Proving that $\Bbb Q$ is dense in $\Bbb R$

Question

I would like a hint for the following:

$\forall \epsilon >0$ $\forall x\in \mathbb{R}$ $\exists y \in \mathbb{Q}$ $:$ $|x-y|< \epsilon$.

The idea I have is to break it into the following cases:

(1) x $< \epsilon$

(2) $x > \epsilon$

(3) x= $\epsilon$

Is there a simpler way to do this?

I'd love a hint.

Answer 1

Since you have the theorem that for any two real numbers $a,b$ with $a < b$, there is a rational number in the interval $(a,b)$, we can argue as follows . . .

Let $\epsilon > 0$, and let $x\in\mathbb{R}$.

Let $y$ be a rational number in the interval $(x-\epsilon,x+\epsilon)$. \begin{align*} &y\in (x-\epsilon,x+\epsilon)\\[4pt] \implies\;&x-\epsilon < y < x+\epsilon\\[4pt] \implies\;&-\epsilon < y-x < \epsilon\\[4pt] \implies\;&|y-x| < \epsilon\\[4pt] \implies\;&|x-y| < \epsilon\\[4pt] \end{align*} as required.

Answer 2

Hint: write $x$ in the decimal form, and take its first $n$ terms as $y$, where $n$ is an integer such that $10^{-n}<\epsilon$.

Answer 3

You can take a number $q \geq 1 $ and subdivide $\mathbb{R}$ in intervals of length $1/q$, so you have $\mathbb{R} = \cup_{i \in \mathbb{Z}} [i/q,(i+1)/q[$, since this union is disjoint then $\exists! i \in \mathbb{Z}$ such that $x \in [i/q,(i+1)/q[$, and since the lenght of the interval is $1/q$ then $|x-i/q| \leq 1/q$, if you take $q$ large enough then $ 1/q < \epsilon$, and $y = i/q$ solves the problem.