What's wrong in this proof of 1=2?

Asked Mar 09 '19 at 20:56

Active Mar 10 '19 at 03:12

Viewed 75 times

Here is the proof:

let a = b
a² = ab
a²-b² = ab-b²
(a-b)(a+b) = b(a-b)
a+b = b
2b = b
2 = 1

Now, of course Problem must lie on the line (a-b)(a+b) = b(a-b), as it seems like multiplying by 0 on both side but does algebraic rule allow that step? Anyway, where is the error in this proof?

edited Mar 10 '19 at 03:12

J. W. Tanner

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asked Mar 09 '19 at 20:56

MessitÖzil

$a-b=0$, you can't go from $(a-b)(a+b)=b(a-b)$ to $a+b=b$ – B.Swan Mar 09 '19 at 20:58
Is there any law that particularly forbid this? – MessitÖzil Mar 09 '19 at 20:59
2

You can see that $1\times 0=2\times 0$, can't you? But that doesn't mean that $1=2$. (And yes, there is a law that particularly forbids this: you can't divide by zero.) – TonyK Mar 09 '19 at 21:00
Yes, dividing by $0$ is not well-defined in general. – lightxbulb Mar 09 '19 at 21:01
$ac=bc \implies a=b$ only if $c$ is not a zero divisor. Think about it, the equation says $0=0$. You cannot do $0=2\times 0=1\times 0 \implies 2=1$. – B.Swan Mar 09 '19 at 21:03

What's wrong in this proof of 1=2?

0 Answers0