Prove Using the Method of Contrapositive

[kmas55]

Prove both by method of contrapositive.
1. If a ≤ b + ε, where ε > 0, then b > a.
2. If 0 ≤ a - b < ε, where ε > 0, then a = b.

I'll start with problem 1.:
p: If a ≤ b + ε, where ε > 0
q: b > a

neg q: b < a
neg p: for some ε' > 0 1/2(a - b), a > b + ε'

define ε' = 1/2(a - b)

I assume the contrapositive goes something like this: p→q, neg q→neg p,

After this, do I start with b < a→ a > b + ε', and then substitute the ε' = 1/2( a - b) into neg p and then try to work it into the form b < a ? However, it seem like I should start b > a and work my way to neg p. But how? Am I on the right path? Thanks

 

[Mark44]

Prove both by method of contrapositive.
1. If a ≤ b + ε, where ε > 0, then b > a.
2. If 0 ≤ a - b < ε, where ε > 0, then a = b.

I'll start with problem 1.:
p: If a ≤ b + ε, where ε > 0
q: b > a

neg q: b < a

The negation of q is "b is not greater than a" I.e., $b \le a$

neg p: for some ε' > 0 1/2(a - b), a > b + ε'

I don't get what you're doing here. Are there any "for all" qualifiers in the original question?
Where did the (1/2)(a - b) come from?

define ε' = 1/2(a - b)

I assume the contrapositive goes something like this: p→q, neg q→neg p,

Yes, you have it right.

After this, do I start with b < a→ a > b + ε', and then substitute the ε' = 1/2( a - b) into neg p and then try to work it into the form b < a ? However, it seem like I should start b > a and work my way to neg p. But how? Am I on the right path? Thanks

 

[kmas55]

yes,,

The negation of q is "b is not greater than a" I.e., $b \le a$
I don't get what you're doing here. Are there any "for all" qualifiers in the original question?

Yes, sorry, instead of "where ε > 0" I should have wrote "for all ε > 0".

 

[Mark44]

yes,,

Yes, sorry, instead of "where ε > 0" I should have wrote "for all ε > 0".

Look up in your textbook or in examples from class how you deal with universal qualifiers in forming the contrapositive.

 

[kmas55]

The negation of q is "b is not greater than a" I.e., $b \le a$
I don't get what you're doing here. Are there any "for all" qualifiers in the original question?
Where did the (1/2)(a - b) come from?
Yes, you have it right.

The (1/2)(a - b) is a hint the 1st problem gives you. I misplaced it in the problem. The "neg p: for some ε' > 0 1/2(a - b), a > b + ε' ", should read " neg p: There is some ε' > 0 , so that a > b + ε'. And then the hint, define ε' = (1/2)(a - b). Sorry, it was late when I typed the original.

Also, thank you for confirming that I'm starting in the right direction. I'll work from there, and then post. Thanks much.