Let be the basic equation.
Then, both sides multiply with .
Move from left to right; and and from right to left.
And we have:
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 🙂
5 thoughts on “1+1 is not 2?”
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. 😛
You have pointed up the keyword. We should never divide something with zero as “1/0” is undefined in Mathematic.
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?
Good observation. Got pandai 🙂
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.