1+1 is not 2?

Let a + b = c be the basic equation.

Then, both sides multiply with (a+b) .
(a + b)^2 = c (a+b) .
a^2 + 2ab +b^2 = ac + bc .

Move ab from left to right; and ac and b^2 from right to left.
a^2 + ab - ac =  bc - b^2 -ab .

Factoring …
a (a+b-c) =  -b (a + b -c)

Removing (a+b-c)
And we have:
a = -b.

Oops. It contradicts with our basic equation.

Obviously something is wrong here.

Guess what is the mistake here? It is a primary school’s question in fact πŸ™‚

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5 thoughts on “1+1 is not 2?

  1. a+b=c. equating both sides to 0, a+b=0, assuming c=0. this is one way. thus a=-b.

    it could go on and on… a+b=1, assuming c=1…. thus a=1-b

    it could go on and on and on.. in the end. if we want an exact answer to a mathematical equation, we would have to make an ass out of you and me. assumptions. πŸ˜›

  2. What you mean is,

    If a=-b, then c=0 and hence (a+b-c)=0. So
    a(a+b-c)=-b(a+b-c), here we can’t cancel the (a+b-c)?

    So it is a contradiction?
    and 1+1=2.

    • Sorry. What I mean is, for any a,b,c \in Real number field, if we state an equation as a+b=c, it implies that a+b+c=0. And, we should not divide zero in the algebra manipulation as 1/0 is undefined in first place.

      Besides, it is a mistake for not stating what field a, b and c are in a first place.

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