• Chais@sh.itjust.works
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    1 year ago

    They’re not prime. By definition primes have two prime factors. 1 and the number itself. 1 is divisible only by 1. 0 has no prime factors.

    • CAPSLOCKFTW@lemmy.ml
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      1 year ago

      Commonly primes are defined as natural numbers greater than 1 that have only trivial divisors. Your definition kinda works, but 1 can be infinitely many prime factors since every number has 1^n with n ∈ ℕ as a prime factor. And your definition is kinda misleading when generalising primes.

      • Chais@sh.itjust.works
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        1 year ago

        Isn’t 1^n just 1? As in not a new number. I’d argue that 1*1==1*1*1. They’re not some subtly different ones. I agree that the concept of primes only becomes useful for natural numbers >1.
        How is my definition misleading?

        • CAPSLOCKFTW@lemmy.ml
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          1 year ago

          It is no new number, though you can add infinitely many ones to the prime factorisation if you want to. In general we don’t append 1 to the prime factorisation because it is trivial.

          In commutative Algebra, a unitary commutative ring can have multiple units (in the multiplicative group of the reals only 1 is a unit, x*1=x, in this ring you have several “ones”). There are elemrnts in these rings which we call prime, because their prime factorisation only contains trivial prime factors, but of course all units of said ring are prime factors. Hence it is a bit quirky to define ordinary primes they way you did, it is not about the amount of prime factors, it is about their properties.

          Edit: also important to know: (ℝ,×), the multiplicative goup of the reals, is a commutative, unitary ring, which happens to have only one unit, so our ordinary primes are a special case of the general prime elements.