This was the question:

Proof that 1 equals 2

I read an equation (you may know it) in a magazine:

a = b

a^2 = a*b

a^2-b^2 = a*b-b^2

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

(a+b) = b

a+a = a

2a = a

2 = 1

I do understand the problem of 2=1, but where and is the equation
incorrect? And most important, why?

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Professor Homunculus' answer:

Good question. I used to crack my brain over that one, too, till someone pointed out to me that at one point you are dividing each side of the equation by 0. That is a big time no-no. You can find out whyon this lesson: "OK, so why can't you divide by 0?").

a = b

multiply both sides by a

a^2 = a*b

subtract b^2 from both sides

a^2-b^2 = a*b-b^2

apply the distributive law to both sides

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

divide both sides by (a-b)

(a+b) = b

substitute all a's for b's (remember, if a = b you can do this)/

a+a = a

regroup the two a's in the left side, and rename it 2a

2a = a

divide both sides by a

2 = 1

The point which I highlighted in red is where you would have to divide each side of
(a+b)(a-b)=b(a-b)
by (a-b) to get
(a+b)=b

But you will notice that if the first part of the proof (a=b) is true, then (a-b) would equal 0.
And, as we know, we can't divide by 0, so from that point on this "proof" that 1=2 is relegated to the non-valid, but interesting, heap of other things which people try to use to prove things which are also false.

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