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 $q<p$,then $q\inα$.
  3. If $p\in α$,then $p<r$ 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.


1 Answer 1


I'll talk about Property 2 first, which boils down to simple casework. Suppose $t\in \alpha'$ and $s\in\mathbb{Q}$ satisfies $s<t$. If $s<0$, then $s\in\{x:x<0\}\subset \alpha'$. If, on the other hand, $0\leq s<t$, then properties of the ordered field $\mathbb{Q}$ imply $0\leq s^2<t^2$. 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 $t^2<p$. 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 $t<r$. We would like to find conditions on $n$ which imply $r^2<p$. For this, note $$r^2=\big(t+\frac{1}{n}\big)^2 = t^2+\frac{2t}{n}+\frac{1}{n^2}.$$ Supposing $r^2<p$ we may rearrange to find $2tn+1<n^2(p-t^2)$, or

$$1<n^2(p-t^2)-2tn = n[n(p-t^2)-2t].$$ 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 $r^2<p$ and the proof is complete.

  • $\begingroup$ 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 $\endgroup$
    – Kylinny
    Dec 6, 2020 at 12:41
  • $\begingroup$ 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 $t<r$. $\endgroup$
    – Glare
    Dec 6, 2020 at 12:46
  • $\begingroup$ Yeah exactly. Sorry about the confusion. @Glare $\endgroup$
    – Kylinny
    Dec 6, 2020 at 12:48
  • $\begingroup$ No problem. I'll edit that bit in my answer as well. $\endgroup$
    – Glare
    Dec 6, 2020 at 12:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .