# Proof of Dedekind cuts.

This is my definition for Dedekind cuts:

A subset α of Q is said to be a cut if:

1. $$α$$ is not empty,$$α\neq \mathbb{Q}$$
2. If $$p \in α,q\in\mathbb{Q}$$,and $$q,then $$q\inα$$.
3. If $$p\in α$$,then $$p for some $$r\inα$$.

I've seen a specific example of Dedekind cut produced by $$\sqrt{2}$$:

$$α={\{p∈Q:p<0\}}∪{\{p∈Q:p≥0 \:\text{ and }\: p^2 <2\}}.$$

To prove this subset is a cut, it needs to be shown it satisfies 1.2.3. For the proof of 3., the author constructed $$r=\frac{2(p+1)}{p+2}$$.

Now I'm working on a more general subset here: $$a' = {\{x\mid x\geq 0 , x^2\leq p\}} \cup {\{x\mid x<0\}}$$. $$p$$ is a positive integer but not a square of integer. I don't know how to prove 2. and 3. and construct such an $$r$$ here.

I'll talk about Property 2 first, which boils down to simple casework. Suppose $$t\in \alpha'$$ and $$s\in\mathbb{Q}$$ satisfies $$s. If $$s<0$$, then $$s\in\{x:x<0\}\subset \alpha'$$. If, on the other hand, $$0\leq s, then properties of the ordered field $$\mathbb{Q}$$ imply $$0\leq s^2. Yet $$t^2\leq p$$, so $$s\in\{x:x\geq 0, x^2\leq p\}\subset \alpha'$$.
Now we'll prove Property 3. Let $$t\in\alpha'$$. If $$t<0$$, then $$r=0$$ works, so suppose $$0\leq t$$ and $$t^2\leq p$$. The square root of such a $$p$$ is irrational, so we actually have $$t^2. To find an $$r$$ that works we'll "rationally perturb" $$t$$; put $$r=t+1/n$$ where $$n$$ is some to-be-determined positive integer. This $$r$$ is rational and satisfies $$t. We would like to find conditions on $$n$$ which imply $$r^2. For this, note $$r^2=\big(t+\frac{1}{n}\big)^2 = t^2+\frac{2t}{n}+\frac{1}{n^2}.$$ Supposing $$r^2 we may rearrange to find $$2tn+1, or
$$1 Hence it is sufficient to choose $$n$$ so large that $$n(p-t^2)-2t>1$$, as this and the fact $$n\geq 1$$ ensure the above inequality is true. Choosing such an $$n$$ is possible by Archimedean property, so $$r=t+1/n$$ has $$r^2 and the proof is complete.
• Sorry, I forgot to mention a condition of $p$. $p$ is a positive integer. Actually I proved that such $p$ wouldn't be the square of a rational number. But I didn't get what you said about prime numbers. @Glare Dec 6, 2020 at 12:41
• Well, so long as $p$ is nonsquare then everything works. The issue I had in mind was something like $p=4$. Then $t=2\in\alpha'$ yet there is no $r\in\alpha'$ with $t<r$. Dec 6, 2020 at 12:46