## Monday, March 06, 2006

### Dedekind Cut

The Dedekind Cut is mathematical construction created by Richard Dedekind to provide a definition for the real numbers.

The Dedekind Cut itself is defined in terms of the rational numbers.

Definition 1 - Dedekind Cut

A Dedekind cut α is defined as the subset of the rational integers Q (ratios of integers) which is less than α.

NOTE: Q is used to present the set of all rational numbers; R is used to represent the set of all real numbers, and Z is used to represent the set of all integers.

Example: π

We could create a Dedekind Cut around π. In this case, we could think of 3, 22/7, -4, etc. as elements of the Dedekind Cut. On the other hand, 32/10, 4, etc. would not be elements of the cut.

Definition 2 - Set of Real Numbers R

The set of real numbers is the set of Dedekind cuts α that have the following properties:

(a) α is not empty

(b) α contains no greatest element

For any element xα, there exists y ∈ α such that x is less than y.

(c) If x,y are rational integers where y is less than x, then x ∈ α → y ∈ α.

With these definitions, we have enough the construct the properties of the real numbers.

Definition 3: Additive Identity: 0

0 is defined as the set { x ∈ Q such that x is less than 0 }

Definition 4: Multiplicative Identity: 1

1 is defined as the set { x ∈ Q such that x is less than 1 }

Definition 5: Addition

α + β is defined as the set { x + y such that x ∈ α, y ∈ β }

Definition 6: Subtraction

α - β is defined as the set { x - y such that x ∈ α, y ∈ β }

Definition 7: Multiplication

α * β is defined as the set { x * y such that x ∈ α, y ∈ β }

Definition 8: Division

α / β is defined as the set { x/y such that β ≠ 0, x ∈ α, y ∈ β }

Definition 9: Irrational Numbers

A real number α is said to be irrational if α ∩ Q does not have a least element.

Now, I will present a proof for a fundamental property of reals using the Dedekind Cut.

Theorem 1: Every real number that is bounded above has a least upper bound.

(1) Let Α be a set of real numbers that is bounded by γ such that α ∈ Α → α ≤ γ

NOTE: Bounded above just means that there is an element γ that is greater or equal to all the elements that make up A.

(2) The union of all the sets that make up Α are themselves a real number because:

(a) Since each real number α is a Dedekind cut, we know that each α is the set of x ∈ Q such that x is less than some real number.

(b) Since each α is not empty, the union of all elements that make up α is not empty.

(c) We know that the union does not have a greatest element since if it did, this element would likewise be the greatest element for whichever real number that it is an element for but this is impossible since by definition, none of the real numbers have a greatest element.

(d) If x ∈ the Union and y is less than x, then y is necessarily an element of the Union, since y would necessarily be an element of the real number that x is an element of.

(3) The union of all sets that make up A is an upper bound for A since every element of A ⊆ Union of all sets of A.

(4) Now, γ is an arbitrary upper bound, so all we need to prove is that A ⊆ γ since this shows that A is necessarily less than or equal to any given upper bound.

(5) But this is easy to prove since x ∈ A → x ∈ γ since we defined each element of A as less than γ.

QED

Theorem 2: For any positive real number ε, there exists a natural number n such that:

0 is less than 1/n is less than ε

Proof:

(1) We can find a natural number n such that:

(n-1) ≤ (1/ε) ≤ n.

(2) Taking the reciprocal for each gives us:

1/(n-1) ≥ ε ≥ 1/n.

QED

Corollary 2.1: Between any two distinct real numbers, there exists a rational number.

Proof:

(1) Let x,y be real numbers such that y is greater than x.

(2) By Theorem 2 above, there exists an integer n such that: 1/n is less than y - x.

(3) Let m = floor(n*x) where floor(n*x) returns the highest integer that is less than n*x.

(4) From this, we know that:

(m/n) ≤ x and x is less than (m+1)/n.

(5) Further, we have:

(m + 1)/n = m/n + 1/n ≤ x + 1/n is less than x + y - x = y.

(6) So that, we have:

x is less than (m+1)/n is less than y.

QED

References

#### 6 comments :

Haroon said...

What would + and - mean exactly? It isn't clear to me.
Thanks.

Larry Freeman said...

That's Definition 5 and Definition 6.

The A + B set is a new set which consists of every combination of a + b where a is an element of A and b is an element of B.

A - B set is a new set which consists of every combination of a - b where a is an element of A and b is an element of B.

-Larry

Anonymous said...

You seem to have a circular arguement here. In definition 1 you define a dedekind cut by a real number but you can't do this as the real numbers are supposed to be defined by dedekind cuts. You need to define dedekind cuts without real numbers and then define real numbers as dedekind cuts.
Also your definition of multiplication is wrong as it allows 100 to be in the set 1*2 as 100=(-1)*(-100).

Larry Freeman said...

I'm not clear on your question.

Definition 1 defines a Dedekind cut in terms of rational numbers.

Definition 2 defines real numbers in terms of Dedekind cut.

100 = (-1)*(100) so I'm not sure what you are saying here. If you add more details, I'll be glad to respond.

Anonymous said...

Then definition 1 makes no sense as you say a dedekind cut is a subset of intergers which are less than that dedekind cut. You can't use a dedekind cut to define a dedekind cut. You also haven't defined what less than means as these are sets you are dealing with not numbers. a < b should be defined as a is a subset of b, but you can't do this until you define dedekind cuts.
The problem with your definition of multiplication is that you say a*b is the set of all products of all rational numbers in the original cuts a and b. So if we multiply 2 and 1 we can take -1 which is in the cut 1 and -100 which is in the cut 2. Then 100=(-1)*(-100) must be in the cut 2*1.
Your definiton of division is also problematic as you only specify that b can not 0 for the cut a/b. but if b>0 then 0 will be in the cut b and so x/0 for x in cut a will be in a/b. But this makes no sense.

Larry Freeman said...

Hi Anonymous,

I agree with your criticisms of the definitions. I'll update the definitions and repost.

Thanks very much for your comments!

-Larry