Fallacious proof/2equals1: Difference between revisions

Revision as of 14:48, 3 August 2006

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.

Revision as of 13:42, 3 August 2006 view source IntrepidMath (talk \| contribs) Administrators 197 edits No edit summary		Revision as of 14:48, 3 August 2006 view source JBL (talk \| contribs) Administrators 2,310 edits mNo edit summary Newer edit →
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	=== Explanation ===		=== Explanation ===
	The trick in this argument is when we divide by <math>a^{2}-ab</math>. Since a=b, <math>a^2-ab</math> ~~equals 0~~, and dividing by zero is illegal.		The trick in this argument is when we divide by <math>a^{2}-ab</math>. Since <math>a=b</math>, <math>a^2-ab = 0</math>, and dividing by zero is illegal.