Prove golden ratio is irrational

Question

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!

As I understand, $p$ divides $q^2$ means $n p = q^2$ for some integer $n$. Suppose $p = 4$ and $n = 25$ then $n p = 100 = 10^2 = q^2$, thus $q = 10$. But $4$ does not divide $10$. — guest, Sep 14 '19 at 21:18
Thanks! Is there a different contradiction that I could make instead at this point? — random_0620, Sep 14 '19 at 21:19
If $d|p-q,q$, then $q=nd,p-q=md$ for some $n,m$, $p=(n+m)d\to d|\gcd(p-q,q)$ — Rushabh Mehta, Sep 14 '19 at 21:26
You have a fine solution but maybe the exercise alludes to $\varphi^2-\varphi-1=0$ can only have $\pm 1$ as rational roots by rational root theorem. — dioid, Sep 14 '19 at 21:35
@DonThousand I see that but it is still not clicking where the contradiction is in your statement. Could I have another hint/step? — random_0620, Sep 14 '19 at 21:52
The bolded statement is true for primes. If $p|q^2$ but $p\not \mid q$ then $p$ has some prime factors that divide into $p$ a higher power than they do into $q$ but all the prime factors $p$ are prime factors of $q$ and their powers in $p$ are no more than twice they are in $q$. — fleablood, Sep 15 '19 at 00:24
Your sketched proof is simply a special case of a common proof of the Rational Root Test. — Bill Dubuque, Sep 15 '19 at 00:56

score 2 · Accepted Answer · answered Sep 15 '19 at 01:03

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.

score 1 · Answer 2 · answered Sep 14 '19 at 21:45

1

Since $\operatorname{gcd}(p,\,q)=1$, by Bézout's lemma integers $x,\,y$ exist with $px+qy=1$. Hence $p^2x^2+2pqxy+q^2y^2=1$. If $p|q^2$, the left-hand side is a multiple of $p$, so $p=1$. But clearly, $1/\varphi\notin\Bbb N$.

answered Sep 14 '19 at 21:45

J.G.

115,835

score 0 · Answer 3 · answered Sep 14 '19 at 21:47

$p\mid q^2$ in general does not imply $p\mid q$; for instance $4\mid2^2$ but $4\not\mid2$.

However, you know that $(p,q)=1$. So $p$ and $q$ have no prime factors in common, so $p$ and $q^2$ certainly don't either. However, this does not imply that $p\mid q$ is false, just that $p=1$.

Go forth from there.

(Aside: Try to avoid using $p$ and $q$ as variables in number-theory-ish proofs. By convention, those variables make readers assume you're referring to prime numbers, which you aren't here.)

score 0 · Answer 4 · answered Sep 14 '19 at 23:50

0

Here's another approach altogether:

From $1/\phi=\phi-1$ (and $\phi\gt0$) it follows that $1\lt\phi\lt2$. Now suppose that $\phi$ is rational. Then it can be written as $p/q$ with a minimal (positive) denominator $q$. From $1\lt p/q\lt 2$ we have $q\lt p\lt2q$. But now

$${1\over\phi}=\phi-1={p\over q}-1={p-q\over q}$$

implies $\phi=q/(p-q)$, and from $q\lt p\lt2q$ we see that $p-q$ is a positive denominator less than $q$. This contradiction shows that $\phi$ is irrational.

answered Sep 14 '19 at 23:50

Barry Cipra

79,832

This is a special case of proof $(3)$ here or, more generally, the conductor ideal approach linked there. The key idea is much clearer when viewed from this more general perspective. – Bill Dubuque Sep 15 '19 at 00:39

Prove golden ratio is irrational

4 Answers4