Proving Dedekind Cuts for D = {x: x \in Q and (x \leq or x^2 < 2)}

  • Thread starter laminatedevildoll
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In summary, D is a Dedekind cut if it is not an empty set or the set of all rational numbers, if for any element in D there exists a larger element in D, and if for any element in D and any smaller element, the smaller element must also be in D. The set D = {x: x \in Q and (x \leq 0 or x^2 < 2)} satisfies these conditions and therefore is a Dedekind cut.
  • #1
laminatedevildoll
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The question:

Show D= {x: x [tex]\in[/tex] Q and (x [tex]\leq[/tex] or x^2 < 2)} is a dedekind cut.


A set D c Q is a Dedekind set if

1)D is not {}, D is not Q
2) if r[tex]\in[/tex] D then there exists a s [tex]\in[/tex] D s.t r<s
3) if r [tex]\in[/tex] D and if s [tex]\leq[/tex] r, then s [tex]\in[/tex] D.

For the first case, D is not an empty set because x is equal to 0 or the sqrt of 2. But, how do I prove case 2,3. Do I have to use addition/multiplication to prove them?
 
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  • #2
I'm not sure I understand your definition of D

is it equivilant to [tex] D=\{x\in Q | x\le 2\}\cup \{x\in Q | x^2 < 2\}[/tex]
Which means [tex]D= \{x\in Q | x\le 2\}[/tex] which seems to contradict 2).

Steven
 
  • #3
snoble said:
I'm not sure I understand your definition of D

is it equivilant to [tex] D=\{x\in Q | x\le 2\}\cup \{x\in Q | x^2 < 2\}[/tex]
Which means [tex]D= \{x\in Q | x\le 2\}[/tex] which seems to contradict 2).

Steven

Sorry, D is actually
D= {x: x [tex]\in[/tex] Q and (x [tex]\leq[/tex] 0 or x^2 < 2)}
 
  • #4
1. Obviously, D is not empty- any negative number is in D. Obviously D is not all rational numbers, 2> 0, 22= 4> 2 so 2 is not in D.

3. if r is in D and s< r then either:
a) r< 0 in which case r is in D or
b) 0< r< s so 0< r2< s2< 2 so r is in D.

2. is the hard one. Obviously if r< 0, we can take s= 0. Ir r> 0, then r2< 2. Take d= 2- s2. Can you show that 0< (r+ d/4)2> 2?
 
  • #5
HallsofIvy said:
1. Obviously, D is not empty- any negative number is in D. Obviously D is not all rational numbers, 2> 0, 22= 4> 2 so 2 is not in D.

3. if r is in D and s< r then either:
a) r< 0 in which case r is in D or
b) 0< r< s so 0< r2< s2< 2 so r is in D.

2. is the hard one. Obviously if r< 0, we can take s= 0. Ir r> 0, then r2< 2. Take d= 2- s2. Can you show that 0< (r+ d/4)2> 2?

0< (r+ d/4)2> 2?
if d= 2- s2
0< r + (2- s2)/4 > 2
Do I let r = sqrt(2) both plus and minus
to show that
0< r > 2, so this will confirm the fact that r<s?
 
  • #6
Sorry, there was a misprint! I mean 0< (r+d/4)2< 2. (not > 2)

Suppose r is the largest number in the set.

It is obvious that (3/2)2= 2.25> 2 so 3/2 is not in this set. It is obvious that 1.42= 1.96 so 1.4 is in this set. Any possible maximum for the set must be greater than or equal to 1.4 and less than 1.5= 3/2: 1.4<= r< 3/2 and so d= 2- r2 must be less than or equal to 2- 1.96= 0.04. (r+ d/4)2= r2+ rd/2+ d2/16 so 2- (r+d/4)2= (2- r2)- rd/2- d2/16. 2- r2= d and since r< 3/2, rd/2< (3/4)d. d/16= d(d/16) and since d< 0.04, d/16< (.04/16)= (.01/4)= 0.0025. That is: 2- (r+d/4)2> d- (3/4)d- 0.0025d= d- (0.7525)d> 0 which means (r+ d/4)2.
r+ d/4 is larger than r but (r+ d/4)2< 2 so r+ d/4 is still in the set contradicting the hypothesis that r is the maximum for the set. Therefore, the set has no maximum.
 
  • #7
For the third proof, do I go onto assume that s is equal or greater than r to prove that it's a contradiction?
 
  • #8
laminatedevildoll said:
For the third proof, do I go onto assume that s is equal or greater than r to prove that it's a contradiction?

No, just prove exactly what's given: if s less than or equal to r, then it must be in the set. That's exactly what I did in my first response:

Suppose r is in this set. There are two possibilities: r< 0 or r2< 2.
(a) If r< 0 and s<= r, then s< 0 so s is in the set.

(b) If 0<= r, r2< 2, and s< r then either
(i) s< 0 so s is in the set
(ii) s>= 0 so 0<= s2< r< 2 and s is in the set.
In any case, if r is in this set and s< r, then s is in the set.
 

FAQ: Proving Dedekind Cuts for D = {x: x \in Q and (x \leq or x^2 < 2)}

What are Dedekind Cuts?

Dedekind Cuts are mathematical objects used in the construction of real numbers. They are a way of defining a real number as the partition of the rational numbers into two sets - one set containing all the rational numbers less than the real number and the other set containing all the rational numbers greater than the real number.

Who is the creator of Dedekind Cuts?

Dedekind Cuts are named after German mathematician Richard Dedekind, who introduced them in his book "Continuity and Irrational Numbers" in 1872.

How are Dedekind Cuts different from Cauchy Sequences?

While both Dedekind Cuts and Cauchy Sequences are methods of constructing real numbers, they differ in their approach. Dedekind Cuts define a real number as a partition of the rational numbers, while Cauchy Sequences define a real number as the limit of a sequence of rational numbers.

What is the significance of Dedekind Cuts in mathematics?

Dedekind Cuts are important in the development of the real number system, as they provide a rigorous definition of real numbers. They also allow for the construction of irrational numbers, which cannot be expressed as a ratio of two integers.

Can Dedekind Cuts be used in other areas of mathematics?

Yes, Dedekind Cuts have applications in various areas of mathematics such as analysis, topology, and algebra. They are also used in other fields such as economics and computer science.

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