Fallacious proof/2equals1

Let $a=b$ .

Then we have

$a^2 = ab$ (since $a=b$ )

$2a^2 - 2ab = a^2 - ab$ (adding $a^2-2ab$ to both sides)

$2(a^2 - ab) = a^2 - ab$ (factoring out a 2 on the LHS)

$2 = 1$ (dividing by $a^2-ab$ )

The trick in this argument is when we divide by $a^{2}-ab$ . Since $a=b$ , $a^2-ab = 0$ , and dividing by zero is illegal.

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