Greater than or equal to
Definition
Greater than or equal to is a relation (specifically a partial ordering) on [ilmath]\mathbb{R} [/ilmath] (and thus [ilmath]\mathbb{Q} [/ilmath], [ilmath]\mathbb{Z} [/ilmath] and [ilmath]\mathbb{N} [/ilmath]).
TODO: Link with ordered integral domain (as that is where the ordering is induced) THE STRUCTURE ON [ilmath]\mathbb{R} [/ilmath] IS VERY IMPORTANT. For example the epsilon form below requires addition, subtraction, so forth
Alternative forms
Epsilon form: [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]
 As [ilmath]\epsilon>0[/ilmath] we see [ilmath]x+\epsilon>0+x=x[/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=yx[/ilmath]
 As [ilmath]x<y[/ilmath] we know [ilmath]0<yx[/ilmath].
 Choose [ilmath]\epsilon:=\frac{yx}{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{yx}{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.
See also
References

