15
$\begingroup$

I'm trying to prove that a certain number is irrational. I've taken a number theory class, so I'm familiar with the proofs that $\sqrt{2}$ and $\sqrt{3}$ are irrational (assume it is rational, square both sides, and then arrive at a contradiction). However, I've tried this method with this number to no avail. This is the number in question:

$\frac{p}{q} = \sqrt{13+\sqrt{52}} - \sqrt{13}$

$\frac{{p}^{2}}{{q}^{2}} = (13+\sqrt{52}) - 2(\sqrt{13+\sqrt{52}})(\sqrt{13}) + 13$

$\frac{{p}^{2}}{{q}^{2}} = 26 + 2\sqrt{13} - 2\sqrt{169 + 13\sqrt{52}}$

Once I get here, it just seems that I've made the question more complicated, or that I'm taking the wrong approach, and have no clue how to move on. Any help would be appreciated. Many thanks.

$\endgroup$
1
  • 7
    $\begingroup$ You may be familiar with the rational root theorem. Find a polynomial with integer coefficients that has that number as a root. $\endgroup$
    – dxiv
    Commented May 17, 2021 at 2:31

6 Answers 6

22
$\begingroup$

Let us assume $m = \sqrt{13 + \sqrt{52} } - \sqrt {13} $ as rational . Next we bring the $\sqrt {13} $ on the other side and then square both sides. This will give : $$ m^2 + 2\sqrt{13} m = \sqrt{52} $$ We rearrange terms and simplify $\sqrt{52} = 2 \sqrt{13} $ and get : $$ m^2 = 2 \sqrt{13} ( 1- m ) $$ Since we have assumed $m$ as rational , by closure $ m^2 $ and $ 2(1-m) $ are both rational but $\sqrt{13} $ is irrational . And it is not possible that we multiply a $\mathbf{non-zero}$ rational and an irrational number to get rational number , hence we arrive at a contradiction . Hence $m$ is irrational .

Based on comments below , we notice that $ m^2=0 $ will imply $m=1$ and similarly $1-m=0$ implies $ m=0 $ , and hence both are not possible .

$\endgroup$
7
  • 5
    $\begingroup$ Your approach is the simplest so far and most suitable for a typical high school curriculum. +1 $\endgroup$
    – Paramanand Singh
    Commented May 17, 2021 at 3:39
  • 3
    $\begingroup$ You should replace $p/q$ with $m$ everywhere. The numbers $p, q$ are not used at all in the argument. $\endgroup$
    – Paramanand Singh
    Commented May 17, 2021 at 3:41
  • 2
    $\begingroup$ @Paramanand Singh Thank you sir ! , I am a high school student , that is why I used that method . $\endgroup$
    – user925963
    Commented May 17, 2021 at 4:06
  • 5
    $\begingroup$ Wish you a great journey of mathematical excitement and enjoyment here! $\endgroup$
    – Paramanand Singh
    Commented May 17, 2021 at 4:07
  • 1
    $\begingroup$ You do need to eliminate the possibility $1-m = 0$ by other means: the correct statement should be that the product of a nonzero rational number and an irrational number cannot be rational. $\endgroup$ Commented May 17, 2021 at 21:47
10
$\begingroup$

Per dxiv's comment, you can find the minimal polynomial of this number to be $x^4-52x^2+104x-52$; I did so with a computer program. By the rational roots theorem, the only possible rational roots of this polynomial are $\pm 1, \pm 2, \pm 4, \pm 13,\pm 26, \pm 52$. But none of these are actually roots of this polynomial and so it has no rational roots, a contradiction.

$\endgroup$
6
  • $\begingroup$ I see. Thank you very much for your help. What computer program did you use to find that polynomial? $\endgroup$
    – martingale
    Commented May 17, 2021 at 3:00
  • 1
    $\begingroup$ It's called Sage Math and is a bit like Python if that's familiar to you $\endgroup$ Commented May 17, 2021 at 3:01
  • $\begingroup$ This is really tedious. Rational root theorem should be used only when it is easy to apply. $\endgroup$
    – Paramanand Singh
    Commented May 17, 2021 at 3:38
  • $\begingroup$ @ParamanandSingh True in general, but this one in particular looks like it was arranged to be easier to carry out. Squaring once $(x+\sqrt{13})^2 = 13 + 2 \sqrt{13}$ and rearranging the terms gives $x^2 = 2(1-x)\sqrt{13}$, then squaring one more time gives the polynomial, which is what the other answer did. $\endgroup$
    – dxiv
    Commented May 17, 2021 at 3:47
  • $\begingroup$ @dxiv: I am not saying about the finding minimal polynomial. I am just highlighting that checking 12 numbers to be root of a 4th degree equation is impractical if such a question comes in a high school exam. On the other hand the minimal polynomial is irreducible by Eisenstein so it can't have any rational root. Rational root theorem should be used for checking whether small numbers like $0,\pm 1,\pm 2 $ are roots or not. $\endgroup$
    – Paramanand Singh
    Commented May 17, 2021 at 3:50
8
$\begingroup$

Alternate way not using Rational root theorem:

We will only use the fact that $\sqrt{13}$ is irrational.

Let, $$a=\sqrt{13+\sqrt{52}} - \sqrt{13}, a>0$$

$$b=\sqrt{13+\sqrt{52}} + \sqrt{13}$$

Suppose that, $a$ is rational. Then we have:

$$b=a+2\sqrt{13}$$ is irrational.

This implies $$b^2=a^2+4a\sqrt {13}+52$$

is also irrational.

Then,

$$\begin{align}(a^2+b^2-52)^2&=(2a^2+4a\sqrt{13})^2\\ &=A+B \sqrt {13}.\end{align}$$

where $A,B\in\mathbb Q^{+}.$

This follows, $$(a^2+b^2-52)^2 \not \in\mathbb Q$$

Finally, we have

$$a^2+b^2=4\sqrt{13}+52$$

$$(a^2+b^2-52)^2 \in\mathbb Q$$

Thus, we get an obvious contradiction.

$\endgroup$
1
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Commented May 18, 2021 at 13:26
8
$\begingroup$

Here is another approach. Consider the number $a=\sqrt{13+\sqrt{52}}$ which is a root of polynomial $$(x^2-13)^2-52$$ or $$x^4-26x^2+117$$ which is irreducible over $\mathbb {Q} [x] $ (by Eisenstein criterion with $p=13$) and therefore $[\mathbb{Q} (a) :\mathbb{Q}] =4$.

Next we observe that $b=\sqrt{13}=(a^2-13)/2$ lies in $\mathbb{Q} (a) $ and $[\mathbb{Q} (b) :\mathbb{Q}] =2$. Hence we get $[\mathbb{Q} (a) :\mathbb{Q} (b)] =2$.

If $a-b$ were rational then this would mean that $a\in\mathbb {Q} (b) $ so that $\mathbb{Q} (a) =\mathbb{Q} (b) $ which contradicts $[\mathbb{Q} (a) :\mathbb{Q} (b)] =2$. Hence $a-b$ is irrational.


This method uses ideas from theory of field extensions and is not the preferred approach for such questions (which are mostly a part of high school curriculum). I took this route because other answers already covered the simpler approaches.

The standard approach of proving some algebraic real number as irrational involves computing the minimal polynomial for that number and verifying that the degree of this polynomial is greater than $1$. This standard approach is also simpler (given in another answer) where we get a minimal polynomial of degree $4$.

$\endgroup$
4
  • 1
    $\begingroup$ Can you please explain the last conclusion a-b is rational. $\endgroup$ Commented May 17, 2021 at 21:51
  • $\begingroup$ @LawrenceMano: updated my answer to add some explanation on that front. $\endgroup$
    – Paramanand Singh
    Commented May 18, 2021 at 2:12
  • $\begingroup$ (+1) Picking a nit: $a\in\mathbb{Q}(b)\implies\mathbb{Q}(a)\subset\mathbb{Q}(b)$ not equality. Of course, we can easily prove the opposite inclusion by noting that $b\in\mathbb{Q}(a)$. However, $\mathbb{Q}(a)\subset\mathbb{Q}(b)$ also contradicts $[\mathbb{Q}(a):\mathbb{Q}(b)]=2$. $\endgroup$
    – robjohn
    Commented May 18, 2021 at 16:28
  • $\begingroup$ Fully agree @robjohn $\endgroup$
    – Paramanand Singh
    Commented May 18, 2021 at 16:56
6
$\begingroup$

$13+\sqrt{52}$ is a root of $x^2-26x+117$, and so $\sqrt{13+\sqrt{52}}$ is a root of $x^4-26x^2+117$. Thus, $\sqrt{13+\sqrt{52}}$ is an algebraic integer.

$\sqrt{13}$ is a root of $x^2-13$ and is therefore an algebraic integer.

This answer shows that a difference of algebraic integers is an algebraic integer. This answer says that a rational algebraic integer is an integer.

Since $\sqrt{x}$ is strictly increasing, $$ \begin{align} 0&=\sqrt{13}-\sqrt{13}\\[15pt] &\lt\sqrt{13+\sqrt{52}}-\sqrt{13}\\[6pt] &=\frac{\sqrt{52}}{\sqrt{13+\sqrt{52}}+\sqrt{13}}\\ &\lt\frac{\sqrt{52}}{\sqrt{13}+\sqrt{13}}\\[9pt] &=1 \end{align} $$ So, $\sqrt{13+\sqrt{52}}-\sqrt{13}$ is an algebraic integer between $0$ and $1$, and is thus, not an integer; therefore, it is not rational.

$\endgroup$
1
  • $\begingroup$ I assume that the downvote is because I am using results about algebraic integers. However, there is nothing in the question or tags that would preclude these topics. Although there are simple computations that can be brought to bear on this particular question, I think that the more approaches that can be applied to a type of problem, the better. $\endgroup$
    – robjohn
    Commented May 23, 2021 at 18:15
4
$\begingroup$

Say $\sqrt{ 13 + \sqrt{52}} = a + b\sqrt{13}$, then $13 + 2\sqrt{13}= (a+ b \sqrt{13})^2$, take the conjugate and get $13 -2\sqrt{13}= (a- b \sqrt{13})^2$, now multiply and get $$117=169- 52 = (a^2 - 13 b^2)^2$$ not possible

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .