# rational number

• Mar 21st 2010, 03:52 PM
elmo
rational number
prove $\displaystyle \sqrt{7}/^{3}\sqrt{15}$ is rational

• Mar 21st 2010, 05:03 PM
chiph588@
Quote:

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

I'm assuming you mean irrational?
• Mar 21st 2010, 06:04 PM
chiph588@
Quote:

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

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

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

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

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

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

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