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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 $n \leq x < n + 1$.

Floris Claassens
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Donlans Donlans
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2 Answers2

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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.

Stefan Egger
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You are assuming that no rational non-negative number $x$ so that $x^2 < 2 < (x + \epsilon)^2$ is true.

$x = 0$ is a rational non-negative number.

So that means, by our assumption, that $0^2 < 2 < (0+\epsilon)^2$ is false.

So $0 < 2 < \epsilon^2$ is false.

$0 < 2$ is true so $2 < \epsilon^2$ is false. So $\epsilon^2 \le 2$.

We have already proven that $\epsilon^2 = 2$ and $\epsilon$ is rational is false. So $\epsilon^2 < 2$.

.....

Now if we let $x = \epsilon$, a non-zero rational number, we have

$\epsilon^2 < 2 < (\epsilon + \epsilon)^2$ is false (by our assumption).

So $\epsilon^2 < 2 < (2\epsilon)^2$ is false.

But we already showed $\epsilon^2 < 2$ is true. So $2 < (2\epsilon)^2$ is false.

So $(2\epsilon)^2 \le 2$. But $2\epsilon$ is rational so $(2\epsilon)^2 \ne 2$.

So $(2\epsilon)^2 < 2$.

And we keep using induction to show if $(n\epsilon)^2 < 2$ then $(n\epsilon)^2 < 2 < ((n+1)\epsilon)^2$ is false so $((n+1)\epsilon)^2 < 2$.

fleablood
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