exhibit a countable set of irrational numbers with justification

Question

So I have been given this problem and I am totally stumped on what to do...everything I have learned says the irrational numbers are uncountable but I am supposed to exhibit a countable set of irrational numbers. This implies that there is a bijection from this countable set (lets call it S) to the natural numbers or a bijection from the natural numbers to this countable set S.

• any thoughts on what I should I do would be appreciated

Answer 1

Take $\{n+\pi \mid n\in \mathbb{N}\}$. The bijection is obvious and each element here is irrational.

Answer 2

Consider $\{\sqrt p \mid p$ is prime$\}$.

Answer 3

Possible alternate.
The definition I know of countable says that a set has to be finite or denumerable.

The set $\{\pi\}$ is countable. In fact, it is finite.