Irrational Square Roots
The square root of n is rational only for perfect squares: a descent proof with no primes, and one by Euclid's lemma.
1 Statement
Theorem 1.1 (Square roots). Let \(n\) be a positive integer. Then \(\sqrt {n}\) is rational if and only if \(n = k^2\) for some integer \(k\). In particular, \(\sqrt 2\), \(\sqrt 3\), \(\sqrt 5\) and \(\sqrt 6\) are irrational.
2 Setting
We use only the integers and the well-ordering principle: every non-empty set of positive integers has a least element. A real number \(x\) is rational if \(x = p/q\) for integers \(p\) and \(q \geq 1\).
3 Idea of the proof
If \(\sqrt n = p/q\), then \(\sqrt n\) times a suitable integer combination of \(p\) and \(q\) produces a representation with a strictly smaller denominator. Choosing the smallest denominator first gives a contradiction. This needs no primes and no unique factorisation.
4 Proof
Lemma 4.1 (Integer part). If \(n\) is not a perfect square, there is an integer \(k\) with \(k < \sqrt n < k + 1\).
Proof. Let \(k\) be the largest integer with \(k^2 \leq n\); it exists because \(0^2 \leq n\) and \(j^2 > n\) for \(j > n\). Then \(k^2 \leq n < (k+1)^2\), and \(k^2 \neq n\) by assumption, so \(k < \sqrt n < k + 1\).
Proof of Theorem 1.1. If \(n = k^2\), then \(\sqrt n = k/1\) is rational. Conversely, suppose \(n\) is not a perfect square but \(\sqrt n\) is rational. Take \(k\) from Lemma 4.1. The set
is non-empty, so by well-ordering it has a least element \(q\). Write \(p = q\sqrt n\), and put
Both \(q'\) and \(p'\) are integers, because \(p' = q\sqrt n \cdot \sqrt n - kq\sqrt n = nq - kp\). Since \(0 < \sqrt n - k < 1\), we have \(0 < q' < q\). So \(q' \in S\) and \(q' < q\), contradicting the minimality of \(q\).
5 A second proof, by prime factorisation
This is standard; it follows from Bézout’s identity. If \(\ell \nmid a\), then \(\gcd (a, \ell ) = 1\), so \(1 = ax + \ell y\) for some integers \(x, y\), and \(b = abx + \ell b y\) is divisible by \(\ell \).
Second proof of Theorem 1.1. Suppose \(\sqrt n = p/q\) in lowest terms, so that \(\gcd (p, q) = 1\) and \(p^2 = nq^2\). If \(q > 1\), let \(\ell \) be a prime dividing \(q\). Then \(\ell \mid p^2\), so \(\ell \mid p\) by Lemma 5.1, contradicting \(\gcd (p, q) = 1\). Hence \(q = 1\) and \(n = p^2\).
6 Remarks
Remark 6.1. The second proof generalises at once. If a rational number \(p/q\) in lowest terms is a root of a monic polynomial \(x^d + c_{d-1}x^{d-1} + \dotsb + c_0\) with integer coefficients, then multiplying by \(q^d\) shows \(q \mid p^d\), so \(q = 1\). A rational root of such a polynomial must therefore be an integer. Taking \(x^d - n\), the number \(\sqrt [d]{n}\) is rational only when \(n\) is a perfect \(d\)th power.