a) Determine if $f$ is injective.Prove your claim
b) Determine if $f$ is surjective. Prove your claim
Answer for (a):
Claim: $f$ is injective.
We will proceed by contrapositive. We know definition of injective is: Let $A,B$ be sets. A function $f:A\to B$ is injective if for every element $(a_1,a_2)\in A$ (with $a_1\neq a_2$), $f(a_1)\neq f(a_2)$. This simply means that different inputs go to different outputs, and one input can not produce two different outputs.
Let $(n,m)∈ \mathbb Z \times\mathbb Z$. We need to show:$n=x$ and $m=y$.
$$f(n,m)=2n+m$$
Let $f(n,m)=f(x,y)$ or $(2n+m)=(2x+y)$
Therefore, $n=x$ and $m=y$.
From this we can see that one element of the domain can only correspond to one value of the codomain. Hence, by the definition of injectivity, we can conclude that $f$ is injective.
I was wondering if this is right or not and how I should approach part b of the question. I know that $f$ is surjective because the codomain is equal to the range but I am not sure about how I can show it in proof format.
Best Answer
It is not injective because $f(0,0)=f(1,-2)$.
It is surjective because any integer $n$ can be written as $f(0,n)$.