Prove that $\sqrt{\sqrt{7}-\sqrt{5}}$ is irrational.

Question

$\sqrt{\sqrt{7}-\sqrt{5}}$

I need to prove this is irrational.

I don't know how.

$\sqrt{\sqrt{7}-\sqrt{5}}$ = $\frac{m}{n}$

$\sqrt{7}-\sqrt{5}$= $(\frac{m}{n})^2$

($\sqrt{7}-\sqrt{5})*n^2$= $m^2$

Answer 1

You can write $(\sqrt{\sqrt7-\sqrt5})^4=7+5-2\sqrt{35}$, we deduce that $\sqrt{\sqrt7-\sqrt5}$ is rational implies that $\sqrt{35}$ is rational.

Write $\sqrt{35}={p\over q}$, where $\gcd(p,q)=1$, $35q^2=p^2$ implies that $5,7$ divides $p$, write $p=35a,p^2=(35)^2a^2=35q^2$ implies that $q^2=35a^2$ and $5,7$ divides $q$ contradiction since $\gcd(p,q)=1$.

Answer 2

Suppose $\sqrt{\sqrt{7}-\sqrt{5}}$ is rational. Then exist rational number $q$ such that:$$\sqrt{\sqrt{7}-\sqrt{5}}=q$$

So $$7-2\sqrt{35}+5= q^4$$

so $$\sqrt{35} = {12-q^4\over 2}$$

We have a contradiction, since on the right side of equation is rational number and on the left is not.

Answer 3

Clearly, it is enough to prove that $\sqrt7-\sqrt5$ is irrational. But it is not hard to see that $\sqrt7-\sqrt5$ is a root of the polynomial $x^4-24x^2+4$. By the rational root theorem, the only rational numbers which can be a root of this polynomial are $\pm1$, $\pm2$, and $\pm4$. Since $\sqrt7-\sqrt5$ is none of them, it is irrational.

Answer 4

Set $x=\sqrt{\sqrt 7-\sqrt5}$. We have $x^2=12-2\sqrt{35}$, whence $(x^2-12)=-2\sqrt{35}$, and squaring we obtain the equation $$x^4-24x^2+4=0 $$ If $x$ is rational, by the rational root theorem, it is an integer which divides the constant term, $4$. As this is a biquadratic equation, we only have to test $1,2$ and $4$. None of these satisfies the equation.