I am trying to show that the golden ratio $\varphi$ is irrational.
Show $\frac{1}{\varphi} =1 – \varphi$
\begin{equation}
\frac{1}{\varphi} = \frac{1}{\frac{\sqrt{5}+1}{2}}=\frac{2}{\sqrt{5}+1}
=\frac{2(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}-1}{2}=\frac{\sqrt{5}-1+1-1}{2}=\frac{\sqrt{5}+1-2}{2}=\frac{\sqrt{5}+1}{2}-\frac{2}{2}=\varphi – 1
\end{equation}
Prove $\varphi$ is irrational
Assume the golden ratio is rational which implies $\varphi=\frac{p}{q}$ where $p,q \in \mathbb{N}$ and gcd($p$,$q$) $= 1$.
Since $\frac{1}{\varphi}=\varphi – 1 \Rightarrow \frac{q}{p}=\frac{p}{q}-1 \Rightarrow \frac{q}{p}=\frac{p-q}{q}\Rightarrow q^2=p(p-q)$. This implies $p$ divides $q^2$.
At this point I would like to say that $p |q^2$ implies $p|q$ however I am unable to verify if this is true. At this point I could easily say that $p|q$ is false because gcd($p$,$q$)=1.
How can I verify the bolded statement? Thanks!
Best Answer
You do have two options.
That not so great and begging the question option:
Show if $\frac 1\varphi = 1-\varphi$ then $\varphi^2 -\varphi + 1 = 0$ and $\varphi =\frac {1 +\sqrt 5}2$ and show that $\sqrt 5$ is irrational.
That's done the "usual" way. If $a^2 = 5b^2$ for integers $a,b$ then if $a$ isnt a multiple of $5$ then $a^2=5b^2$ either. SO $a$ is a multiple of $5$ and $a =5a'$ and $5a'^2 = b^2$ and by the same argument $b$ is a multiple of $5$ so $\sqrt 5 = \frac ab$ where $a,b$ are integers in lowest terms is impossible.
But I call that "begging the question" because it doesn't really have anything to do with the golden ratio aspect; just something about boring old square roots of integers.
Other option:
Let $\varphi =\frac pq$ where $p,q\in \mathbb Z$ and in lowest terms, and $\frac qp = 1-\frac pq$.
Then $q^2 = p(q- p)$. Now $p,q$ are in lowest terms. So any factor $n$ of $q$ so that $n|q$ can not have any factor with $p$ at all. So $n|q^2$ means $n|p(q-p)$ but $n$ has nothing in common with $p$ so $n|q-p$. But $n|q$ so if $n$ also divides $q-p$ then $n|q -(q-p)=p$ but that's a contradiction.... unless $n=1$ is the only factor of $q$. But that means $q = 1$ and ... that just doesn't work. $1 = p(1-p)$ has no integer solutions.
That requires a fair number of assumptions about numbers and factors; namely that all numbers have indivisible prime factors.
YOu can do a well ordering type argument:
$\frac ab$ where $a=1; b=1$ is not a pair of integers where $\frac ba =1-\frac ab$. Let $p,q$ be such that $q$ is the smallest possible positive integer where there is any integer $p'$ where $\frac q{p'} = 1-\frac {p'}q$ and $p$ is the least possible $p'$ integer where that is true. Then $\frac qp = \frac {q -p}q$ and $\frac q{q-p} = \frac pq = \frac {q-(q-p)}q = 1-\frac {q-p}q$ and $q$ and $q-p$ are also such numbers. But $q-p < q$ and that is a contradiction.