# rational number

• March 21st 2010, 04:52 PM
elmo
rational number
prove $\sqrt{7}/^{3}\sqrt{15}$ is rational

• March 21st 2010, 06:03 PM
chiph588@
Quote:

Originally Posted by elmo
prove $\sqrt{7}/^{3}\sqrt{15}$ is rational

I'm assuming you mean irrational?
• March 21st 2010, 07:04 PM
chiph588@
Quote:

Originally Posted by elmo
prove $\sqrt{7}/^{3}\sqrt{15}$ is $\color{red}\text{ir}$ $\text{rational}$.

$a = \frac{\sqrt{7}}{\sqrt[3]{15}} = \sqrt[6]{\frac{7^3}{15^2}}$

Assume $a=\frac{p}{q}$, where $(p,q)=1$.

Then $a^6 = \frac{p^6}{q^6}$.
So $7^3\cdot q^6 = 15^2\cdot p^6$.

This means $7^3\mid p^6 \implies 7\mid p$ since $7$ is prime.
So $15^2\cdot p^6 =15^2\cdot (7k)^6 = 7^6\cdot 15^2\cdot k^6 = 7^3\cdot q^6\implies 7^3\cdot 15^2\cdot k^6 = q^6$.

Therefore $7 \mid q$ for the same reasons as above.
But this means $7\mid (p,q)$ which is a contradiction since we were under the assumption that $(p,q)=1$.

Therefore $\frac{\sqrt{7}}{\sqrt[3]{15}}$ is irrational.