1. ## Range of f

let f:NxN->N be defined by f(x,y)=(1/2)(x+y)(x+y-1)-(1-y)

Is f one-to-one? Does f map (NxN) to N or onto N (meaning are all values in N in the range of f or is the range a subset of N)?

Can I get a hint on how to go about answering either of those questions. I know what each of the questions mean, but I have no idea how to answer them.

2. Originally Posted by billa
let f:NxN->N be defined by f(x,y)=(1/2)(x+y)(x+y-1)-(1-y)

Is f one-to-one? Does f map (NxN) to N or onto N (meaning are all values in N in the range of f or is the range a subset of N)?

Can I get a hint on how to go about answering either of those questions. I know what each of the questions mean, but I have no idea how to answer them.
For the first, does $\displaystyle f(x,y) = f(x',y') \Rightarrow (x,y) = (x',y')$? For the second, put $\displaystyle y=1$. Then $\displaystyle f(x,1) = \frac{1}{2}x(x+1)$ whose range is....?

3. Originally Posted by Sampras
For the first, does $\displaystyle f(x,y) = f(x',y') \Rightarrow (x,y) = (x',y')$?
It is not 1-1. There are two pairs that map to zero - $\displaystyle (-1,1)$ and $\displaystyle (0,1)$.

4. Originally Posted by Swlabr
It is not 1-1. There are two pairs that map to zero - $\displaystyle (-1,1)$ and $\displaystyle (0,1)$.
I think since f:NxN -> N, (-1,1) is not in the domain, so it might still be one-to-one

For the first, does
I set f(x,y)=f(x',y') and couldn't show that it implies (x,y)=(x',y')

$\displaystyle f(x,y)=f(x',y') \implies x^2+y^2+2xy-x-3y=(x')^2+(y')^2+2x'y'-x'-3y'$

how can i tell that there aren't different pairs such that both of these are equal

For the second, put . Then whose range is....?
I am still working on this