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RE: [PrimeNumbers] Non-prime numbers

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  • Jon Perry
    1+2+3 is a sum, using sum as a noun to describe a sequence of mathematical symbols, esp. using the + sign. You may continue to argue that it is 2 sums, and
    Message 1 of 16 , Apr 21, 2002
      1+2+3 is a sum, using sum as a noun to describe a sequence of mathematical
      symbols, esp. using the + sign.

      You may continue to argue that it is 2 sums, and ignore it's abelian
      properties if you must, but you know, I think most people stopped listening
      to you a while ago, so I doubt it will make any difference.

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths
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      -----Original Message-----
      From: Phil Carmody [mailto:thefatphil@...]
      Sent: 20 April 2002 10:54
      To: primenumbers
      Subject: RE: [PrimeNumbers] Non-prime numbers


      --- Jon Perry <perry@...> wrote:
      > Yes, but all I argued was that the + symbol that we all know and
      > love to
      > little bits requires 2 (two) operands.
      >
      > It was you who decided that once they had thought of the, admitedly
      > immense,
      > idea of invoking a sum{} operator, that the + operator that we all
      > grew up
      > with was now defunct in purpose, and should immediately be
      > superceded
      > (superseded for our American friends).
      >
      > As to my claim that 1+2+3 is a single sum, I didn't.

      That's not my memory of the exchange. I rampantly delete e-mails, and
      almost never keep mails I send, so there's no evidence either way.

      > I actually said that 1+2+3 is permissible as a sum

      'a' sum? How many would that be then?
      The version of the language (English) that I use marks the definite
      article 'a' with the number 'one'.

      You know, I don't think my memory was flawed at all.

      [rewinding]
      > I actually said that 1+2+3 is permissible as a sum
      > using only the +
      > operator.

      When applied twice.
      So is "1" using zero applications of the operator.
      Why can't you count to zero? It's less effort than you seem to think?

      > The abelian nature of the formula was purely an accident.

      Abelian is irrelevant.
      I think you mean /associative/ rather than abelian, an instead of
      /purely an accident/ you mean /absolutely necessary in order to be
      unambiguous/.

      Phil

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