# Proof that $\sqrt[p]{n}$ is irrational if $n$ is not a perfect pth power [duplicate]

i'm using Courant's book for self study, i would like to know if my proof that $$\sqrt[p]{n}$$ is irrational if $$n$$ is not a perfect pth power. Also would appreciate if someone do know where i can look for the solutions.

If $$n$$ is not a perfect pth power i can express it in terms of $$n^{pm + 1}$$. Assuming that $$\sqrt[p]{n^{pm+1}}$$ is rational i would have:

$$\sqrt[p]{n^{pm+1}} = \frac{k}{j} \\ n^{pm + 1} = \frac{k^p}{j^p} \\ n^{pm + 1}j^p = k^p \\ n^mn^{\frac{1}{p}}j = k \\ j^p = \frac{k^p}{n^{pm+1}} \\ j = \frac{k}{n^mn^{\frac{1}{p}}}$$

Which is a contradiction because they do have a common factor.

• It seems to me like circular logic! What you ended up with is just a rearrangement of the first equality/assumption! – Anastassis Kapetanakis Sep 11 at 18:24
• Your first step is confusing and wrong. You are thinking that you can write $n = s^{pm+1}$ for some integer $s$ if $n$ is not a $p$th power, but this is wrong. (Try $n=2$ and $p=3$.) I don't know that book, so I can't guess what technique you're supposed to try. Has he proved the Fundamental Theorem of Arithmetic, perhaps? – Ted Shifrin Sep 11 at 18:24
• @TedShifrin sorry about that, that's one of my first tries to proof something. He didn't yet proved the FTA. – Elias Sep 11 at 19:31

We will prove the contrapositive statement. Suppose that $$\sqrt[p]{n}$$ is rational i.e $$~ \sqrt[p]{n}=\frac{a}{b}, ~ b \ne 0$$. Assume that $$gcd(a,b)=1$$, so $$gcd(a^{p},b^{p})=1.$$ Now, $$n = \frac{a^{p}}{b^{p}}$$, but $$gcd(a^{p},b^{p})=1$$ and $$n$$ is an integer implies that $$b^{p}=1$$. So, $$n=a^{p}$$.
$$a_0 x^n + \cdots a_n = 0$$ a polynomial equation where $$a_i \in \mathbb Z$$. Then if $$x$$ is a rational root of this equation it must be of the form $$x=\frac{r}{s}$$ where $$r$$ is a divisor of $$a_n$$ and $$s$$ a divisor of $$a_0$$. Now consider the equation $$x^p-n=0.$$ You would have that $$r^p=n$$ which is a contradiction .