## Saturday, April 05, 2008

### Let a and b be equal non-zero quantities

$a = b \,$

Multiply through by a

$a^2 = ab \,$

Subtract $b^2 \,$

$a^2 - b^2 = ab - b^2 \,$

Factor both sides

$(a - b)(a + b) = b(a - b) \,$

Divide out $(a - b) \,$

$a + b = b \,$

Observing that $a = b \,$

$b + b = b \,$

Combine like terms on the left

$2b = b \,$

Divide by the non-zero b

$2 = 1 \,$

The fallacy is in line 5: the progression from line 4 to line 5 involves division by (ab), which is zero since a equals b. Since division by zero is undefined, the argument is invalid. Deriving that the only possible solution for lines 5, 6, and 7, namely that a = b = 0, this flaw is evident again in line 7, where one must divide by b (0) in order to produce the fallacy (not to mention that the only possible solution denies the original premise that a and b are nonzero). A similar invalid proof would be to say that 2(0) = 1(0) (which is true) therefore, by dividing by zero, 2 = 1.