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Thread: order of reals

  1. #1
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    order of reals

    1. Prove that < is an order for for all real numbers.
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    Quote Originally Posted by princess08 View Post
    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"
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    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.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by princess08 View Post
    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 "$\displaystyle \le$", 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")
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