The Irrationality of $\sqrt{2}$




Some basic facts

  1. Definition: A number $X$ is rational if it can be written as a quotient of two whole numbers: $X=A/B$.

    If $X$ is rational then, by reducing to lowest terms, we can be assume that $A$ and $B$ have no common factor. In particular, if $X$ is rational, we can assume that $X=A/B$ where $A$ and $B$ are not both even.

  2. If $N$ is even, its square $N^{2}$ is also even, while if $N$ is odd, its square is odd. This shows that if the square of a number is even, it must have been even.

So now we come to the proof. Suppose that $\sqrt{2}$ is rational. We will show that this leads to an absurdity, so can not be true.




If $\sqrt{2}$ is rational, we can write:MATHWe can now square both sides of this equation, obtaining: MATHThus, since $A^{2}$ is twice a number, $A^{2}$ is even, so we can conclude A is even, and we can write: $A=2M$ for some whole number $M$. Thus, we can rewrite our equation: MATHNow, we can divide through by 2, obtaining: MATHBut, since $B^{2}$ is even, $B$ is even also. Thus, we have seen that both $A$ and $B$ turn out to be even. This contradicts the fact that we chose $A$ and $B$ so as not to have any common factors.

Thus, the only assumption we made, namely that $\sqrt{2}$ is rational, leads to the contradiction that $A$ and $B$ have no common factors, yet both are even. Thus, $\sqrt{2}$ can not be rational.

Q.E.D. (= Quod Erat Demonstrondum = Which Was to be Proved)