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The question is: prove $x=\sqrt{3} - \sqrt{2}$ is not rational.

I can "prove" the above (ie. I saw the answer in my book) but can't quite understand it. $x = \sqrt{3} - \sqrt{2}$, $x+\sqrt{2}=\sqrt{3}$, $(x + \sqrt{2})^2 = 3$, $x^2+2\sqrt{2}x+2 = 3$, $\sqrt{2} = \frac{3-x^2-2}{2x}$.

This is a contradiction as $\sqrt{2}$ is not rational. Ok, I understand it's a contradiction, but it contradicts what?

I mean this proof didn't start with "let's assume [...]" so I don't know is the assumptions that is being contradicted. This is probably a very basic question, but can someone please explain where in the proof did they assume $\sqrt{3} - \sqrt{2}$ is rational?

Thanks!

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    $\begingroup$ If $x=\sqrt 3 - \sqrt 2$ were rational, then they show that $\sqrt 2$ would be rational (as it is a rational expression in $x$). $\endgroup$ – lulu Jun 17 at 10:12
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    $\begingroup$ Indeed, without the assumption that $x$ is rational, you cannot conclude that $\frac{3-x^2-2}{2x}$ is rational $\endgroup$ – Hagen von Eitzen Jun 17 at 10:15
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The assumption is right at the start: assume that $x=\sqrt{3}-\sqrt{2}$ is rational. Then follow through the algebraic manipulations that you've listed until you reach $$ \sqrt{2} = \frac{3-x^2-2}{2x}$$

Since $x$ is assumed to be rational, both the numerator and denominator of this fraction must be rational, which means that $\sqrt{2}$ is rational. And that's the contradiction.

(This assumes, of course, that you've seen or worked out the standard proof that $\sqrt{2}$ is irrational)

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  • $\begingroup$ Please allow me a dummy question... The text in my book never says "assume that $x=\sqrt{3} - \sqrt{2}$ is rational - I don't know where this came from. So, if I were to say assume $x = \sqrt{3} - \sqrt{2}$ is a real number (which it is), then the contradiction would show x isn't real (of course not). My point is that the (correct) assumption is crucial here (or so I think). I'm sorry to be a pain, I just want to understand this: how do you know the assumption is that the difference is rational?? $\endgroup$ – TheVoiceInMyHead Jun 17 at 12:34
  • $\begingroup$ @TheVoiceInMyHead You'd probably need to quote the full text from your book for us to be able to judge whether it was presented well or not I'm afraid. However, if the question is "prove that $x$ is irrational" then (standard) proof by contradiction would be to assume that $x$ is rational and work from there. Assuming $x$ is not real isn't motivated and you'd need to explain why you made that step as part of your proof. (If the question was "prove $x$ is purely imaginary" then "assume $x$ has non-zero real part" is a reasonable start for proof by contradiction) $\endgroup$ – postmortes Jun 17 at 12:38
  • $\begingroup$ Thanks for your time on this. To be honest it isn't quite clear to me but I think it's my book's fault. It doesn't state the assumption anywhere. I also don't quite understand why assuming the difference is rational matters here. I mean, If we were to assume the difference is rational, equal the difference to some a/b and then get to a contradiction - this I would understand. $\endgroup$ – TheVoiceInMyHead Jun 17 at 13:47
  • $\begingroup$ @TheVoiceInMyHead It sounds like your book could clarify its last step better, as it sounds to me like the last step is "now if we assume that $x$ is rational we have proven $\sqrt{2}$ is rational, which is impossible so $x$ must be irrational too". But if it doesn't explicitly say that then there's good reason for you to be a little puzzled. However it is common as you advance through mathematics to write down fewer and fewer of the "obvious" steps, so it's very good that you're taking the time to think about this and figure it out :) $\endgroup$ – postmortes Jun 17 at 13:51
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Also, after writing $$2\sqrt2x=1-x^2$$ we need to say that $x\neq0$(otherwise, $0=1$, which is impossible).

Thus, indeed, $$\sqrt{2}=\frac{1-x^2}{2x},$$ which is a contradiction because we assumed before that $x\in\mathbb Q$.

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