I am having a little trouble understanding this proof.
For every rational number $\varepsilon > 0$, there exists a
non-negative rational number $x$ such that $x^2 < 2 < (x + \varepsilon)^2$.
Proof. Let $\varepsilon> 0$ be rational. Suppose for sake of contradiction that
there is no non-negative rational number $x$ for which $x^2 < 2 < (x+\varepsilon)^2$.
This means that whenever $x$ is non-negative and $x^2 < 2$, we must also
have $(x + \varepsilon)^2 < 2$. Since $0^2 < 2$, we thus have $\varepsilon^2 < 2$, which then implies $(2\varepsilon)^2 < 2$,
and indeed a simple induction shows that $(n\varepsilon)^2 < 2$ for every natural
number n. But, by Proposition 4.4.1 we can find an integer $n$ such that
$n > 2/\varepsilon$, which implies that $n\varepsilon > 2$, which implies that $(n\varepsilon)2 > 4 > 2$,
contradicting the claim that $(n\varepsilon)2 < 2$ for all natural numbers $n$. This
contradiction gives the proof.
What I specifically don't understand is why $0^2$ implies that $\varepsilon^2 < 2$, and why this implies that $(2\varepsilon)^2 < 2$, I understand the rest of the proof except those two parts, I'd be very grateful if someone could enlighten me.
note: Proposition 4.4.1 asserts that between two integers $n$ and $n + 1$, you can always find a rational number $x$ such that $n \leq x < n + 1$.
Best Answer
Your assumption is that if $x$ is non-negative and $x^2 < 2$ then also $(x+\varepsilon)^2 < 2$. Ok, now take $x=0$: $0$ is nonnegative and $0^2 < 2$. Therefore, the assumption gives us that $(0+\varepsilon)^2 < 2$, which means that $\varepsilon^2 < 2$. Now we can use our assumption again for $x=\varepsilon$ and therefore we get $(\varepsilon + \varepsilon)^2 < 2$, which means $(2\varepsilon)^2 < 2$ and so on.