Irrational Numbers

Could someone check my work please? The Problem is: Prove that the cube root of any irrational number is an irrational number.


I let $x$ be some irrational number. Then I tried to use proof by contradiction. I assumed that $x^{\frac{1}{3}} = \frac{p}{q}$, where $p$ and $q$ are integers in lowest terms. Therefore $xq^3 = p^3$ and since p and q are in lowest terms this equation implies that $p|x$, however $x$ is an irrational number and so it's impossible for it to be divisible by another integer. This is a contradiction. Our initial assumption that $x^\frac{1}{3}$ was rational must therefore be false and so the cube root of $x$ must be irrational.

Therefore $xq^3 = p^3$ and since p and q are in lowest terms this equation implies that $p|x$

You can consider a prime factorisation of x though and use the fact that if a is prime a|b^n if and only if a|b.

So in your working you should have x is prime and x|p or some prime factor of x divides p.

This is not true. The equation implies x and q^3 divides p^3. Work from there.

You can just observe that p^3 and q^3 are both rational, and that irrational mult by (non-zero) rational must be irrational.

So from xq^3 = p^3, we can say that p^3 is irrational and similarly q^3 is irrational as well. So is it fine to say that since p and q were assumed to be integers a product of integers can't result in an irrational number, which is the contradiction we need?

No.

You assumed p,q are integers with q being non-zero. Hence, p^3 and q^3 are rational. But we have xq^3=p^3 where x is irrational.

If you can’t see the contradiction here, it might be good to prove a simple lemma of rational*irrational=irrational

Tbh you should see the problem just from x=p^3/q^3

So p^3 and q^3 are both rational. But xq^3 = p^3 is irrational *rational, so that implies p^3 is irrational, which is a contradiction. (Why don't we say that p^3 and q^3 are integers as p and q are both integers, does it matter? Or is that just because every integer is a rational number and it's better to refer to rational numbers in this case?)

With regards to seeing it from x = p^3/q^3, do you mean that the LHS is irrational, whereas the RHS is a rational number divided by another rational number i.e. a rational number, so that is also a contradiction?

x is irrational, not sure why he needs to assume its prime to deduce that the cube root is irrational. It doesn’t have prime factors.

First part: yes, you can say they are integers but it’s better to use rational/irrational since that’s what we’re looking at and comparing.