What is the range of the function g: ZxZ --> ZxZ given by g(m,n)=(m-n,m+n)?

[geskekj]

Homework Statement

Find the range of the function g: ZxZ --> ZxZ given by g(m,n)=(m-n,m+n). Hints: First recall that if f: A ---> B then Range (f)={b e B such that there exists an A in A with b=f(a). Second, if you claim that some set C is the range, then you must show that i) C is a subset Range(f) and ii) range (f) is a subset of C. Both i and ii are required to conclude that C=range(f).

Homework Equations

none

The Attempt at a Solution

I am think that the range is all integers from -infinity to +infinity.

I know that:
* ZxZ denotes a set of ordered pairs.
* m,n exist in Z
* range (g) exists in Z

I don't know where to go with this!

 

[Dick]

The range is all pairs (p,q) such that (p,q)=(m+n,m-n) for integers m and n. Solve for m and n in terms of p and q. Now rethink your guess that the range is all integer pairs. E.g. is (1,2) in the range?

 

[geskekj]

So we figured out that p=q. So our new guess for the range is all integers such that p=q for the ordered pair (p,q). If we say that, is that enough to justify our answer?

 

[Dick]

How did you figure out p=q? p=m+n, q=m-n for the domain value (m,n) and ask yourself when m and n are both integers. That's two equations in two unknowns. Solve them. Your new guess is worse than the last.

 

[Dick]

Actually, let's go back. You never answered my question. Is (1,2) in the range? If you answer that it will give you a big clue in a non-abstract sense.

 

[geskekj]

(1,2) is not in the range. When substituted in the equations for p and q the results for m and n are not integers.
After looking at your first response to our terrible answer, we looked that problem and decided that the range is (p,q) when p and q are any even integers.
**I am working on this a group of people and the more we talk about it, the more confused we get.**

 

[Dick]

If you solved those equations you should have gotten m=(p+q)/2 and n=(p-q)/2. Now think about pairs (p,q) where both p and q are odd.