Thread: 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

Can you show that can not contain any negative numbers?

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.

Originally Posted by mathgirl1188

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 , 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 and and .

Then by definition.
Now, using the closure of under addition, can you get a number which will cancel to get 0 by addition?

Originally Posted by mathgirl1188

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.