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The question is: Proove that $\left(\sqrt[3] \frac{q^2-1}{qx}\right)$ is irrational if x is irrational and nonzero and q is a rational number that is not 0 or 1.
I started my proof with: To get a contradiction, suppose that $\left(\sqrt[3] \frac{q^2-1}{qx}\right)$ is rational. Therefore $\left(\sqrt[3] \frac{q^2-1}{qx}\right)$ is in the set of rational numbers. So $\left(\sqrt[3] \frac{q^2-1}{qx}\right)$ = r. I know I now have to proove that x is irrational, but how?
Are these steps correct so far?

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  • $\begingroup$ It's an expression not an equation. $\endgroup$ – Deepak Feb 28 at 8:28
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From $$r=\frac{c}{d}=\sqrt[3]{\frac{q^2-1}{qx}}.$$ we obtain $$\frac{c^3}{d^3}=\frac{q^2-1}{qx}\Leftrightarrow\frac{c^3q}{d^3(q^2-1)}=\frac{1}{x}.$$ The LHS is rational, since $q,c,d$ are rational. Thus, $1/x$ is rational and $x$ is rational, contradiction. Note that $c^3q$ is an integer and $d^3(q^2-1)$ is a non-zero integer by assumption.

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  • $\begingroup$ Is it wrong if I say that x = $\left( \frac{q^3-1}{qr^3}\right)$ because r hasn't been proved rational? $\endgroup$ – Name Feb 28 at 8:40
  • $\begingroup$ q^2-1* I was hoping I could say that x was equal to that where q,1, and r is rational, but I realized that I couldn't prove r was rational. So I would be better off with this? $\endgroup$ – Name Feb 28 at 8:41
  • $\begingroup$ The assumption to obtain a contradiction is that $r$ is rational. So you can do that but should write a line about why $r$ is non-zero. $\endgroup$ – James Feb 28 at 8:42
  • $\begingroup$ Then the final proof would be : Restate what is written in the question, do algebra to solve for x, explain that r is rational and not 0, and then say that x equals the rational number, a contradiction. And would this be correct so I don't have to do it the way you said in your question? $\endgroup$ – Name Feb 28 at 8:45
  • $\begingroup$ That's what you can do. However, what I wrote in my answer is just the algebra to solve for $x$. The only difference I can see in what you wrote is that you keep $r$ as it is and don't write it as $c/d$ as I did. $\endgroup$ – James Feb 28 at 8:50
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You can continue by solving for $x$ in $\sqrt[3]{\frac{q^2-1}{qx}} = r$ and get a contradiction once you look at what you obtain.

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Let $c=\left(\sqrt[3] \frac{q^2-1}{qx}\right)$

Say $c$ is rational meaning we can express $c = \frac mn$, where $m$ and $n$ are integers.

Similarly since $q$ is rational we can write $q = \frac ab$, where $a$ and $b$ are distinct integers with $a \neq 0$.

You can do the algebra (including a cubing step) and express $x$ in terms of the other variables. You should easily be able to reach the required contradiction showing $x$ as a rational expression of two integer expressions.

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