MHB How do you find the image of the function?

[WannaBe]

What is the image of the function f : Z*N -> R ; f(a,b)= a/b?

I know the answer is Q (rational numbers) but I don't know how to find it.

 

[Plato2]

What is the image of the function f : Z*N -> R ; f(a,b)= a/b?
I know the answer is Q (rational numbers) but I don't know how to find it.

You surely know that any rational number is the ratio of two integers.
Note that any rational can be written as an integer divided by a positive integer.
Here let $$\mathbb{N}=\mathbb{Z}^+$$.

 

[HallsofIvy]

Technically, you don't "find the image" of a function, you find the image of a set by a function: the image of set A, by function f, is \{ y| y= f(x), x\in A\}.

As Plato said, the set of rational numbers is defined as the set of all fractions, \frac{a}{b} with a any integer, b any positive integer (taking the denominator from the positive integers let's us assign the sign of the fraction to the numerator and voids division by 0).

Now, if y is any rational number, there exist integer, m, and positive integer, n, such that y= m/n. That is precisely f(m, n) so every rational number is in the image.

Conversely, for any pair, (m, n), m an integer, n a positive integer, f(m,n)= m/n is a rational number so the image is precisely the set of rational numbers.

(This function is NOT "one-to-one", f(2, 4)= f(1, 2), but that is not relevant to this problem.)