Proof that 2=1: Difference between revisions

Revision as of 20:53, 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.

<size=200>Thus, this prove is false.</size>

Revision as of 23:58, 2 February 2020 view source Shurong.ge (talk \| contribs) 155 edits →Error ← Older edit		Revision as of 20:53, 3 February 2020 view source Kevinmathz (talk \| contribs) editor 341 edits →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>