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It's been proven that √7 and √2 are irrational.

However, I am not sure how to go about proving that √7 - √2. Is it an acceptable proof to just solve the equation which would prove/disprove the equation or as should the proof be done as a contrapositive, similar to how √7 and √2 are proven to be irrational.

What would a valid proof/disproof of irrationality look like in this case?

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marked as duplicate by Hans Lundmark, Chase Ryan Taylor, Parcly Taxel, The Phenotype, Arnaud Mortier Feb 8 '18 at 13:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ One way is to find the equation with integer coefficients that $\,\sqrt{7}-\sqrt{2}\,$ satisfies, then prove that it has no rational roots. $\endgroup$ – dxiv Feb 8 '18 at 5:42
  • $\begingroup$ not quite sure what you mean by integer coefficients, could you provide an example? $\endgroup$ – Jessica Tiberio Feb 8 '18 at 5:44
  • $\begingroup$ Please use MathJax to format your posts. $\endgroup$ – Chase Ryan Taylor Feb 8 '18 at 6:15
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Suppose $\sqrt{7}-\sqrt{2}$ were rational; that is, suppose $$\sqrt{7}-\sqrt{2}=\frac{a}{b}, $$ where $\text{gcd}(a,b)=1$. Multiply both sides of the equation by $\sqrt{7}+\sqrt{2}$ to obtain $$ 5=7-2=(\sqrt{7}-\sqrt{2})(\sqrt{7}+\sqrt{2}) = \frac{a}{b}(\sqrt{7}+\sqrt{2}). $$ Since $\frac{5b}{a}\in\mathbb{Q}$, $\sqrt{7}+\sqrt{2}$ is also a rational number. Since the sum of two rational numbers is rational, $$ (\sqrt{7}-\sqrt{2}) + (\sqrt{7}+\sqrt{2}) = 2\sqrt{7} $$ is rational. So $\sqrt{7}$ is rational. This is a contradiction.

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Hint:  suppose $\,\sqrt{7}-\sqrt{2}\,$ were rational, then so would be $\,\dfrac{5}{\sqrt{7}-\sqrt{2}}=\sqrt{7}+\sqrt{2}\,$, then so would be their difference $\,2 \sqrt{2}\,$.


[ EDIT ] Following up on the previous comment: let $\,x=\sqrt{7}-\sqrt{2}\,$, then $\,x^2=9-2\sqrt{14}\,$, then $(x^2-9)^2=4 \cdot 14 \iff x^4 - 18 x^2 + 25 = 0\,$. But the latter equation has no rational roots, since by the rational root theorem the only such roots could be $\,\pm1, \pm5, \pm25\,$ which none work.

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    $\begingroup$ Not really a hint :) $\endgroup$ – user223391 Feb 8 '18 at 5:46
  • $\begingroup$ Your alt hint is not really a hint either :) $\endgroup$ – user223391 Feb 8 '18 at 5:49
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    $\begingroup$ @ZacharySelk Right about the alt, just dropped the "hint" misnomer. The first one still leaves at least something to work out, though ;-) $\endgroup$ – dxiv Feb 8 '18 at 5:51
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Suppose $\sqrt{a}-\sqrt{b} = r $ is rational.

Squaring this, $a+b-2\sqrt{ab} = r^2$, so $\sqrt{ab}$ is rational.

If $\sqrt{ab}$ is irrational, this can not hold.

Therefore, if $\sqrt{ab}$ is irrational, so is $\sqrt{a}-\sqrt{b}$.

Since $\sqrt{14}$ is irrational, so is $\sqrt{7}-\sqrt{2}$.

Note that this works for $\sqrt{a}+\sqrt{b}$ also.

Note 2: There are many proofs here that if $n$ is not a perfect square then $\sqrt{n}$ is irrational.

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