Make integers constitute a field

[msbell1]

Homework Statement

This question consists of three parts, the first two of which I have answered:
a) Is the set of all positive integers a field? (positive indicates greater than or equal to 0, and ordinary definitions of addition and multiplication are being used)
No. There is no additive inverse for any element other than 0. Also, there is not a multiplicative inverse for any element in the set of positive integers other than 1.

b) What about the set of all integers?
Almost, but no. Again, there is not a multiplicative inverse for most elements.

c) Can the answers to these questions be changed by re-defining addition or multiplication (or both)?

Maybe I'm not imaginative enough to answer this. My problem is that I'm not sure how I can change addition or multiplication and still have the operations satisfy the axioms that define a field. For instance, does the multiplicative inverse of a always have to be 1/a? I guess it does, since a(1/a) must equal 1. In that case, I am tempted to answer this question by saying "no". Is this the right answer?

Homework Equations

Axioms for a field:
to every pair of scalars a and b, there is a scalar a + b such that
a + b = b + a
a + (b + c) = (a + b) + c
a + 0 = a
a + -a = 0
ab = ba
a(bc) = (ab)c
a1 = a
a(1/a) = 1

The Attempt at a Solution

See above.

Thank you very much for the help! For anyone interested, this is problem 2 in Finite Dimensional Vector Spaces by Halmos, which I am currently trying to get through on my own, although the going is slow so far (I am stuck on page 2).

 

[micromass]

The answer is yes. But you'll need to come up with a weird addition and multiplication.

Let $$f:\mathbb{Z}\rightarrow \mathbb{Q}$$ be a bijection.
Define addition of a and b as

$$a\oplus b=f^{-1}(f(a)+f(b))$$

(where the + is just the regular addition in Q). Similarly, we define multiplication as

$$ab=f^{-1}(f(a)f(b))$$

 

[msbell1]

Thanks! I kind of see it now. Definitely I can see how these still satisfy commutativity and associativity for both addition and multiplication.

Now if I focus on addition, I guess I need to define some scalar, I will call it 0*, such that a $$\oplus$$ 0* = a (which implies that f(0*) = 0 where 0 is the usual 0), and some scalar that I will call a- such that a $$\oplus$$ a- = 0*. I guess I can convince myself that this will work, although if I try to draw a graphical representation of the bijection, I'm not really sure what that would look like (has to be 1 to 1, but also f(x) must have a negative value corresponding to every positive value)

Focusing on multiplication:

I define 1* such that
a$$\otimes$$1* = a (so f(1*) must equal the regular old 1), and define a^ such that
a$$\otimes$$a^ = 1*

I guess that will work. But I'm not sure if I have the right idea or not. Not that I need to say what kind of transformation (or bijection--I don't know if I can use those words interchangeably or not) is used, but if I try to picture it, it seems like (for the case of part a, which deals with the positive integers) I need to have f(0) = -$$\infty$$ or +$$\infty$$, and then f($$\infty$$) = +$$\infty$$ or -$$\infty$$ (so that there is an element such that f(0*) = regular old 0, and so that there is a value of f(a-) such that, when added to f(a), I get 0*.

Anyway, this is a lot of writing, and I guess my main concern is that I'm not quite sure that I see how the multiplicative inverse (and maybe the additive inverse) axioms are obeyed by this bijection. Thanks a lot for the help!

 

[micromass]

Well, take an $$a\in \mathbb{Z}$$. Then $$f(a)\in \mathbb{Q}$$ has an additive inverse -f(a). The additive inverse of a is then $$f^{-1}(-f(a))$$.
The same thing with multiplication.