Help with onto functions...?

Question

Given that Z is the set of all real integers, determine whether f: Z * Z -> Z is onto if

1) f(m,n) = 2m - n

2) f(m,n) = m^2 - n^2

3) f(m,n) = m + n + 1

I am thinking that 1 is onto, 2 is not onto, and 3 is onto. Is that correct? Thanks...

Answer 1

1 and 3 are onto.

to prove 2 is you have to take three seperate cases..

Case 1, m is odd, n is even:
let m = 2x ... for every x is an integer (m is even)
let n = 2p+1... for every p i an integer (n Is odd)

F(m,n) = (2x)^2 - (2p-1)^2
= 4x^2 - 4p^2 + 4p - 1
= 2(2(x^2 - p^2 +p)) - 1
^^this is an odd number in the form of 2x - 1


Case 2, both even:
m = 2s
n = 2p
F(m,n) = 4s^2 - 4p^2
= 2(2(s^2 - p^2))
^^this is an even number in the form of 2x


Now that there is both an even and odd possibility for f(there is no need for case 3), therefor m^2 - n^2 can fulfill the set of all real integers.

functions 1, 2, and 3 are onto

Answer 2

I believe "onto" means that the result returned by the function is part of the set specified. All three functions specified can result in any integer(negative,zero or positive). Each possible combination of values of "m" or "n" results in an integer as long as "m" and "n" are members of the set of integers. All three functions can produce the same result with different inputs. By that I mean that none of the functions have an inverse.

All three functions are ONTO the integers.

Answer 3

You're right...