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We have to prove that $\dfrac{\sqrt{8}}{\sqrt{7}}$ is irrational(try not to use the Rational Root Theorem)

At first,we prove that the expression is not an integer.

$\dfrac{\sqrt{8}}{\sqrt{7}}=\sqrt{\dfrac{8}{7}}<2$

and hence is not an integer,since the only square integer less than $2$ is $1$ and the above expression is greater than $1$. Now,let us assume that $$\dfrac{\sqrt{8}}{\sqrt{7}}=\dfrac{p}{q}$$

with $p$ and $q$ coprime.Then, $$\dfrac{\sqrt{8}+\sqrt{7}}{\sqrt{8}-\sqrt{7}}=\dfrac{p+q}{p-q}$$ using componendo-divideno.But,$$\dfrac{\sqrt{8}+\sqrt{7}}{\sqrt{8}-\sqrt{7}}=(\sqrt{8}+\sqrt{7})^2=15+2\sqrt{56}$$ which is irrational,thus arriving at a contradiction.However,I find my proof to be unnecessarily big.So I want some help in finding a more succinct proof.I tried to proceed the in the usual manner(like the proof of irrationality of $\sqrt{2}$) but I do not arrive at any contradiction.Some help will be appreciated.

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    $\begingroup$ Another approach is to realize that a real number $x$ is irrational if and only if $7x$ is irrational. Thus to prove $\sqrt8/\sqrt7$ is irrational, it suffices to prove that $7\sqrt8/\sqrt7 = \sqrt{56}$ is irrational. $\endgroup$ – Greg Martin Feb 6 '14 at 8:17
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The usual proof works. We have $8q^2=7p^2$, so $7$ divides $q$, so $7$ divides $p$.

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  • $\begingroup$ Oops yes,can't believe I missed that.Thanks for helping. $\endgroup$ – rah4927 Feb 6 '14 at 7:52
  • $\begingroup$ You are welcome. It might have been easier if you did not have the componendo et dividendo tool, which used to be a standard item in school algebra, but is now infrequently taught. $\endgroup$ – André Nicolas Feb 6 '14 at 7:59
  • $\begingroup$ Indeed, we are working with linear independence here so that much junk above is unnecessary. $\endgroup$ – Balarka Sen Feb 6 '14 at 10:19
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Hint $\,\ r\in\Bbb Q,\,\ r^2 = 8/7\,\Rightarrow\,(7r)^2=56\,\overset{\large \rm Rat\ Root\ Test}\Rightarrow\,7r\in\Bbb Z\,\Rightarrow\,56 = $ perfect square $\,\Rightarrow\Leftarrow$

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