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This bothers me for a while:

Proof $a = b$ $$a=b$$ $$b=a$$ $$a+b=b+a$$ $$a+b=a+b$$ $$0 = 0$$

This looks like a proof for $a=b$, but it shouldn't work like this. But i can't put my finger on why it's wrong.

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if you start from $0=0$ you can't assume that a=b and cancel a and b in step 2. And if you're trying to show a=b and you start with a=b and accept things as is then you've only shown a=b when a=b=0 not when a,b can be anything. If you started with 0=0 everything is good up until step 4 when for no reason $a+b=b+a$ turns into $b=a$. Also there are no justifications for any of these steps. – Eleven-Eleven Apr 22 '13 at 13:26
This (among other things) proves that $a=b$ assuming $a=b$. There is a simpler way to do this: just note that $a=b$. – Dejan Govc Apr 22 '13 at 14:00
More simply: $\ a = b\:\Rightarrow\: 0 = 0\:$ by multiplying both sides by $\,0.\:$ Because that operation is not invertible, you cannot reverse the implication, i.e. the converse is not true. However, it is true that $\:a = b\iff ac = bc\:$ if $\:c \ne 0\:$ in a field (or if $\:c\:$ is cancellable in a ring). – Math Gems Apr 22 '13 at 14:06
up vote 22 down vote accepted

This merely proves that if $a=b$ then $0=0$. Try proving it the other way around and you'll see that you can't reverse the steps.

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+1: Gets at the heart of the question. – Neal Apr 22 '13 at 13:28
Yes, all of the responses get at aspects of the difficulty with this "proof". It is not "reversible" exactly because it does follow that if a number is equal to itself (a = b and b = a), then 0 must equal 0. It does not in turn follow that a = b = 0 for all values of $a$. – RecklessReckoner Apr 22 '13 at 13:35
@RecklessReckoner I don't think the deduction is that $a=b=0$, only that $a=b$. There is an intuitive technique for checking identities to manipulate them until you get a "true" identity. But, as you say, that only works when the steps are reversible. – Thomas Andrews Apr 22 '13 at 13:38
if we don't reverse it and assume that a=b you can't go from $a+b=a+b$ to $0=0$ unless we've already said that $a=b=0$ – Eleven-Eleven Apr 22 '13 at 13:41
No, we get that by adding $-(a+b)$ to both sides. @ChristopherErnst – Thomas Andrews Apr 22 '13 at 13:42

If you reverse the steps, there is no viable rule or theory that lets you go from $$a+b=b+a$$ to $$b=a$$

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Well put. Just because $3+2=2+3$, does not mean $2=3$ !! – goblin Aug 1 '13 at 2:28

There's nothing wrong with this. You started by assuming that $a=b$, so your proof only holds if, indeed, $a$ is equal to $b$.

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Make the proof complete by adding either $\;\Rightarrow\;$, $\;\Leftarrow\;$, or $\;\Leftrightarrow\;$ inbetween each two expressions. Then you will see that you have proved $\;a = b \;\Rightarrow\; 0 = 0\;$.

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It starts off with what it attempts to prove.

Prove $a=b$, is not supposed to start off with $\text{Let }a=b$.

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