If $f: \mathbb N \rightarrow \mathbb N \times \mathbb N $ such as $f(n)=(n,n+1)$ Is it surjective and/or injective? If $f: \mathbb N \rightarrow \mathbb N \times \mathbb N $ such as $f(n)=(n,n+1)$ Is it surjective and/or injective?
I know that it is surjective $\Leftrightarrow \forall (a,b) \in \mathbb N \times \mathbb N \exists c \in \mathbb N:f(c)=f(a,b)$
It is obviously injective because if $(n,n+1)=(m,m+1) \rightarrow n=m$
I can see that it is not surjective but do not know how to prove it, can I get some help?

Consider $(1,1)\in\mathbb{N}\times\mathbb{N}$. Suppose for contradiction that there exists $n\in\mathbb{N}$ with $f(n)=(n,n+1)=(1,1)$. Then reading the first entry, we get $n=1$. Reading the second entry, we get $n+1=1\implies n=0$. Clearly we can't have $n=1$ and $n=0$ at the same time. Contradiction. Hence $f$ is not surjective.
