# Proving that 3 + 2 = 0

Assume A + B = C, and assume A = 3 and B = 2.

Multiply both sides of the equation A + B = C by (A + B). We obtain A² + 2AB + B² = C(A + B)

Rearranging the terms we have A² + AB - AC = - AB - B² + BC

Factoring out (A + B - C), we have A(A + B - C) = - B(A + B - C)

Dividing both sides by (A + B - C), that is, dividing by zero, we get A = - B, or A + B = 0, which is evidently absurd.

I found this delouse paradox from http://www.paradoxes.co.uk

## Leave a comment