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?  
 A: 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 $\mathbb R$.
A: 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.
A: 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.
