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Is zero even or odd?

Discussion in 'Electronic Design' started by Gactimus, Dec 20, 2004.

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  1. vonroach

    vonroach Guest

    You rely on a computer? How slipshod.
  2. vonroach

    vonroach Guest

    None of which have been accepted.
  3. So you think the polynomial 2 x^2 + 3 x^1 + 4 x^0 is undefined at x=0 ?
  4. Dave Seaman

    Dave Seaman Guest

    It is no more unjustified than your false claim that I "just want to win
    the argument, come what may." I do not base my facts on my opinions, and
    I strongly resent any implication to the contrary. I have endeavored to
    stick to the mathematics in this discussion.

    It seems that there have been interpretation issues on both sides, and
    each of us has reacted by drawing unjustified conclusions. For my part,
    I am sorry that I spoke in haste. However, I hope you will recognize
    that you are not without blame.
  5. Dave Seaman

    Dave Seaman Guest

    0 factorial is an empty product, and therefore equal to 1.
    0^0 is an empty product, and therefore equal to 1.
    The same argument also shows that empty sums are equal to 0.
  6. Dave Seaman

    Dave Seaman Guest

    i^0 = exp(0*log(i)) = exp(0) = 1,

    where log(i) is multivalued (= i*pi/2 + 2*n*pi*i), but it doesn't
    particularly matter which value you choose. And (-i)^0 = 1 by similar
  7. John Fields

    John Fields Guest


    1/0 is a small number?


    y = lim x->0 ---

    Then it seems to me like y gets pretty big when x gets pretty small!
  8. John Fields

    John Fields Guest

  9. So (k*0)/0 is not equal to k*(0/0), then?
    What a pity!
    To defend your point, you have to abandon the associative law of

  10. Doesn't matter; you can find a delta for every epsilon.
    If there was a proof, one which doesn't depend on a contrived meaning
    for exponentiation, I'd agree with you. But then, there wouldn't have
    been such a controversy if there was such a proof.
  11. Dave Seaman

    Dave Seaman Guest

    I rely on the mathematical arguments I have previously posted with references
    to recognized authoritative authors (Suppes, Lang, and the sci.math FAQ).
  12. Dave Seaman

    Dave Seaman Guest

    Not true. In fact, you cannot find a single delta that works for *any*
    epsilon > 0. I'll let you choose any epsilon you like. You still lose.
    This is not a contrived meaning of exponentiation. It's the standard
    definition for a^b where a and b are cardinal numbers, and this is a
    necessary first step before extending the definition to more general

    Consider 2^3, for example. It's the cardinality of the set of mappings
    from the set { 0, 1, 2 } to the set { 0, 1 }, which is 8.

    Besides, there is also the definition from algebra, in which x^n is
    defined whenever x is a member of a monoid M and n is a natural number.
    In particular, x^0 = e, the identity in M.
  13. George Cox

    George Cox Guest

    You're confused. If y = 1/x then y gets pretty big when x gets pretty

    But if y = lim x->0 --- then y doesn't exist.
  14. John Fields

    John Fields Guest

    Obviously, but no matter.


    If my point is that 0/0 = 1, and if k*(0/0)=k then 0/0 _must_ be
    equal to 1, otherwise k would not be equal to k. That, I believe,
    proves my point, my point being:

    y = lim x->0 --- = 1
  15. John Fields

    John Fields Guest

  16. No, that is circular reasoning:
    "If 0/0 = 1 the 0/0 must be equal to 1".

    If dividing by zero yields no result (i.e. it is undefined), then both
    (k*0)/0 and k*(0/0) produce the same, i.e. they're both undefined.
    But they're not equal any more,
    and (a*b)/c = a*(b/c) by the associative law.

    I am trying to show to you that _assuming_ 0/0=1 (to be 100% clear,
    assuming that 0/0=1 is _always_ true) leads to you having to abandon the
    generality of that law.

    If by associative law, you find that (k*0)/0=k*(0/0), that proves that
    0/0 can't be 1,
    because the equation simplifies to 1=k.

    So it's either that law or 0/0=1, but not both.

  17. George Cox

    George Cox Guest

    Write what? That 1/x gets big as x gets small?

    1/x -> +infinity as x -> +0

    1/x -> -infinity as x -> -0.
  18. George Cox

    George Cox Guest

    This is true, but it is true because

    y = lim x->0 --- = lim x->0 1 = 1.

    The "1" on the left of the equals sign is there because the two x's

    It is not always the case that

    lim x -> A f(x) = f(A).

    So the true statement

    y = lim x->0 --- = 1

    does not allow you to conclude that 0/0 = 1.
  19. keith

    keith Guest

    Of course. We had no issues with discontinuities either.
    That's about when I took it. (actually 33 years).
    Which makes no practical sense. Engineers like the practical, even though
    it's infinite. ;-)
  20. John Fields

    John Fields Guest

    But _does_ allow me to conclude that any other number, no matter how
    small, and no matter what its sign will, when divided by itself, give
    a quotient of 1?

    OK, look at it this way:

    These two number lines are behind a screen, so the numbers can't be
    seen, but arranged in such a way that when a number is picked from
    one, the same number will be picked from the other and those two
    numbers will be divided and the quotient presented for viewing.

    -1 0 1

    -1 0 1

    I fail to see why, at one infinitesimal spot on the line(s), the
    division becomes intractable and the machine refuses to output a 1.
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