Prove that the set of nonnegative numbers is denumerable

Question

I need to show through a proof that the set of nonnegative numbers is denumerable

I know a set is denumerable if its members or elements can be put into an order and counted. I am supposed to show this through a proof as well. I was considering using Induction possibly but I'm not sure if this is a viable or feasible method to use. Any help is great appreciated

Answer 1

Hint – Your set is $A=\mathbb N\cup\{0\}$. How would you proceed, using this?

Answer 2

The set of non-negative numbers: I assume you talk of non-negative integers. The set would be $\{0,1,2,3...\}$.

Every element of the set can be associated with a natural number as $n-1$ where $n$ is the natural number.

That means: $0$ is associated with $1$, $1$ is associated with $2$ and so on. When every element of the given set can be associated with a natural number, It is denumerable (countably infinite).

Answer 3

Every subset of a denumerable set is either denumerable or finite. In particular , your set is an infinite subset of set of all integers and hence denumerable.