The discussion revolves around a theorem involving real numbers x, y, and ε, specifically addressing the implications of the statement "if x ≤ y + ε for every ε > 0, then x ≤ y." Participants are exploring the contrapositive proof of this theorem and questioning the nature of ε in this context.
The discussion is ongoing, with participants raising questions about the correctness of statements made regarding ε and the implications of the contrapositive. Some guidance has been offered regarding the interpretation of quantifiers and the conditions necessary for the proof.
There is a focus on the definitions and properties of ε, particularly in relation to the inequalities involved. Participants are examining the implications of negating the original statement and how it affects the proof structure.
trebolian said: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> 0"?trebolian said: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?
HallsofIvy said:I don't know what you mean by "the equality of \epsilon". Are you referring to the in equality "\epsilon> 0"?
In any case your first statement is incorrect. If x> y then there exist an infinite number of positive \epsilon such that x> y+ \epsilon. x> y implies x-y> 0. Take \epsilon to be any positive number less than x- y.