# Famous Theorems of Mathematics/√2 is irrational

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

## ProofEdit

Assume for the sake of contradiction that ${\displaystyle {\sqrt {2}}\in \mathbb {Q} }$. Hence ${\displaystyle {\sqrt {2}}={\frac {a}{b}}}$ holds for some a and b that are coprime.

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

Since the left-hand side of the equation is divisible by 2, then so must the right-hand side, i.e., ${\displaystyle 2|a^{2}}$. Since 2 is prime, we must have that ${\displaystyle 2|a}$.

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

Dividing both sides with 2 yields ${\displaystyle b^{2}=2A^{2}\!\,}$, and using similar arguments as above, we conclude that ${\displaystyle 2|b}$. However, we assumed that ${\displaystyle {\sqrt {2}}={\frac {a}{b}}}$ such that that a and b were coprime, and have now found that ${\displaystyle 2|a}$ and ${\displaystyle 2|b}$; a contradiction.

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

## Another ProofEdit

The following reductio ad absurdum argument is less well-known. It uses the additional information √2 > 1.

1. Assume that √2 is a rational number. This would mean that there exist integers m and n with n ≠ 0 such that m/n = √2.
2. Then √2 can also be written as an irreducible fraction m/n with positive integers, because √2 > 0.
3. Then ${\displaystyle {\sqrt {2}}={\frac {{\sqrt {2}}\cdot n({\sqrt {2}}-1)}{n({\sqrt {2}}-1)}}={\frac {2n-{\sqrt {2}}n}{{\sqrt {2}}n-n}}={\frac {2n-m}{m-n}}}$, because ${\displaystyle {\sqrt {2}}n=m}$.
4. Since √2 > 1, it follows that m > n, which in turn implies that m > 2nm.
5. So the fraction m/n for √2, which according to (2) is already in lowest terms, is represented by (3) in strictly lower terms. This is a contradiction, so the assumption that √2 is rational must be false.

Similarly, assume an isosceles right triangle whose leg and hypotenuse have respective integer lengths n and m. By the Pythagorean theorem, the ratio m/n equals √2. It is possible to construct by a classic compass and straightedge construction a smaller isosceles right triangle whose leg and hypotenuse have respective lengths m − n and 2n − m. That construction proves the irrationality of √2 by the kind of method that was employed by ancient Greek geometers.

## NotesEdit

• 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.