The Golden Ratio is Irrational
Three proofs that phi is irrational: descent from phi = 1 + 1/phi, the rational root test, and its endless continued fraction.
1 Statement
Theorem 1.1 (Irrationality of \(\varphi \)). The golden ratio \(\varphi = \dfrac {1 + \sqrt 5}{2}\) is irrational.
2 Setting
The number \(\varphi \) is the positive root of
Since \(2 < \sqrt 5 < 3\), we have \(1 < \varphi < 2\).
3 Idea of the proof
The relation \(\varphi = 1 + 1/\varphi \) is self-similar: removing a unit from \(\varphi \) and inverting gives \(\varphi \) back. A fraction cannot behave like this, because each step strictly shrinks its denominator.
4 First proof: descent
Proof. Suppose \(\varphi \) is rational. Among all ways of writing \(\varphi = p/q\) with integers \(p, q \geq 1\), choose one with \(q\) least, by well-ordering. By (2.1),
Since \(1 < p/q < 2\), we have \(0 < p - q < q\). This writes \(\varphi \) with the positive denominator \(p - q < q\), contradicting the choice of \(q\).
5 Second proof: rational roots
Lemma 5.1 (Rational root test). If \(p/q\) is in lowest terms and is a root of \(a x^2 + b x + c\) with integer coefficients, then \(p \mid c\) and \(q \mid a\).
Proof. Multiplying \(a(p/q)^2 + b(p/q) + c = 0\) by \(q^2\) gives \(ap^2 + bpq + cq^2 = 0\). So \(p \mid cq^2\). Since \(\gcd (p, q) = 1\), also \(\gcd (p, q^2) = 1\), so Bézout gives \(1 = px + q^2y\) and \(c = pcx + cq^2y\) is divisible by \(p\). Similarly, \(q \mid ap^2\) gives \(q \mid a\).
Second proof of Theorem 1.1. By Lemma 5.1, a rational root of \(x^2 - x - 1\) has \(p \mid 1\) and \(q \mid 1\), so it is \(\pm 1\). But \(1 - 1 - 1 = -1\) and \(1 + 1 - 1 = 1\), so neither is a root.
6 Third proof: continued fractions
For \(x > 0\), the continued fraction algorithm sets \(x_0 = x\), records \(a_j = \lfloor x_j \rfloor \), and continues with \(x_{j+1} = 1/(x_j - a_j)\), stopping if \(x_j\) is an integer.
Lemma 6.1 (Rationals terminate). If \(x\) is rational, the algorithm stops after finitely many steps.
Proof. If \(x_j = p/q\) with \(q \geq 1\) is not an integer, write \(p = a_j q + r\) with \(0 < r < q\). Then \(x_{j+1} = q/r\) has denominator \(r < q\). Denominators are positive integers that strictly decrease, so the process stops.
Third proof of Theorem 1.1. For \(x_0 = \varphi \): \(a_0 = 1\) because \(1 < \varphi < 2\), and \(x_1 = 1/(\varphi - 1) = \varphi \) by (2.1). So \(x_j = \varphi \) for every \(j\), and the algorithm never stops:
By Lemma 6.1, \(\varphi \) is irrational.
7 Remarks
Remark 7.1. The three proofs are one proof in different clothing. The step \(p/q \mapsto q/(p-q)\) in the first proof is one step of the continued fraction algorithm, and it is the Euclidean algorithm run on \((p, q)\).
Remark 7.2. Truncating the continued fraction gives \(1, 2, \frac 32, \frac 53, \frac 85, \dotsc \), the ratios \(F_{j+1}/F_j\) of consecutive Fibonacci numbers. If \(r_j = F_{j+1}/F_j\), then \(r_{j+1} = 1 + 1/r_j\), the same recursion as (2.1). One can show \(\abs {r_j - \varphi } < 1/F_j^2\), so these fractions approximate \(\varphi \), but only slowly: every partial quotient is \(1\), the smallest possible.