1. ## order of reals

1. Prove that ‘<’ is an order for for all real numbers.

2. Originally Posted by princess08
1. Prove that ‘<’ is an order for for all real numbers.
it may help if you tell us what kind of axioms you are using and how your text/class/professor defines "order"

3. Defn.
From the introductory lectures, an ordered set is a set S with a relation ‘<’ which satisfies
two properties:
1. (Trichotomy property) for any two elements
a, b is an element of S exactly one of the following hold a < b, a = b, or b < a.

2. (Transitive property) for any three elements
a, b, c is an element ofS, if a < b and b < c, then a < c.

In this case the relation ‘
<’ is called an order.

4. Originally Posted by princess08
Defn.
From the introductory lectures, an ordered set is a set S with a relation ‘<’ which satisfies
two properties:
1. (Trichotomy property) for any two elements
a, b is an element of S exactly one of the following hold a < b, a = b, or b < a.

2. (Transitive property) for any three elements
a, b, c is an element ofS, if a < b and b < c, then a < c.

In this case the relation ‘
<’ is called an order.
ok, so you want to prove these two things for the relation "<" on the real numbers. (by the way, if it has to work for all real numbers, shouldn't we be considering " $\le$", also, what do you know about real numbers? are we allowed to talk about "negative" and "positive" here? the proof here is subtle, since " $\le$" is defined on the real numbers to fit these conditions, so it might seem like begging the question if we use concepts that we aren't allowed to)

so, (1) Let $a,b \in \mathbb{R}$. Is it true that $a = b$, or $a < b$ or $b < a$? how would you show that? does considering $(a - b)$ help?

and (2) Let $a,b,c \in \mathbb{R}$, and assume that $a < b$ and $b < c$ according to the relation of "<" on the reals. then that would mean that $a - b < 0$ and $b - c < 0$. what about their sum? (the reals are closed under addition, thus we can use the idea of "sum")