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I've just started learning about relations and now I'm at partial order relations and total order relations; essentially, I'm trying to convey that I'm very much a beginner to this relations stuff.

My textbook includes the following remark:

The usual order relation on the real line $ \Bbb R $ is a total order relation.

A little later in my textbook is the following exercise (a portion of it, anyways):

Let $A$={1,2}.
List all the partial order relations on A. (The usual order relation on A is $R$={$\mathsf (1,2) \cup E)$}, where $E$={$(1,1),(2,2)$} is the relation of equality on A. ...)

The textbook also refers to a usual order relation again, somewhere later, so I would like to know what is meant by the term. I tried to infer something from the exercise whose excerpt I included, but I still don't understand. I went to mathworld.com and I didn't find anything there.

Andrés E. Caicedo
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Jordan Miller
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    The usual order relation is the ordering you would assume someone was talking about if the issue of different possible orders had never come up. So $2$ is less than $5,$ $e^2$ is less than ${\pi}^{2},$ $-3$ is less than $-\frac{1}{2},$ etc. – Dave L. Renfro Aug 17 '15 at 20:09
  • The "usual" ordering is the ordering as you are familiar with it. For instance $n \gt n-1$ in the whole numbers, and the following rules are satisified: $a \lt b, c \gt 0 \implies ac \lt bc$, multiplying by a $-1$ flips the inequality, and we also have $a \lt b, c \in \Bbb{R} \implies a + c \lt b + c$. Prove those, using your intuition about real numbers. – Daniel Donnelly Aug 17 '15 at 20:10
  • @Enjoys Math: How come you didn't put this as an answer? Anyways...then, based on what you wrote, one might say: S is a relation on $ \Bbb R $ defined by $x R y$ iff there exists a real number $c$ such that if $a < B$, then $a + c = b + c$; and S is the usual order relation on $ \Bbb R $. – Jordan Miller Aug 17 '15 at 20:36

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The "usual order relation" is a creation of laziness. It's the order relation that your intuition comes up with first when you have to define an order relation on the space it's attached to.

So for $\Bbb R$, it's the ordering of numbers, for $\{1,2\}$ it's the only "sensible" ordering "up to renaming of the elements" (i.e., $1 < 2$). I hope that clears the air.

Lord_Farin
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The usual ordering on $\Bbb{R}$ can be uniquely defined as the total order $\lt$ on $\Bbb{R}$ that satisfies the following two properties:

(Let $a,b,c \in \Bbb{R}$)

1) $c \gt 0 \iff c$ is positive.

2) $a \lt b, \ \ 0 \lt c \implies ac \lt bc$

Proof that this defines uniquely $\lt$: Suppose that $\lt'$ also satisfies these properties and is a total order on $\Bbb{R}$. Then if $a \lt b$, assume that $b \lt' a$. Then we have $0 \lt' a-b$, or by (1) that $a-b$ is positive. Then let $c = a-b \gt' 0$. This connects us to $\lt$ by (1) since we can also then write $c = a-b \gt 0$. Now we need one more defining property to complete the proof:

3) $a \lt b \implies a + c \lt b + c$

$\square$.

Daniel Donnelly
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