6/2 * 2) + (6/2 * 1)

x = 6 [(1/2 * 2/6)+(1/2 * 1/6)]

x^(2) = [ 6 (1/2 * 2/6)+(1/2 * 1/6)]^2

log x^2 = log [ 6 (1/2 * 2/6)+(1/2 * 1/6)]^2

2 log x = 2 log [ 6 (1/2 * 2/6)+(1/2 * 1/6)]

log x = 2 log [ 6 (1/2 * 2/6)+(1/2 * 1/6)] / 2

therefore

x = 10 ^ (2 log {[ 6 (1/2 * 2/6)+(1/2 * 1/6)] / 2})

Of course.

LOL jk it's:

6/2 (2 + 1) = x

x = (6/2 * 2) + (6/2 * 1)

x = 6 [(1/2 * 2/6)+(1/2 * 1/6)]

x^(2) = [ 6 (1/2 * 2/6)+(1/2 * 1/6)]^2

log x^2 = log [ 6 (1/2 * 2/6)+(1/2 * 1/6)]^2

2 log x = 2 log [ 6 (1/2 * 2/6)+(1/2 * 1/6)]

log x = 2 log [ 6 (1/2 * 2/6)+(1/2 * 1/6)] / 2

therefore

x = 10 ^ (2 log {[ 6 (1/2 * 2/6)+(1/2 * 1/6)] / 2})

Of course.

Super Dave

Oct 11th, 2012, 02:07 PM

HOWEVER - I will admit that the formatting of the question in the thread title is ridiculous and could lead to issues. I can forgive people wanting to distribute the 2 into the parentheses, which would give you 6 divided by 6, which is non-fucking 1.

But this question is "6÷2 x (2+1)" and that has only one answer per order of operations, which is fucking 9.

It is poorly written and designed to cause ad nauseam arguments on the internets. Mission accomplished.

But this question is "6÷2 x (2+1)" and that has only one answer per order of operations, which is fucking 9.

It is poorly written and designed to cause ad nauseam arguments on the internets. Mission accomplished.

ElusiveChanteuse

Oct 11th, 2012, 02:08 PM

:spit:

KournikovaFan91

Oct 11th, 2012, 02:18 PM

I got 1 but I suck at maths :lol: I see its 9 now.

Novichok

Oct 11th, 2012, 02:48 PM

Please tell me you are trolling. You solved both equations you set up wrong.

Half of the posts in this thread are the reasons why Romney cannot be allowed to win the election.

LOL. It's really not that serious. You get 1 or 9 depending on how you interpret the syntax of the question. Choosing 1 as the answer is not indicative of poor math skills. It's only indicative of not knowing the accepted order of operations.

Half of the posts in this thread are the reasons why Romney cannot be allowed to win the election.

LOL. It's really not that serious. You get 1 or 9 depending on how you interpret the syntax of the question. Choosing 1 as the answer is not indicative of poor math skills. It's only indicative of not knowing the accepted order of operations.

Kuilli

Oct 11th, 2012, 03:07 PM

All I know is if you want to get into a decent graduate school via the GRE, you better answer this question as fucking 9, otherwise you're going to Arizona State.

So the Arizona State didn't work out that well, eh? ;)

So the Arizona State didn't work out that well, eh? ;)

Cajka

Oct 11th, 2012, 03:12 PM

It's not clearer.

Visually, it is. But, of course, it's the same thing.

Visually, it is. But, of course, it's the same thing.

SVK

Oct 11th, 2012, 04:53 PM

It´s 9.

I was confused with this too when I saw this for the first time and when they write it as 6/2(2+1) it is even more confusing, because you don´t know if it´s fraction or not. The problem is that 6÷2(2+1) is not the same as 6/2(2+1) as a fraction, those are two completely different things with 2 different results and that´s why people mess this up.

Let´s say you have 10÷5*3. When you have 2 operations with the same priority the another rule is that you need to count from left to right. You got the result of 6. If you count it as a fraction 10/5*3 the result is of course different (10/15).

The other thing is that if you want to make a fraction from x÷y*z then it is x*z/y(!) and not x/y*z...Example - 10÷5*3 = 10*3/5 as a fraction = 30/5 = 6.

If it would be 10÷(5*3) so x÷y then it´s different case and you can write it as 10/5*3.

We have 6÷2（2+1） what is x(6)÷y(2)*z(2+1) so if you want to make a fraction of it you need to count 6(2+1)/2 = 6*3/2 = 18/2 = 9.

If you want to count it in an original form then 6÷2（2+1） = 6÷2*3 = 3*3 = 9.

The another problem might be that people don´t see "*" between 2 and (2+1) and you might think (2+1) is fixed to 2 what is not true of course x÷y*z is the same as x÷yz.

Result 1 would be if the original thing would be in this form - 6÷[2（2+1）]. the extra brackets would make it as x÷y and you could count is as 6/2(2+1) as a fraction what equals 6/2*3 = 6/6 = 1 (the "/" sign is used as a fraction, because i don´t know how to write it properly, but it does not equal "÷", therefore I can´t count it from left to right in this case, because 2*3 is a denominator).

Simply - 6÷2（2+1） what equals 9 is not the same thing as 6/2(2+1) as a fraction what equals 1. 6÷[2(2+1)] would equal 6/2(2+1) as a fraction.

I was confused with this too when I saw this for the first time and when they write it as 6/2(2+1) it is even more confusing, because you don´t know if it´s fraction or not. The problem is that 6÷2(2+1) is not the same as 6/2(2+1) as a fraction, those are two completely different things with 2 different results and that´s why people mess this up.

Let´s say you have 10÷5*3. When you have 2 operations with the same priority the another rule is that you need to count from left to right. You got the result of 6. If you count it as a fraction 10/5*3 the result is of course different (10/15).

The other thing is that if you want to make a fraction from x÷y*z then it is x*z/y(!) and not x/y*z...Example - 10÷5*3 = 10*3/5 as a fraction = 30/5 = 6.

If it would be 10÷(5*3) so x÷y then it´s different case and you can write it as 10/5*3.

We have 6÷2（2+1） what is x(6)÷y(2)*z(2+1) so if you want to make a fraction of it you need to count 6(2+1)/2 = 6*3/2 = 18/2 = 9.

If you want to count it in an original form then 6÷2（2+1） = 6÷2*3 = 3*3 = 9.

The another problem might be that people don´t see "*" between 2 and (2+1) and you might think (2+1) is fixed to 2 what is not true of course x÷y*z is the same as x÷yz.

Result 1 would be if the original thing would be in this form - 6÷[2（2+1）]. the extra brackets would make it as x÷y and you could count is as 6/2(2+1) as a fraction what equals 6/2*3 = 6/6 = 1 (the "/" sign is used as a fraction, because i don´t know how to write it properly, but it does not equal "÷", therefore I can´t count it from left to right in this case, because 2*3 is a denominator).

Simply - 6÷2（2+1） what equals 9 is not the same thing as 6/2(2+1) as a fraction what equals 1. 6÷[2(2+1)] would equal 6/2(2+1) as a fraction.

égalité

Oct 11th, 2012, 05:00 PM

It´s 9.

I was confused with this too when I saw this for the first time and when they write it as 6/2(2+1) it is even more confusing, because you don´t know if it´s fraction or not. The problem is that 6÷2(2+1) is not the same as 6/2(2+1) as a fraction, those are two completely different things with 2 different results and that´s why people mess this up.

Let´s say you have 10÷5*3. When you have 2 functions with the same priority the another rule is that you need to count from left to right. You got the result of 6. If you count it as a fraction 10/5*3 the result is of course different (10/15).

The other thing is that if you want to make a fraction from x÷y*z then it is x*z/y(!) and not x/y*z...Example - 10÷5*3 = 10*3/5 as a fraction = 30/5 = 6.

If it would be 10÷(5*3) so x÷y then it´s different case and you can write it as 10/5*3.

We have 6÷2（2+1） what is x(6)÷y(2)*z(2+1) so if you want to make a fraction of it you need to count 6(2+1)/2 = 6*3/2 = 18/2 = 9.

If you want to count it in an original form then 6÷2（2+1） = 6÷2*3 = 3*3 = 9.

The another problem might be that people don´t see "*" between 2 and (2+1) and you might think (2+1) is fixed to 2 what is not true of course x÷y*z is the same as x÷yz.

Result 1 would be if the original thing would be in this form - 6÷[2（2+1）]. the extra brackets would make it as x÷y and you could count is as 6/2(2+1) as a fraction what equals 6/2*3 = 6/6 = 1 (the "/" sign is used as a fraction, because i don´t know how to write it properly, but it does not equal "÷", therefore I can´t count it from left to right in this case, because 2*3 is a denominator).

Simply - 6÷2（2+1） what equals 9 is not the same thing as 6/2(2+1) as a fraction what equals 1. 6÷[2(2+1)] would equal 6/2(2+1) as a fraction.

Nnnoooo those two things are exactly the same. 6/2(2+1) does not mean 6÷[2(2+1)]. ÷ and / mean the same thing.

I was confused with this too when I saw this for the first time and when they write it as 6/2(2+1) it is even more confusing, because you don´t know if it´s fraction or not. The problem is that 6÷2(2+1) is not the same as 6/2(2+1) as a fraction, those are two completely different things with 2 different results and that´s why people mess this up.

Let´s say you have 10÷5*3. When you have 2 functions with the same priority the another rule is that you need to count from left to right. You got the result of 6. If you count it as a fraction 10/5*3 the result is of course different (10/15).

The other thing is that if you want to make a fraction from x÷y*z then it is x*z/y(!) and not x/y*z...Example - 10÷5*3 = 10*3/5 as a fraction = 30/5 = 6.

If it would be 10÷(5*3) so x÷y then it´s different case and you can write it as 10/5*3.

We have 6÷2（2+1） what is x(6)÷y(2)*z(2+1) so if you want to make a fraction of it you need to count 6(2+1)/2 = 6*3/2 = 18/2 = 9.

If you want to count it in an original form then 6÷2（2+1） = 6÷2*3 = 3*3 = 9.

The another problem might be that people don´t see "*" between 2 and (2+1) and you might think (2+1) is fixed to 2 what is not true of course x÷y*z is the same as x÷yz.

Result 1 would be if the original thing would be in this form - 6÷[2（2+1）]. the extra brackets would make it as x÷y and you could count is as 6/2(2+1) as a fraction what equals 6/2*3 = 6/6 = 1 (the "/" sign is used as a fraction, because i don´t know how to write it properly, but it does not equal "÷", therefore I can´t count it from left to right in this case, because 2*3 is a denominator).

Simply - 6÷2（2+1） what equals 9 is not the same thing as 6/2(2+1) as a fraction what equals 1. 6÷[2(2+1)] would equal 6/2(2+1) as a fraction.

Nnnoooo those two things are exactly the same. 6/2(2+1) does not mean 6÷[2(2+1)]. ÷ and / mean the same thing.

SVK

Oct 11th, 2012, 05:10 PM

Nnnoooo those two things are exactly the same. 6/2(2+1) does not mean 6÷[2(2+1)]. ÷ and / mean the same thing.

It´s very difficult to write fractions properly here, and I used / as the line between numerator and denominator and not as a ÷ so there is a difference...count the example 10÷5*3 I used, when you have it in this form you must count from left to right, that´s simply a rule just like other rules in maths - you got a result of 6. and now count it as a fraction 10/5*3 where the / is just a line between numerator and denominator - denominator would be 15 and the whole thing would look like 10/15 what is of course differnt result as 6.

It´s very difficult to write fractions properly here, and I used / as the line between numerator and denominator and not as a ÷ so there is a difference...count the example 10÷5*3 I used, when you have it in this form you must count from left to right, that´s simply a rule just like other rules in maths - you got a result of 6. and now count it as a fraction 10/5*3 where the / is just a line between numerator and denominator - denominator would be 15 and the whole thing would look like 10/15 what is of course differnt result as 6.

SVK

Oct 11th, 2012, 05:51 PM

Nnnoooo those two things are exactly the same. 6/2(2+1) does not mean 6÷[2(2+1)]. ÷ and / mean the same thing.

However, I got your point, because / is very often used as ÷ in PC and LinLin ruined this thing a bit:p because he wrote it as 6÷2(2+1) what has a clear answer. You have this PEMDAS thing, so I copied what it´s written there:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

So 6÷2(2+1) = 6÷2*3 = 3*3 = 9.

Just like any other x÷yz example.

But, if it would be written as 6/2(2+1) as it´s originally written all over internet then the fun would begin. What the / sign would mean? Is it 6÷2(2+1) or is it a fraction where 6 is numerator and 2(2+1) is denominator? As I showed on the previous example x÷yz is a completely different thing than x(as numerator) / yz (as a denominator), just like 10÷5*3 is different than 10(numerator)/5*3(denominator) - you got two different results as both of these cases are completely different things.

However, I got your point, because / is very often used as ÷ in PC and LinLin ruined this thing a bit:p because he wrote it as 6÷2(2+1) what has a clear answer. You have this PEMDAS thing, so I copied what it´s written there:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

So 6÷2(2+1) = 6÷2*3 = 3*3 = 9.

Just like any other x÷yz example.

But, if it would be written as 6/2(2+1) as it´s originally written all over internet then the fun would begin. What the / sign would mean? Is it 6÷2(2+1) or is it a fraction where 6 is numerator and 2(2+1) is denominator? As I showed on the previous example x÷yz is a completely different thing than x(as numerator) / yz (as a denominator), just like 10÷5*3 is different than 10(numerator)/5*3(denominator) - you got two different results as both of these cases are completely different things.

silverwhite

Oct 11th, 2012, 06:59 PM

Visually, it is. But, of course, it's the same thing.

That's subjective. IMO, it's visually clearer without an extra set of brackets :lol:

That's subjective. IMO, it's visually clearer without an extra set of brackets :lol:

Wigglytuff

Oct 11th, 2012, 07:53 PM

Nnnoooo those two things are exactly the same. 6/2(2+1) does not mean 6÷[2(2+1)]. ÷ and / mean the same thing.

Yeah those are two different equations.

6/2(2+1) is 3x(3)

Is 9

But 6/[2(2+1)] is

6/[2(3)] is

6/[2x(3)] is

6/[6] is

6/6 is

1

Yeah those are two different equations.

6/2(2+1) is 3x(3)

Is 9

But 6/[2(2+1)] is

6/[2(3)] is

6/[2x(3)] is

6/[6] is

6/6 is

1

Inger67

Oct 11th, 2012, 09:29 PM

LOL. It's really not that serious. You get 1 or 9 depending on how you interpret the syntax of the question. Choosing 1 as the answer is not indicative of poor math skills. It's only indicative of not knowing the accepted order of operations.

Last time I checked order of operations is a part of math skills? :shrug:

Last time I checked order of operations is a part of math skills? :shrug:

Wigglytuff

Oct 11th, 2012, 09:39 PM

I think people are making more complicated than it needs to be. If you look at the topic title it is clearly NOT a fraction.

Patrick345

Oct 11th, 2012, 10:07 PM

It is a math problem that appeared on facebook some time ago, some people argue to death and defriend each other arguing which is the correct answer for it, 9 or 1. Scandalous.

This. It almost led to WW III. :lol::lol:

This. It almost led to WW III. :lol::lol:

Novichok

Oct 11th, 2012, 11:11 PM

Last time I checked order of operations is a part of math skills? :shrug:

No, it's a useful piece of knowledge to help one utilize their math skills. But it's completely arbitrary (in this instance).

No, it's a useful piece of knowledge to help one utilize their math skills. But it's completely arbitrary (in this instance).

égalité

Oct 12th, 2012, 12:43 AM

No, it's a useful piece of knowledge to help one utilize their math skills. But it's completely arbitrary (in this instance).

Order of operations is arbitrary in that it's contrived, but it's contrived in such a way so as to make it the best way of writing arithmetic. By "best" I mean that it leaves no room for ambiguity, and anyone who thinks the answer is 1 is missing that point.

Order of operations is arbitrary in that it's contrived, but it's contrived in such a way so as to make it the best way of writing arithmetic. By "best" I mean that it leaves no room for ambiguity, and anyone who thinks the answer is 1 is missing that point.

Novichok

Oct 12th, 2012, 01:06 AM

Order of operations is arbitrary in that it's contrived, but it's contrived in such a way so as to make it the best way of writing arithmetic. By "best" I mean that it leaves no room for ambiguity, and anyone who thinks the answer is 1 is missing that point.

Yes, the rules get rid of the ambiguity. But is there a non-arbitrary way to justify: 6÷2（2+1） as (6÷2)(2+1) rather than 6÷(2(2+1))? I don't see how one is necessarily better than the other. I also don't see how interpreting the question as 6÷(2(2+1)) is indicative of poor math skills. More plausible, it's indicative of being ignorant of convention.

Yes, the rules get rid of the ambiguity. But is there a non-arbitrary way to justify: 6÷2（2+1） as (6÷2)(2+1) rather than 6÷(2(2+1))? I don't see how one is necessarily better than the other. I also don't see how interpreting the question as 6÷(2(2+1)) is indicative of poor math skills. More plausible, it's indicative of being ignorant of convention.

Cajka

Oct 12th, 2012, 01:58 AM

That's subjective. IMO, it's visually clearer without an extra set of brackets :lol:

Perhaps. But if there were brackets, those poor people probably wouldn't defriend each other on fb. :tears:

Perhaps. But if there were brackets, those poor people probably wouldn't defriend each other on fb. :tears:

égalité

Oct 12th, 2012, 02:48 AM

Yes, the rules get rid of the ambiguity. But is there a non-arbitrary way to justify: 6÷2（2+1） as (6÷2)(2+1) rather than 6÷(2(2+1))? I don't see how one is necessarily better than the other. I also don't see how interpreting the question as 6÷(2(2+1)) is indicative of poor math skills. More plausible, it's indicative of being ignorant of convention.

Of course there is. Mathematicians don't establish rules like these unless they're essential. Here's why it matters for the problem in this thread:

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d.

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

Of course there is. Mathematicians don't establish rules like these unless they're essential. Here's why it matters for the problem in this thread:

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d.

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

Hurley

Oct 12th, 2012, 03:02 AM

Of course there is. Mathematicians don't establish rules like these unless they're essential. Here's why it matters for the problem in this thread:

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d.

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

God, shit like this makes me want to take all of my pants off.

http://i1112.photobucket.com/albums/k494/tbenner12/blanche.gif

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d.

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

God, shit like this makes me want to take all of my pants off.

http://i1112.photobucket.com/albums/k494/tbenner12/blanche.gif

égalité

Oct 12th, 2012, 03:06 AM

gifs like that make me want to take my pants off as well :hearts:

Hurley

Oct 12th, 2012, 03:09 AM

Blanche Devereaux is a handsome woman.

Her answer is 9 too.

Her answer is 9 too.

égalité

Oct 12th, 2012, 03:12 AM

Rose's answer is 12 :oh:

Hurley

Oct 12th, 2012, 03:13 AM

http://media.tumblr.com/tumblr_lfvt66ej9v1qc03he.gif

young_gunner913

Oct 12th, 2012, 03:14 AM

http://media.tumblr.com/tumblr_lfvt66ej9v1qc03he.gif

:bigcry: GOAT moment.

:bigcry: GOAT moment.

Hurley

Oct 12th, 2012, 03:15 AM

:bigcry: GOAT moment.

I don't remember this one at all. :help:

I don't remember this one at all. :help:

young_gunner913

Oct 12th, 2012, 03:19 AM

I don't remember this one at all. :help:

It's the one where they get robbed and Blanche thinks they stole her mother's jewelry and she thought she hid it in the flour and comes back from the kitchen with flour all over her and sobs:

"They got my mama's jewels!"

"Well I see they didn't get your cocaine." - Dorothy

"Oh my God! Blanche has cocaine?!" - Rose

It's the one where they get robbed and Blanche thinks they stole her mother's jewelry and she thought she hid it in the flour and comes back from the kitchen with flour all over her and sobs:

"They got my mama's jewels!"

"Well I see they didn't get your cocaine." - Dorothy

"Oh my God! Blanche has cocaine?!" - Rose

égalité

Oct 12th, 2012, 03:22 AM

Dorothy: Uh, tell me, Rose, um... Ah-ha ha ha!... Did they ever shoot a herring out of a cannon?

Rose: Only once. But they shot him into a tree. After that no other herring would do it.

When I heard this my life was over.

Rose: Only once. But they shot him into a tree. After that no other herring would do it.

When I heard this my life was over.

young_gunner913

Oct 12th, 2012, 03:26 AM

When I heard this my life was over.

:crying2:

iRrGCWUuQbM

:07 - :12 KILLS me.

:crying2:

iRrGCWUuQbM

:07 - :12 KILLS me.

Hurley

Oct 12th, 2012, 03:27 AM

When I heard this my life was over.

That was so well-delivered. Dorothy laughing in the middle, her and Blanche cracking up and clomping their feet after Rose responded totally stone-faced. NEO-REALISM.

That was so well-delivered. Dorothy laughing in the middle, her and Blanche cracking up and clomping their feet after Rose responded totally stone-faced. NEO-REALISM.

Hurley

Oct 12th, 2012, 03:28 AM

:crying2:

iRrGCWUuQbM

:07 - :12 KILLS me.

BEST. EPISODE. EVER.

iRrGCWUuQbM

:07 - :12 KILLS me.

BEST. EPISODE. EVER.

LeRoy.

Oct 12th, 2012, 03:30 AM

^ Get a room and stop spamming this thread. :yawn:

9. If you answered 1, I hope its because you never took Algebra in school.

9. If you answered 1, I hope its because you never took Algebra in school.

Hurley

Oct 12th, 2012, 03:34 AM

Apropos of nothing, if you don't know the difference between "its" and "it's," I hope it's because you're functionally retarded and no one likes you.

jameshazza

Oct 12th, 2012, 03:36 AM

BODMAS.

Division comes before Multiplication.

So 9.

Division comes before Multiplication.

So 9.

Hurley

Oct 12th, 2012, 03:38 AM

Division comes before Multiplication

Sigh. Read the thread. :facepalm:

You got lucky this time because the division sign comes first left-to-right. You would have been dead wrong had those signs been reversed.

Sigh. Read the thread. :facepalm:

You got lucky this time because the division sign comes first left-to-right. You would have been dead wrong had those signs been reversed.

LeRoy.

Oct 12th, 2012, 03:40 AM

Apropos of nothing, if you don't know the difference between "its" and "it's," I hope it's because you're functionally retarded and no one likes you.

Oooh a racist grammar police? Go to Becca's board and bitch about me with your other 5 personalities.

Oooh a racist grammar police? Go to Becca's board and bitch about me with your other 5 personalities.

jameshazza

Oct 12th, 2012, 03:41 AM

Sigh. Read the thread. :facepalm:

You got lucky this time because the division sign comes first left-to-right. You would have been dead wrong had those signs been reversed.

Ain't nobody got time for that.

You got lucky this time because the division sign comes first left-to-right. You would have been dead wrong had those signs been reversed.

Ain't nobody got time for that.

Hurley

Oct 12th, 2012, 03:42 AM

Ain't nobody got time for that.

Ah, touché.

Ah, touché.

Novichok

Oct 12th, 2012, 03:44 AM

Of course there is. Mathematicians don't establish rules like these unless they're essential. Here's why it matters for the problem in this thread:

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d.

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

I'm not sure if I'm understand where you're getting at. Help me if you have time. :tape:

You supplied semantics for the expression and then by associativity you showed that (ab/c)d=a(b/cd). Then you gave an interpretation to these two (equal by associativity) expressions. And since (abd)/c is not equal to ab/(cd), this shows that multiplication becomes non-associative.

But why would we expect (abd)/c and ab/(cd) to be equal? You showed that they were equal by associativity based on the first interpretation that you provided. Then you applied another interpretation ("interpreting...in the same manner that is used to achieve an answer of 1"). Is the identity relation still true if the semantics change? Shouldn't (ab/c)d interpreted as ((ab)/c)d if we interpret in the same manner that is used to achieve an answer of 1?

You're probably right; It's not at all obvious that it's arbitrary in the way that I meant it.

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d.

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

I'm not sure if I'm understand where you're getting at. Help me if you have time. :tape:

You supplied semantics for the expression and then by associativity you showed that (ab/c)d=a(b/cd). Then you gave an interpretation to these two (equal by associativity) expressions. And since (abd)/c is not equal to ab/(cd), this shows that multiplication becomes non-associative.

But why would we expect (abd)/c and ab/(cd) to be equal? You showed that they were equal by associativity based on the first interpretation that you provided. Then you applied another interpretation ("interpreting...in the same manner that is used to achieve an answer of 1"). Is the identity relation still true if the semantics change? Shouldn't (ab/c)d interpreted as ((ab)/c)d if we interpret in the same manner that is used to achieve an answer of 1?

You're probably right; It's not at all obvious that it's arbitrary in the way that I meant it.

égalité

Oct 12th, 2012, 03:55 AM

I'm not sure if I'm understand where you're getting at. Help me if you have time. :tape:

You supplied semantics for the expression and then you by associativity you showed that (ab/c)d=a(b/cd). Then you gave an interpretation to these two (equal by associativity) expressions. And since (abd)/c is not equal to ab/(cd), this shows that multiplication becomes non-associative.

But why would we expect (abd)/c and ab/(cd) to be equal? You showed that they were equal by associativity based on the first interpretation that you provided. Then you applied another interpretation ("interpreting...in the same manner that is used to achieve an answer of 1"). Is the identity relation still true if the semantics change? Shouldn't (ab/c)d interpreted as ((ab)/c)d if we interpret in the same manner that is used to achieve an answer of 1?

You're probably right; It's not at all obvious that it's arbitrary in the way that I meant it.

I didn't show they were equal. I said that they SHOULD be equal based on the fact that multiplication of real numbers is associative. But if you rearrange the terms on the left hand side, you get (ab/c)d=d(ab/c)=(dab)/c=(abd)/c (justified by the fact that multiplication is also commutative). If you rearrange the terms on the right hand side, according to the flawed order of operations being used in this thread, you get a(b/cd)=a(b/(cd))=(ab)/(cd). So the two things that were supposed to be equal are not equal. You can certainly find numbers that make (abd)/c=(ab)/(cd) true, like if they're all equal to 1, for example, but this is ALWAYS supposed to be true, not just sometimes.

It's a proof by contradiction. :D Assume that order of operations works in the "6/2(2+1)=1" way, derive a mathematical absurdity, conclude that your assumption was wrong.

You supplied semantics for the expression and then you by associativity you showed that (ab/c)d=a(b/cd). Then you gave an interpretation to these two (equal by associativity) expressions. And since (abd)/c is not equal to ab/(cd), this shows that multiplication becomes non-associative.

But why would we expect (abd)/c and ab/(cd) to be equal? You showed that they were equal by associativity based on the first interpretation that you provided. Then you applied another interpretation ("interpreting...in the same manner that is used to achieve an answer of 1"). Is the identity relation still true if the semantics change? Shouldn't (ab/c)d interpreted as ((ab)/c)d if we interpret in the same manner that is used to achieve an answer of 1?

You're probably right; It's not at all obvious that it's arbitrary in the way that I meant it.

I didn't show they were equal. I said that they SHOULD be equal based on the fact that multiplication of real numbers is associative. But if you rearrange the terms on the left hand side, you get (ab/c)d=d(ab/c)=(dab)/c=(abd)/c (justified by the fact that multiplication is also commutative). If you rearrange the terms on the right hand side, according to the flawed order of operations being used in this thread, you get a(b/cd)=a(b/(cd))=(ab)/(cd). So the two things that were supposed to be equal are not equal. You can certainly find numbers that make (abd)/c=(ab)/(cd) true, like if they're all equal to 1, for example, but this is ALWAYS supposed to be true, not just sometimes.

It's a proof by contradiction. :D Assume that order of operations works in the "6/2(2+1)=1" way, derive a mathematical absurdity, conclude that your assumption was wrong.

Otlichno

Oct 12th, 2012, 04:07 AM

If it's 9, then there should be parentheses around the 6/2.

In this case it is 6/(2(2+1)), which is obviously 1. :shrug:

In this case it is 6/(2(2+1)), which is obviously 1. :shrug:

Pvt. Kovalenko

Oct 12th, 2012, 04:09 AM

It's 9 :shrug:

6/2 * (2+1) = 3 * (3) = 9

or

6/2 * (2+1) = 3 * (2+1) = (distributing the 3) = (3*2 + 3*1) = (6+3) = 9

Now, IF you're a calculator, then, probably this equation will end up in 1

6/2 * (2+1) = 3 * (3) = 9

or

6/2 * (2+1) = 3 * (2+1) = (distributing the 3) = (3*2 + 3*1) = (6+3) = 9

Now, IF you're a calculator, then, probably this equation will end up in 1

Otlichno

Oct 12th, 2012, 04:19 AM

It's 9 :shrug:

6/2 * (2+1) = 3 * (3) = 9

or

6/2 * (2+1) = 3 * (2+1) = (distributing the 3) = (3*2 + 3*1) = (6+3) = 9

Now, IF you're a calculator, then, probably this equation will end up in 1

But you're using the wrong equation.

6/2(2+1) not 6/2 * (2+1)

This is all about division, not order of operations.

6/2 * (2+1) = 3 * (3) = 9

or

6/2 * (2+1) = 3 * (2+1) = (distributing the 3) = (3*2 + 3*1) = (6+3) = 9

Now, IF you're a calculator, then, probably this equation will end up in 1

But you're using the wrong equation.

6/2(2+1) not 6/2 * (2+1)

This is all about division, not order of operations.

égalité

Oct 12th, 2012, 04:21 AM

But you're using the wrong equation.

6/2(2+1) not 6/2 * (2+1)

This is all about division, not order of operations.

These are the same. I actually provided a proof of this on the previous page. :lol:

6/2(2+1) not 6/2 * (2+1)

This is all about division, not order of operations.

These are the same. I actually provided a proof of this on the previous page. :lol:

Novichok

Oct 12th, 2012, 04:22 AM

I didn't show they were equal. I said that they SHOULD be equal based on the fact that multiplication of real numbers is associative. But if you rearrange the terms on the left hand side, you get (ab/c)d=d(ab/c)=(dab)/c=(abd)/c (justified by the fact that multiplication is also commutative). If you rearrange the terms on the right hand side, according to the flawed order of operations being used in this thread, you get a(b/cd)=a(b/(cd))=(ab)/(cd). So the two things that were supposed to be equal are not equal. You can certainly find numbers that make (abd)/c=(ab)/(cd) true, like if they're all equal to 1, for example, but this is ALWAYS supposed to be true, not just sometimes.

It's a proof by contradiction. :D Assume that order of operations works in the "6/2(2+1)=1" way, derive a mathematical absurdity, conclude that your assumption was wrong.

Ok. :D I'm just confused. :o

Isn't the identity relation (ab/c)d=d(ab/c)=(dab)/c=(abd)/c dependent on the semantics?

If we change the semantics, why should we expect the identity relation to be true?

It's a proof by contradiction. :D Assume that order of operations works in the "6/2(2+1)=1" way, derive a mathematical absurdity, conclude that your assumption was wrong.

Ok. :D I'm just confused. :o

Isn't the identity relation (ab/c)d=d(ab/c)=(dab)/c=(abd)/c dependent on the semantics?

If we change the semantics, why should we expect the identity relation to be true?

égalité

Oct 12th, 2012, 04:29 AM

Ok. :D I'm just confused. :o

Isn't the identity relation (ab/c)d=d(ab/c)=(dab)/c=(abd)/c dependent on the semantics?

If we change the semantics, why should we expect the identity relation to be true?

No, all the steps there are true in both versions of order of operations in question.

I'm not totally sure what you mean by "dependent on the semantics!"

Isn't the identity relation (ab/c)d=d(ab/c)=(dab)/c=(abd)/c dependent on the semantics?

If we change the semantics, why should we expect the identity relation to be true?

No, all the steps there are true in both versions of order of operations in question.

I'm not totally sure what you mean by "dependent on the semantics!"

Otlichno

Oct 12th, 2012, 04:33 AM

These are the same. I actually provided a proof of this on the previous page. :lol:

Oh, okay. :o :smash:

Thanks for that. :D

Oh, okay. :o :smash:

Thanks for that. :D

PhilePhile

Oct 12th, 2012, 04:37 AM

Of course there is. Mathematicians don't establish rules like these unless they're essential. Here's why it matters for the problem in this thread:

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d. A x (B/C) x D

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

Fixed.

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d. A x (B/C) x D

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

Fixed.

égalité

Oct 12th, 2012, 04:38 AM

Yeah, that's what I said. :spit:

Novichok

Oct 12th, 2012, 04:43 AM

No, all the steps there are true in both versions of order of operations in question.

I'm not totally sure what you mean by "dependent on the semantics!"

Nevermind. I just realized where I was mistaken. Your proof is flawless. :angel:

I'm not totally sure what you mean by "dependent on the semantics!"

Nevermind. I just realized where I was mistaken. Your proof is flawless. :angel:

Sam L

Oct 12th, 2012, 10:16 AM

I said 9 but I can see how someone can get 1.

Can someone tell me why it's not 1 again?

Thanks. I can't be bothered reading 13 pages of this.

Can someone tell me why it's not 1 again?

Thanks. I can't be bothered reading 13 pages of this.

Sam L

Oct 12th, 2012, 10:35 AM

I thought the best way to prove that 9 was right was this:

2(2+1) = 2 x (2+1) just like 2(3) = 2 x 3. Right?

Therefore, it is 6 ÷ 2 x 3 = 9

To say the answer is 1 would mean the question is written 6÷(2(2+1)). Where do the extra brackets come from? Where?

2(2+1) = 2 x (2+1) just like 2(3) = 2 x 3. Right?

Therefore, it is 6 ÷ 2 x 3 = 9

To say the answer is 1 would mean the question is written 6÷(2(2+1)). Where do the extra brackets come from? Where?

Sam L

Oct 12th, 2012, 11:31 AM

OMG there are Youtube clips on this:

JPe1aBW_YCg

JPe1aBW_YCg

Super Dave

Oct 12th, 2012, 02:11 PM

http://www.myfacewhen.net/uploads/316-make-it-stop.jpg

JJ Expres

Oct 12th, 2012, 02:24 PM

http://www.wolframalpha.com/input/?i=6%2F2%282%2B1%29

just use Wolfram, however you type it with or without * it gives the same answer, it's 9...

just use Wolfram, however you type it with or without * it gives the same answer, it's 9...

zigga

Oct 12th, 2012, 08:30 PM

This is too easy :)

Here is one more interesting math problem: how much is 0.9 periodic (so 0.9999...)?

I just find it amusing that it can be proven that it equals 1 :)

Here is one more interesting math problem: how much is 0.9 periodic (so 0.9999...)?

I just find it amusing that it can be proven that it equals 1 :)

pov

Oct 12th, 2012, 08:50 PM

hmmm . my original posted answer was 1. Based on multiplication before division.

Rewriting though 6x1/2x3 =9 so :shrug:

However according to Wolfram Alpha "implied multiplication without parentheses precedes division" so that makes the answer 1.

The overall issue here is a poorly written problem but given its parameters the correct reading, by the accepted standards I know and those I've re-checked, is 6÷ (2（2+1）)

Rewriting though 6x1/2x3 =9 so :shrug:

However according to Wolfram Alpha "implied multiplication without parentheses precedes division" so that makes the answer 1.

The overall issue here is a poorly written problem but given its parameters the correct reading, by the accepted standards I know and those I've re-checked, is 6÷ (2（2+1）)

pov

Oct 12th, 2012, 09:03 PM

one more source:

American Mathematical Society (AMS)

"multiplication indicated by juxtaposition is carried out before

division." Thus, in general, for any variables a, b and c, we would

have a/bc = a/(bc) (assuming, of course, that b and c are nonzero).

American Mathematical Society (AMS)

"multiplication indicated by juxtaposition is carried out before

division." Thus, in general, for any variables a, b and c, we would

have a/bc = a/(bc) (assuming, of course, that b and c are nonzero).

Nicolás89

Oct 12th, 2012, 09:05 PM

Lol It is 9.

delicatecutter

Oct 12th, 2012, 09:16 PM

Jesus Christ on a cross, egalite is a mathematics professional. Take his word for it.

pov

Oct 12th, 2012, 11:15 PM

Of course there is. Mathematicians don't establish rules like these unless they're essential. Here's why it matters for the problem in this thread:

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d.

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

Perhaps you should contact the AMS (see above) since according to them a/bc is to be interpreted as a/(bc)

Either way, your "proof" is flawed. There is nothing in the problem (or either answer) that countermands the associative property. I'll let you figure it out. ;)

Multiplication of real numbers is associative: (ab)c=a(bc).

Take four nonzero integers a,b,c,d and multiply: a times b/c times d.

By associativity, (ab/c)d=a(b/cd).

But interpreting these expressions in the same manner that is used to achieve an answer of 1, we get (ab/c)d=(abd)/c, and a(b/cd)=ab/(cd). These are not equal. If we mess around with order of operations, multiplication becomes non-associative. :lol:

Perhaps you should contact the AMS (see above) since according to them a/bc is to be interpreted as a/(bc)

Either way, your "proof" is flawed. There is nothing in the problem (or either answer) that countermands the associative property. I'll let you figure it out. ;)

JJ Expres

Oct 12th, 2012, 11:24 PM

This is too easy :)

Here is one more interesting math problem: how much is 0.9 periodic (so 0.9999...)?

I just find it amusing that it can be proven that it equals 1 :)

x=0.99999999999999999999999999999....

10x=9.99999999999999999999999999999......

10x-x=9

9x=9

x=1

what's amusing about that :S

Here is one more interesting math problem: how much is 0.9 periodic (so 0.9999...)?

I just find it amusing that it can be proven that it equals 1 :)

x=0.99999999999999999999999999999....

10x=9.99999999999999999999999999999......

10x-x=9

9x=9

x=1

what's amusing about that :S

Sammo

Oct 12th, 2012, 11:38 PM

It's 9. Trust me, I'm a 1st grade Chemical Engineering student

Miracle Worker

Oct 12th, 2012, 11:59 PM

WTF is （? If （ = () the result should be 9.

6:2x(2+1)= 3x3=9

6:2x(2+1)= 3x3=9

Miracle Worker

Oct 13th, 2012, 12:06 AM

Perhaps you should contact the AMS (see above) since according to them a/bc is to be interpreted as a/(bc)

Either way, your "proof" is flawed. There is nothing in the problem (or either answer) that countermands the associative property. I'll let you figure it out. ;)

:eek: It seems maths changed a lot in last few years :eek:

a/bc it's not the same as a/(bc).

Either way, your "proof" is flawed. There is nothing in the problem (or either answer) that countermands the associative property. I'll let you figure it out. ;)

:eek: It seems maths changed a lot in last few years :eek:

a/bc it's not the same as a/(bc).

SilverPersian

Oct 13th, 2012, 12:06 AM

Multiplication indicated by juxtaposition is carried out before

division - American Mathematical Society (AMS)

This is the crux of the problem resolved then :shrug:

division - American Mathematical Society (AMS)

This is the crux of the problem resolved then :shrug:

Cajka

Oct 13th, 2012, 12:37 AM

:eek: It seems maths changed a lot in last few years :eek:

a/bc it's not the same as a/(bc).

It is. But what we have here is a/b(c), not a/(bc)

a/bc it's not the same as a/(bc).

It is. But what we have here is a/b(c), not a/(bc)

Wigglytuff

Oct 13th, 2012, 12:50 AM

hmmm . my original posted answer was 1. Based on multiplication before division.

Rewriting though 6x1/2x3 =9 so :shrug:

However according to Wolfram Alpha "implied multiplication without parentheses precedes division" so that makes the answer 1.

The overall issue here is a poorly written problem but given its parameters the correct reading, by the accepted standards I know and those I've re-checked, is 6÷ (2（2+1）)

You. Can . Not. Add. An. Extra. Parenthesis. The. Equation. Is. Not. A. Fraction.

Even writing it out as simply as that you still won't get it.

Six divided by two multiplied by (two plus one)

Six divided by two multiplied by (three)

Three multiplied by three is nine.

It. Is. Not. Rocket. Science.

$9 says you still don't get it.

Rewriting though 6x1/2x3 =9 so :shrug:

However according to Wolfram Alpha "implied multiplication without parentheses precedes division" so that makes the answer 1.

The overall issue here is a poorly written problem but given its parameters the correct reading, by the accepted standards I know and those I've re-checked, is 6÷ (2（2+1）)

You. Can . Not. Add. An. Extra. Parenthesis. The. Equation. Is. Not. A. Fraction.

Even writing it out as simply as that you still won't get it.

Six divided by two multiplied by (two plus one)

Six divided by two multiplied by (three)

Three multiplied by three is nine.

It. Is. Not. Rocket. Science.

$9 says you still don't get it.

Wigglytuff

Oct 13th, 2012, 12:52 AM

It is. But what we have here is a/b(c), not a/(bc)

You are correct.

You are correct.

Wigglytuff

Oct 13th, 2012, 01:03 AM

You. Can . Not. Add. An. Extra. Parenthesis. The. Equation. Is. Not. A. Fraction.

SilverPersian

Oct 13th, 2012, 01:03 AM

You. Can . Not. Add. An. Extra. Parenthesis. The. Equation. Is. Not. A. Fraction.

Even writing it out as simply as that you still won't get it.

Six divided by two multiplied by (two plus one)

Six divided by two multiplied by (three)

Three multiplied by three is nine.

It. Is. Not. Rocket. Science.

$9 says you still don't get it.

Firstly, calm down.

Secondly, you didn't deal with the crux of his post. He provided two quotes from mathematical authorities suggesting that implied multiplication trumps division in the order of operations. If you have authoritative source material contradicting this then I would be interested to see it, but until then I think I'll take the opinion of the American Mathematical Society :shrug:

Even writing it out as simply as that you still won't get it.

Six divided by two multiplied by (two plus one)

Six divided by two multiplied by (three)

Three multiplied by three is nine.

It. Is. Not. Rocket. Science.

$9 says you still don't get it.

Firstly, calm down.

Secondly, you didn't deal with the crux of his post. He provided two quotes from mathematical authorities suggesting that implied multiplication trumps division in the order of operations. If you have authoritative source material contradicting this then I would be interested to see it, but until then I think I'll take the opinion of the American Mathematical Society :shrug:

pov

Oct 13th, 2012, 01:07 AM

:eek: It seems maths changed a lot in last few years :eek:

a/bc it's not the same as a/(bc).

The whole point and the reason this "problem" (more like a brain-teaser) was first posted to the net is that a/bc is ambiguous. It can be taken to mean a/(bc) or (a/b)c. Both are correct and the case can be made for either. However the "standard setting bodies" I checked suggest read a/bc as a/(bc) which is the way I know it.

a/bc it's not the same as a/(bc).

The whole point and the reason this "problem" (more like a brain-teaser) was first posted to the net is that a/bc is ambiguous. It can be taken to mean a/(bc) or (a/b)c. Both are correct and the case can be made for either. However the "standard setting bodies" I checked suggest read a/bc as a/(bc) which is the way I know it.

Wigglytuff

Oct 13th, 2012, 01:12 AM

It is not ambiguous the topic title clearly uses the division symbol. Simple as that.

Sam L

Oct 13th, 2012, 01:14 AM

The whole point and the reason this "problem" (more like a brain-teaser) was first posted to the net is that a/bc is ambiguous. It can be taken to mean a/(bc) or (a/b)c. Both are correct and the case can be made for either. However the "standard setting bodies" I checked suggest read a/bc as a/(bc) which is the way I know it.

Yes but that is a fraction. This is a division sign.

You're talking about 6/2(2+1) which is not the same as 6÷2（2+1）.

Yes but that is a fraction. This is a division sign.

You're talking about 6/2(2+1) which is not the same as 6÷2（2+1）.

NoppaNoppa

Oct 13th, 2012, 01:16 AM

Looks 9

SilverPersian

Oct 13th, 2012, 01:17 AM

~Multiplication indicated by juxtaposition is carried out before division~

It doesn't seem to need to be a fraction.

It doesn't seem to need to be a fraction.

Wigglytuff

Oct 13th, 2012, 01:20 AM

Firstly, calm down.

Secondly, you didn't deal with the crux of his post. He provided two quotes from mathematical authorities suggesting that implied multiplication trumps division in the order of operations. If you have authoritative source material contradicting this then I would be interested to see it, but until then I think I'll take the opinion of the American Mathematical Society :shrug:

I'm calm, I am trying to use as small word as I can without any ambiguity so he can understand.

Multiplication DOES trump division, but not parenthesis. And NOT the order of LEFT TO RIGHT. You can't add extra parenthesis or ignore existing ones.

That's why I had to spell it out in small words.

P

E

M

D

A

S

You have to do the one existing parenthesis first.

Secondly, you didn't deal with the crux of his post. He provided two quotes from mathematical authorities suggesting that implied multiplication trumps division in the order of operations. If you have authoritative source material contradicting this then I would be interested to see it, but until then I think I'll take the opinion of the American Mathematical Society :shrug:

I'm calm, I am trying to use as small word as I can without any ambiguity so he can understand.

Multiplication DOES trump division, but not parenthesis. And NOT the order of LEFT TO RIGHT. You can't add extra parenthesis or ignore existing ones.

That's why I had to spell it out in small words.

P

E

M

D

A

S

You have to do the one existing parenthesis first.

Wigglytuff

Oct 13th, 2012, 01:22 AM

Yes but that is a fraction. This is a division sign.

You're talking about 6/2(2+1) which is not the same as 6÷2（2+1）.

You are correct.

I tried. I tried to write it out, count it out and even use words. I still could not explain it.

Even google says its 9

http://www.google.com/search?q=6÷2（2%2B1）&ie=UTF-8&oe=UTF-8&hl=en&client=safari

You're talking about 6/2(2+1) which is not the same as 6÷2（2+1）.

You are correct.

I tried. I tried to write it out, count it out and even use words. I still could not explain it.

Even google says its 9

http://www.google.com/search?q=6÷2（2%2B1）&ie=UTF-8&oe=UTF-8&hl=en&client=safari

Hurley

Oct 13th, 2012, 01:24 AM

Stop being uppity when you're wrong about some stuff.

Multiplication DOES trump division

No it doesn't. M and D (like A and S) are done simultaneously left to right.

This has only been said 7x5÷1+10000^2 times in the thread if you bothered to read and comprehend it.

Multiplication DOES trump division

No it doesn't. M and D (like A and S) are done simultaneously left to right.

This has only been said 7x5÷1+10000^2 times in the thread if you bothered to read and comprehend it.

Hurley

Oct 13th, 2012, 01:28 AM

Secondly, you didn't deal with the crux of his post.

Right, she didn't. If she had done it as she said, her answer would NOT be 9. It would be 1. :facepalm:

But the answer is 9.

Right, she didn't. If she had done it as she said, her answer would NOT be 9. It would be 1. :facepalm:

But the answer is 9.

pov

Oct 13th, 2012, 01:28 AM

You. Can . Not. Add. An. Extra. Parenthesis. The. Equation. Is. Not. A. Fraction.

Even writing it out as simply as that you still won't get it.

Six divided by two multiplied by (two plus one)

Six divided by two multiplied by (three)

Three multiplied by three is nine.

It. Is. Not. Rocket. Science.

$9 says you still don't get it.

- The forward slash is used as a divisor. Technically any division that has a modulus ( and some claim that even without) is a fraction.

- The division comes before multiplication. Generally they are given equal precedence but because of the occasionally grey-area guidelines are offered on which to do first and when.

- Lastly one thing that you said that I think most here will laugh at you about is "You. Can . Not. Add. An. Extra. Parenthesis." (I won't even go on about the misspelling) Adding parentheses for clarification is standard procedure.

Even writing it out as simply as that you still won't get it.

Six divided by two multiplied by (two plus one)

Six divided by two multiplied by (three)

Three multiplied by three is nine.

It. Is. Not. Rocket. Science.

$9 says you still don't get it.

- The forward slash is used as a divisor. Technically any division that has a modulus ( and some claim that even without) is a fraction.

- The division comes before multiplication. Generally they are given equal precedence but because of the occasionally grey-area guidelines are offered on which to do first and when.

- Lastly one thing that you said that I think most here will laugh at you about is "You. Can . Not. Add. An. Extra. Parenthesis." (I won't even go on about the misspelling) Adding parentheses for clarification is standard procedure.

Wigglytuff

Oct 13th, 2012, 01:30 AM

Stop being uppity when you're wrong about some stuff.

No it doesn't. M and D (like A and S) are done simultaneously left to right.

This has only been said 7x5÷1+10000^2 times in the thread if you bothered to read and comprehend it.

Errr did you read my edited post (I made the mistake of taking it for granted that everyone knew you had to go left to right first. ). And nothing should be taken for granted here.

No it doesn't. M and D (like A and S) are done simultaneously left to right.

This has only been said 7x5÷1+10000^2 times in the thread if you bothered to read and comprehend it.

Errr did you read my edited post (I made the mistake of taking it for granted that everyone knew you had to go left to right first. ). And nothing should be taken for granted here.

Wigglytuff

Oct 13th, 2012, 01:33 AM

Right, she didn't. If she had done it as she said, her answer would NOT be 9. It would be 1. :facepalm:

But the answer is 9.

Try again bob. Again, I made the mistake, and apparently a big one, of assuming folks took it got granted that you have to go from left to right first, clearly I gave you too much credit and for that I am sorry.

But the answer is 9.

Try again bob. Again, I made the mistake, and apparently a big one, of assuming folks took it got granted that you have to go from left to right first, clearly I gave you too much credit and for that I am sorry.

Hurley

Oct 13th, 2012, 01:33 AM

Errr did you read my edited post (I made the mistake of taking it for granted that everyone knew you had to go left to right first. ). And nothing should be taken for granted here.

Yes. You said (as I quoted) multiplication trumps division. If that is true then you would have done:

6÷2（2+1）--> 6÷2（3) --> 6÷6 --> 1

...because you would have done 2 times 3 before 6÷2. :shrug:

Yes. You said (as I quoted) multiplication trumps division. If that is true then you would have done:

6÷2（2+1）--> 6÷2（3) --> 6÷6 --> 1

...because you would have done 2 times 3 before 6÷2. :shrug:

Hurley

Oct 13th, 2012, 01:35 AM

Multiplication DOES trump division, but not parenthesis. And NOT the order of LEFT TO RIGHT.

Ah, now I see this.

So what you actually mean to say is:

Multiplication in fact does NOT trump division. EVER. They are done simultaneously.

So you mean the opposite of what you say but that's cool.

Ah, now I see this.

So what you actually mean to say is:

Multiplication in fact does NOT trump division. EVER. They are done simultaneously.

So you mean the opposite of what you say but that's cool.

Wigglytuff

Oct 13th, 2012, 01:40 AM

Ah, now I see this.

So what you actually mean to say is:

Multiplication in fact does NOT trump division. EVER. They are done simultaneously.

So you mean the opposite of what you say but that's cool.

I does, if and only if, all other rules apply and it does not trump left to right.

So what you actually mean to say is:

Multiplication in fact does NOT trump division. EVER. They are done simultaneously.

So you mean the opposite of what you say but that's cool.

I does, if and only if, all other rules apply and it does not trump left to right.

Wigglytuff

Oct 13th, 2012, 01:42 AM

Ah, now I see this.

So what you actually mean to say is:

Multiplication in fact does NOT trump division. EVER. They are done simultaneously.

So you mean the opposite of what you say but that's cool.

Check it out. You might learn something.

http://en.wikibooks.org/wiki/Arithmetic/Order_of_Operations

So what you actually mean to say is:

Multiplication in fact does NOT trump division. EVER. They are done simultaneously.

So you mean the opposite of what you say but that's cool.

Check it out. You might learn something.

http://en.wikibooks.org/wiki/Arithmetic/Order_of_Operations

Nicolás89

Oct 13th, 2012, 01:42 AM

BTW to everyone the juxtaposition rule is used in problems where we work with variables and since we are only working with numbers here, we use pemdas. The problem's syntaxis is just confusing because it is mixing literal operations with juxtaposition in a problem where there isn't a variable.

pov

Oct 13th, 2012, 01:46 AM

There Will Never not Be Left To Right Coming Into Play.

Yes there will. As it often does. ;)

Yes there will. As it often does. ;)

Hurley

Oct 13th, 2012, 01:48 AM

Yes there will. As it often does. ;)

Provide an example of this instance and I will retract everything except calling Wigglytuff dumb, because, I mean we've all been here for years and we all know she is.

Provide an example of this instance and I will retract everything except calling Wigglytuff dumb, because, I mean we've all been here for years and we all know she is.

pov

Oct 13th, 2012, 01:59 AM

For all those still reading this thread.

6/2(3)

Three possible criteria:

A - Since M and D have equal precedence - invoke left to right.

B - In this case division takes precedence over multiplication.

C - In this case multiplication takes precedence over division.

I'm saying C. The AMS is saying C. Others say A. And that's it.

To Wigglytuff,

All the insults will not change the fact that you specifically offer nothing to back up your viewpoint. Vehement assertions don't mean much other than show a desperation to be right. And now you may get in a few more choice phrases since the point has been made and those who still don't get it probably just don't want to.

6/2(3)

Three possible criteria:

A - Since M and D have equal precedence - invoke left to right.

B - In this case division takes precedence over multiplication.

C - In this case multiplication takes precedence over division.

I'm saying C. The AMS is saying C. Others say A. And that's it.

To Wigglytuff,

All the insults will not change the fact that you specifically offer nothing to back up your viewpoint. Vehement assertions don't mean much other than show a desperation to be right. And now you may get in a few more choice phrases since the point has been made and those who still don't get it probably just don't want to.

Hurley

Oct 13th, 2012, 02:03 AM

For all those still reading this thread.

6/2(3)

Three possible criteria:

A - Since M and D have equal precedence - invoke left to right.

B - In this case division takes precedence over multiplication.

C - In this case multiplication takes precedence over division.

I'm saying C. The AMS is saying C. Others say A. And that's it.

So...your example is not a specific example. Your example is "this is what I believe." OK, that's fine. But:

1. Well...we've been telling you otherwise. And thus

2. Left to right ALWAYS comes into play, ergo multiplication NEVER takes preference over division, and therefore Wigglytuff, again, doesn't know what she's talking about.

Someone wants to convince me otherwise, go right ahead. :shrug:

6/2(3)

Three possible criteria:

A - Since M and D have equal precedence - invoke left to right.

B - In this case division takes precedence over multiplication.

C - In this case multiplication takes precedence over division.

I'm saying C. The AMS is saying C. Others say A. And that's it.

So...your example is not a specific example. Your example is "this is what I believe." OK, that's fine. But:

1. Well...we've been telling you otherwise. And thus

2. Left to right ALWAYS comes into play, ergo multiplication NEVER takes preference over division, and therefore Wigglytuff, again, doesn't know what she's talking about.

Someone wants to convince me otherwise, go right ahead. :shrug:

Nicolás89

Oct 13th, 2012, 02:09 AM

For all those still reading this thread.

6/2(3)

Three possible criteria:

A - Since M and D have equal precedence - invoke left to right.

B - In this case division takes precedence over multiplication.

C - In this case multiplication takes precedence over division.

I'm saying C. The AMS is saying C. Others say A. And that's it.

Juxtaposition is used with variables, there's no variable here. :shrug:

Again, the problem's syntaxis is just wrong and confusing.

6/2(3)

Three possible criteria:

A - Since M and D have equal precedence - invoke left to right.

B - In this case division takes precedence over multiplication.

C - In this case multiplication takes precedence over division.

I'm saying C. The AMS is saying C. Others say A. And that's it.

Juxtaposition is used with variables, there's no variable here. :shrug:

Again, the problem's syntaxis is just wrong and confusing.

SilverPersian

Oct 13th, 2012, 02:13 AM

^But variables are just numbers in disguise, so shouldn't the same rules still apply? (Not a maths guy so I could be wrong about this).

Nicolás89

Oct 13th, 2012, 02:16 AM

By convention we say that 2(2+1) = 2 x (2+1) but that's not true because you can't infer a multiplication just with numbers otherwise two numbers put together like 10 could be read 1x0 and that's wrong. The parentheses only purpose is to separate the operations but when we calculate whats inside of them is useless, 2(3) isn't an operation 2 x 3 is. That's why I think the problem's syntaxis is wrong.

ampers&

Oct 13th, 2012, 02:32 AM

http://www.reactiongifs.com/wp-content/uploads/2012/09/cYuvmwqI2Tv5rxP.gif

The only correct answer to this question. You're welcome.

The only correct answer to this question. You're welcome.

delicatecutter

Oct 13th, 2012, 02:37 AM

Ryan, please close the thread! :sobbing:

Cajka

Oct 13th, 2012, 02:42 AM

Ryan, please close the thread! :sobbing:

He won't. Evil Ryan there somewhere laughing at us all. :sobbing:

He won't. Evil Ryan there somewhere laughing at us all. :sobbing:

Hurley

Oct 13th, 2012, 02:52 AM

This thread can't be closed until it approaches determining who is greater, Serena or Maria.

Vartan

Oct 13th, 2012, 02:56 AM

Maria.

Hurley

Oct 13th, 2012, 02:59 AM

Because M comes before S in the alphabet left to right? BUT DID YOU THINK ABOUT DIFFERENT ALPHABETS.

Does clay trump grass. What does the SEAMS say about it.

Does clay trump grass. What does the SEAMS say about it.

delicatecutter

Oct 13th, 2012, 03:06 AM

Well Vika is the greatest as the Majors are equal and we have to go in order starting with January.

Novichok

Oct 13th, 2012, 03:07 AM

Perhaps you should contact the AMS (see above) since according to them a/bc is to be interpreted as a/(bc)

Either way, your "proof" is flawed. There is nothing in the problem (or either answer) that countermands the associative property. I'll let you figure it out. ;)

Here is what I thought last night:

(ab/c)d = a(b/cd) by associativity (based on one interpretation of the expressions)

(ab/c)d=(abd)/c and a(b/cd)=ab/(cd)(based on another interpretation of the expressions). These are not equal so there's a mathematical absurdity.

His error is that he thinks that (ab/c)d = a(b/cd) (based on the 2nd interpretation). This isn't true. ((ab)/c)d does not equal a(b/(cd))

If we're consistent with the interpretations, the mathematical absurdity doesn't arise.

Either way, your "proof" is flawed. There is nothing in the problem (or either answer) that countermands the associative property. I'll let you figure it out. ;)

Here is what I thought last night:

(ab/c)d = a(b/cd) by associativity (based on one interpretation of the expressions)

(ab/c)d=(abd)/c and a(b/cd)=ab/(cd)(based on another interpretation of the expressions). These are not equal so there's a mathematical absurdity.

His error is that he thinks that (ab/c)d = a(b/cd) (based on the 2nd interpretation). This isn't true. ((ab)/c)d does not equal a(b/(cd))

If we're consistent with the interpretations, the mathematical absurdity doesn't arise.

Hurley

Oct 13th, 2012, 03:07 AM

I read somewhere else that Ivanovic > Vika though.

Hurley

Oct 13th, 2012, 03:08 AM

Wait, are people questioning egalite????

LOOOOOOOOOOOOOOOOL okay.

LOOOOOOOOOOOOOOOOL okay.

Novichok

Oct 13th, 2012, 03:14 AM

Wait, are people questioning egalite????

LOOOOOOOOOOOOOOOOL okay.

Yes, he's a professional mathematician. Yes, he knows more mathematics than anyone on this forum. But that doesn't entail that his argument is sound.

LOOOOOOOOOOOOOOOOL okay.

Yes, he's a professional mathematician. Yes, he knows more mathematics than anyone on this forum. But that doesn't entail that his argument is sound.

Nicolás89

Oct 13th, 2012, 03:42 AM

Here is what I thought last night:

(ab/c)d = a(b/cd) by associativity (based on one interpretation of the expressions)

(ab/c)d=(abd)/c and a(b/cd)=ab/(cd)(based on another interpretation of the expressions). These are not equal so there's a mathematical absurdity.

His error is that he thinks that (ab/c)d = a(b/cd) (based on the 2nd interpretation). This isn't true. ((ab)/c)d does not equal a(b/(cd))

If we're consistent with the interpretations, the mathematical absurdity doesn't arise.

Yep I thought the same as you.

Sent from my LG-E510g using VerticalSports.Com App

(ab/c)d = a(b/cd) by associativity (based on one interpretation of the expressions)

(ab/c)d=(abd)/c and a(b/cd)=ab/(cd)(based on another interpretation of the expressions). These are not equal so there's a mathematical absurdity.

His error is that he thinks that (ab/c)d = a(b/cd) (based on the 2nd interpretation). This isn't true. ((ab)/c)d does not equal a(b/(cd))

If we're consistent with the interpretations, the mathematical absurdity doesn't arise.

Yep I thought the same as you.

Sent from my LG-E510g using VerticalSports.Com App

PhilePhile

Oct 13th, 2012, 03:49 AM

There are different "fields" of mathematics. Here is one of them, dramatized, called game theory:

From the movie A beautiful Mind, "Adam Smith needs revision ... that's the only way we all get laid."

CemLiSI5ox8

From the movie A beautiful Mind, "Adam Smith needs revision ... that's the only way we all get laid."

CemLiSI5ox8

Vincey!

Oct 13th, 2012, 03:54 AM

Technically can't that be written as 6÷2（2+1）= 6 ÷(2x2 + 2x1) = 1?

Then again I'm not sure. :lol:

you do this in algebra but not in numeric equations. :shrug: A parenthesis is there to imply that the result of the equation in that parenthesis is affected by what follows and what is before it.

Then again I'm not sure. :lol:

you do this in algebra but not in numeric equations. :shrug: A parenthesis is there to imply that the result of the equation in that parenthesis is affected by what follows and what is before it.

Sam L

Oct 13th, 2012, 05:45 AM

Technically can't that be written as 6÷2（2+1）= 6 ÷(2x2 + 2x1) = 1?

Then again I'm not sure. :lol:

That's what someone I know said. But if you calculate it that way, wouldn't it become 6 ÷ 4 + 2? Without brackets? :confused: Because once you calculate the brackets, they go. That's what I remember. Maybe a mathpert can help me out.

Then again I'm not sure. :lol:

That's what someone I know said. But if you calculate it that way, wouldn't it become 6 ÷ 4 + 2? Without brackets? :confused: Because once you calculate the brackets, they go. That's what I remember. Maybe a mathpert can help me out.

égalité

Oct 13th, 2012, 06:01 AM

Perhaps you should contact the AMS (see above) since according to them a/bc is to be interpreted as a/(bc)

Either way, your "proof" is flawed. There is nothing in the problem (or either answer) that countermands the associative property. I'll let you figure it out. ;)

The source you posted is basically a "letter to the editor," isn't it? :lol:

So when you multiply an integer a by a rational number b/c, you can write it as ab/c or as a*b/c, but you can't write it both as b/c*a and as b/ca? How does that make sense?

Either way, your "proof" is flawed. There is nothing in the problem (or either answer) that countermands the associative property. I'll let you figure it out. ;)

The source you posted is basically a "letter to the editor," isn't it? :lol:

So when you multiply an integer a by a rational number b/c, you can write it as ab/c or as a*b/c, but you can't write it both as b/c*a and as b/ca? How does that make sense?

Morning Morgan

Oct 13th, 2012, 06:03 AM

Ok, there is one correct answer only, but only because some people made a decision about the convention on the order of operations. The convention is supposedly more natural because of the left to right prioritizing of operations, but it was a choice nonetheless. They could have come up with MDEPAS instead of PEDMAS and made everyone's life absolutely miserable with its stupid non-intuitiveness, but that would have given a different "correct" answer. Can we just leave the argument at that please?

Hurley

Oct 13th, 2012, 06:18 AM

Well...yeah. But, as my father says, if the queen had nuts, she'd be the king. :shrug:

égalité

Oct 13th, 2012, 06:23 AM

Here is what I thought last night:

(ab/c)d = a(b/cd) by associativity (based on one interpretation of the expressions)

(ab/c)d=(abd)/c and a(b/cd)=ab/(cd)(based on another interpretation of the expressions). These are not equal so there's a mathematical absurdity.

His error is that he thinks that (ab/c)d = a(b/cd) (based on the 2nd interpretation). This isn't true. ((ab)/c)d does not equal a(b/(cd))

If we're consistent with the interpretations, the mathematical absurdity doesn't arise.

No, this isn't based on either interpretation. This is the definition of associativity when the operation of multiplication is indicated by juxtaposition. If you and pov don't like my argument, put an extra set of parentheses around the d. Then it's the exact same situation at what's in the problem and my argument doesn't change. :lol:

(ab/c)d = a(b/cd) by associativity (based on one interpretation of the expressions)

(ab/c)d=(abd)/c and a(b/cd)=ab/(cd)(based on another interpretation of the expressions). These are not equal so there's a mathematical absurdity.

His error is that he thinks that (ab/c)d = a(b/cd) (based on the 2nd interpretation). This isn't true. ((ab)/c)d does not equal a(b/(cd))

If we're consistent with the interpretations, the mathematical absurdity doesn't arise.

No, this isn't based on either interpretation. This is the definition of associativity when the operation of multiplication is indicated by juxtaposition. If you and pov don't like my argument, put an extra set of parentheses around the d. Then it's the exact same situation at what's in the problem and my argument doesn't change. :lol:

Novichok

Oct 13th, 2012, 06:48 AM

No, this isn't based on either interpretation. This is the definition of associativity when the operation of multiplication is indicated by juxtaposition. If you and pov don't like my argument, put an extra set of parentheses around the d. Then it's the exact same situation at what's in the problem and my argument doesn't change. :lol:

Yeah, my mistake.

But (ab/c)d = a(b/cd) is false* if we use the interpretation of 6÷2（2+1）=1. That interpretation of the expressions would give us ((ab)/c)d = a(b/(cd)). But these are not equal.

So the fact that (ab/c)d=(abd)/c and a(b/cd)=ab/(cd) using the "6÷2（2+1）=1 interpretation" doesn't give us a mathematical absurdity.

*It's true if we use the interpretation that gives us an answer of 9 though. (a(b/c))d=a((b/c)d).

Yeah, my mistake.

But (ab/c)d = a(b/cd) is false* if we use the interpretation of 6÷2（2+1）=1. That interpretation of the expressions would give us ((ab)/c)d = a(b/(cd)). But these are not equal.

So the fact that (ab/c)d=(abd)/c and a(b/cd)=ab/(cd) using the "6÷2（2+1）=1 interpretation" doesn't give us a mathematical absurdity.

*It's true if we use the interpretation that gives us an answer of 9 though. (a(b/c))d=a((b/c)d).

Miracle Worker

Oct 13th, 2012, 10:17 AM

Ok. I think I understand what's going on here.

We just don't know how this should look like.

It can be:

6:2*(2+1) =9

Or

6: (2*(2+1)) =1

I guess this whole discussion is about which version was chosen by OP?

And a/bc is not the same as a/(bc).

We just don't know how this should look like.

It can be:

6:2*(2+1) =9

Or

6: (2*(2+1)) =1

I guess this whole discussion is about which version was chosen by OP?

And a/bc is not the same as a/(bc).

debby

Oct 13th, 2012, 03:58 PM

Omg this thread is gold :rolls:

http://media.tumblr.com/tumblr_m0a00iEKyZ1qbbg8c.gif

Can't believe some answers in that thread, and I had a terrible grade au baccalauréat de maths :oh: (ok not terrible, but still bad)

http://media.tumblr.com/tumblr_m0a00iEKyZ1qbbg8c.gif

Can't believe some answers in that thread, and I had a terrible grade au baccalauréat de maths :oh: (ok not terrible, but still bad)

debby

Oct 13th, 2012, 04:00 PM

Ok. I think I understand what's going on here.

We just don't know how this should look like.

It can be:

6:2*(2+1) =9

Or

6:(2*(2+1)) =1

I guess this whole discussion is about which version was chosen by OP?

And a/bc is not the same as a/(bc).

Yup.

No one adds a parenthesis just because it looks cute. If there is an extra parenthesis, it's for the order of resolution. So it obviously affects the result.

So if we go by the version chosen by the OP, it's clearly 9. Period. I don't understand why some argue this.

We just don't know how this should look like.

It can be:

6:2*(2+1) =9

Or

6:(2*(2+1)) =1

I guess this whole discussion is about which version was chosen by OP?

And a/bc is not the same as a/(bc).

Yup.

No one adds a parenthesis just because it looks cute. If there is an extra parenthesis, it's for the order of resolution. So it obviously affects the result.

So if we go by the version chosen by the OP, it's clearly 9. Period. I don't understand why some argue this.

Sammo

Oct 13th, 2012, 04:06 PM

Ah, now I see this.

So what you actually mean to say is:

Multiplication in fact does NOT trump division. EVER. They are done simultaneously.

So you mean the opposite of what you say but that's cool.

They aren't done simultaneously, they're done from left to right.

So what you actually mean to say is:

Multiplication in fact does NOT trump division. EVER. They are done simultaneously.

So you mean the opposite of what you say but that's cool.

They aren't done simultaneously, they're done from left to right.

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