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1) Prove that number irrational $\sqrt{7-\sqrt{2}}$

I created a polynomial $x=\sqrt{7-\sqrt{2}}$ so

$P(x)=x^4-14x^2+47$ and since $47$ is prime we check $P(x)$ for $ {1,-1,47,-47}$ and since all of them are $P(x)\neq0$ it means our number is irrational.

Is my prof OK ?

2) Decide if the number $\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$ is rational or irrational. I don't know how to tackle this one. I'd be grateful for hints

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  • $\begingroup$ The approach in your first proof is correct, though it could be a bit more polished: $P(x)$ certainly has this number as a root, and then you apply the Rational Root Theorem. As for the second, think about how you created the polynomial in 1) so that it'd include that root. $\endgroup$ – Semiclassical Jul 31 '14 at 14:20
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    $\begingroup$ 1) That's the correct idea, but check your numbers. $\endgroup$ – Empy2 Jul 31 '14 at 14:20
  • $\begingroup$ 2) Since there are four square-roots, I suspect the polynomial is going to be degree $2^4=16$! $\endgroup$ – Empy2 Jul 31 '14 at 14:23
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    $\begingroup$ It should be $P(x) \equiv x^4 - 14 x^2 \color{red}{+ 47}$ and not $-47$ there. $\endgroup$ – hjpotter92 Jul 31 '14 at 16:23
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    $\begingroup$ If you square your number, you get 7 - sqrt (2) which is irrational. Squaring a rational produces a rational, so you must have squared an irrational number. $\endgroup$ – gnasher729 Jul 31 '14 at 22:20
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For 2).

If $p=\sqrt{\sqrt 5+3}+\sqrt{\sqrt 5-2}$ is a rational number, then $$\begin{align}p-\sqrt{\sqrt5+3}=\sqrt{\sqrt5-2}&\Rightarrow p^2-2p\sqrt{\sqrt5+3}+\sqrt5+3=\sqrt5-2\\&\Rightarrow 2p\sqrt{\sqrt5+3}=p^2+5\\&\Rightarrow 4p^2(\sqrt 5+3)=(p^2+5)^2\\&\Rightarrow \sqrt5=\frac{(p^2+5)^2}{4p^2}-3\end{align}$$ implies that $\sqrt 5$ is a rational number. This is a contradiction.

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  • $\begingroup$ Very smooth! I am curious - can the irrationality of all algebraic irrationals, no matter how complex the number, be demonstrated by this "inductive" method (given the irrationality of all prime roots is a given) or are there any exceptions? $\endgroup$ – Just_a_fool Jul 31 '14 at 14:54
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    $\begingroup$ @Just_a_fool The reason that the trick works here is that the 'inner' roots are the same, so after two squaring steps you're left with a single square root. OTOH, the method that you used for 1) (with the rational root theorem) can be used as an algorithm to determine the rationality or irrationality of an arbitrary algebraic expression (since the equation gives you a finite number of possible values to test against and all roots can be computed to sufficient precision). OTOH, there's a lot of subtlety in this; look up the 'sum-of-square-roots' problem in computational geometry for details.. $\endgroup$ – Steven Stadnicki Jul 31 '14 at 17:06
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You don't need all that! If $\sqrt{7-\sqrt 2} = \dfrac{p}{q}$ is rational, then $\sqrt 2 = 7 - \dfrac{p^2}{q^2}$is rational. Which it isn't.

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