Should we or should we not take $1$ as a prime number?

Question

I think I know that there were times in the past when it was convenient to look at a number $1$ as a prime number, and, as far as I can remember, even then it was dependent on who we ask is it prime or not?

This question is primarily here because in some of the questions I posted in the last few days I wanted that we for the purpose of the question consider $1$ as prime, or because of some similar reason.

So, let us see why, in my opinion, we should take $1$ as prime, and why we should not.

First, the definition of prime numbers as seen on Wikipedia:

A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.

If we interpret here this greater as $n \in \mathbb N \setminus \{1\}$, in other words, as strictly greater, then clearly $1$ cannot be prime because it is, by definition, not taken into consideration.

If we change the definition and remove part of the sentence, the part greater than 1, then we have this definition:

A prime number (or a prime) is a natural number that has no positive divisors other than 1 and itself.

Then we could say that $1$ is prime, because he has no other positive divisors other than 1 and itself, and he has two divisors which coincide and are both equal to $1$ (1 and itself).

But do we stumble upon some problem if we declare $1$ as prime?

Yes we do, but all these problems (that I know of) are not so serious.

First, we should rephrase fundamental theorem of arithmetic which says that:

Every positive integer $n>1$ can be represented in exactly one way as a product of prime powers.

If we would take $1$ as prime then the fundamental theorem of arithmetic could be stated as:

Every positive integer $n>1$ can be represented in exactly one way as a product of prime powers in such a way that in the factorization of $n$ powers of number $1$ are absent.

This is because we have $1^a=1^b$ for $a \neq b$ so if we want unique factorization then powers of $1$ must be absent from the factorization.

But do we gain something positive if we declare $1$ as prime number?

I think we do.

For example, the Goldbach conjecture states:

Every even integer greater than 2 can be expressed as the sum of two primes.

If we would declare $1$ as prime we could shorten the Goldbach conjecture:

Every even natural number can be expressed as the sum of two primes.

Of course, this is because $1+1=2$ so now even two has a representation as a sum of two primes.

But in this case we would gain even more. Now, when $1$ is taken as a prime number, then we have slightly greater chance that every even natural number can be expressed as a sum of two primes because now we can consider sums of the form $2n=(2n-1)+1$ and we cannot consider them when we do not take $1$ as a prime.

It seems (but I could be wrong), that it is the matter of agreement between mathematicians should we or should we not take $1$ as a prime number.

So, I have a few questions:

1) Is there any serious problem or inconsistency inside of mathematics if we take $1$ as a prime number (other than just re-writing the theorems/lemmas/conjectures/definitions in a different way)?

2) Should we or should we not take $1$ as a prime number?

3) Can you give some other examples of conjectures for which it would be better to take $1$ as a prime?

4) What are your thoughts on this?

@lhf I would not call it a duplicate if only because of the length of this and length of that question. — Farewell, May 03 '16 at 02:38
Well, $1$ would be the only prime with the property that $1=1\cdot 1 \cdot \dotso$ which would change how youd talk about factorization of an arbitrary prime.... — Nap D. Lover, May 03 '16 at 02:41
@Farewell Sometimes answers to short questions solve long questions. — , May 03 '16 at 07:36
Or of http://math.stackexchange.com/questions/1058414/primality-of-number-1 — Mr. Brooks, May 03 '16 at 21:14

Robert Soupe · Answer 1 · 2016-05-04T00:28:50.537

1) Is there any serious problem or inconsistency inside of mathematics if we take 1 as a prime number (other than just re-writing the theorems/lemmas/conjectures/definitions in a different way)?

No, no serious problem, only a bunch of minor but unnecessary inconveniences.

2) Should we or should we not take 1 as a prime number?

We should not. Unless we like making things more difficult than they need to be.

3) Can you give some other examples of conjectures for which it would be better to take 1 as a prime?

Well, no, but for constructing magic prime squares it's occasionally useful to accept 1 as a prime.

4) What are your thoughts on this?

The prime numbers come first and our theorems about them second. It's like in chemistry: a metal is a metal regardless of whether or not we recognize it as such. And so in mathematics, 1 and $-1$ have always been units, long before we recognized them as such.

We're in agreement that 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ... are all primes, right? Just checking. What characteristics do they have in common? Does 1 share all those characteristics? No, it does not. In fact, there are also many characteristics 1 does not share with the composite numbers either. Therefore 1 is neither prime nor composite, but a unit.

That we recognize this does not change the nature of the number. 1 has always been a unit. It's quite a different question from whether Y is a vowel or a consonant. All of us English speakers could get together and decide it one way or the other. But if there is a civilization in some distant world, they may too at first consider 1 to be a prime number but gradually come to realize that it is not. I venture to guess the Vulcans realized this a lot sooner than the Klingons, wink, wink.

I think this is important. A surprising number of people don't know about units, numbers that divide $1$. Together with primes and composites, they partition the algebraic integers over a field, but a lot of people don't seem to know that the third category exists. — Stella Biderman, Jun 01 '16 at 18:50
In our current teach-to-the-test educational climate, teachers are probably quite stressed out enough about getting kids to understand the difference between primes and composites, so algebraic integers and fields, that's for the genius children only. — Robert Soupe, Jun 02 '16 at 02:56

score 1 · Answer 2 · answered May 03 '16 at 02:40

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In particular the little theorem of Fermat and Wilson's theorem would be valid because $a^{1-1}\equiv 1\pmod 1$ and $(1-1)!\equiv -1\pmod 1$ (since $a\equiv 0\pmod 1$ for all integer $a$).

On the other hand $\mathbb Z/(1\cdot \mathbb Z)=\mathbb Z$ is not a field so, according to the corresponding well known theorem, $1$ is not a prime.

answered May 03 '16 at 02:40

Piquito

29,594

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I think that $\mathbb{Z}/(1\cdot \mathbb{Z})\cong {*}$, which could be considered a field if we don't require the additive and multiplicative identities to be distinct. – Eric May 03 '16 at 02:47
@Eric: "Because the identity condition is generally required to be different for addition and multiplication, every field must have at least two elements" (Wolfram MathWorld). Regards. – Piquito May 03 '16 at 11:22
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@Piquito if we are willing to consider $1$ as prime, I think dropping that condition on a field is probably not a bad idea either (to make the full theory self-consistent). – Cameron Williams May 03 '16 at 21:17
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@Piquito, I was mostly trying to correct your claim that $\mathbb{Z}/(1\cdot\mathbb{Z})\cong\mathbb{Z}$. The notion of prime number is related to the more general notion of a prime ideal. Normally we require prime ideals to be proper--that is we typically do not consider the ideal $(1)$ to be prime (as it is the whole ring itself). We could 'decide' to let $(1)$ be prime and this would be the same thing as allowing a field with one element as Cameron Williams pointed out. – Eric May 03 '16 at 22:27

score 1 · Answer 3 · answered May 03 '16 at 21:15

I have answered an extremely similar question before. I will edit my old answer to better suit this question.

I am so old I was taught that definition, that it's enough for a prime number to be divisible only by $1$ and itself. But my children and grandchildren were taught that a prime number must have exactly two distinct divisors among the positive integers. When I first heard about this redefinition, I thought it made sense.

Because, you see, there were some things that bothered me about $1$ being a prime number, some things that didn't quite make sense to me. For example, consider the sequence $p^n$ where $n$ runs over the positive integers. $p^n$ gets larger as $n$ gets larger. But $1^n = 1$ no matter what $n$ is. In this way, $1$ is different from the prime numbers. But it is also different from the composite numbers.

There is at least one property $1$ shares with some composite numbers: $\sqrt{1} \in \mathbb{Q}$. But $\sqrt[n]{1} = 1$ no matter what $n$ is. Compare a number like $4$, see that while $\sqrt{4} = 2$, $\sqrt[3]{4} \not\in \mathbb{Q}$.

As a schoolchild I wasn't able to express these misgivings I had about $1$ being a prime number. But it was clear to me that $1$ differs from the prime numbers in a much more fundamental way than $2$ differs from the odd primes.

Actually, the only thing that I have to add to my old answer is that this stuff about unique factorization is a red herring.

score 1 · Answer 4 · answered May 03 '16 at 21:39

I think this is the most natural way to define a prime number:

An integer $n$ is prime if it has exactly two integer divisors, up to equivalence.

If you don't like thinking about equivalence, we can also say:

A positive integer $n$ is prime if it has exactly two positive integer divisors.

This is analogous to many other concepts in mathematics:

A group $G$ is called simple if it has exactly two normal subgroups. The trivial group is not simple.
An $R$-module $M$ is called simple if it has exactly two submodules. The zero module is not simple.
A unital commutative ring $R$ is a field if it has exactly two ideals. The zero ring is not a field.
A unital commutative ring is an integral domain if it has exactly two annihilators. The zero ring is not an integral domain.
A topological space is irreducible if there are exactly two possibilities for the closure of an open set. The empty space is not irreducible.

The reason that these definitions are so popular, rather than their "at most two" versions, is that generally the "exactly one" version gives objects with completely different behavior from the "exactly two" version. They are often in some way degenerate, or give us some sort of non-uniqueness (e.g. if the empty space is irreducible, we can no longer hope to uniquely decompose any space into irreducibles).

So maybe we should think of the integers this way: the units $\pm 1$ have "zero complexity", the primes $\pm 2, \pm 3,\ldots$ have "minimal non-zero or irreducible complexity", the remaining non-zero integers have "reducible complexity", and $0$ has... well, I'm not sure of a good name, but it's somehow infinitely complex because it's divisible by everything.

I think there may be situational reasons to consider $1$ a prime, and Conway has argued that $-1$ should sometimes be considered a prime. But I think the conceptual reasons to exclude $1$ from the list of primes are overwhelming. It is nothing like the usual prime numbers, and a definition that makes it appear so is probably not a great definition.

Should we or should we not take $1$ as a prime number?

4 Answers4