Rational and irrational numbers

Question

Consider $x$ a rational number. Let $\epsilon \geq 0$ be the minimal value such that $x + \epsilon$ is irrational, and let also $\gamma > 0$ be the minimal value such that $x+\gamma$ is rational. How to compute $\gamma/\epsilon$?

Answer 1

Simply: $\lim\limits_{\varepsilon\downarrow0}(x+\varepsilon)=x$, and it was given that $x$ is irrational.

Update after drastic revision of the question: There are no such minimal values, in either of the two cases. If some positive number $x+\varepsilon$ is irrational, then $x+\varepsilon$ is not the minimal value for which that holds, because $x+\dfrac\varepsilon2$ is also irrational. To prove this, suppose it were rational. Then $x+\dfrac\varepsilon2 = \dfrac m n$ for some integers $m$ and $n$. It follows that $\dfrac\varepsilon2=\dfrac m n - x$ and so $\dfrac\varepsilon2$, being a difference between two rational numbers, is rational. And therefore so is $\varepsilon$, so $x+\varepsilon$ must be rational, and we have a contradiction.

It's even simpler if $x+\gamma$ is rational. Since $x$ is rational, and $x+\gamma$ is rational, we have $\gamma = \left(x+\gamma\right)-x$, a difference of two rational numbers, which is therefore rational. Since $\gamma$ is thus proved to be rational, so is $\gamma/2$, and then so is $x+\gamma/2$.

So in either of the two cases, every instance fails to be minimal; there's always a smaller one. There are no such minimal numbers $\gamma$ and $\varepsilon$.

Answer 2

$\epsilon$ doesn't exist! And even if it did, $\gamma$ wouldn't exist!

Theorem: If $a$ and $b$ are real numbers with $a<b$, then there exists a rational number $c$ such that $a<c<b$ and there exists an irrational number $d$ such that $a<d<b$.