Let $f: Z \times Z \to Z \times Z$ defined like this: $f(x,y) = (x+y, x-y)$
Prove that $f$ is injective, and not surjective.
For injectivity I did that:
Let $(a,b) \in Z\times Z$ and $(c,d) \in Z\times Z$ such that: $f(a,b) = f(c,d)$
and let's show that: $(a,b) = (c,d)$
$$(a+b,a-b) = (c+d,c-d)$$
$$a+b = c+d $$ and
$$a-b = c-d$$
and we end up with: $a=c$ and $b=d$, that concludes that: $(a,b) = (c,d)$ which proves that $f$ is injective.
Can someone show me how to prove that $f$ is NOT surjective? I couldn't do it by myself.
Thanks.
Best Answer
It is not surjective. For example $(1,0)$ dont have a source. Assume that $$(x+y,x-y)=(1,0).$$ Then $x=y$ and then $2x=1$ which have no solutions over $\mathbb{Z}$.
By the way over $\mathbb{R}$ it is surjective since the corresponding matrix is invertible.