View source for Greater than or equal to/Epsilon form

<noinclude>
==Statement==
* {{M|1=x\ge y\iff\forall\epsilon>0[x+\epsilon>y]}}
==Proof==</noinclude>
{{M|1=x\ge y\implies\forall\epsilon>0[x+\epsilon>y]}}
* Let {{M|\epsilon > 0}} be given
** As {{M|\epsilon>0}} we see {{M|1=x+\epsilon>0+x=x}}
*** But by hypothesis {{M|x\ge y}}
** So {{M|1=x+\epsilon>x\ge y}}
** Thus {{M|x+\epsilon>y}}
* This completes this part of the proof.
{{M|1=\forall\epsilon>0[x+\epsilon>y]\implies x \ge y}} (this will be a proof by [[contrapositive]])
* We will show: {{M|x<y\implies\exists\epsilon>0[x+\epsilon < y]}} {{Warning|I wrongly negated {{M|>}}, it should be {{M|\le}} not {{M|<}} - in light of this I might be able to get away with {{M|1=\epsilon=y-x}}}}
** As {{M|x<y}} we know {{M|0<y-x}}.
** Choose {{M|1=\epsilon:=\frac{y-x}{2} }} (which we may do for both {{M|\mathbb{R} }} and {{M|\mathbb{Q} }})
** Now {{M|1=x+\epsilon=\frac{2x}{2}+\frac{y-x}{2}=\frac{x+y}{2} }}
*** But by hypothesis {{M|x<y}} so {{M|1=x+y<y+y=2y}}, so:
** {{M|1=x+\epsilon=\frac{x+y}{2}<\frac{2y}{2}=y}}
* We have shown {{M|\exists\epsilon >0[x+\epsilon<y]}}
This completes this part of the proof.
{{Todo|Fix warning. Note that {{M|x+\epsilon < y\implies x+\epsilon \le y}} so this content isn't wrong, but it requires multiplication by {{M|\frac{1}{2} }} which you cannot do in the [[ring]] {{M|\mathbb{Z} }} for example.}}
<noinclude>
==References==
<references/>
{{Theorem Of|Order Theory|Elementary Set Theory|Real Analysis}}
</noinclude>