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Suppose $\sqrt{2} + \sqrt{n}$ is rational,

so it can be represented like so: $\sqrt{2} + \sqrt{n} = {x \over y}$

then $\sqrt{2} = {{x \over y} - \sqrt{n}}$

so $\sqrt{2} = {x - y\sqrt{n}\over y}$

but $\sqrt{2}$ is irrational, which leads to a contradiction.

Does this proof make sense?

edit: fixed some mistakes

goldenotaste
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    What is the contradiction actually? More often than not $\sqrt{n}$ isn't rational either, so you haven't deduced an equation such that one side is rational and the other irrational. – Jyrki Lahtonen Oct 12 '19 at 21:36
  • Also you have sign error almost immediately. – Arturo Magidin Oct 12 '19 at 21:37
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    Do reuse the ideas described here! I like to think have a thread dedicated to the irrationalit of $\sqrt2+\sqrt3$, but it's too late an hour for me to search properly, so that other one has to do. – Jyrki Lahtonen Oct 12 '19 at 21:38

3 Answers3

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At the last step in your proof the contradiction is not evident: why is ${x - y\sqrt{n}\over y}$ a rational number?

My hint: if $\sqrt{2} + \sqrt{n}=r$ is a rational number then it is positive and by squaring we get $$n=(\sqrt{n})^2=(r-\sqrt{2})^2=r^2-2r\sqrt{2}+2.$$ Any contradiction here?

Robert Z
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The polynomial $$\left(x^2-\left(\sqrt{2}-\sqrt{n}\right)^2\right) \left(x^2-\left(\sqrt{2}+\sqrt{n}\right)^2\right) = x^4-2(n+2)\,x^2+(n-2)^2$$ has $\sqrt2+\sqrt n$ as one of its roots. Any rational divisor would have to be an integer factor of $(n-2)^2$ by the rational roots theorem.

But $\sqrt 2+\sqrt n$ is not an integer, indeed, if $\sqrt 2+\sqrt n=a\in\mathbb Z$, then $n=a^2+2-2a\sqrt 2$, implying that $2a\sqrt 2\in\mathbb Z$.

Luke Collins
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3

The sum or product of rational numbers is rational.

If $\sqrt 2 + \sqrt n$ is rational, then so is $\frac{2 - n}{\sqrt 2 + \sqrt n} = \sqrt 2 - \sqrt n$.

Then $\frac{\left(\sqrt 2 + \sqrt n\right) + \left(\sqrt 2 - \sqrt n\right)}{2} = \sqrt 2$ is rational.

David Diaz
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