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I haven't seen really a straightforward proof towards this question. All of them regarding this topic focus on the fact that $x+y$ can be rational even if x and y are irrational because you could set y as the negative of x, but there isn't really anything about if both are positive and have no minus signs "inside" the variable (say 10-sqrt2 is not valid).

Gerard L.
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3 Answers3

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You may take $\sqrt 2$ and $10-\sqrt 2$, they are both positive and irrational and their sum is $10$

Momo
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  • Right, but this doesn't answer my question. I'm asking to prove that if both x and y are positive (without any minus signs inside of y). – Gerard L. Feb 03 '17 at 01:27
  • @GerardL. $10-\sqrt{2}$ is certainly positive. If you really want to get rid of the $-$ minus sign, just write it as $$\cfrac{98}{10+\sqrt{2}},$$ – dxiv Feb 03 '17 at 01:30
  • @GerardL. It doesn't make sense to say "no minus signs inside $y$". $y$ is a perfectly respectable irrational number. You could write it as $98/(10 + \sqrt{2})$ with no minus signs if you wish. – Ethan Bolker Feb 03 '17 at 01:31
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    maybe $\log 2$ and $\log(e/2)$? – Momo Feb 03 '17 at 01:33
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Not true.

$x = \sqrt{2}, y = 2-\sqrt{2}$.

marty cohen
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Counterexample: let $x \in \mathbb{R}^+ \setminus \mathbb{Q}$ be an arbitrary positive irrational, and define $y=\lceil x \rceil - x$. Then $y$ is itself a strictly positive irrational (why?) and $x+y=\lceil x \rceil \in \mathbb{Z^+}$.

dxiv
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