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    Her lack of situational awareness is her dominant trait.
    • CommentAuthorloreman
    • CommentTimeMar 10th 2021
    Posted By: AngusI was much chipper than loreman!

    I'm not very chipper at all, presently
    + => * = > power => factorial ...

    all through simple counting up.

    The Moessner Miracle
    Is the universe geometric or algebraic?

    I enjoyed this, not least because Grassman and Clifford get mentions. Towards the end he tries to tie multiplication and addition to the scattering of elementary particles, and makes the comment that distributivity reveals a fundamental asymmetry between multiplication and addition, in that the former distributes and the latter does not.

    I thought I'd take a look at what's going on here. The multiplicative case is
    a.(b+c) = a.b + a.c
    and if we try this for addition by merely exchanging symbols, we get
    a + (b.c) != (a+b).(a+c)

    i.e for the equality to hold, we require
    a + b.c = a.(a+b+c) +b.c
    which is satisfied when
    a + b + c = 1

    A rather nice symmetrical condition.
    What it implies is mysterious to me.
    I have no geometrical insight about it!
    To recap, the discovery is that 3 entities distribute additively when their sum is unity.

    Which is not what the man on the podium said.

    Trust, but verify.
    • CommentTimeJul 28th 2021
    Your definition of distributive is weird. I would interpret the thing as

    a+(b+c)!=a+b +a+c
    if a=b=1 we get 3!=4
    as stated
    The multiplicative case is
    a.(b+c) = a.b + a.c
    and if we try this for addition by merely exchanging symbols, we get
    a + (b.c) != (a+b).(a+c)
    Seems self-evident. Recall we're examining the relationship between '+' and '.' and so exchanging symbols is the natural operation here.
    • CommentTimeJul 28th 2021 edited
    You are just swapping x and +. That's not distribution. Distribution says that doing someting F to individual things is the same as doing it to a group.

    a F b + a F c = a F (b+ c)

    If F is x it's true if F is + it's not.
    omg i proved something else then!!
    • CommentTimeJul 28th 2021
    Posted By: Andrew Palfreymanomg i proved something else then!!

    Seems a bit trivial.

    a + b.c = a F(a,b,c....z) + b.c

    iff F(a,b,c...z) =1
    It's not iff. The condition is merely sufficient.

    Trivial? Depends whether or not you're interested in the symmetry between addition and multiplication, I suppose.
    • CommentTimeJul 28th 2021
    Cancelling the second term we get

    a = a F(a,b,c....z)

    iff F(a,b,c...z) =1

    That seems pretty trivial to me.

    I think it's a necessary condition as well as sufficient. Give me an example where it is true with
    F(a,b,c...z) !=1
    Yes, you're right. Thanks.

    Anent this asymmetry between addition and multiplication, I had a look at sums equalling products.
    For 2 variables it's trivial that a = b / (b-1), b = a / (a-1).
    For any number of variables it generalises to
    a = s' / (p' - 1)
    where s' is the sum excluding a, and p' is the product excluding a.

    Kind of curious (albeit again trivial to prove).
    • CommentTimeJul 28th 2021
    That is kind of neat!
    Well, there's two results about exchanging the operations. For sure there are a bunch more. It's weird that, so late in life, I've seen these for the very first time.

    ETA I really should try to generalise that first result.