I have looked around at the questions about Dedekind cuts and still have some questions.

For example,

1) Why is $\{r \in \mathbb{Q}: r^2 < 2 \}$ not a Dedekind cut and yet $\big( 0^{\ast}:=\{r \in \mathbb{Q}: r < 0\} \big) \cup \{r \in \mathbb{Q}: r \geq 0 \textrm{ and } r^2 < 2 \}$ is a Dedekind cut (for representing $\sqrt{2}$)?

I can verify the union is a Dedekind cut for $\sqrt{2}$ by drawing an open interval, $(-\infty,0)$, union-ed with the open interval $[0,\sqrt{2})$ to produce what I think is the definition of a Dedekind cut "by the lower half", $(-\infty,\sqrt{2})$.

The first part - I use the definition:

(a) Cut $A \neq \varnothing$ and $A \neq \mathbb{Q}$.

(b) If $r \in A$ and $s \in \mathbb{Q}$ and $s < r$, then $s \in A$.

(c) A contains no largest rational

Doesn't the first "cut" satisfy all these? Because can't I just write it as $\{r \in \mathbb{Q}: r < \sqrt{2} \}$? Then (a) is satisfied, (b) is satisfied, and (c) is true because you can get arbitrarily close to $\sqrt{2}$.

It seems that in both $\{r \in \mathbb{Q}: r^2 < 2 \}$ and $\{r \in \mathbb{Q}: r^3 < 2\}$, the suprema of each are irrational....?

2) For $\{r \in \mathbb{Q}: r^3 < 2 \}$ is a Dedekind cut - can't I just say that it is because it is the set $\{r \in \mathbb{Q}: r < 2^{1/3}\}$ and $r^3 - 2 = 0$ can only have rational roots that are multiples of $\frac{r\,a_0}{s\,a_n}: \pm 1, \pm 2$?

The separation is on an irrational number and that this cut represents an irrational number?

3) Is it fair to say that a rational is represented by the cut, $\{(-\infty,r),[r,+\infty)\}$; and an irrational is represented by the cut, $\{(-\infty,i),(i,+\infty)\}$?

Thank you all for the help!

  • 1
    $\begingroup$ For (1), with your $A$ we have $0 \in A$ and yet $-2 \notin A$, since $4$ is not less than $2$. $\endgroup$ Commented Feb 6, 2012 at 2:36
  • $\begingroup$ I clearly see what you are talking about! Totally missed that.... ;) $\endgroup$
    – nate
    Commented Feb 6, 2012 at 2:40
  • $\begingroup$ One last question though... Is my **3)$$ correct, if I am to still use the definition of a cut by each "half"? $\endgroup$
    – nate
    Commented Feb 6, 2012 at 2:52
  • $\begingroup$ Well, I think it's true that a cut $A$ as defined above represents a rational number if and only if the set $A$ has a least upper bound (the corresponding rational number) in $\mathbf Q$. And once you specify the ordering on cuts you can make sense of your (3). But early on it doesn't make sense to say, "Well, $\sqrt{2}$ is represented by $(-\infty, \sqrt{2})$." Because it isn't clear what those last two things are. $\endgroup$ Commented Feb 6, 2012 at 3:28
  • $\begingroup$ Also, it seems like it would be good to "cheat" early on: look at the graph of $x^2$ and see what kind of region on the $x$-axis a condition like $x^2 < 2$ gives you. Similarly for $x^3$. That might catch some of the more obvious errors. $\endgroup$ Commented Feb 6, 2012 at 3:39

3 Answers 3


For (1): As Dylan Moreland points out in a comment, the problems lies in (b). It's not true that a rational smaller than an element of your $S$ is also in $S$.

For (2) and (3), I have the same comment in both: A common and understandable confusion when being first exposed to Dedekind cuts is that you have to completely forget that real numbers exist. It doesn't really make sense to say that a cut is on an irrational number, since there are no such thing as rational numbers when you're defining Dedekind cuts. (Of course, most of the things you're saying in this respect are "morally correct" in the sense that they end up being true once you've defined real numbers, but in the long run, this is circular.)

Similarly, for your $2^{1/3}$ Dedekind cut, it may be helpful while you're going through this to never ever use notations like $2^{1/3}$ or $\sqrt[3]{2}$, except possibly as shorthand for the Dedekind cut $S$ you describe (and even then, only once you've defined multiplication of cuts and checked that $S^3=2$), since in the rational world, this number does not exist. So no, you can't just say it's that set, because you don't know that set is a Dedekind cut. You need to prove that it has no maximal element, and if you think about it, checking that there is no rational number whose cube equals 2 is a little shy of that stronger conclusion.


If $r$ is in the left part of the cut, and $s$ is a rational such that $s<r$, then $y$ must be in the left part of the cut. That is part of the definition of cut, it is your condition (b).

If you look at the rationals $r$ such that $r^2<2$, note that for example $-17$ is not in. But for example $1$ is in which violates part of the definition of cut.

Added: For $\sqrt[3]{2}$, it so happens that the rationals $r$ such that $r^3<2$ satisfy your condition (b), since the function $x^3$ is increasing on the rationals, and goes to $-\infty$ as $x$ goes to $-\infty$. The function $x^2$ has neither of these properties.

Your characterization of "rational" and "irrational" cuts near the end sounds reasonable enough. The main trouble with it is that the standard meaning of say $(-\infty,5)$ is the set of all reals $x$ such that $x<5$. However, a cut is a certain set of rationals. Also, we use cuts to define the reals. Before the reals are defined, the expression $(-\infty,\sqrt{2})$ has no official meaning.

Please note that there are some inessential differences between cuts as defined by various people. Some definitions allow a largest number for the "left" part, and not for the "right" part. Some, following Dedekind's original definition, mention explicitly both parts, so a cut is an ordered pair $(L,U)$ such that neither is empty, together they make up the rationals, and so on.

  • $\begingroup$ So this clarifies why say $\{r \in \mathbb{Q}: r^3 < 2\}$ is a cut but not $\{r \in \mathbb{Q}: r^2 < 2 \}$. Thanks! $\endgroup$
    – nate
    Commented Feb 6, 2012 at 2:46
  • $\begingroup$ @nate: Yes, but the whole thing is a technicality, the "imortant" part of the cut is the general area where it changes from in to out. $\endgroup$ Commented Feb 6, 2012 at 3:00

A Dedekind cut is a partition of the rational numbers $$\mathbb{Q} = L \cup U,$$ so that for all $x\in L$ and $y\in U$, we have $x < y$.

This should answer your question about $\{x\in \mathbb{Q}| x^2 < 2\}$.


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