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Prove that $\sqrt{2} + \sqrt{7}$ is irrational using contradiction.

I would appreciate if you could go over the proof in as much detail as possible. Thanks in advance.

MJD
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nocomment
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    Do you know the rational root theorem? – J. W. Tanner Feb 05 '20 at 22:32
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    Have you seen the proof that $\sqrt 2$ is irrational? – Ross Millikan Feb 05 '20 at 22:32
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    "I would appreciate if you could go over the proof in as much detail as possible". So where is the proof then? – imranfat Feb 05 '20 at 22:33
  • @J.W.Tanner Unfortunately not. – nocomment Feb 05 '20 at 22:36
  • @RossMillikan I have seen the proof for $ \sqrt{2} $, and I wonder whether I'd have to prove that first in order for this proof to be complete. Meaning, whether I must prove that $ \sqrt{2} $ is irrational before I've proved this one. – nocomment Feb 05 '20 at 22:39
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    @nocomment: see here, and note that $\sqrt2+\sqrt7$ is a root of $x^4-18x^2+25$ – J. W. Tanner Feb 05 '20 at 22:43
  • There are several approaches. The linked question has a general proof. You would usually start with $\sqrt 2$, then note the same works for any square root of a prime, then go on. The rational root theorem is another way-find a polynomial that $\sqrt 2 + \sqrt 7$ satisfies and show it has no rational roots. – Ross Millikan Feb 05 '20 at 22:43
  • Note that $(\sqrt{7}+\sqrt{2})(\sqrt{7}-\sqrt{2})=7-2=5$. Assume $\sqrt{7}+\sqrt{2}$ is rational. Then so is $\frac{5}{\sqrt{7}+\sqrt{2}}=\sqrt{7}-\sqrt{2}$. If both $\sqrt{7}+\sqrt{2}$ and $\sqrt{7}-\sqrt{2}$ are rational, then so is their sum (and difference) implying that $2\sqrt{7}$ is rational. You should be able to readily show that is impossible, so the assumption that $\sqrt{7}+\sqrt{2}$ is rational must be incorrect. – Keith Backman Feb 06 '20 at 02:36

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