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Before this is marked as a duplicate I just want to say that I've already read a similar thread, where the original poster asked how they would prove that $\sqrt 2 + \sqrt 5 + \sqrt 7$ is an irrational number. I've read the answers to that thread and I couldn't really "apply"/use them in my situation.

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Exercise added for reference. Post was edited because it was referring to a particular case (where n was assigned a value such as 6).

I'm trying to reopen the topic because I am more curious as to how you'd solve this type of exercise in the general form (like in the image).

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    $\begingroup$ These problems are all the same. Square to show that $\sqrt {30}+\sqrt{35}+\sqrt{42}\in \mathbb Q$ Now square again to get $A\sqrt {30}+B\sqrt{35}+C\sqrt{42}\in \mathbb Q$ for suitable integers $A,B,C$. Use those two equations to eliminate one radical. $\endgroup$
    – lulu
    Commented Oct 4, 2020 at 17:14
  • $\begingroup$ thanks very much @lulu :) $\endgroup$
    – cezar
    Commented Oct 4, 2020 at 17:16
  • $\begingroup$ I think you can do this by showing that $x^2-5$ is irreducible Over $\mathbb Q [\sqrt 7, \sqrt 6]$, or that the relevant field extension has degree $2$. I seem to remember an answer to a question which took this approach and showed how it applied to the general case of adjoining "coprime" square roots. $\endgroup$
    – Mark Bennet
    Commented Oct 4, 2020 at 20:42

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If $$r-\sqrt6=\sqrt5+\sqrt7,$$ where $r$ is a rational number, we obtain: $$r^2+6-2r\sqrt6=12+2\sqrt{35},$$ which gives $$(r^2-6)^2=24r^2+8r\sqrt{210}+140$$ and since $r=0$ is impossible, we obtain a contradiction: $$\sqrt{210}=\frac{r^4-36r^2-104}{8r}\in\mathbb Q.$$

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As @lulu noted, if $r\in\Bbb Q$ then $\sqrt{30}+\sqrt{35}+\sqrt{42}\in\Bbb Q$ and, squaring again, $7\sqrt{30}+6\sqrt{35}+5\sqrt{42}\in\Bbb Q$. So $\sqrt{42}-\sqrt{30}\in\Bbb R$; squaring again, $\sqrt{35}\in\Bbb Q$, a contradiction.

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Suppose $a\sqrt {n-1}+b\sqrt n +c\sqrt {n+1}=r$ where $a,b,c,r$ are rational, then $$\left(r-b\sqrt n\right)^2=\left (a\sqrt {n-1}+c\sqrt {n+1}\right)^2$$ or $$r^2+b^2n-2br\sqrt n=a^2(n-1)+c^2(n+1)+2ac\sqrt{n^2-1}$$ or $$q\sqrt {n^2-1}=s+t\sqrt n$$

Where $q,s,t\in \mathbb Q$

Square again, and you get a rational expression for $\sqrt n$, and this is only possible if $n$ is a square.

As I noted in the comments there, each additional square root is either obviously contained in the extension of $\mathbb Q$ generated by the previous square roots, or generates an extension of order $2$. I am sure that this has been explained in an answer before, but I can't find the link. Essentially this codifies the theory in a general context and reduces the need for specific computations.

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Just muck. Isolate radicals one by one to side; square and keep going to eliminate radicals one by one.

If $\sqrt 5 + \sqrt 6 + \sqrt 7 = r\in \mathbb Q$ then

$\sqrt 6 + \sqrt 7 = r- \sqrt 5$ then square both side

$6 + 7 + 2\sqrt{6*7} = r^2 - 2r \sqrt 5 + 5$.

$2\sqrt {6*7} + 2r \sqrt 5= r^2+5-6 -7$ then square both sides

$2^2*6*7 + 2^2r^2*5 + 2*2*2r\sqrt{5*6*7} = (r^2 + 5 -6-7)^2$

Thus $\sqrt {5*6*7} = \frac {(r^2 + 5 -6-7)^2-2^2*6*7-2^2r^2*5}{2*2*2r} \in \mathbb Q$.

And as $5*6*7$ is not a perfect square that is a contradiction.

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