Proof $t$ is irrational $ t = a-bs $ , Given $a$ and $b$ are rational numbers, $b \neq 0$ and $s$ is irrational. Hence show that $(\sqrt3-1)/(\sqrt3+1)$ is irrational

  • $\begingroup$ Can you simplify $(\sqrt3-1)/(\sqrt3+1)$? $\endgroup$ – Lord Shark the Unknown Nov 17 '18 at 12:52
  • $\begingroup$ The problem is in two parts, and it's not clear to me whether you are saying that you have already solved the first part or not. $\endgroup$ – TonyK Nov 17 '18 at 12:55
  • $\begingroup$ Try to prove that $t$ is irrational by contradiction i.e. assume that $t$ was a rational number. $\endgroup$ – clark Nov 17 '18 at 13:00
  • $\begingroup$ @TonyK I got stuck not sure. $t=a-bs$ I assumed t to be rational i.e $p/q$ where $p$ and $q$ member of integers and $p\neq 0$ thus $p/(a-bs)=q$ So p is divisible by $a-bs $ but don't know how to find a divisor for $q$ to disqualify $t$ as rational $\endgroup$ – Eric kioko Nov 17 '18 at 14:15
  • $\begingroup$ You have not used the first part to prove the second part. Rather, you have used the same method of proof. But you don't need to go through that again! You can just put $a=2$, $b=1$, and $s=\sqrt 3$, and apply what you proved in the first part. $\endgroup$ – TonyK Nov 17 '18 at 15:23

For the first part: in a comment you have solved for $s$ in terms of $a$, $b$ and $t$. You know the first two are rational. What could you say about $s$ if $t$ were rational too?

You have started the second part correctly. Now it's in a form where you can apply the first part.

  • $\begingroup$ Thanks for the eye opener, I have attempted check to see if I'm right. $\endgroup$ – Eric kioko Nov 17 '18 at 15:14
  • $\begingroup$ WIth your eyes open you have a pretty much correct solution. You need the fact that the rationals are closed under division, too, which you use but don't state. For part 2 you have given a direct proof based on the idea from part 1, but didn't explicitly use that result, so the "hence" in the question hasn't been addressed. You could edit the question to remove the solution and then post the solution as an answer to your own question. That's allowed here, and it will take the question off the unanswered queue. $\endgroup$ – Ethan Bolker Nov 17 '18 at 15:18

For the first part

$a/b -s = t/b$


Assuming t was rational would mean $ s $ is rational since rational numbers are closed under addition and subtraction.

This contradicts s being irrational. Therefore t is irrational.

The second part $(\surd3-1)/(\surd3+1) * (\surd3-1)/(\surd3-1) $


substituting $2-\surd3$

in $t=a-bs$



Therefore $t=2-\surd3$ is irrational too.

  • $\begingroup$ You can avoid the checks: $2-\sqrt{3}$ is irrational because $s=\sqrt{3}$ is irrational and you can take $a=2$, $b=1$. $\endgroup$ – egreg Nov 17 '18 at 16:56
  • $\begingroup$ @egreg Thanks, that's how its supposed to be done. I'm gonna edit my answer to reflect that. $\endgroup$ – Eric kioko Nov 21 '18 at 18:53

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