Contrapositive Proof of Theorem: x > y → x > y+ε

[trebolian]

Homework Statement

Theorem: Let x,y,ε be ℝ. If x≤ y+ε  for every ε > 0 then x ≤ y.

Write the above as a logic statement and prove it using contrapositive proof.


The attempt at a solution

The contrapositive statement x > y → x > y+ε is only true if ε < 0. Does a contrapositive proof negate the equality of ε?

 

[cris(c)]

Homework Statement

Theorem: Let x,y,ε be ℝ. If x≤ y+ε  for every ε > 0 then x ≤ y.

Write the above as a logic statement and prove it using contrapositive proof.


The attempt at a solution

The contrapositive statement x > y → x > y+ε is only true if ε < 0. Does a contrapositive proof negate the equality of ε?

It should. The negation of x≤ y+ε  for every ε > 0 requires X>y+e for some e<0.

 

[HallsofIvy]

Homework Statement

Theorem: Let x,y,ε be ℝ. If x≤ y+ε for every ε > 0 then x ≤ y.

Write the above as a logic statement and prove it using contrapositive proof.The attempt at a solution

The contrapositive statement x > y → x > y+ε is only true if ε < 0. Does a contrapositive proof negate the equality of ε?

I don't know what you mean by "the equality of \epsilon". Are you referring to the in equality "\epsilon&gt; 0"?

In any case your first statement is incorrect. If x> y then there exist an infinite number of positive \epsilon such that x&gt; y+ \epsilon. x> y implies x-y> 0. Take \epsilon to be any positive number less than x- y.

 

[trebolian]

I don't know what you mean by "the equality of \epsilon". Are you referring to the in equality "\epsilon&gt; 0"?

In any case your first statement is incorrect. If x> y then there exist an infinite number of positive \epsilon such that x&gt; y+ \epsilon. x> y implies x-y> 0. Take \epsilon to be any positive number less than x- y.

HallsofIvy, are you saying that the contrapositive of "for every ε > 0..." is actually "there exists an ε > 0 such that..."?

I would have thought that the contrapositive should be "there exists an ε < 0 such that...", i.e switch the inequality as well as the universal/existential quantifier

 

[cris(c)]

you can't negate saying that you need an epsilon greater than zero. The negation must be done looking for some nonnegative epsilon. Any will do it, in particular epsilon=y-x.

 

[cris(c)]

A mistake in my previous post. Indeed, to prove ~Q implies ~P you have to show that for some e>0, x > y → x > y+ε, since negating Q means that there is at least one e>0 such that ~Q is true.