# 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!)

Please help! And Thank You!

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

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

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

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

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

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

This is a binary function defined by:

$f(n, m) = nm$

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

It can't be one-to-one because $\mathbb{Z}^2$ and $\mathbb{Z}$ are equinumerous, so the range of the function is also a subset of $\mathbb{Z}^2$, 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!)
$z \in \mathbb{Z}\; \Rightarrow \;f(1,z) = z$
It is clearly onto. Then the range is $\mathbb{Z}$

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

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!