Proof by contradiction that an expression is irrational

The question is: Proove that $$\left(\sqrt \frac{q^2-1}{qx}\right)$$ 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 $$\left(\sqrt \frac{q^2-1}{qx}\right)$$ is rational. Therefore $$\left(\sqrt \frac{q^2-1}{qx}\right)$$ is in the set of rational numbers. So $$\left(\sqrt \frac{q^2-1}{qx}\right)$$ = r. I know I now have to proove that x is irrational, but how?
Are these steps correct so far?

• It's an expression not an equation. – Deepak Feb 28 at 8:28

From $$r=\frac{c}{d}=\sqrt{\frac{q^2-1}{qx}}.$$ 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.

• Is it wrong if I say that x = $\left( \frac{q^3-1}{qr^3}\right)$ because r hasn't been proved rational? – Name Feb 28 at 8:40
• 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? – Name Feb 28 at 8:41
• 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. – James Feb 28 at 8:42
• 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? – Name Feb 28 at 8:45
• 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. – James Feb 28 at 8:50

You can continue by solving for $$x$$ in $$\sqrt{\frac{q^2-1}{qx}} = r$$ and get a contradiction once you look at what you obtain.

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

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.