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, 03:43 AM

#181

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

#182

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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.



Oct 12th, 2012, 09:35 AM

#183

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Join Date: Oct 2001
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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?



Oct 12th, 2012, 10:31 AM

#184

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Re: 6÷2（2+1）=？



Oct 12th, 2012, 01:11 PM

#185

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

#187

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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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Oct 12th, 2012, 07:50 PM

#188

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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 rechecked, is 6÷ (2（2+1）)
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Oct 12th, 2012, 08:03 PM

#189

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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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Oct 12th, 2012, 08:05 PM

#190

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Re: 6÷2（2+1）=？
Lol It is 9.
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Oct 12th, 2012, 08:16 PM

#191

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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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Oct 12th, 2012, 10:15 PM

#192

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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.

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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Oct 12th, 2012, 10:24 PM

#193

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Posts: 3,461

Re: 6÷2（2+1）=？
Quote:
Originally Posted by zigga
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......
10xx=9
9x=9
x=1
what's amusing about that :S



Oct 12th, 2012, 10:38 PM

#194

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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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Oct 12th, 2012, 10:59 PM

#195

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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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