1. Prove that ‘<’ is an order for for all real numbers.
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 elementsa, 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 "$\displaystyle \le$", also, what do you know about real numbers? are we allowed to talk about "negative" and "positive" here? the proof here is subtle, since "$\displaystyle \le$" 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 $\displaystyle a,b \in \mathbb{R}$. Is it true that $\displaystyle a = b$, or $\displaystyle a < b$ or $\displaystyle b < a$? how would you show that? does considering $\displaystyle (a - b)$ help?
and (2) Let $\displaystyle a,b,c \in \mathbb{R}$, and assume that $\displaystyle a < b$ and $\displaystyle b < c$ according to the relation of "<" on the reals. then that would mean that $\displaystyle a - b < 0$ and $\displaystyle b - c < 0$. what about their sum? (the reals are closed under addition, thus we can use the idea of "sum")