6÷2(2+1)=? - Page 12 - TennisForum.com

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1. 40 26.85%
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post #166 of 355 (permalink) Old Oct 12th, 2012, 02:38 AM
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Re: 6÷2(2+1)=?

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Division comes before Multiplication
Sigh. Read the thread.

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.

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post #167 of 355 (permalink) Old Oct 12th, 2012, 02:40 AM
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Re: 6÷2(2+1)=?

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post #168 of 355 (permalink) Old Oct 12th, 2012, 02:41 AM
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Re: 6÷2(2+1)=?

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Sigh. Read the thread.

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.

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post #169 of 355 (permalink) Old Oct 12th, 2012, 02:42 AM
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Re: 6÷2(2+1)=?

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Ain't nobody got time for that.
Ah, touché.

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post #170 of 355 (permalink) Old Oct 12th, 2012, 02:44 AM
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Re: 6÷2(2+1)=?

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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.
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 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.

Last edited by Novichok; Oct 12th, 2012 at 02:49 AM.
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post #171 of 355 (permalink) Old Oct 12th, 2012, 02:55 AM
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Re: 6÷2(2+1)=?

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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 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. 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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post #172 of 355 (permalink) Old Oct 12th, 2012, 03:07 AM
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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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post #173 of 355 (permalink) Old Oct 12th, 2012, 03:09 AM
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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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post #174 of 355 (permalink) Old Oct 12th, 2012, 03:19 AM
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Re: 6÷2(2+1)=?

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Originally Posted by Pvt. Kovalenko View Post
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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post #175 of 355 (permalink) Old Oct 12th, 2012, 03:21 AM
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Re: 6÷2(2+1)=?

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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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post #176 of 355 (permalink) Old Oct 12th, 2012, 03:22 AM
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Re: 6÷2(2+1)=?

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Originally Posted by égalité View Post
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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post #177 of 355 (permalink) Old Oct 12th, 2012, 03:29 AM
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Re: 6÷2(2+1)=?

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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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post #178 of 355 (permalink) Old Oct 12th, 2012, 03:33 AM
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Re: 6÷2(2+1)=?

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These are the same. I actually provided a proof of this on the previous page.
Oh, okay.

Thanks for that.
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post #179 of 355 (permalink) Old Oct 12th, 2012, 03:37 AM
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Re: 6÷2(2+1)=?

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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. 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.

Fixed.
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post #180 of 355 (permalink) Old Oct 12th, 2012, 03:38 AM
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Re: 6÷2(2+1)=?

Yeah, that's what I said.

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