# 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.

## Historical NotesEdit

First mentions of this proof claimed it was authored by Greek mathematicians of Pythagorean school, and indeed the first formal proof was presented later in Euclid's Elements. However, most of contemporaries (as well as many of today's historians of mathematics) were of the opinion that Pythagoreans themselves have borrowed almost all of their mathematics (including this proof) from Egyptian sources. Alas, destruction of the Great Library of Alexandria wiped out almost all the then extant scientific texts of that civilization so it is very likely this will forever remain a mystery.

## 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.