# Where is wrong in this proof [duplicate]

Suppose $a=b$.

Multiplying by $a$ on both sides gives $a^2 = ab$. Then we subtract $b^2$ on both sides, and get

$a^2-b^2 = ab-b^2$.

Obviously, $(a-b)(a+b) = b(a-b)$, so dividing by $a - b$, we find

$a+b = b$.

Now, suppose $a=b=1$. Then $1=2$ :)

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## marked as duplicate by Ishan Banerjee, azimut, Julian Kuelshammer, Davide Giraudo, AmzotiMay 12 '13 at 12:18

You divide by $a-b$. This is possible and justified only if $a-b\ne0$. So the proof would be wrong in general because you might have $a-b=0$ unless you proof otherwise (or treat the case specially). In this particular case, you assume right away that $a=b$, hence $a-b=0$. – Hagen von Eitzen May 12 '13 at 7:15
I would be amazed if this wasn't a duplicate, but I can't find a version of this question because I'm not sure what keywords to search for. – Qiaochu Yuan May 12 '13 at 8:31

The mistake is that from the fact that $$xy = zy$$ you cannot conclude that $x=z$. You can only conclude that $x=z$, if $y \neq 0$, i.e., $$xy = zy \implies x=z \color{red}{\text{ or } y=0}$$

In your case, you had $$(a-b)(a+b) = b(a-b)$$ Since $a = b$, we have $a-b = 0$ and hence you cannot cancel $a-b$ from both sides, i.e., from the fact that $$(a-b)(a+b) = b(a-b)$$ we can only conclude that $$a+b = b \color{red}{\text{ or } a-b=0}$$ The $\color{red}{\text{later}}$ is what is valid in your case, since you started with the assumption $a=b$, i.e., $a-b=0$.

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yes.that's right – Mehdi Khademloo May 12 '13 at 7:20

By a simple example:

if( A = 0 ) A * 5 = A * 7 and by dividing by A we have 5 = 7.

Then we can dividing by A if A not equal to zero.

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by omit this fact we can proof any of the number is equal to any of the other number – Mehdi Khademloo May 12 '13 at 13:35