Proving a Function is OntoIs f: ZxZ->Z an onto function?

[alman9898]

Homework Statement

Determine whether f: ZxZ->Z is onto if

(a) f(m, n) = 2m-n
(b) f(m, n) = n^2 - m^2
(c) f(m, n) = m + n + 1
(d) f(m, n) = |m| - |n|
(e) f(m, n) = m^2 - 4

Homework Equations

A function is onto if for every y in the codomain there is at least one X in the domain st f(x) = y

The Attempt at a Solution

I have no idea how to start proving this. I know I'm supposed to prove that for every integer y, there is at least 1 (m, n) pair, but that requires solving for another variable m or n doesn't it? This leaves a lot of m,n to test for yes?

For example, I can go about the first one by setting m = (y+n)/2 and if y = 0 and n is odd then m is not an integer. However, I know from reading ahead that the first one is onto.

For the last one I can do

sqrt(y+4) = m
If y = 1, then m is not an integer and so y = 1 is not in the range

Am I supposed to prove the cases where m is odd, n is even, both are even, both are odd, m is even, n is odd?

 

[praharmitra]

find what the range is. Clearly, it is the set of integers. Thus for every integer a, we should have f(x, y) = a

for the first part 2m - n = a => 2m = a + n. Since m and n are also integers. find how many such pairs satisfy. In fact, don't find all pairs, just prove that atleast one answer exists...

Remember, that a is an integer and can be negative

 

[alman9898]

Heres a try at the first one

m = (a+n)/2 is an integer only if m is even which => (a+n) must be even.

(a+n) is even if a & n are both even or both odd which covers a for both even and odd integers

 

[praharmitra]

Heres a try at the first one

m = (a+n)/2 is an integer only if m is even which => (a+n) must be even.

(a+n) is even if a & n are both even or both odd which covers a for both even and odd integers

ok, why should m be even? a + n should be even, because it is being divided by two, and the result must be an integer. However, m need not be.

9 = (11 + 7)/2

 

[HallsofIvy]

Heres a try at the first one

m = (a+n)/2 is an integer only if m is even which => (a+n) must be even.

(a+n) is even if a & n are both even or both odd which covers a for both even and odd integers

No, m does not have to be even, only a+ n does.

For a any integer, 2m-n= a implies m= (a+n)/2. Okay, if a is even, say a= 2k, take n= 0 so m= a/2= k: 2k- 0= 2k= a. If a is odd, say a= 2k+1, take n= 1 so m= (2k+ 2)/2= k+1; 2(k+1)- 1= 2k+ 2- 1= 2k+ 1= a. For any a, there exist m and n such that 2m- n= a. The function is "onto" Z.

(Note: this function is not "one to one". f(3,5)= 2(3)- 5= 1 and f(2, 3)= 2(2)- 3= 1.)