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My professor in class only did such a proof by contradiction. She would assume $r$ is rational, and contradict this in the end by showing that it cannot be, but I think a direct is cleaner. For my HW, I've done this:

Prove: $\sqrt{5 - \sqrt{3}}$ is not rational

Proof:

$\sqrt{5 - \sqrt{3}}$ solves the polynomial $x^4 - 10x^2 + 22 = 0$

by the rational zeroes theorem, all rational solutions to the equation possible are $\{\pm22,\pm11,\pm2,\pm1\}$

plugging these in to the polynomial, we see none of these are solutions.

therefore, since there are no rational solutions to the equation, which $\sqrt{5 - \sqrt{3}}$ solves, then $\sqrt{5 - \sqrt{3}}$ cannot be rational.

user129393192
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  • Seems correct to me. – Abezhiko Oct 08 '23 at 19:47
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    Curious: How did professor show irrational via contradiction? [And yes your solution correct] – coffeemath Oct 08 '23 at 19:48
  • I will outline @coffeemath : for $\sqrt{2}$, argue by contradiction, assume is rational, then $\sqrt{2} = \frac{p}{q}$ where $p, q$ relatively prime, in integers, and q not 0. $\sqrt{2}$ is solution to $x^2 - 2 = 0$, possible rational solutions is $\frac{p}{q} \in {\pm1, \pm2}$ and these don't satisfy, with set not containing solutions to the polynomial equation, which is a contradiction because $\frac{p}{q}$ should be a representation for $\sqrt{2}$, therefore our assumption is false and $\sqrt{2}$ is not in $\mathbb{Q}$. I can see her way being useful if a rational solution does exist. – user129393192 Oct 08 '23 at 21:24
  • Because if a rational does exist, then you show that $\sqrt{2}$ is not equal to that, and it must be if it were to be rational. That's my reasoning at least, but I'm not entirely certain. – user129393192 Oct 08 '23 at 21:25
  • @user129393192 What you have is for $\sqrt{2}$ however your number was more complicated, $\sqrt{5-\sqrt{3}}.$ – coffeemath Oct 08 '23 at 21:37
  • I gave you my teacher's proof she did in class @coffeemath – user129393192 Oct 08 '23 at 21:42
  • The one I gave was for my HW – user129393192 Oct 08 '23 at 21:42
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    @user129393192 Both your proof and your teacher's proof use the rational root theorem. Each proof begins with a polynomial p having integer coefficients which has the number x (whose irrationality is to be shown) as a solution to p=0. Then one lists possible rational roots and notes none works. So your proof is really the same idea as your teacher's, but for different numbers to show irrational. Either both or neither proof is "by contradiction". – coffeemath Oct 08 '23 at 22:09
  • Let us continue this discussion in chat. – coffeemath Oct 09 '23 at 02:28

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