If $p$ is prime and any integer $k>1$, then $p^{\frac 1k} $ is irrational. Prove this by assuming $p^{\frac 1k}$ rational

Question

I've tried setting $p^{\dfrac 1k}= \dfrac a b$, and then raising $p^{\dfrac 1k}$ to the $k^\text{th}$ power, but I'm stuck.

where did you get stuck? Because your idea is good! – sranthrop Oct 09 '13 at 16:34 — sranthrop, Oct 09 '13 at 16:34
http://math.stackexchange.com/a/324729/25928 – Baby Dragon Oct 09 '13 at 16:42 — Baby Dragon, Oct 09 '13 at 16:42
Remember $p\mid nm\implies p\mid n$ or $p\mid m$. – Pedro Oct 09 '13 at 16:46 — Pedro, Oct 09 '13 at 16:46

score 4 · Answer 1 · answered Oct 09 '13 at 16:34

4

Without loss of generality, assume that $(a,b)=1$. If $a^k = p b^k$ then what can you say about $a$ (and then, using $k \geqslant 2$, about $b$)?

answered Oct 09 '13 at 16:34

njguliyev

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score 0 · Accepted Answer · answered Oct 09 '13 at 16:35

let $p^{\frac1k}$ be rational then $p^{\frac1k}=\frac ab$ with $(a,b)=1$ or $p=\frac {a^{k}}{b^{k}}$,

since p is a prime hence integer we have $b=1$ and $p=a^k$, and if $k>1$ then p will be divisible by more then two numbers, hence it will not be prime, a contradiction, therefore $p^{\frac1k}$ is irrational for $k>1$ and p be a prime.

score 0 · Answer 3 · answered Oct 09 '13 at 16:48

Assuming $p^{1/k}=\frac{a}{b}$is in lowest terms. Taking $k$-th powers and rewriting gives: $b^kp=a^k$. By the funamental theorem of arithmetic the prime factorization of any positive integer is unique. Now note that the number of times $p$ appears on the left hand side will be more than the number of times it is on the right hand side. This violates this uniqueness.

If $p$ is prime and any integer $k>1$, then $p^{\frac 1k} $ is irrational. Prove this by assuming $p^{\frac 1k}$ rational

3 Answers3