View Poll Results: ？

1.


40 
26.85% 
9.


99 
66.44% 
Others.(Please specify.)


2 
1.34% 
Hard to say.


8 
5.37% 
Voters: 149. You may not vote on this poll 


Oct 12th, 2012, 02:38 AM

#166

Enemy of Art
Join Date: Sep 2001
Location: location, location
Posts: 13,068

Re: 6÷2（2+1）=？
Quote:
Originally Posted by jameshazza
Division comes before Multiplication

Sigh. Read the thread.
You got lucky this time because the division sign comes first lefttoright. You would have been dead wrong had those signs been reversed.



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Oct 12th, 2012, 02:40 AM

#167

Senior Member
Join Date: Nov 2004
Posts: 15,919

Re: 6÷2（2+1）=？
Quote:
Originally Posted by Hurley
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.



Oct 12th, 2012, 02:41 AM

#168

Senior Member
Join Date: Jan 2012
Posts: 11,796

Re: 6÷2（2+1）=？
Quote:
Originally Posted by Hurley
Sigh. Read the thread.
You got lucky this time because the division sign comes first lefttoright. You would have been dead wrong had those signs been reversed.

Ain't nobody got time for that.
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Oct 12th, 2012, 02:42 AM

#169

Enemy of Art
Join Date: Sep 2001
Location: location, location
Posts: 13,068

Re: 6÷2（2+1）=？
Quote:
Originally Posted by jameshazza
Ain't nobody got time for that.

Ah, touché.



Oct 12th, 2012, 02:44 AM

#170

Senior Member
Join Date: Jan 2011
Location: Bliss
Posts: 10,309

Re: 6÷2（2+1）=？
Quote:
Originally Posted by égalité
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 nonassociative.

I'm not sure if I'm understand where you're getting at. Help me if you have time.
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 nonassociative.
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.
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Last edited by Novichok : Oct 12th, 2012 at 02:49 AM.



Oct 12th, 2012, 02:55 AM

#171

Senior Member
Join Date: Jun 2006
Location: Hingistan
Posts: 11,976

Re: 6÷2（2+1）=？
Quote:
Originally Posted by Novichok
I'm not sure if I'm understand where you're getting at. Help me if you have time.
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 nonassociative.
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. Assume that order of operations works in the "6/2(2+1)=1" way, derive a mathematical absurdity, conclude that your assumption was wrong.
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Oct 12th, 2012, 03:07 AM

#172

Splendid#man
Join Date: May 2009
Posts: 21,928

Re: 6÷2（2+1）=？
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.
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Oct 12th, 2012, 03:09 AM

#173

Senior Member
Join Date: Dec 2011
Location: Intentionally Lost
Posts: 1,639

Re: 6÷2（2+1）=？
It's 9
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
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Oct 12th, 2012, 03:19 AM

#174

Splendid#man
Join Date: May 2009
Posts: 21,928

Re: 6÷2（2+1）=？
Quote:
Originally Posted by Pvt. Kovalenko
It's 9
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.
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Oct 12th, 2012, 03:21 AM

#175

Senior Member
Join Date: Jun 2006
Location: Hingistan
Posts: 11,976

Re: 6÷2（2+1）=？
Quote:
Originally Posted by Otlichno
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.
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Oct 12th, 2012, 03:22 AM

#176

Senior Member
Join Date: Jan 2011
Location: Bliss
Posts: 10,309

Re: 6÷2（2+1）=？
Quote:
Originally Posted by égalité
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. 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. I'm just confused.
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?
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Oct 12th, 2012, 03:29 AM

#177

Senior Member
Join Date: Jun 2006
Location: Hingistan
Posts: 11,976

Re: 6÷2（2+1）=？
Quote:
Originally Posted by Novichok
Ok. I'm just confused.
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!"
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Oct 12th, 2012, 03:33 AM

#178

Splendid#man
Join Date: May 2009
Posts: 21,928

Re: 6÷2（2+1）=？
Quote:
Originally Posted by égalité
These are the same. I actually provided a proof of this on the previous page.

Oh, okay.
Thanks for that.
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Oct 12th, 2012, 03:37 AM

#179

Senior Member
Join Date: Sep 2009
Posts: 2,470

Re: 6÷2（2+1）=？
Quote:
Originally Posted by égalité
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 nonassociative.

Fixed.



Oct 12th, 2012, 03:38 AM

#180

Senior Member
Join Date: Jun 2006
Location: Hingistan
Posts: 11,976

Re: 6÷2（2+1）=？
Yeah, that's what I said.
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