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Is $\frac{\sqrt7}{\sqrt[3]{15}}$ rational or irrational? Prove it.

I am having a hard time with this question. So far what I did was say, assume it's rational, then $$\frac{\sqrt7}{\sqrt[3]{15}}=\frac{x}{y} \Rightarrow \sqrt{7}y=x\sqrt[3]{15}$$ I then showed the product of a rational number and an irrational number is irrational so the expression above is irrational on the left and right side. I can't get to a contradiction to prove it's irrational so I'm currently thinking it is rational but I don't know how to go about proving it.

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    $\begingroup$ Take the $6^{\textrm{th}}$ power of each side and then argue as in standard proofs that $\sqrt{2}$ is irrational. $\endgroup$ – aes Feb 27 '15 at 1:25
  • $\begingroup$ We can assume $x$ and $y$ are relatively prime. Take the sixth power. We have $7^3 y^6=15^2x^6$. Argue that $7$ divides $x$ and therefore $7$ divides $y$ contradiction. Or argue that $3$ divides $y$ and therefore $x$. $\endgroup$ – André Nicolas Feb 27 '15 at 1:27
  • $\begingroup$ So it is irrational? I just want to be sure I'm right. It contradicts the unique factorization into primes so it must be irrational, right? $\endgroup$ – user212744 Feb 27 '15 at 1:34
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Factoring $\,7^{\,\large \color{}{3}} y^{\large 6}\!=15^{\large 2}x^{\large\color{}{6}}$ into primes, $\,7\,$ has odd power $\,3\!+\!6j\,$ on LHS, but even $\,6k\,$ on RHS

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    $\begingroup$ Note that the argument depends crucially on both the existence and uniqueness of prime factorizations, i.e. the Fundamental Theorem of Arithmetic. $\endgroup$ – Bill Dubuque Feb 27 '15 at 2:42
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Suppose the number is rational, let $$ \frac{\sqrt{7}}{\sqrt[3]{15}} = \frac{x}{y} $$ where $\gcd(x, y) = 1$, then $$ \sqrt{7}y = \sqrt[3]{15} x \implies 7^3 y^6 = 225 x^6 $$ So $$ 7^3 | x^6 $$ which implies $ 7| x $.

Let $x = 7m, \; m \in \mathbb N$. Then we have $$ 7^3 y^6 = 225 \times 7^6 m^6 $$ which means $$ y^6 = 225 \times 7^3 m^6 $$ once again we have $$ 7^3 | y^6 $$ which implies $ 7| y $. And this is contradictory to the coprimality between $x$ and $y$.

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If $x$ is rational then so is $x^3$, and hence so is $\frac{15}{7}x^3$.

Applying this to $$x=\frac{\sqrt7}{\root3\of{15}}$$ shows that if $x$ is rational then so is $\sqrt7$. But I expect you already know that this is not the case. Therefore $x$ is irrational.

(If you are not allowed to assume that $\sqrt7$ is irrational, it is pretty easy to prove.)

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Hint: This is the sixth root of $\dfrac{343}{225}$.

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hint:$7^3y^6 = 15^2x^6 \Rightarrow \left(\dfrac{x}{y}\right)^6 = r \Rightarrow t^6 - r = 0 \Rightarrow pt^6 - q = 0$. Use Eisenstein's criteria.

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