Here are four proofs of the irrationality of the square root of 11. Can you think of another proof?

#### Unique factorization

Suppose that the square root of 11 is rational. Then we can express the square root of 11 as a fraction of integers.

\[\sqrt{11} = \frac{a}{b}\]

Therefore, \(a^2 = 11b^2\).

Now, 11 occurs an even number of times in the prime factorization of \(a^2\), but it occurs an odd number of times in the prime factorization of \(11b^2\). This is a contradiction, since prime factorizations are unique.

See James Tanton’s essay for a more detailed explanation.

#### Euclid’s lemma

Euclid’s lemma states that if a prime number divides the product of two numbers, then it divides one or both of the factors.

Suppose that the square root of 11 is rational. Then we can express the square root of 11 as a fraction of integers \(a/b\). We may assume that \(a\) and \(b\) have no common factor other than 1.

As before, we write \(a^2 = 11b^2\). Since 11 is a prime number and it divides \(a^2\), Euclid’s lemma implies that 11 also divides \(a\).

Let \(a = 11c\). Then \((11c)^2 = 11b^2\), hence \(11c^2 = b^2\). This time, Euclid’s lemma implies that 11 divides \(b\). This contradicts the assumption that \(a\) and \(b\) have no common factor.

#### Rational root theorem

The rational root theorem states that if \(x\) is a rational root of a polynomial with integer coefficients, then \(x = p/q\), where \(p\) divides the constant term of the polynomial and \(q\) divides the leading coefficient.

The square root of 11 is a root of the polynomial \(x^2 – 11\). According to the rational root theorem, the only possible rational roots are \(\pm 1\) or \(\pm 11\). Since none of these are equal to the square root of 11, we conclude that the square root of 11 is irrational.

#### Infinite descent

Suppose that the square root of 11 is rational. Then there exists a positive integer \(n\) so that \(m := n\sqrt{11}\) is an integer. Let \(n\) be the smallest positive integer with this property.

Then

\[(m – 3n)\sqrt{11} = (n\sqrt{11}-3n)\sqrt{11} = 11n-3n\sqrt{11} = 11n-3m\]

which is also an integer.

On the other hand, \(0 < m - 3n < n\), since \(3 < \frac{m}{n} < 4\). This is a contradiction, because we assumed that \(n\) is the smallest positive integer so that \(n\sqrt{11}\) is an integer.

#### BONUS: Continued fraction

The square root of 11 has an infinite continued fraction representation.

\[

\sqrt{11} = [3; \overline{3, 6}]

= 3 + \cfrac{1}{3

+ \cfrac{1}{6

+ \cfrac{1}{3

+ \cfrac{1}{6 + \cdots} } } }

\]

But the continued fraction representation of a rational number is finite. Therefore, the square root of 11 is irrational.

It’s fun to distinguish between the proofs that work equally well for any prime number replacing 11, and the proofs that don’t.

Theodorus proved that the square roots of primes up to 17 are irrational, and people like to argue about why he stopped there. It must have something to do with the spiral of Theodorus.

Very interesting.