1
$\begingroup$

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).

  • 2
    How about $5-\sqrt 2$ and $\sqrt{2}$ their sum is rational. – 2017-02-03
  • 0
    its not $$ $$ $$ $$ – 2017-02-03
  • 0
    "no minus sign" in the number is quite a ludicrous demand – 2017-02-03

3 Answers 3

5

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

  • 0
    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). – 2017-02-03
  • 0
    @GerardL. $10-\sqrt{2}$ is certainly positive. If you really want to get rid of the $-$ minus sign, just write it as $$\cfrac{98}{10+\sqrt{2}}\,$$ – 2017-02-03
  • 0
    @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. – 2017-02-03
  • 1
    maybe $\log 2$ and $\log(e/2)$? – 2017-02-03
4

Not true.

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

3

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 $x+y=\lceil x \rceil \in \mathbb{Z^+}$.