Greater than or equal to/Epsilon form

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  • [ilmath]x\ge y\iff\forall\epsilon>0[x+\epsilon>y][/ilmath]


[ilmath]x\ge y\implies\forall\epsilon>0[x+\epsilon>y][/ilmath]

  • Let [ilmath]\epsilon > 0[/ilmath] be given
    • As [ilmath]\epsilon>0[/ilmath] we see [ilmath]x+\epsilon>0+x=x[/ilmath]
      • But by hypothesis [ilmath]x\ge y[/ilmath]
    • So [ilmath]x+\epsilon>x\ge y[/ilmath]
    • Thus [ilmath]x+\epsilon>y[/ilmath]
  • This completes this part of the proof.

[ilmath]\forall\epsilon>0[x+\epsilon>y]\implies x \ge y[/ilmath] (this will be a proof by contrapositive)

  • We will show: [ilmath]x<y\implies\exists\epsilon>0[x+\epsilon < y][/ilmath] Warning:I wrongly negated [ilmath]>[/ilmath], it should be [ilmath]\le[/ilmath] not [ilmath]<[/ilmath] - in light of this I might be able to get away with [ilmath]\epsilon=y-x[/ilmath]
    • As [ilmath]x<y[/ilmath] we know [ilmath]0<y-x[/ilmath].
    • Choose [ilmath]\epsilon:=\frac{y-x}{2}[/ilmath] (which we may do for both [ilmath]\mathbb{R} [/ilmath] and [ilmath]\mathbb{Q} [/ilmath])
    • Now [ilmath]x+\epsilon=\frac{2x}{2}+\frac{y-x}{2}=\frac{x+y}{2}[/ilmath]
      • But by hypothesis [ilmath]x<y[/ilmath] so [ilmath]x+y<y+y=2y[/ilmath], so:
    • [ilmath]x+\epsilon=\frac{x+y}{2}<\frac{2y}{2}=y[/ilmath]
  • We have shown [ilmath]\exists\epsilon >0[x+\epsilon<y][/ilmath]

This completes this part of the proof.

TODO: Fix warning. Note that [ilmath]x+\epsilon < y\implies x+\epsilon \le y[/ilmath] so this content isn't wrong, but it requires multiplication by [ilmath]\frac{1}{2} [/ilmath] which you cannot do in the ring [ilmath]\mathbb{Z} [/ilmath] for example.