Greater than or equal to

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I made this page just so I could document the epsilon form

Definition

Greater than or equal to is a relation (specifically a partial ordering) on $\mathbb{R}$ (and thus $\mathbb{Q}$, $\mathbb{Z}$ and $\mathbb{N}$).

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TODO: Link with ordered integral domain (as that is where the ordering is induced) THE STRUCTURE ON $\mathbb{R}$ IS VERY IMPORTANT. For example the epsilon form below requires addition, subtraction, so forth

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Alternative forms

[Expand]

Epsilon form: $x\ge y\iff\forall\epsilon>0[x+\epsilon>y]$

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

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

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

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

This completes this part of the proof.

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

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See also

• Less than or equal to
  • Strictly less than
• Strictly greater than
• Lower bound
  • Infimum
• Upper bound
  • Supremum