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

Question

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.

Answer 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 .

Answer 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.

Answer 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.

Answer 4

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 $4$.

Answer 5

$13+\sqrt{52}$ 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.

Answer 6

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