MHB How do you find the image of the function?

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The image of the function f: Z*N -> R, defined as f(a,b) = a/b, is the set of rational numbers Q. This is because any rational number can be expressed as a ratio of an integer (a) and a positive integer (b). The function maps each pair of integers (m, n) to a rational number m/n, confirming that every rational number is included in the image. Additionally, the function is not one-to-one, as different pairs can yield the same rational number, but this does not affect the overall image. Thus, the image of the function is precisely the set of rational numbers.
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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.
 
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WannaBe said:
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 $$\mathbb{N}=\mathbb{Z}^+$$.
 
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.)
 
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