# Math Help - Abstract Algebra II

1. ## 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).

2. Can you show that $D^p$ can not contain any negative numbers?

3. 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.

4. 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 $a \in \mathcal{D}^p$, 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?

5. 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.