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Proof: if x≤y+ε for every ε>0 then x≤y

by samsun2024
Tags: ε, ε>0, proof, x≤y
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samsun2024
#1
Apr8-12, 10:30 AM
P: 5
let x,y,ε in ℝ.
if x≤y+ε for every ε>0 then x≤y

hints: use proof by contrapositive .

i try to proof it, and end up showing that....
if x+ε≤y for every ε>0 then x≤y
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Dickfore
#2
Apr8-12, 10:34 AM
P: 3,014
Suppose, [itex]x > y[/itex]. Then, take [itex]\epsilon = 2 (x - y)[/itex]. Is the first inequality satisfied?
oleador
#3
Apr19-12, 09:25 PM
P: 19
Quote Quote by Dickfore View Post
Suppose, [itex]x > y[/itex]. Then, take [itex]\epsilon = 2 (x - y)[/itex]. Is the first inequality satisfied?
The contrapositive is [itex]x>y \Rightarrow x>y+ε[/itex]
[itex]\epsilon = 2 (x - y)[/itex] would not work:
[itex]x>y+ε \Rightarrow x>y+2 (x - y) \Rightarrow -x>-y[/itex], a contradiction unless [itex]x=y[/itex].
[itex]\epsilon = (x - y)/2[/itex] would work though.

DonAntonio
#4
Apr19-12, 10:51 PM
P: 606
Proof: if x≤y+ε for every ε>0 then x≤y

Quote Quote by oleador View Post
The contrapositive is [itex]x>y \Rightarrow x>y+ε[/itex]


*** No, it is not. The contrapositive is [itex]x>y\Longrightarrow x\nleq y+\epsilon[/itex] , for some [itex]\epsilon > 0[/itex]

DonAntonio



[itex]\epsilon = 2 (x - y)[/itex] would not work:
[itex]x>y+ε \Rightarrow x>y+2 (x - y) \Rightarrow -x>-y[/itex], a contradiction unless [itex]x=y[/itex].
[itex]\epsilon = (x - y)/2[/itex] would work though.
....
oleador
#5
Apr19-12, 11:10 PM
P: 19
True. Confused [itex]\forallε>0[x≤y+ε]\Rightarrow x≤y[/itex] with [itex]\forallε>0[x≤y+ε\Rightarrow x≤y][/itex].
The former is true.

This, however, does not change my conclusion. [itex]ε=2(x−y)[/itex] doesn't work, while [itex]ε=(x−y)/2[/itex] does.


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