$\sqrt p: p $ is a prime number is countable or uncountable?

Question

$\sqrt p: p $ is a prime number is uncountable

Set of primes is countable and set of irrational is uncountable.

$ \sqrt p$ is an irrational number as p is not perfect square. So given set would be subset of irrationals. Further if i make a function $f(p) = \sqrt p$ would that make a bijective? if so, then countable right? pls clarify

Answer 1

Let $P = \{p:p \textrm{ is prime}\}$. Then $P$ is countable.

Let $P' = \left\{\sqrt p:p \textrm{ is prime}\right\}$. There is a bijection $f: P' \rightarrow P$, namely $f(x) = x^2$. Therefore $P'$ has the same cardinality as $P$.

Answer 2

It's countable. To convince you that I'm not lying, I'll find you a bijection from $ P:=\{\sqrt{p}$ | $p$ is a prime$\} $ to a subset of natural numbers.

Certainly all primes are natural numbers, so the set $C:=\{ p $ | p is a prime$ \}$ is a subset of natural numbers.

Now, it's easily shown that mapping $\sqrt p$ to $p$ gives a bijection between $P$ and $C$.

Answer 3

Let $P$ denote the set of primes and define a set $\sqrt{P}$ by declaring that $\sqrt{p}$ $\in \sqrt{P}$ $\iff$ $p\in P$. We aim to show that $\sqrt{P}$ is countable. The set of primes is clearly countably infinite, since it is a subset of the natural numbers. This means that we may find a bijection between $P$ and $\mathbb{N}$. To this end, to show that $\sqrt{P}$ is countable, it suffices to show that there exists an injection from $\sqrt{P}$ to $P$.

Define a map $f: \sqrt{P}\rightarrow P$ by the assignment $f(\sqrt{p})=p$. This map is clearly injective.

Note that if $A$ is uncountable, then a subset $B\subseteq A$ does not need to be uncountable. Just consider a subset of $A$ with only one element. In fact, a nice exercise would be to show that every infinite subset contains a countably infinite subset.