For all natural numbers $n$, if $n$ is odd, then $\sqrt{15^n}$ is irrational.

Question

How can I show that for all natural numbers $n$, if $n$ is odd, then $\sqrt{15^n}$ is irrational? I have tried to use a proof by contradiction to no avail. I have gotten decently far with a proof by contraposition, where I assume that $\sqrt{15^n}$ is rational and prove that $n$ is even.

For integers $x$ and $y$:

\begin{align*} \sqrt{15^n} &= \frac{x}{y} \\ 15^n &= \frac{x^2}{y^2} \\ 15 &= \frac{x^{\frac{2}{n}}}{y^{\frac{2}{n}}} \end{align*}

From here, I don't know how I would isolate the $n$ to prove that it is even. Any help would be greatly appreciated.

Answer 1

I'd start by saying that if $n$ is the odd natural numbers then it can be represented as: $$n=2k+1\quad\forall k\in\mathbb{N}_0$$

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Secondly, $$15=5\cdot3\rightarrow \sqrt{15^n}=5^{n/2}3^{n/2}$$

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now substitute in our form for $n$ and we get: $$5^k3^k\sqrt{5}\sqrt{3}$$ or: $$15^k\sqrt{15}$$ then we know that: $$15^k\in\mathbb{Q}\forall k,\sqrt{15}\notin \mathbb{Q}$$ and so: $$15^k\sqrt{15}\notin\mathbb{Q}$$

Answer 2

If $n=2k+1$ for some $k \in \mathbb{Z}$, then $$ \sqrt{15^n} = 15^{(2k+1)/2} = 15^k \sqrt{15}. $$ So if $\sqrt{15^n} \in \mathbb{Q}$, then $\sqrt{15} \in \mathbb{Q}$, as well. Can you show $\sqrt{15}$ is irrational?