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Math Help - Abstract Algebra II

  1. #1
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    Abstract Algebra II

    Let D denote the field of real numbers, and let D^p be any subset of D that satisfies the conditions of the definition here:
    (1)closure under addition if a,b exits in D^P, then a + b exist D^P
    (2)closure under multiplication if a, b exist in D^p, then ab exist in D^p
    (3)law of trichotomy if a exist in D, then exactly one of the following is true: a = 0, a exist in D^p or -a exist in D^p

    Prove that D^p = (0, infinity)(In other words, the usual set
    of positive real numbers is the only subset of D = R that can serve as D^p).

    Please help! We are stuck
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  2. #2
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    Can you show that D^p can not contain any negative numbers?
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  3. #3
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    This is what we have so far, because we are trying to prove that (0, infinity) is a subset of D^P and D^p is a subset of (0, infinity).

    Part 1:

    If a exists (0, infinity)

    Then according to the Definition:

    a + a = 2a, which must be positive because 2 times a positive number is a positive number.

    Additionally, a * a = a^2, which is obviously positive because any number squared is positive.

    Part 2:

    We proved in Part 1 that a exists in (0, infinity). Now, we are letting a belong to D^p.

    Thus, by the last part of the definition, we see that if a exists D^p, then exactly one of the following is true:

    a=0, a belongs to D^p, or -a belongs to D^p.

    Since we proved that a belongs to D^p. This means that only a can belong to D^p.

    Which means that a is not 0 or negative.
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  4. #4
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    Quote Originally Posted by mathgirl1188 View Post
    This is what we have so far, because we are trying to prove that (0, infinity) is a subset of D^P and D^p is a subset of (0, infinity).

    Part 1:

    If a exists (0, infinity)

    Then according to the Definition:

    a + a = 2a, which must be positive because 2 times a positive number is a positive number.

    Additionally, a * a = a^2, which is obviously positive because any number squared is positive.

    Part 2:

    We proved in Part 1 that a exists in (0, infinity). Now, we are letting a belong to D^p. Pay attention that this is not necessarily the same a!

    Thus, by the last part of the definition, we see that if a exists D^p, then exactly one of the following is true:

    a=0, a belongs to D^p, or -a belongs to D^p.

    Since we proved that a belongs to D^p. This means that only a can belong to D^p. You assumed(!) that a \in \mathcal{D}^p, not proved! You don't know anything about this a!

    Which means that a is not 0 or negative.
    Here is what I would do.

    Assume by contradiction that a \in \mathbb{R} and a<0 and a \in \mathcal{D}^p.

    Then a^2 \in \mathcal{D}^p by definition.
    Now, using the closure of \mathcal{D}^p under addition, can you get a number which will cancel a^2 to get 0 by addition?
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  5. #5
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    Quote Originally Posted by mathgirl1188 View Post
    This is what we have so far, because we are trying to prove that (0, infinity) is a subset of D^P and D^p is a subset of (0, infinity).

    Part 1:

    If a exists (0, infinity)

    Then according to the Definition:

    a + a = 2a, which must be positive because 2 times a positive number is a positive number.

    Additionally, a * a = a^2, which is obviously positive because any number squared is positive.

    Part 2:

    We proved in Part 1 that a exists in (0, infinity).
    NO, you didn't! You started Part 1 with "if a exists in (0, infinity)" as hypothesis, not conclusion. I thought you were going to show that a was in D^p but you did not. you simply showed that 2a and a^2 were positive. I don't see what that has to do with a being in D^p.


    Now, we are letting a belong to D^p.

    Thus, by the last part of the definition, we see that if a exists D^p, then exactly one of the following is true:

    a=0, a belongs to D^p, or -a belongs to D^p.

    Since we proved that a belongs to D^p. This means that only a can belong to D^p.

    Which means that a is not 0 or negative.
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