Proving that $\sqrt{13+\sqrt{52}} - \sqrt{13}$ is irrational.

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.

Solution 1:

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 .

Solution 2:

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.

Solution 3:

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.