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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.

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    $\begingroup$ Yes that's fine! $\endgroup$ – gimusi Aug 31 '18 at 13:45
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    $\begingroup$ There's no need for the "$-\{0\}$" in "$1\in\mathbb{Q}-\{0\}$" to show that $1+\sqrt5$ is irrational, but there is a need for it in $r\in\mathbb{Q}-\{0\}\implies rx\not\in\mathbb{Q}$. $\endgroup$ – Barry Cipra Aug 31 '18 at 13:50
  • $\begingroup$ There's a short proof in Wikipedia, and essentially the same proof is in this answer. I dimly recall reading somewhere that the irrationality of $\phi$ was known to the ancient Greeks before the irrationality of $\sqrt{2}$, but I can't find a reference right now, and my ancient memory may be playing tricks. $\endgroup$ – Calum Gilhooley Aug 31 '18 at 15:32
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Another approach:

Continued fractions are finite for rationals. Now, try to find the continued fraction representation for $\frac{1+\sqrt{5}}{2}\big($you may use the fact that it satisfies the equation $x^2-x-1=0\big)$, it will be $[1;1 ,1,1,\cdots] $, which has infinitely many $1$. Conidering the contrapositive statement of the statement above, if a number has infinitely many terms in it's cont. fraction representation then it is not rational. Hence, $\frac{1+\sqrt{5}}{2} $ is not rational or irrational.


Another one:

Let, $\frac{1+\sqrt{5}}{2}=\frac{p}{q},p,q\in\mathbb{Z}$, then $\sqrt{5}=\frac{2p-q}{q}$. Now, we can prove that $\sqrt{5}$ is irrational using this argument:

$\sqrt{5}$ satisfies the monic polynomial $x^2-5=0$, hence, if it is an rational algebraic number it need to be an integer. But, $2<\sqrt{5}<3$ (as, $4<5<9$). Hence, $\sqrt{5}$ is irrational.

Now come back to the equation $\sqrt{5}=\frac{2p-q}{q}$. As, $p,q\in\mathbb{Z}$, $\frac{2p-q}{q}\in\mathbb{Q}$, as, both $2p-q,q\in\mathbb{Z}$. But, we have shown that $\sqrt{5}$ is irrational. Hence, contradiction!

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  • $\begingroup$ In the second proof you show that $\sqrt{5}$ is irrational, assume that it is rational and then show a contradiction? What? $\endgroup$ – LionCoder Sep 2 '18 at 11:22
  • $\begingroup$ en.wikipedia.org/wiki/Algebraic_number $\endgroup$ – tarit goswami Sep 2 '18 at 13:49
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    $\begingroup$ But $\sqrt{5}$ is algebraic $\endgroup$ – LionCoder Sep 3 '18 at 8:14
  • $\begingroup$ Ok, I will fix my words. It is algebraic but not rational. $\endgroup$ – tarit goswami Sep 3 '18 at 8:47
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    $\begingroup$ @LionCoder It is algebraic, but if it is rational algebraic then it need to be a integer, as the minimal polynomial is monic. But, as $2<\sqrt{5}<3$, means it lies between two consecutive integers, so it is not a integer. Hence, $\sqrt{5}$ is not a rational algebraic number. Thank you for finding the error :) $\endgroup$ – tarit goswami Sep 3 '18 at 8:51
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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.

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    $\begingroup$ Reminded me of this xkcd $\endgroup$ – rafa11111 Aug 31 '18 at 14:12
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    $\begingroup$ @rafa11111 LoL. I actually recently came across the mentioned theorem in a previous question a couple of days ago (I had forgotten about its existence) and went for it :) $\endgroup$ – b00n heT Aug 31 '18 at 15:32
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    $\begingroup$ Or, you could combine this with the other argument: if $x = \frac{a}{b}$ in lowest terms then $a^2-ab-b^2 = 0$. So, if $p$ is prime and $p \mid b$, that implies $p \mid a^2$ so $p \mid a$, giving a contradiction. And similarly if $p \mid b$. So, $a,b=\pm 1$ which is ridiculous. $\endgroup$ – Daniel Schepler Sep 6 '18 at 0:58
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Yes, this is basically how you prove that $\sqrt n$ is irrational. In general the argument is that, if $n$ is not a square it has at least prime factor $p$ of odd exponent in the factorization of $n$, but then if $a^2=b^2 n$, and so the exponent of $p$ is odd on the right but even on the left, contradiction. This is just a generalization of your argument.

About the way you proved that if $p|n^2$ then $p|n$ for prime $p$ via FTA, not only is this not an inelegant method, I think it's basically how anyone who knows number theory would do it, and it's probably the method that most clearly explains "why" the theorem is true. Whatever simpler method you know for the case $p=2$, it's probably either just this method in disguise, or it involves a property of the number $2$ that really doesn't generalize to other primes.

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Proposition 1: Let $a$ and $b$ be two nonnegative integers satisfying

$\tag 1 ab + b^2 = a^2$

Then both $a$ and $b$ must be equal to zero.

Proof:

Note that if $a \gt 0$ then the only way $\text{(1)}$ can hold is if $b \gt 0$, and similarly, if $b \gt 0$ then $a \gt 0$. So to get a contradiction, assume we have two positive integers satisfying $\text{(1)}$.

It is easy to see that both $a$ and $b$ would necessarily have to be strictly greater than $1$ in this scenario. Moreover, since $ab + b^2 = (a + b) b$, using the Fundamental theorem of arithmetic and $\text{(1)}$ , we have to conclude that $b$ and $a$ can't be relatively prime.

We can certainly restrict our attention to solutions of $\text{(1)}$ where the sum of $a$ and $b$ is minimal. But if $d$ is a common factor so that $a = da^{'}$ and $b = d b^{'}$, then

$\quad ab + b^2 = a^2 \text{ iff } d a^{'} d b^{'} + {(d b^{'})}^2 = {(d a^{'})}^2$

We can factor out $d^2$ in the above equation giving us another 'descending solution' $a^{'}$ and $b^{'}$, contradicting our choice of $a$ and $b$. $\quad \blacksquare$

Proposition 2: If $a$ and $b$ are real numbers satisfying

$a > b > 0 \text{ and } \frac {a+b}{a}=\frac {a}{b}$

then $\frac {a}{b}$ must be an irrational number.

Proof:

To get a contradiction, assume to the contrary that $\frac {a}{b} = \frac {b^{'}}{a^{'}}$ with integers $a^{'}$ and $b^{'}$. Using algebra, we can show that $a^{'} b^{'} + {a^{'}}^2 = {b^{'}}^2$, but by proposition 1 that is impossible. $\quad \blacksquare$


In the above we show that if solutions can be found in the context of proposition 2, then the ratio $\frac{a}{b}$ is irrational. It wasn't necessary to find the existence of solutions or to discuss why the ratio $\frac{a}{b}$ is unique (no reason to examine $\sqrt 5$).

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The book A History Of Mathematics, by Boyer and Merzbach, suggests that the Golden Ratio $\psi$ may have been the first number known to be irrational. They present a proof that plausibly could have occurred 25 centuries ago:

By contradiction, suppose $\psi =A/B$ where $A, B \in \Bbb N$ and $B$ is as small as possible.

We have $1=\psi^2-\psi\implies 1/\psi=\psi -1 \implies \psi=1/(\psi - 1).$ So $\psi=A/B=1/(A/B-1)=B/(A-B).$

Now $B>A-B>0$ because $2B>A>B$ because $2>A/B=\psi>1.$

So the denominator in $B/(A-B)$ is a positive integer, smaller than the smallest possible, which is $B,$ which is absurd.

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