Why if $a =b$ then $a = 0$ is not a correct statement Bogus Claim: If $a$ and $b$ are two equal real numbers, then $a = 0$
$a = b$
$a^2 = ab$
$a^2 - b^2 = ab - b^2$
$(a-b)(a+b) = (a-b)b$
$a + b = b$
$a = 0$
I found this in my proof handouts, and correct me if I'm wrong,but is it wrong because after line 4 we divide both sides by $(a - b)$ which would be $0$ if $a = b$ ?

*

*You take the equation $(a-b)(a+b)=(a-b)b$

*You divide each side by $(a-b)$

*You get the equation $a + b = b$
But $a=b\implies(a-b)=0$, which means that step #2 is illegal.
