Theorem proof: If, for every ? > 0, a ? b + ? , then a ? b.

[jsyc9301]

I've been working on doing a proof for this theorem and I was hoping that someone could help me ensure that I am performing this correctly. It's been a while since I've taken math. Thanks in advance.

Real Analysis Theorem:
If, for every ? > 0, a ? b + ? , then a ? b .

Proof:
Let p ? for every ? > 0, a ? b + ? and q ? a ? b.

Therefore p -> q

I believe that the next step to proving this would be to form the negations ~p and ~q.

This would mean that ~p ? there exists ? > 0, a > b + ? and ~q ? a > b

I'm not sure of the next step, but I think the next logical step would be to write a contrapositive statement in the form of ~q -> ~p

The contrapositive statement is below:
If, a>b, then there exists ? > 0, a > b + ? .

After this step, I get lost. I assume I am to prove the contrapositive statement by letting ? = (a-b)/2 > 0 . But I'm not sure. So if anyone could guide me to what I should be doing here, I would be very grateful.

 

[pka]

Re: Theorem Help Please

Suppose that \(\displaystyle a>b\) then let \(\displaystyle \varepsilon = \frac{a-b}{2}\).
It is easy to get a contradiction from there.

 

[jsyc9301]

Re: Theorem Help Please

Apparently I'm getting to old for this kind of math logic Just had an epiphany as to what I was doing. Thanks pka, your simple observation made me realize what I'm doing. I should have the final proof up here by tonight if you guys want to check my work.

 

[jsyc9301]

Re: Theorem Help Please

Still a little confused apparently as I can't finish the proof. I got some insight from an acquaintance, but I am not quite understanding what they meant.

They said that Because a > b this means that there is a Delta > 0 for which a - b = Delta, we have a contradiction, because we can take Epsilon = Delta/2 :

a - b = Delta

a <= b + Delta/2

b+Delta <= b + Delta/2

Delta < Delta/2 and this is a contradiction

It seems like there would be a much simpler explanation to this all.

Am I supposed to be choosing a value for Epsilon or choosing values for a and b?

 

[pka]

Re: Theorem Help Please

\(\displaystyle a > b\, \Rightarrow \,a < b + \left( {a - b} \right) = a\, \Rightarrow \,a < a\)

 

[jsyc9301]

Re: Theorem Help Please

Here is the final proof. Can someone help me edit out the unnecessary stuff to simplify the proof? Thanks again for all the help.

Real Analysis Theorem:
If, for every ? > 0, a ? b + ? , then a ? b .

Proof:
Let p ? for every ? > 0, a ? b + ? and q ? a ? b.
Therefore p -> q
Let ~p ? there exists ? > 0, a > b + ? and ~q ? a > b
If, a>b, then there exists ? > 0, a > b + ?
Suppose that a > b then let ? = (a-b)/2 > 0
So a > b + ?
a > b + (a-b)/2 > 0
a – b > (a-b)/2 > 0
a – b > (a-b)/2 > 0
Let a=4, b =2
4 – 2 > (4-2)/2 > 0
2 > 2/2 > 0
2 > 1 > 0

Therefore, if a > b, then there exists ? > 0, a > b + ?. Thus, if, for every ? > 0, a ? b + ? , then a ? b .

 

[daon]

Re: Theorem Help Please

Here is the final proof. Can someone help me edit out the unnecessary stuff to simplify the proof? Thanks again for all the help.

Real Analysis Theorem:
If, for every ? > 0, a ? b + ? , then a ? b .

Proof:
Let p ? for every ? > 0, a ? b + ? and q ? a ? b.
Therefore p -> q
Let ~p ? there exists ? > 0, a > b + ? and ~q ? a > b
If, a>b, then there exists ? > 0, a > b + ?
Suppose that a > b then let ? = (a-b)/2 > 0
So a > b + ?
a > b + (a-b)/2 > 0
a – b > (a-b)/2 > 0
a – b > (a-b)/2 > 0
Let a=4, b =2
4 – 2 > (4-2)/2 > 0
2 > 2/2 > 0
2 > 1 > 0

Therefore, if a > b, then there exists ? > 0, a > b + ?. Thus, if, for every ? > 0, a ? b + ? , then a ? b .

a and b should remain arbitrary.

If you wish to prove his by contrapositive then suppose a>b. let ?=(a-b)/2. Note ?>0. Then a = b+2?. a>b+? iff b+2?>b+? iff 2? > ? iff ? > 0.

 

[jsyc9301]

Re: Theorem Help Please

Haha, yes, I didn't do so well on this proof. I haven't taken any math logic classes in a long time so my brain is still catching up. But I did learn a lot from my mistakes and I think I'm beginning to understand the basics of proofs. I do appreciate all your tutoring though and hints.