# Are irrational numbers prime? [closed]

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

## closed as unclear what you're asking by Shailesh, Leucippus, Morgan Rodgers, Juniven, GNUSupporter 8964民主女神 地下教會Mar 19 '17 at 3:25

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• script8man Do you mean in last comment that 1.5 is $3^1 \times 2^{-1}$? – coffeemath Mar 19 '17 at 1:30
• script8man I just meant the numerator of 1.5 should be 3 and denominator 2, but you had 2 in numerator. – coffeemath Mar 19 '17 at 1:33
• This question would be more interesting in literal sense when it comes about structure of prime numbers and composition of irrationals. – usiro Mar 19 '17 at 1:48
• Take number 2 as an example. It is an even number with one of the properties of primes. In similar way irrational numbers can inherit structure of primes since all numbers can be thought as polynomials. I have to study it first to say more - it is just my first feeling. – usiro Mar 19 '17 at 2:13
• I voted to close this question while thinking about prime elements in a more abstract and general sense, which gives us more insight into the primality of an element and saves time from ambiguous discussions, in a ring $R$. So what in which ring we're talking about? Unlike the usual prime numbers, irrational numbers live outside $\Bbb Q$. I think it's more sensible to discuss in $\Bbb R$, which is a field, so it's a principal ideal, so every prime ideal is a maximal ideal, which doesn't exist in $\Bbb R$. Thus there no prime ideal in $\Bbb R$, so the answer is no. – GNUSupporter 8964民主女神 地下教會 Mar 19 '17 at 3:35

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.
• I think that any definition of a prime must specify that a prime must be an integer. Otherwise you arrive at things like $\pi$ isn't divisible by itself. – mrnovice Mar 19 '17 at 1:31
• "Only described by themselves" is ultimately meaningless. Let $a= \pi -e$ then $e=\pi - a$ describes $e$ without using $e$. Or if $b =\pi/3e$ then $e =3*\pi b$. Is that "prime" or not? – fleablood Mar 19 '17 at 3:27