The discussion centers on the contrapositive proof of the theorem stating that if \( x \leq y + \epsilon \) for every \( \epsilon > 0 \), then \( x \leq y \). Participants clarify that the contrapositive statement \( x > y \rightarrow x > y + \epsilon \) holds true only if \( \epsilon < 0 \). The key conclusion is that the negation of the statement requires demonstrating the existence of a positive \( \epsilon \) that satisfies the inequality, specifically \( \epsilon = y - x \), which is crucial for the proof.
PREREQUISITESMathematics students, educators, and anyone interested in formal logic and proof techniques, particularly in the context of real analysis and inequalities.
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.