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

  1. Apr 8, 2012 #1
    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
  2. jcsd
  3. Apr 8, 2012 #2
    Suppose, [itex]x > y[/itex]. Then, take [itex]\epsilon = 2 (x - y)[/itex]. Is the first inequality satisfied?
  4. Apr 19, 2012 #3
    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.
  5. Apr 19, 2012 #4
  6. Apr 19, 2012 #5
    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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