This is an example on how to design proofs. Another one is needed for definitions and axioms.

Irrationality of the square root of 2

The square root of 2 is irrational, ${\displaystyle {\sqrt {2}}\notin \mathbb {Q} }$

Proof

This is a proof by contradiction, so we assume that ${\displaystyle {\sqrt {2}}\in \mathbb {Q} }$ and hence ${\displaystyle {\sqrt {2}}={\frac {a}{b}}}$ for some a, b that are coprime.

This implies that ${\displaystyle 2={\frac {a^{2}}{b^{2}}}}$. Rewriting this gives ${\displaystyle 2b^{2}=a^{2}\!\,}$.

Since ${\displaystyle b^{2}\in \mathbb {Z} }$, we have that ${\displaystyle 2|a^{2}}$. Since 2 is prime, 2 must be one of the prime factors of ${\displaystyle a^{2}}$, which are also the prime factors of ${\displaystyle a}$, thus, ${\displaystyle 2|a}$.

So we may substitute a with ${\displaystyle 2k,k\in \mathbb {Z} }$, and we have that ${\displaystyle 2b^{2}=4k^{2}\!\,}$.

Dividing both sides with 2 yields ${\displaystyle b^{2}=2k^{2}\!\,}$, and using similar arguments as above, we conclude that ${\displaystyle 2|b}$.

Here we have a contradiction; we assumed that a and b were coprime, but we have that ${\displaystyle 2|a}$ and ${\displaystyle 2|b}$.

Hence, the assumption was false, and ${\displaystyle {\sqrt {2}}}$ cannot be written as a rational number. Hence, it is irrational.

Notes

• As a generalization one can show that the square root of every prime number is irrational.
• Another way to prove the same result is to show that ${\displaystyle x^{2}-2}$ is an irreducible polynomial in the field of rationals using Eisenstein's criterion.