## Dedekind Cut

this isn't homework this is from a book im reading for fun

L is the set of all negative rationals and zero and all numbers whose square is less than 2. R is the set of all numbers whose squares are greater than 2. this defines$\sqrt{2}$

so thats cool i understand that, the irrational number is the interstitial region between the two sets.

what i don't understand is how applying the dedekind to the reals you get reals. maybe by analogy in that if you apply it to the rationals and the cut is inside one of the sets then the cut defines a rational number?

x is real and has a square less than 2. this defines an L class with no largest member and an R class with smallest member $\sqrt{2}$

when we cut the rationals we say that if L is everything with a square less than 2 then that set has no greatest number and conversely or w/e R is the set of all numbers with squares greater than 2 but this set has no bottom.

what is different in the definition of the cut in the reals that makes it include the square root of 2 in R?

edit

ahh i understand wow i'm dumb it makes sense that if we are now including irrationals in field(right word?) then sqrt(2) is included in one of the real sets. and since its square 2 it should be in the set that does not contain the numbers whose squares are less than 2

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 Quote by ice109 this isn't homework this is from a book im reading for fun L is the set of all negative rationals and zero and all numbers whose square is less than 2. R is the set of all numbers whose squares are greater than 2. this defines$\sqrt{2}$ so thats cool i understand that, the irrational number is the interstitial region between the two sets.
I would consider that poorly phrased. Using the Dedekind cut definition, the irrational number $\sqrt{2}$ is the union of the set of all negative rational numbers and the set of all rational numbers, the set L. As long as you are talking about rationals only, there is no "interstitial" region. And, by the way, part of the set R you define is in set L.\
But basically, you are right- the set of all rational numbers is not "complete"- there are holes in it. The Dedekind cut is a way of "filling" those holes.

[/quote]what i don't understand is how applying the dedekind to the reals you get reals. maybe by analogy in that if you apply it to the rationals and the cut is inside one of the sets then the cut defines a rational number?[/quote]
The Dedekind cut is a real number. You then prove that the set of all Dedekind cuts has all properties you associate with real numbers. In particular, if r is rational number, then the set {x| x is rational, x< r} is the "real number" representing the rational number r.

[/quote]x is real and has a square less than 2. this defines an L class with no largest member and an R class with smallest member $\sqrt{2}$[/quote]
No, it doesn't. Remember that the purpose of this is to DEFINE the real numbers. Until you have done that there is NO $\sqrt{2}$. Defining L as you have above and R as the set of positive rational numbers whose square is larger than 2, you have partition the rational numbers into a set, L, that has no largest member and a set, R, that hs no smallest member. In particular, L is a set having an upper bound but no least upper bound and R is a set having a lower bound but no greatest lower bound.
Oh, wait, you are talking about sets of real numbers, not just sets of rational numbers. Sorry.

 when we cut the rationals we say that if L is everything with a square less than 2 then that set has no greatest number and conversely or w/e R is the set of all numbers with squares greater than 2 but this set has no bottom.
You are being just a bit sloppy. L is "everything with a square less than 2" and all negative rationals. R is "all positive rational numbers with square larger than 2"

 what is different in the definition of the cut in the reals that makes it include the square root of 2 in R?
The "least upper bound property". If we think of L (still a set of rationals) as a subset of R, then, since it has an upper bound, 2, it must have a least upper bound so that either L contains a largest member or R contains a smallest. In this particular case, $\sqrt{2}$ is the least upper bound and is in R.

 edit ahh i understand wow i'm dumb it makes sense that if we are now including irrationals in field(right word?) then sqrt(2) is included in one of the real sets. and since its square 2 it should be in the set that does not contain the numbers whose squares are less than 2
Pretty much. The point of the Dedekind cut is that the set, L, above is the irrational number $\sqrt{2}$. The rational numbers themselves are a subset of the real numbers in the way I described above: the rational number r corresponds to the set {x| x is rational, x< r}. By the way, that is the definition Dedekind used- it is only the set you call "L" and doesn't mention "R" except for the requirement that L not be all rational numbers:
A "Dedekind cut" is a set or rational numbers satisfying
a) The set is not empty; there exist a least one rational number in it
b) The set is not all rationals; there exist at least one rational not in it
c) If a< b and b is in the set then so is a
d) The set has no largest member

If A and B are Dedekind cuts, then A< B if and only if A is a proper subset of B.

The real point of the Dedekind cut definition it that it makes it easy to prove one of the "defining" properties of the real numbers- the Least Upper Bound property that I mentioned before.
Let A be a non-empty set of real numbers, having upper bound b. Since a "real number" is a Dedekind cut, a set, A is a collection of sets. Let X be the union of all cuts in A.

Now show that X is itself a cut:
a) Since A is non-empty, it contains at least one cut. Since cuts are non-empty, the union contains at least one rational number.
b) Since A has b as upper bound, a< b for every member of A so all rational number a in some member of A are contained in b. But there exist some rational number that is not in b and so not in the union of A= x.
c) Suppose b is in x, a< b. Since x= union of A, b is in some member of A. Since a< b, a is also in that cut and so in the union: in A.
d) Suppose x were the largest member of x. Since it is in x, it must be a rational number in some member of A, say y. Since x is the largest member of y, it must be larger than any other member of y (since they are also in X) and so is the largest member of y, contradicting the fact that y is a cut.

It's easy to show that X is an upper bound on A: X is the union of all members of A so every member of A is a subset.

Finally show that there cannot exist a cut that contains every member of A and is properly contained in X.