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How do you find the image of the function?

WannaBe

New member
Oct 4, 2013
11
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 dont know how to find it.
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
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 dont 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 \(\displaystyle \mathbb{N}=\mathbb{Z}^+\).
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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 [tex]\{ y| y= f(x), x\in A\}[/tex].

As Plato said, the set of rational numbers is defined as the set of all fractions, [tex]\frac{a}{b}[/tex] with a any integer, b any positive integer (taking the denominator from the positive integers lets 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.)