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I am trying to learn some discrete mathematics alongside my arts course to try and expand my knowledge. I have this question:

Prove that if $x$ is irrational then $\frac{x+1}{x-1}$ is irrational.

My working is as follows and I'm not sure if this is a valid way of proving the theorem? (If indeed it is called a theorem, I've not yet fully grasped the lingo.

Proof by contraposition, so if

$\frac{x+1}{x-1}$ is rational then $x$ is rational

If $x$ is rational then $x = \frac{a}{b}$ where a, b are integers

substitute $x = \frac{a}{b}$ into $\frac{x+1}{x-1}$

to get

$\frac{\frac{a}{b}+1}{\frac{a}{b}-1}$

simplify

$\frac{\frac{a+b}{b}}{\frac{a-b}{b}}$

multiply the numerator and denominator by $\frac{b}{b}$ which is equivalent to multiplying by 1

to get

$\frac{a+b}{a-b}$

As a,b are integers, and an integer over an integer will yield a rational number, this proves the initial statement by contraposition?

The thing is, I don't think this addresses the fact the denominator can't be $0$?

Apologies, this is probably a rubbish question and I've probably not used this maths formatting language properly :[

Thanks for your help

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  • $\begingroup$ What you actually showed is: "If $x$ is rational (and $\ne1$) then $\frac{x+1}{x-1}$ is rational" Or equivalently "If $\frac{x+1}{x-1}$ is irrational then $x$ is irrational". $\endgroup$ – Hagen von Eitzen Oct 17 '18 at 6:12
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No, it is not correct. What you actually did was to prove that if $x$ is rational, then $\frac{x+1}{x-1}$ is rational too. A proof by contradiction would consist in assuming that $\frac{x+1}{x-1}$ is rational and to deduce from that that $x$ is rational too. That's not hard, since$$x=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}.$$

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  • $\begingroup$ Hi, thank you so much :) ! $\endgroup$ – potatonotjacket Oct 17 '18 at 15:05
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No need to be so apologetic! You can start by assuming that $x \neq 1$ and $\frac {x+1} {x-1}$ is rational. Then your argument shows that $x$ is rational. You can now conclude that if $\frac {x+1} {x-1}$ is rational then either $x=1$ or $x$ is rational, hence rational in all cases. This takes care of the problem of denominator becoming $0$.

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  • $\begingroup$ Thank you so much! $\endgroup$ – potatonotjacket Oct 17 '18 at 15:05

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