6÷2（2+1）=？ - Page 13 - TennisForum.com
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Old Oct 12th, 2012, 03:43 AM
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Re: 6÷2（2+1）=？

Quote:
Originally Posted by égalité View Post
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.
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Old Oct 12th, 2012, 09:16 AM
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Re: 6÷2（2+1）=？

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.

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Old Oct 12th, 2012, 09:35 AM
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Re: 6÷2（2+1）=？

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?

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Old Oct 12th, 2012, 10:31 AM
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Re: 6÷2（2+1）=？

OMG there are Youtube clips on this:

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Old Oct 12th, 2012, 01:11 PM
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Re: 6÷2（2+1）=？

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Old Oct 12th, 2012, 01:24 PM
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Re: 6÷2（2+1）=？

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...

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Old Oct 12th, 2012, 07:30 PM
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Re: 6÷2（2+1）=？

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

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Old Oct 12th, 2012, 07:50 PM
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Re: 6÷2（2+1）=？

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

Rewriting though 6x1/2x3 =9 so

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）)

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Old Oct 12th, 2012, 08:03 PM
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Re: 6÷2（2+1）=？

one more source:
American Mathematical Society (AMS)

Quote:
"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).

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Old Oct 12th, 2012, 08:05 PM
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Re: 6÷2（2+1）=？

Lol It is 9.

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Old Oct 12th, 2012, 08:16 PM
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Re: 6÷2（2+1）=？

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

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Old Oct 12th, 2012, 10:15 PM
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Re: 6÷2（2+1）=？

Quote:
Originally Posted by égalité View Post
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.
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.

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Old Oct 12th, 2012, 10:24 PM
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Re: 6÷2（2+1）=？

Quote:
Originally Posted by zigga View Post
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

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Old Oct 12th, 2012, 10:38 PM
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Re: 6÷2（2+1）=？

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

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Old Oct 12th, 2012, 10:59 PM
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Re: 6÷2（2+1）=？

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

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

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