Real Between Rationals

Question

Let $x$ be a real number. Show that, for any $\varepsilon>0$, there exist two rationals $q$ and $q'$ such that $q<x<q'$ and $|q-q'|<\varepsilon$ How should I approach this prove?

Answer 1

If you are allowed to use the property that the rational numbers are dense in $\mathbb{R}$, and it sounds like you are from the comments, then:

$x - \frac{\epsilon}{2}$ and $x + \frac{\epsilon}{2}$ are both real numbers, so find $q, q'$ rational such that $$x - \frac{\epsilon}{2} < q < x < q' < x + \frac{\epsilon}{2}.$$ Then $|q' - q| < |x + \frac{\epsilon}{2} - (x - \frac{\epsilon}{2})| = \epsilon$.

Answer 2

Let $x=x_0,x_1x_2x_3...$ be the decimal representation of $x>0$. Given $\varepsilon>0$, take $n$ such that $\dfrac{2}{10^n}<\varepsilon$. Than the rationals $q=\sum_{k=0}^n\dfrac{x_k}{10^k}-\dfrac{1}{10^n}$ and $q'=\sum_{k=0}^n\dfrac{x_k}{10^k}+\dfrac{1}{10^n}$, satisfies the conditions desired. For $x<0$, make a similar aproach.

Answer 3

You say in comments that you know that between any two real numbers there is a rational number.

So apply that to the the two real numbers $x$ and $x+\dfrac\varepsilon2$, and then again to the two real numbers $x$ and $x-\dfrac\varepsilon2$.

Answer 4

Let $B(x-\epsilon/2, \epsilon/2)$ and $B(x+\epsilon/2, \epsilon/2)$ be open balls. By density of the rationals, there is a rational in each of these balls, and it is trivial to check that they are less than $\epsilon$ apart.