Division by Zero Fallacy

A "proof" that $2 = 1$ :

0	Let $a = b$	Given
1	$a \times a = a \times b$	Multiply both sides by $a$
2	$a^{2} = a b$
3	$a^{2} - b^{2} = a b - b^{2}$	Subtract $b^{2}$ from both sides
4	$a^{2} - b^{2} = a b - b b$
5	$(a + b) (a - b) = (a - b) b$	Factor
6	$(a + b) (a - b) = (a - b) b$	Divide both sides by $(a - b)$
7	$a + b = b$
8	$b + b = b$	Substitute $a = b$ (Premise)
9	$2 b = 1 b$
10	$2 b = 1 b$	Divide both sides by $b$
11	$2 = 1$	Q.E.D.
The error occurs in step 6, where the equality is divided by $(a - b)$ . Since $a = b$ , $(a - b) = 0$ , and we cannot divide by 0 (for this reason and many others).