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This is a Tutorial sheet problem for my module in college for Introduction to analysis. The lecturer isn't the best and tends to not fully explain things and he gets most his solutions wrong too. We have never been showed how to do this proof, and any help would be very much appreciated.

The question is:

Let $p$ be a prime number and $k ∈ \mathbb{N}$ with $k ≥ 2$. Prove that there does not exist $q ∈ \mathbb{Q}$ such that $q^k = p$.

I know that $\mathbb{Q}$ is a set of rational numbers, I think.

I started with $a,b∈\mathbb{Q}$ where either $a=0$ or $b=0$ which then gives us either $a+b=a$ or $a+b=b$. We know from this that $a+b∈\mathbb{Q}$.

It's the next part I'm struggling with. An example given was that $\sqrt[k]{p}$ is not a rational number.

Christian E. Ramirez
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ADinar
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  • Usually, $\mathbb Q$ denotes the set of rational numbers. But then, of course we can take $q=\frac pk$ and get a rational number. You should seek clarification from your instructor, but I really can't imagine someone using $\mathbb Q$ to denote the irrationals. – lulu Feb 09 '23 at 20:05
  • sorry i mean rational not irrational ill edit it thanks – ADinar Feb 09 '23 at 20:07
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    @MatthewTowers Ah, that makes sense. In that case, the argument is analogous to the standard proof that $\sqrt 2\not \in \mathbb Q$. – lulu Feb 09 '23 at 20:09
  • If you wrote q- a rational number, as a ratio in lowest terms between two integers m, n, then those integers are relatively prime. But $q^k=p$, meaning p is a prime factor of $m^k$, or p is a factor of m. This leads to p being a factor of n, which contradicts that m,n are relatively prime. – WindSoul Feb 09 '23 at 20:21
  • @lulu i think it does. I'll try proving it now. thanks, it means a lot. – ADinar Feb 09 '23 at 20:32

1 Answers1

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$$q\in \mathbb Q\Rightarrow \exists m,n \in \mathbb Z, n \ne 0, \text{m,n relatively prime}:q=\frac{m}{n}$$

$$\begin{align}q^k=p\Rightarrow \left(\frac{m}{n}\right)^k=p&\Rightarrow m^k=p\times n^k\\&\Rightarrow \underbrace{m\equiv 0\mod p}_{\text{p is a factor of m}}\\&\Rightarrow n\equiv 0\mod p\\&\Rightarrow \text{m,n share common factor p}\\ &\Rightarrow \text{m,n are not relatively prime}\end{align}$$

$$\therefore q^k=p\Rightarrow q \notin \mathbb Q$$

WindSoul
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