I'm trying to prove that $\frac{1 + \sqrt{5}}{2}$, the "Golden ratio," is irrational. I have only been able to do so thus far by taking as given (which I haven't had any trouble proving) exercise 1.1. in Rudin:
Lemma. (Taken as proved) If $r \neq 0$ is rational and $x$ is irrational, then $r + x$ is irrational and $rx$ is irrational.
I'd be very interested if someone knows of an alternate approach. Since the proof I have written was far from elegant, I'd appreciate any critiques on it as well. Here's what I came up with:
Proof. Let's first establish that $\sqrt{5}$ is irrational. Assume to the contrary that $\sqrt{5} \in \mathbb{Q}$. Then, $\exists m, n \in \mathbb{Z}, \left(\sqrt{5} = \frac{m}{n} \wedge n \neq 0\right)$. Without loss of generality, let $m$ and $n$ be coprime. Squaring both sides and algebraically rearranging this equation yields
\begin{align}
5n^2 = m^2,
\end{align}
in which case $5 \mid m^2$. We can therefore deduce that $5 \mid m$. (Side note: the only explanation I was able to come up with, which I will leave as a conceptual argument for the moment, is the fundamental theorem of arithmetic: if we attempt to establish the contrapositive, that $5$ does not divide $m$ means that for all $p_i$ in the prime factorization of $m$, $p_i \neq 5$. In squaring $m$, we double the exponents, but $5$ is still not a factor, and because it's prime it can't be generated from any of the other factors. So, $5$ also doesn't divide $m^2$. This is far from as elegant as establishing, say, that if $m$ is odd, $m^2$ is also odd. If there is a better way to establish this fact, though, I'd be very interested in hearing it.)
So, since $5 \mid m$, we can write $m = 5a$ for some $a \in \mathbb{Z}$. Substituting into our equation gives
\begin{align}
5n^2 = (5a)^2 = 25a^2,
\end{align}
and simplifying gives
\begin{align}
n^2 = 5a^2,
\end{align}
so $5 \mid n^2$ and thus $5 \mid n$, a contradiction, as we assumed $m$ and $n$ were coprime. Thus, $5$ is irrational.
From here, since $1 \in \mathbb{Q} – \{0\}$, we can use the fact that $r + x$ is irrational with $r = 1$ and $x = \sqrt{5}$ to deduce that $1 + \sqrt{5}$ is irrational. Similarly, using the fact that $rx$ is irrational, we can set $x = 1 + \sqrt{5}$ and $r = \frac{1}{2}$ to deduce that $rx = \frac{1 + \sqrt{5}}{2}$ is irrational, which is our goal.
How does this look? I'd be very interested in any critiques of this or alternate methods of proof.
Thanks.
Best Answer
Another completely different approach: One can easily see (by squaring the golden ratio) that it satisfies the quadratic equation $x^2-x-1=0$. Using the Rational Root Theorem we concude that it must be irrational.