Proof that we can find rational numbers arbitrarily close to $\sqrt{2}$: direct approach.

Question

The statement of the proposition:

Proposition. For every rational number $\epsilon > 0$, there exists a non-negative rational number $x$ such that $x^{2} < 2 < (x+\epsilon)^2$.

The most common approach to proving the proposition is by using contradiction (1,2).

My question is: is it possible to prove the proposition directly? More concretely, is it possible to find a function $f: \mathbb Q^+\rightarrow \mathbb Q^+$ such that for arbitrary positive rational $\epsilon$, we have

$$f(\epsilon)^2 < 2 < (f(\epsilon) + \epsilon)^2 $$

?

Answer 1

Define $f(\varepsilon)$ to be the truncation of $\sqrt{2}$ to $n$ decimal places, where $10^{-n} \leq \varepsilon$ is the nearest power of $10$ from below.

This ensures that $$f(\varepsilon) < \sqrt{2} < f(\varepsilon) + 10^{-n} \leq f(\varepsilon) + \varepsilon$$

To illustrate, if $\varepsilon=0.2$ then $n=1$ and the inequality reads $$1.4 < \sqrt{2} < 1.5 \leq 1.6$$

Answer 2

Take $$\epsilon\left\lfloor\frac{\sqrt2}\epsilon\right\rfloor.$$ This rational is less than $\epsilon$ away from $\sqrt2$.

Answer 3

Using the fact that between any two real numbers there exists a rational number, given any rational $\varepsilon$ such that $4\sqrt{2}>\varepsilon>0, \exists x \in \mathbb{Q}$ such that $x \in (\sqrt{2} - \frac{\varepsilon}{2}, \sqrt{2})$, which gives: $x^2 < 2$ and $(x+\varepsilon)^2 > 2.$