How significant is the fact 1 isn't a prime number? What will happen if it is? What areas of Mathematics are affected by changing the fact? I know why and how 1 isn't a prime. My question is how significant is the fact.

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    $\begingroup$ Since $1^n = 1$, there is no longer a unique prime factorization of any integer, and pretty much the whole of number theory will fall apart unless you change "prime" to "prime $\ne 1$" virtually everywhere. $\endgroup$
    – alephzero
    Feb 18, 2019 at 16:53
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    $\begingroup$ It will make theorems more annoying to state. $\endgroup$
    – Randall
    Feb 18, 2019 at 16:54
  • $\begingroup$ See the answers in Why is $1$ not a prime number? $\endgroup$ Feb 18, 2019 at 16:55
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    $\begingroup$ It's analogous to asking what will change if we allow $0$ to be positive (as the French do). Many theorems using "positive" would need to be updated to remain correct. $\endgroup$ Feb 18, 2019 at 17:10
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    $\begingroup$ @usiro I can't make any sense of your remarks. In any case notions such as irreducible, prime and unit are independent of geometry. $\endgroup$ Feb 18, 2019 at 21:27

2 Answers 2


That wouldn't be a disaster, but it would add some very annoying things. For example, instead of the uniqueness in the fundamental theorem of arithmetic we would have "uniqueness up to finitely many multiplications of $1$".

  • $\begingroup$ We can have infinitely many 1's too, right? $\endgroup$ Feb 18, 2019 at 17:02
  • $\begingroup$ Actually yes, we can. But even if we want to say that every integer can be represented as a finite product of primes it would still not be a unique decomposition. $\endgroup$
    – Mark
    Feb 18, 2019 at 17:04
  • $\begingroup$ There wouldn't be any uniqueness.Got your point. $\endgroup$ Feb 18, 2019 at 17:05
  • $\begingroup$ @Krishna No, infinite products don't make sense in general (one needs additional ring structure to make sense of them). $\endgroup$ Feb 18, 2019 at 17:16
  • $\begingroup$ additional ring structure? $\endgroup$ Feb 18, 2019 at 19:27

We need factorisation of positive integers to be unique in many contexts, and this fails if we take $1$ to be prime, since then $6,$ for example, would factorise in the infinity of ways $$2×3=2×3×\underbrace{1×1×\cdots×1}_{n},$$ where $n$ is a nonnegative integer. This happens for the other positive integers too.

Actually, $1$ is different from the other positive integers in that it is neither composite nor prime. This distinguished position follows from the fact that $1$ is the multiplicative identity in $\mathbf Z;$ that is, for any integer $m,$ we have that $$1×m=m×1=m.$$


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