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Fallacious proof/2equals1

Revision as of 13:42, 3 August 2006 by IntrepidMath (talk | contribs)
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2 = 1

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 $a^2-ab$)

Explanation

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

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