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I'm trying to evaluate the following claim:

$$ \sqrt{2} + \sqrt{n} $$ is irrational.

This is what I tried:

Proof by contrapositive: Suppose $$ r = \sqrt{2} + \sqrt{n} $$ and r is rational. Then $$ \frac{m}{l} = \sqrt{2} + \sqrt{n} $$ $$\frac{m^2}{l^2} = 2 + 2\sqrt{2n} + n $$

I'm not sure where to proceed or if I'm even heading in the right direction.

Could anyone give me a tip?

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    $\begingroup$ Can you assume $\sqrt{2}$ is irrational? Also I believe you are doing contradiction, not contrapositive $\endgroup$ – graydad Sep 10 '14 at 4:13
  • $\begingroup$ I think so. Right? $\endgroup$ – Pandamonium Sep 10 '14 at 4:15
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    $\begingroup$ I think you mean by contradiction, I don't believe you can take the contrapositive of a nonconditional statement. $\endgroup$ – IAmNoOne Sep 10 '14 at 4:15
  • $\begingroup$ Yeah, it's a natural number. Sorry for not including that. $\endgroup$ – Pandamonium Sep 10 '14 at 4:21
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I don't think this is going in the right direction, because $\sqrt{2n}$ could very well be an integer. (EDIT: There is actually a proof building on this in the answer by shooting-squirrel.)

I would say this. Let $r = \sqrt{n} + \sqrt{2}$, and let $s = \sqrt{n} - \sqrt{2}$. Then we have $rs = n - 2$ and $(1/2)(r - s) = \sqrt{2}$.

Now assume for a contradiction that $r$ is rational. Since $s = (n-2)/r$, the number $s$ must also be rational. And for that reason, the number $(1/2)(r - s)$ too must be rational. But this is $\sqrt{2}$, which we know to be irrational, a contradiction.

Hence $r$ must be irrational.

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  • $\begingroup$ Could you explain how s = (n-2)/r must also be rational? Thanks in advance. $\endgroup$ – Pandamonium Sep 10 '14 at 4:55
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    $\begingroup$ It's an integer divided by a nonzero rational number. $\endgroup$ – Dave Sep 10 '14 at 4:56
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Case 1: $n$ is of form $2u^2$ , $u$ being an integer

Then our number is equal to $\sqrt 2 (u+1)$, clearly this is irrational

Otherwise $2n$ is not a square, hence you get contradiction in the last identity you wrote.(since $\sqrt {2n}$ is irrational)

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  • $\begingroup$ A matter of personal taste: I find that statements like "clearly..." and "it is obvious that..." come off as condescending and avoid them, especially in teaching/helping. $\endgroup$ – Omnomnomnom Sep 10 '14 at 4:34
  • $\begingroup$ Tipically, I'm very open to questions... $\endgroup$ – shooting-squirrel Sep 10 '14 at 4:42
  • $\begingroup$ I think words like "clearly" are okay. They orient the reader somewhat as to the relative difficulty of a claim. In other words, the author is informing them of how difficult it is to establish that fact in comparison with other facts that appear in their text, or texts of the same level of difficulty. For example, if the reader is searching for the reason the claim is true, and they've been told it's obvious, they won't wander off into complex and meandering arguments. $\endgroup$ – Dave Sep 10 '14 at 4:48
  • $\begingroup$ @Dave a fair point, never thought of it that way. $\endgroup$ – Omnomnomnom Sep 10 '14 at 5:02
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Dear Alex solve your equation for $\sqrt{n}$. That is $r-\sqrt{2}=\sqrt{n}$ then by squaring the equation and solving for $\sqrt{2}$we have $\sqrt{2}=\frac{n-2-r^2}{2r}.$ A contradiction

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