# Thread: Determining whether a function is 1-to-1 &/or onto

1. ## Determining whether a function is 1-to-1 &/or onto

For purposes of the problem, let Z=the set of all integers. Here goes:

"Define f: Z x Z --> Z by f(n,m)=nm"

I found f to NOT be 1-to-1. Now I need to determine whether or not it is onto (i.e. ran(f)=Z). If it is not onto I need to determine the range. I have an inkling it is onto but I have no idea to show that (nor how to determine the range!)

2. Let's dissect this function really quickly, first we define the set of integers:

$\displaystyle \mathbb{Z} = \{..., -3, -2, -1, 0, 1, 2, 3, ...\}$

Ok, now we will define the Cartesian Product of itself, which I will call $\displaystyle \mathbb{Z}^2$:

$\displaystyle \mathbb{Z}^2 = \{..., (-3, -3), (-3, -2), (-3, -1), ..., (3, -3), (3, -2), ..., (3, 1), (3, 0), ...\}$

So, now we have what we need to define our function, which is:

$\displaystyle f : \mathbb{Z}^2 \to \mathbb{Z}$

This is a binary function defined by:

$\displaystyle f(n, m) = nm$

So, the range consists of all the products of the integers which is a subset of $\displaystyle \mathbb{Z}$, and the domain consists of any pair of integers.

It can't be one-to-one because $\displaystyle \mathbb{Z}^2$ and $\displaystyle \mathbb{Z}$ are equinumerous, so the range of the function is also a subset of $\displaystyle \mathbb{Z}^2$, and can therefore cover the entire range, and so it is onto.

3. Originally Posted by dolphinlover
For purposes of the problem, let Z=the set of all integers. Here goes:

"Define f: Z x Z --> Z by f(n,m)=nm"

I found f to NOT be 1-to-1. Now I need to determine whether or not it is onto (i.e. ran(f)=Z). If it is not onto I need to determine the range. I have an inkling it is onto but I have no idea to show that (nor how to determine the range!)
$\displaystyle z \in \mathbb{Z}\; \Rightarrow \;f(1,z) = z$
It is clearly onto. Then the range is $\displaystyle \mathbb{Z}$

4. Originally Posted by Plato
$\displaystyle z \in \mathbb{Z}\; \Rightarrow \;f(1,z) = z$
It is clearly onto. Then the range is $\displaystyle \mathbb{Z}$

It has been a while since I've been to the forum. What you stated was exactly what I used to show it was onto. Thanks!