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Novel Multiplication and Division by odd squares

  1. Dec 1, 2012 #1
    let [tex]A_i[/tex] be an odd integer, [tex]s_i[/tex] be the square of [tex]a_i[/tex] and [tex]t_i[/tex] be the triangular number, [tex](s_i -1)/8[/tex]. Same for [tex]a_j , s_j, t_j, etc[/tex]. Define Multiplication of [tex]n X A_i , etc[/tex] to be n * s_i - t_j and division to be the reverse of this process. I found that

    n X A_i X A_j X A_k = n X A_k X A_j X A_i = n X A_j X A_k X A_i etc.

    for instance ((((4 * 9 - 1)*49 - 6)*25 -3) + 1) / 9 = (4*25-3)*49-6 = (4*49-6) * 25 - 3 = B

    8*4-1 = 31 and 8*B - 1 = 31*25*49

    Is there a simple way to prove this general result?
  2. jcsd
  3. Dec 1, 2012 #2
    I can barely make this out, since you didn't format it.
    If I'm reading it right, this just follows from the commutativity of multiplication.
  4. Dec 1, 2012 #3
    Yes, removing all unnecessary notation, the result your are looking for is the following:

    Let k be a real number. For real numbers n and x, define n[itex]\circ[/itex]x=nx+k(x-1). Then (n[itex]\circ[/itex]x)[itex]\circ[/itex]y=n[itex]\circ[/itex](xy).

    The proof of this is trivial, and your result is the special case of k=-1/8, and where nXa=n[itex]\circ[/itex]a2.
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