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

[laminatedevildoll]

The question:

Show D= {x: x $$\in$$ Q and (x $$\leq$$ 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$$\in$$ D then there exists a s $$\in$$ D s.t r<s
3) if r $$\in$$ D and if s $$\leq$$ r, then s $$\in$$ 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?

 

[snoble]

I'm not sure I understand your definition of D

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

Steven

 

[laminatedevildoll]

I'm not sure I understand your definition of D

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

Steven

Sorry, D is actually
D= {x: x $$\in$$ Q and (x $$\leq$$ 0 or x^2 < 2)}

 

[HallsofIvy]

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?

 

[laminatedevildoll]

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?

 

[HallsofIvy]

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.}

 

[laminatedevildoll]

For the third proof, do I go onto assume that s is equal or greater than r to prove that it's a contradiction?

 

[HallsofIvy]

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.