# Famous Theorems of Mathematics/Proof style

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.