- #1

scorpius1782

- 107

- 0

## Homework Statement

I'm having a lot of problems with this class of thinking I understand something then getting a grade back that proves otherwise. I'd like someone to check my proof of this:

Claim: ##f:\mathbb{N}^2 \rightarrow \mathbb{N}^2 by f(x,y) = (x+y, 2x-3y)## is one to one.

## Homework Equations

## The Attempt at a Solution

Using the definition of one to one: ## f:A\rightarrow B## is one-to-one, i.e. ##\forall x,y \in A, f(x) = f(y) \rightarrow x = y##

Proof: Let x, y, w, z ##\in \mathbb{Z}##

I'll take the two elements (that is x+y and 2x-3y) separately in order to prove one to one.

In order for ##f(x,y)=g(w,z)## then ##x+y=w+z## therefore x=w or x=z and y=w or y=z. So it is clear that by itself this part of the function is not one to one.

Now for ##2x-3y## we have ##f(x,y)=g(w,z)##. Then ##2x-3y=2w-3z##. However here we do not have the possibility for x to equal w or z and y cannot equal w or z. It must be one or the other as ##2x-3y\neq2y-3x## and ##2w-3z\neq2z-3w##. Except if, and only if, ##x=y=w=z=0##. But this fits in the definition of one to one, regardless.

Finally, from the above constraints ##w=x## and ##y=z##. Therefore ##x, y, w, z \in \mathbb{Z}, f(x,y)=f(w,z) \implies (x,y)=(w,z)##