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 $a = b$ ?