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I need help understanding the following proof:

Prove that $\Bbb Q$ is dense in $\Bbb R$, i.e.

$$\forall \varepsilon>0,\forall x\in\Bbb R, \ (x-\varepsilon,x+\varepsilon)\land \Bbb Q\neq \emptyset.$$

Let's assume the opposite:

$\exists\varepsilon>0, \exists x\in\Bbb R, \forall q\in\Bbb Q, (q\leq x-\varepsilon)\lor (x+\varepsilon\leq q)$.

Let $q_1\leq x-\varepsilon$ and $x+\varepsilon\leq q_2$. Let $n\in\Bbb N$ such that $2n\varepsilon>q_2-q_1$ and let $\alpha_k=q_1+\frac{k}{n}(q_2-q_1)\in\Bbb Q (k=0,1,...,n-1,n)$. Let $1\leq k_0

I'm completely lost. Can someone tell me what's the idea behind this proof?

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    What's the first thing you don't understand? – 2017-02-19
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    @NoahSchweber why we chose $n$ such that $2n\varepsilon>q_2-q_1$ – 2017-02-20

1 Answers 1

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The idea is that if you have a hole of diameter $2\epsilon$ and a frog whose jumping length is $\frac{1}{n}$ (rational) and less than the diameter of the hole $\left(\frac{1}{n}<2\epsilon\right)$, and the frog is initially located in origin and starts jumping in the direction of the hole, it will eventually end up in the hole.