Famous Theorems of Mathematics/√2 is irrational

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

ProofEdit

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

This implies that 2 = \frac{a^2}{b^2}. Rewriting this gives 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., 2 | a^2 . Since 2 is prime, we must have that 2 | a .

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

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

Therefore, the assumption was false, and  \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 \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 \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 x^2-2 is an irreducible polynomial in the field of rationals using Eisenstein's criterion.
Last modified on 14 December 2012, at 09:42