# 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 ðŸ™‚

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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. ðŸ˜›

2. Bravo.

You have pointed up the keyword. We should never divide something with zero as “1/0” is undefined in Mathematic.

3. Liang says:

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.

1. Good observation. Got pandai ðŸ™‚

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.