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Is zero even or odd?

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

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  1. John Fields

    John Fields Guest

  2. ISU WINS!

    ISU WINS! Guest

    Dont ever say "must". Thatw as inthe Novice exam!
  3. Guest

    Sigh. Forget machines. Is the definition of division clear to you?
    the result of the division a/b is such a number c that c*b = a.
    That's *all* there is to it. So, the result of 0/0 is a number c such
    that c*0 = 0. Can you tell me what c is?

    OK, here is a little exercise for you. At x = 0, sin(x) is also zero.
    And, so is any positive power of either x or sin(x). So, using the
    Taylor expansion for sin(x), evaluate the following limits:

    1) lim_x->0 sin(x)/x
    2) lim_x->0 sin(x)/x^2
    3) lim_x->0 (sin(x))^2/x

    Mati Meron | "When you argue with a fool,
    | chances are he is doing just the same"
  4. Guest

    In such case you may conclude that the specific limit evaluated is 1.
    Not that 0/0 = 1.

    Mati Meron | "When you argue with a fool,
    | chances are he is doing just the same"
  5. Since this was (cross)posted -for reasons that I can not imagine- also
    to clpmisc, to bring it _slightly_ more on topic:

    $ perl -le 'print 0**0'

    BTW: my good 'ol faithful HP-28s tells me the same.

    PS: the abondance of crankery this thread is revealing is tending to
    make me feel homesick of my sci.math days... (has JSH pop out with an
    FLT-based proof about 0^0 being... <whatever>?!?)

  6. [/QUOTE]

    The point is that you can't cancel them when x may be 0, because
    division by 0 is undefined. However, in the lim x->0 case x approaches
    0, but never really equals zero, so the cancellation may be performed.

    It is desirable for mathematics not to allow two contradictory
    statements such as 0/0=1 and 0/0=2 to be true at the same time; and your
    resoning allows these statements to be true in some cases (we did this
    in the Ohm's law episode, where you never said my reasoning was wrong,
    it was merely not the thing you wanted to prove).

    0/0=1 and 0/0=2 leads to 1=2, which is undesirable.

    lim x->0 x/x = 1 and lim x->0 2x/x = 2 doesn't have this problem.
    It is downright fine and dandy, because
    2 lim x->0 x/x = lim x->0 2x/x

    It is more inconvenient to write, so maybe you could agree to rephrasing
    that 0/0=1 where 0/0 is shorthand for lim x->0 x/x ?
    (You would decidedly be in a minority if you used that shorthand).

  7. You eventually lower the voltage to 0V.
    That's what I achieved with the short.
    I am trying to illustrate that I can make a point that 0/0=2.
    You cannot discard my point without adding extra information about your
    set of conditions.

    This extra information is not present in the 0/0 term, but it _is_
    explicitly written in lim x->0 x/x and lim x->0 2x/x , respectively.

    You want 1 ohm, that's what you bring into the computation. You're
    setting everything up so that 1 ohm results, which means it's circular

    If you hadn't set everything up that way, the 0V/0A measurement would
    leave you stumped as to the value of the resistor, and 0/0=??? then.

    Again, the set of values that are divided by itself is x/x for all x in
    R\{0}, and lim x->0 x/x = 1 as well.

    You are doing the limit argument nicely.
    Now you arrived at x=1 not by dividing by 0 outright, but by taking
    larger values and going closer to 0. I would explicitly write what you
    did, so your last formula would change to
    x = lim --- = 1
    r->0 r

    This is mathematical shorthand for your reasoning that any nonzero value
    plugged into r/r is 1, so it makes sense to take r/r=1 for r=0 as well.

    You misunderstood my point. My point is that you have to arrive at x/x
    or r/r before you plug in the zero. Conmsider: if you had x=2r/r, put in
    r=0 to get x=2*0/0 and use 2*0=0 to simplify to x=0/0, then the
    numerator is algebraically 0=2r and the denominator is algebraically
    0=r, and then
    x = lim ---- = 2
    r->0 r
    which would lead you to conclude that 0/0=2, in this case. You set your
    case up so that x=r/r, i.e. numerator and denominator are algebraically
    the same *before* you plug in the zero.

    This is not true for R=E/I, which is why you can't make a measurement in
    the E=0, I=0 case (unless you have extra information).

    You seem to grasp that so well. You can certainly measure 0V and 0A, but
    you can't determine R. For what reason do not grasp the analogy that 0/0
    represents the same condition in mathematics?

  8. vonroach

    vonroach Guest

    Otherwise? Accept your opinion? Crazy George is really crazy!
  9. vonroach

    vonroach Guest

    No 0^0 is meaningless .
  10. vonroach

    vonroach Guest

    Ah, another subject where you are poorly informed.
  11. vonroach

    vonroach Guest

    What a profound observation. How many years have you ben working on
    this concept? I suppose k*k= k^2 will be next? Have you carefully
    assured yourself that 1+1=2?
  12. vonroach

    vonroach Guest

    Parentheses and exponents are dealt with first.
  13. vonroach

    vonroach Guest

    Both are meaningless and undefined.
  14. vonroach

    vonroach Guest

    What a pity, you blunder on your very first point.
  15. vonroach

    vonroach Guest

    No it is crap with no evidence of any reasoning.
  16. Dave Seaman

    Dave Seaman Guest

    Your rants are meaningless, since you never include supporting arguments and
    you never answer contrary arguments, even when documented with authoritative
  17. vonroach

    vonroach Guest

    No it allows you to conclude that it is meaningless.
  18. vonroach

    vonroach Guest

    Chuckle....0/0 is meaningless and you are wrong and bordering on
  19. vonroach

    vonroach Guest

    0/n where n=any number except 0 = 0. You seem to be trying to make
    something hard out of a triviality.
  20. vonroach

    vonroach Guest

    But, but Mati, nobody is arguing with you.
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