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Prove that $\sqrt{2} + \sqrt{n}$ is irrational $\forall \ n \in \mathbb{N}$.

I have tried though contradiction but can't seem to come up with an answer

I did this where for $\frac ab$, $a \in \mathbb{N}, b \in \mathbb{Z}$ and in reduce form, and $0 \not \in \mathbb{N}$

$$\sqrt 2 + \sqrt n = \frac ab$$ $$\frac {2+n}{\sqrt 2 - \sqrt n} = \frac ab$$ $$\frac {(2+n)(\sqrt 2 + \sqrt n)}{2 + n} = \frac ab$$ $$\frac {(2+n)(a)}{(2-n)(b)}=\frac ab$$ $$\frac {2+n}{2-n}=1$$ $$2+n=2-n$$ $$2+2n=2$$ $$n=0$$ $$n \in \mathbb{N} \therefore n \not =0$$ $$\therefore \sqrt{2} + \sqrt{n} \not \in \mathbb{Q}$$

However this seems clearly flawed as even if $n = 0$ it still would be irrational, and that it subsitiuting $\sqrt 2 + \sqrt n$ for $\frac ab$ seems pretty incorrect.

More just a desperate attempt, pretty confident its not even close.

cmk
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Dakota
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2 Answers2

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Suppose $\sqrt{2}+\sqrt{n}\in\Bbb Q$. Since the ratio of rationals is rational, $\frac{2-n}{\sqrt{2}+\sqrt{n}}=\sqrt{2}-\sqrt{n}\in\Bbb Q$. Averaging, $\sqrt2\in\Bbb Q$, which contradicts a well-known result.

J.G.
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  • So basically whats happening is you are claiming that for $\frac {2-n}{\sqrt 2 + \sqrt n} \in \mathbb{Q}$ both the numerator and denominator have to be rational, So assuming that both are rational you take $\sqrt 2 - \sqrt n \in \mathbb {Q}$, then that is only true if the average of the two radicals, which in this case is $\sqrt 2$ is rational, thus contridicting. I don't really follow why the average of the two radicals is $\sqrt 2$ though. – Dakota Oct 12 '19 at 04:41
  • @Dakota Because their sum is $2\sqrt{2}$. – J.G. Oct 12 '19 at 06:30
  • I still don't quite follow. If $n = 2$ then I understand $2\sqrt2$ but why for all n? – Dakota Oct 12 '19 at 16:47
  • @Dakota Summing $\sqrt{2}\pm\sqrt{n}$ gives an $n$-independent result. – J.G. Oct 12 '19 at 16:50
  • So we sum $\sqrt 2 + \sqrt n$ and $\sqrt 2 - \sqrt n$ given $2\sqrt 2$. So is the reasoning that if $\sqrt 2 + \sqrt n$ and $\sqrt 2 - \sqrt n$ are $\in \mathbb{Q}$ then the average of them must be $\in \mathbb{Q}$? – Dakota Oct 12 '19 at 23:00
  • @Dakota Yes, that's right. – J.G. Oct 13 '19 at 07:12
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The numerator in your second line should be $(\sqrt{2} + \sqrt{n})(\sqrt{2} - \sqrt{n}) = 2 - n$. ... and then your fourth line is a tautology.

Eric Towers
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