Difference between revisions of "Greater than or equal to"
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'''[[Greater than or equal to/Epsilon form|Epsilon form]]:''' {{M|1=x\ge y\iff\forall\epsilon>0[x+\epsilon>y]}} | '''[[Greater than or equal to/Epsilon form|Epsilon form]]:''' {{M|1=x\ge y\iff\forall\epsilon>0[x+\epsilon>y]}} | ||
{{Begin Inline Proof}} | {{Begin Inline Proof}} | ||
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{{Definition|Real Analysis|Elementary Set Theory|Order Theory}} | {{Definition|Real Analysis|Elementary Set Theory|Order Theory}} | ||
Revision as of 15:08, 9 April 2016
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)
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=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.
See also
References
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