• CompassRed@discuss.tchncs.de
    585·
    1 year ago

    2 may be the only even prime - that is it’s the only prime divisible by 2 - but 3 is the only prime divisible by 3 and 5 is the only prime divisible by 5, so I fail to see how this is unique.

  • EatBorekYouWreck@lemmy.worldEnglish
    442·
    1 year ago

    Even vs odd numbers are not as important as we think they are. We could do the same to any other prime number. 2 is the only even prime (meaning it is divisible by 2) 3 is the only number divisible by 3. 5 is the only prime divisible by 5. When you think about the definition of prime numbers, this is a trivial conclusion.

    Tldr: be mindful of your conventions.

    • alvvayson@lemmy.world
      153·
      1 year ago

      Yes, but not really.

      With 2, the natural numbers divide into equal halves. One of which we call odd and the other even. And we use this property a lot in math.

      If you do it with 3, then one group is going to be a third and the other two thirds (ignore that both sets are infinite, you may assume a continuous finite subset of the natural numbers for this argument).

      And this imbalance only gets worse with bigger primes.

      So yes, 2 is special. It is the first and smallest prime and it is the number that primarily underlies concepts such as balance, symmetry, duplication and equality.

      • EatBorekYouWreck@lemmy.worldEnglish
        8·
        1 year ago

        But why would you divide the numbers to two sets? It is reasonable for when considering 2, but if you really want to generalize, for 3 you’d need to divide the numbers to three sets. One that divide by 3, one that has remainder of 1 and one that has remainder of 2. This way you have 3 symmetric sets of numbers and you can give them special names and find their special properties and assign importance to them. This can also be done for 5 with 5 symmetric sets, 7, 11, and any other prime number.

  • AssortedBiscuits [they/them]@hexbear.netEnglish
    27·
    1 year ago

    The meme works better if it’s 1 instead of 2. 1 is mostly not considered a prime number because a bunch of theorems like the fundamental theorem of arithmetic would have to be reworked to say “prime numbers greater than 1.” However, just because 1 is not a prime number doesn’t mean it’s a composite number, so 1 is a number that is neither prime nor composite.

    • MyFeetOwnMySoul
      131·
      1 year ago

      🚨 NERD ALERT🚨

      Go define a vector space, nerd.

      Go compute the p value of you being cool

      Go integrate f(x)= 1/x on the domain (-1,1)

      This is meme-ville population: me

      Take a hike.

      • EatBorekYouWreck@lemmy.worldEnglish
        3·
        1 year ago

        • let V be you mom’s vagina, a vector space over the field of pubes. We define my d as a vector such that d is in V. Thus my dick is in your mom’s vagina.

        • In this vector space p values are not defined, but I can assure you that my pp is > 9000.

        • The integral of f(x)=1/x from -1 to 1 does not converge, just like how your father is never coming back from buying milk. The principal value of that integral tho is 0, just like the amount of hugs you got as a kid.

        • math is cool, you just too stupid to get it.

    • Aesthesiaphilia@kbin.social
      5·
      1 year ago

      Pretty sure that when we plug in a correction factor for the relative age of the Fediverse userbase, “today’s lucky 10,000” becomes more like “today’s lucky 10 million”

    • Chais@sh.itjust.works
      35·
      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
        182·
        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
          101·
          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
            3·
            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.

            • T0Keh@feddit.de
              1·
              1 year ago

              There is multiple things wrong here.

              1. 1 is not a prime number because it is a unit and hence by definition excluded from being a prime.

              2. You probably don’t mean units but identity elements:

              • A unit is an element that has a multiplicative inverse
              • An identity element is an element 1 such that 1x =x1 = x for all x in your ring

              There are more units in R than just 1, take for example -1(unless your ring has characteristic 2 in which case thi argument not always works; however for the case of real numbers this is not relevant). But there is always just one identity element, so there is at most one “1” in any ring. Indeed suppose you have two identities e,f. Then e = ef = f because e,f both are identities.

              1. The property “their prime factorisaton only contains trivial prime factors” is a circular definition as this requires knowledge about “being prime”. A prime (in Z) is normally defined as an irreducible element, i.e. p is a prime number if p=ab implies that either a or b is a unit (which is exactly the property of only having the factors 1 and p itself (up to a unit)).

              2. (R,×) is not a ring (at least not in a way I am aware of) and not even a group (unless you exclude 0).

              3. What are those “general prime elements”? Do you mean prime elements in a ring (or irreducible elements?)? Or something completely different?

              • CAPSLOCKFTW@lemmy.ml
                1·
                1 year ago

                You’re mostly right, i misremembered some stuff. My phone keyboard or my client were not capable of adding a small + to the R. With general prime elements I meant prime elements in a ring. But regarding 3.: Not all reducible elements are prime nor vice versa.

                • T0Keh@feddit.de
                  1·
                  1 year ago

                  That’s why I wrote prime number instead of prime element to not add more confusion. I know that in general prime and irreducible are not equivalent.