Are irrational numbers prime?

Question

I don't mean in the literal sense, but from the perspective of not having divisible integers

Answer 1

Treating "prime" as meaning "not divisible by any integer other than $1$ and possibly itself", we have to decide what "divisible" means. Usually, we say that $n$ is divisible by $a$ if there is an integer $b$ so that $ab = n$. Under that interpretation, irrational numbers are certainly "prime", but so are all non-integers; remember that if $a$ and $b$ are both integers, so is $ab$.

However, we could also define "prime" as "divisible only by $1$ and itself". Then, for example, $\pi$ cannot be prime - while it isn't divisible by any integers other than $1$, it isn't divisible by itself either (under our definition of "divisible" above).

EDIT: You recently commented an observation that suggests another definition. If we count, for example, $3^{-1}$ for the purposes of divisibility, rational numbers are nonprime. So let's use this definition: $n$ is divisible by $a$ if there is some integer power of an integer $b$ so that $ab = n$. So, for example, $3/2$ is divisible by $3$ because $2^{-1}$ is an integer power of an integer. Under that definition, rational numbers are not prime, but irrational numbers are.

Unfortunately, then we have that integers are not prime. For example, $2$ is "divisible" by $4$, because $2 = 4 \cdot 2^{-1}$. So this definition of divisibility doesn't respect the usual one, unlike the previous candidate.