Prove that the golden ratio is irrational by contradiction

Question

I am struggling to see where the contradiction lies in my proof. In a previous example, $1/\phi = \phi-1$ where $\phi$ is the golden ratio $\frac{\sqrt{5} + 1}{2}$.

Since I am proving by contradiction, I started out by assuming that $ϕ$ is rational. Then, by definition, there exists $a,b$ such that $\phi = a/b$. After some simple calculations and using the result shown from my previous example, I found that $\phi= b/(a-b)$. I also know that $b < a$ from directly calculating the ratio.

I know there is a contradiction in the result $ϕ = b/(a-b)$ but I cannot see it. Any help would be appreciated.

Answer 1

Here's one idea that works directly without showing anything about $\sqrt 5$:

We know $\varphi > 1$ so if it is rational, we could write $$ \varphi = \frac{a}{b},$$ where $a > b > 0$ are integers and $\operatorname{gcd}(a,b) = 1$. Then using the relation $\frac{1}{\varphi} = \varphi - 1$ gives $$ \frac{b}{a} = \frac{a - b}{b},$$ which is a contradiction since $\operatorname{gcd}(a,b) = 1$ by construction and $a > b$ (it would be a further reduction of a fraction that we already chose to be completely reduced).

Answer 2

HINT: $$\sqrt{5}=\frac{2a-b}{b}$$ is a contradiction, since the numbers $a,b$ are rational and $$\frac{2a-b}{b}$$ is also rational and $\sqrt{5}$ is irrational

Answer 3

If $\phi=\frac ab$ with $\gcd(a,b)=1$ then $\frac 1{\phi}=\phi-1\implies\frac ba=\frac ab-1\implies b^2=a^2-ab=a(a-b)$

This means $b\mid a(a-b)$ but since $\gcd(a,b)=1$ then $b\mid a-b\implies b\mid a$ which is a contradiction.

Answer 4

Though Nitin showed an excellent solution (much better than mine), I've already typed the routine solution approach. So I'm posting it anyway:

If it is rational, we can write it in the form of the fraction $m/n$ where $m$ and $n$ are integers and have no common factors – always, otherwise it is not rational. So we have $$\frac{\sqrt5+1}{2}=\frac{m}{n}$$ $$\sqrt5=\frac{2m}{n}-1$$ But $\frac{2m}{n}-1$ is rational because $\frac{m}{n}$ is rational. Therefore $\sqrt5$ is rational. But it is not. We have a contradition, so the whole number must be irrational. q.e.d.

If you want to prove that $\sqrt5$ is irrational. Do the same: if it is rational, we can write it in the form of the fraction $m/n$ where $m$ and $n$ are integers and have no common factors (We can factor out any common factors). So we have $$\sqrt5=\frac{m}{n}$$ $$5=\frac{m^2}{n^2}$$ $$5n^2=m^2$$ Therefore $m$ is divisible by $5$ and we can re-rewrite it as $m=5k$. So we have $$5n^2=25k^2$$ $$n^2=5k^2$$ Therefore $n$ is also divisible by 5 and we can re-rewrite it as $n=5p$. Therefore, the fraction $\frac{m}{n}$ can be simplified because the numerator and denominator have a common factor of $5$, which contradicts our assumption that there will be no common factors. This cannot be, so $\sqrt5$ cannot be rational. Therefore it’s irrational. q.e.d.

Answer 5

Suppose $\phi$ is rational. From $\frac{1}{\phi} = \phi - 1$, we see that $\phi$ satisfies the polynomial $\phi^2 - \phi - 1 = 0$. By the rational root theorem , $\phi = p/q$ where $p | -1$ and $q | 1$, forcing $\phi = \pm 1$. But neither of these is a root, contradicting that $\phi$ is rational. Therefore, $\phi$ is not rational.

Answer 6

Another way:

Let´s assume that φ is rational.

a/b=φ is completely reduced (we can do this when it is rational)

b<a (by definition)

a/b = (1+ √5)/2 < ( 1 + √9) / 2 = 2 → a < 2 b → a-b < b

From 1/φ = φ - 1

→ b/a = a/b -1 = (a - b)/b

b/a is completely reduced but is equal to another fraction with both a smaller denominator and a smaller numerator.

This is a contradiction, so φ is not rational.

I explained a bit more in :
https://www.valgetal.com/physics/Mathematics/Golden%20Ratio/Golden%20Ratio.htm