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Bogus Claim: If $a$ and $b$ are two equal real numbers, then $a = 0$

$a = b$

$a^2 = ab$

$a^2 - b^2 = ab - b^2$

$(a-b)(a+b) = (a-b)b$

$a + b = b$

$a = 0$

I found this in my proof handouts, and correct me if I'm wrong,but is it wrong because after line 4 we divide both sides by $(a - b)$ which would be $0$ if $a = b$ ?

kub0x
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2 Answers2

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You are dividing the 4th line by $a-b$, which is zero by your assumption

Canardini
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  1. You take the equation $(a-b)(a+b)=(a-b)b$

  2. You divide each side by $(a-b)$

  3. You get the equation $a + b = b$

But $a=b\implies(a-b)=0$, which means that step #2 is illegal.

barak manos
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