# Dedekind cuts, can it have an irrational member?

From what I know, Dedekind cuts are partitions of rational numbers which have L and R classes used to define a real number. From what I understood, it's not necessary for the L or R classes to have an irrational member; but from this book that I'm reading (Methods of Mathematical Physics by The Jeffreys) it says:

"'X is a real and has a square less than 2' defines an L class with no largest member and an R class with smallest member $\sqrt{2}$." But $\sqrt{2}$ is an irrational number.

Can anyone explain this to me, is this just a mistake from the author or my misconception?

Dedekind cuts in the set of rational numbers are partitions of the set of rational numbers into two sets, $L$ and $R$, having certain properties. By definition $L$ and $R$ are subsets of $\Bbb Q$, so neither of them can contain an irrational number: it’s not just not necessary for either of them to contain an irrational number, it’s impossible. However, it appears that Jeffreys is talking about Dedekind cuts in the set of real numbers (‘$X$ is real and ...’), in which case both $L$ and $R$ necessarily contain lots of irrational numbers.

• So what he means by 'X is real' is all the possible elements of the L and R set (which are real numbers in this case)? – Transcedental Number Oct 29 '12 at 9:31
• @TranscedentalNumber: He’s saying that if you set $L=\{x\in\Bbb R:x^2<2\}$ and $R=\Bbb R\setminus L$, you get a Dedekind cut in $\Bbb R$ in which $L$ has no largest member, and $\min R=\sqrt2$. – Brian M. Scott Oct 29 '12 at 9:34
• Yes, I understand that, I think it's just confusing because the book also defined the Dedekind cuts in terms of rationals. – Transcedental Number Oct 29 '12 at 9:36
• @TranscedentalNumber: I’m going to guess that it did that when defining/constructing the reals. Once that’s been done, you can still get some extra insight by looking at Dedekind cuts in the reals. (Mind you, I’ve not seen the book, so I can’t say whether what he’s doing here is actually useful!) – Brian M. Scott Oct 29 '12 at 9:39

What the excerpt is saying that you are taking cuts as partitions of $\mathbb R$. The $L$ part is those whose square is strictly less than $2$, or they are negative; and $R$ is those whose square is at least $2$ and they are positive.

In $\mathbb R$ we already have $\sqrt2$ so we have that $\sqrt2\in R$ and it is the minimal element there.

Do note, however, that it is common to define $\mathbb R$ as Dedekind-cuts of the rationals, but here we do it over $\mathbb R$.

• If you aren't doing it over the rationals, is that still a 'Dedekind cut'? – Transcedental Number Oct 29 '12 at 9:34
• @TranscedentalNumber: Yes. You can talk about Dedekind cuts in any linearly ordered set. – Brian M. Scott Oct 29 '12 at 9:34
• @Brian M. Scott: Thank you, this helped me a lot. – Transcedental Number Oct 29 '12 at 9:38
• @TranscedentalNumber: You’re welcome; glad to help. – Brian M. Scott Oct 29 '12 at 9:39

GH Hardy in "Pure Mathematics" cites Dedekind's Theorem as follows:

If the real numbers are divided into two classes $L$ and $R$ in such a way that

(i) every number belongs to one or other of the two classes

(ii) each class contains at least one member

(iii) any member of $L$ is less than any member of $R$

then there is a number $\alpha$, which has the property that all the numbers less than it belong to $L$ and all the numbers greater than it to $R$. The number $\alpha$ itself may belong to either class.

The real numbers have previously been defined as sections of the rationals, and this theorem is an expression of the completeness of the real numbers. The whole discussion of this in Chapter 1 of Hardy illuminates the relationships involved. The theorem, or something equivalent, is needed to work with the real numbers if they are defined as sections of the rationals.