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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.

Elias
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  • 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 '19 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 '19 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 '19 at 19:31

2 Answers2

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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, $n=a^{p}$.

Manoj Kumar
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Let be

$$ a_0 x^n + \cdots a_n = 0 $$ 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 .

Luca Goldoni Ph.D.
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