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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.

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  • $\begingroup$ You could start by proving that $\sqrt{5}$ is irrational, and then use the fact that sum and product of a nonzero rational and irrational number are irrational. $\endgroup$ – mechanodroid Sep 13 '17 at 16:35
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    $\begingroup$ Another possible approach: $\phi$ is a root of $x^2 - x - 1$, but the only possible rational roots would be $\pm 1$, for example by using Gauss's lemma on factorization of polynomials in $\mathbb{Z}[x]$ - and those obviously aren't roots. $\endgroup$ – Daniel Schepler Sep 13 '17 at 19:15
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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).

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  • $\begingroup$ Excellent solution! $\endgroup$ – Ken Draco Sep 13 '17 at 17:19
  • $\begingroup$ Because the GCD of the two denominators would have to be greater than 1 if both a and b were rational? $\endgroup$ – CJ Dennis Sep 14 '17 at 6:18
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    $\begingroup$ @CJDennis no, but because $\operatorname{gcd}(a,b) = \operatorname{gcd}(b-a,b)$ whence both fractions are in reduced form which is unique, i.e., $a=b$, contradiction. $\endgroup$ – yo' Sep 14 '17 at 10:13
  • $\begingroup$ Sorry for my ignorance but I am struggling to see why this is a contradiction. We assumed $a\over b$ was fully reduced and then showed ${a-b}\over b$ was fully reduced... how is this a contradiction? $\endgroup$ – Crystal Dec 1 '19 at 23:13
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    $\begingroup$ @Thomas: one can think of it like so - by assumption, a > b so let c = a - b. Then we have b / a = c / b. Now, since all are integers, and a > b, we should be able to reduce the fraction on the left side to obtain exactly the fraction on the right be cancelling some terms; that is: b = kc and a = kb. But by assumption gcd(a, b) is 1. $\endgroup$ – Nitin Jan 9 '20 at 9:17
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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

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  • $\begingroup$ This really helps! Thank you very much $\endgroup$ – Michelle Drolet Sep 13 '17 at 16:41
  • $\begingroup$ ok you are welcome $\endgroup$ – Dr. Sonnhard Graubner Sep 13 '17 at 16:41
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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.

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  • $\begingroup$ For the uninitiated, what is that vertical line that you use in your conclusion? $\endgroup$ – Dancrumb Sep 14 '17 at 13:57
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    $\begingroup$ it means "divides" $\endgroup$ – zwim Sep 14 '17 at 14:03
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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.

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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.

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