Can you show that can not contain any negative numbers?
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
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.
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.