Final answer:
The function f(m, n) = 2m - n is onto, meaning it maps the domain Z \times Z to the entire codomain Z. By fixing m and adjusting n, one can achieve any integer value, hence every element in Z can be reached by the function, proving its surjectivity.
Step-by-step explanation:
To determine whether the function f: Z \times Z \rightarrow Z defined by f(m, n) = 2m - n is onto, we must establish if every element b in the codomain Z (set of all integers) has at least one preimage in the domain Z \times Z (set of all ordered pairs of integers). An onto function (also called a surjective function) maps the domain onto the entire codomain.
The question asks, for a given integer, can we find integers m and n such that the function equates to that integer? The formula f(m, n) = 2m - n suggests that varying m and n allows us to achieve any integer value. For instance, to achieve a desired b, we could fix m and adjust n until the equation is satisfied.
However, to prove that f is onto formally, we must show that for any integer b, there exist integers m and n such that f(m, n) = b. Let's select m equal to b and n equal to m, thus n = b. Then we have f(b, b) = 2b - b = b, indicating that for any b, we can find values m and n (in this case, both equal to b) that satisfy the function. Therefore, the function is indeed onto.