Difference between revisions of "Proof that 2=1"

Revision as of 20:54, 3 February 2020

1) $a = b$ . Given.

2) $a^2 = ab$ . Multiply both sides by a.

3) $a^2-b^2 = ab-b^2$ . Subtract $b^2$ from both sides.

4) $(a+b)(a-b) = b(a-b)$ . Factor both sides.

5) $(a+b) = b$ . Divide both sides by $(a-b)$

6) $a+a = a$ . Substitute $a$ for $b$ .

7) $2a = a$ . Addition.

8) $2 = 1$ . Divide both sides by $a$ .

Usually, if a proof proves a statement that is clearly false, the proof has probably divided by zero in some way.

In this case, the quantity of $a-b$ is $0$ as $a = b$ , since one cannot divide by zero, the proof is incorrect from that point on.

Thus, this proof is false.

Revision as of 20:54, 3 February 2020 (view source) Kevinmathz (talk \| contribs) (→‎Error) ← Older edit		Revision as of 20:54, 3 February 2020 (view source) Kevinmathz (talk \| contribs) (→‎Error) Newer edit →
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	In this case, the quantity of <math>a-b</math> is <math>0</math> as <math>a = b</math>, since one cannot divide by zero, the proof is incorrect from that point on.		In this case, the quantity of <math>a-b</math> is <math>0</math> as <math>a = b</math>, since one cannot divide by zero, the proof is incorrect from that point on.

−	~~[size=200]~~Thus, this ~~prove~~ is false.[/~~size]~~	+	<b>Thus, this proof is false.</b>