zero divided by zero

Discussion in 'Undergraduate Math' started by Turco, Jan 27, 2006.

  1. Turco

    Turco Guest

    Is there anything special about 0/0, I mean is that any different from
    1/0? I know they are both indeterminations, but I would like to know
    if 0/0 is different from k/0 in any way... thx
     
    Turco, Jan 27, 2006
    #1
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  2. Turco

    majamin Guest

    There *is* a system of arithmetic in which mathematicians do allow such
    things as (0*infinity) and (infinity + infinity), but there it is
    mearly a custom of convenience and not of much fundamental importance.
    Firstly, 0/0 does not have a meaning if one considers it as the
    operation 0*(0)^(-1). That is, zero multiplied by the inverse of zero.
    Mathematicians cringe when they see something like this because
    *usually* real numbers (*ahem*) have unique inverses. That is, say,
    0.25 is the unique inverse of 4. Hence, we cringe, because zero times
    *any* number is zero (weird, right? ;-) )Consider this:

    (1) a = b
    (2) a - b = 0
    (3) (a-b)/(a-b) = 0
    (4) 1 = 0

    Obviously, we cannot allow division by zero (line (3) ) for the sake of
    consistency (and sanity?).Now, as for 1/0 and k/0 as you mentioned, the
    same problem creeps up: what number do you multiply 0 by to get a
    non-zero number k?

    If you are interested numbers the idea of the 'limit' and '0/0' and
    'infinity/infinity' are important in earlier stages of analysis (read
    'calculus'), especially if you know what a derivative is, etc. Refer to
    L'Hospital's rule here:

    http://mathworld.wolfram.com/LHospitalsRule.html

    Good luck,

    M.M.
     
    majamin, Jan 28, 2006
    #2
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  3. Turco

    Turco Guest

    thank you.
     
    Turco, Jan 28, 2006
    #3
  4. On 27 Jan 2006 13:45:05 -0800, Turco <>
    wrote in
    In the context of arithmetic of real numbers, both are
    undefined. In general, if x and y are real numbers, we
    define x/y to be the unique real number z such that x = yz,
    if such a real number exists; if not, x/y is undefined. (It
    isn't actually indeterminate; that's a slightly different
    concept that doesn't really apply here.) 1/0 is therefore
    undefined because there is no real number z such that 1 =
    0*z; 0/0, on the other hand, is undefined because every real
    number z satisfies the equation 0 = 0*z, and therefore there
    is no *unique* real number that does so.


    There is another context in which there *is* a very
    important difference between two things that are often
    symbolized by '0/0' and '1/0', though this is very sloppy
    notation, and the context doesn't actually involve any
    attempt to divide by 0. Suppose that you're considering the
    quotient of two functions, f(x)/g(x), and you're interested
    in what happens to it as x gets very close to some number c.
    You may find, for instance, that f(x) approaches 1, while
    g(x) approaches 2; in this case the quotient f(x)/g(x)
    approaches 1/2, and we know this as soon as we know that
    f(x) approaches 1 and g(x) approaches 2.

    Or you may find that f(x) approaches 1, while g(x)
    approaches 0. This is a very different kettle of fish: if
    you divide numbers that are very close to 1 by numbers that
    are very close to 0, you get quotients that are very large
    in absolute value (i.e., either very large in the positive
    direction, or very 'large' in the negative direction).

    For instance, consider f(x) = x, g(x) = x - 1, and c = 1.
    Then f(1.01)/g(1.01) = 1.01/0.01 = 101, f(1.001)/g(1.001) =
    1.001/0.001 = 1001, f(1.0001)/g(1.0001) = 1.0001/0.0001 =
    10001, etc., while f(0.99)/g(0.99) = 0.99/(-0.01) = -99,
    f(0.999)/g(0.999) = 0.999/(-0.001) = -999,
    f(0.9999)/g(0.9999) = 0.9999/(-0.0001) = -9999 etc. Clearly
    f(x)/g(x) isn't settling down near any particular number as
    x gets closer and closer to 1: for x just a tiny bit bigger
    than 1, f(x)/g(x) is a huge positive number, while for x
    just a tiny bit less than 1, f(x)/g(x) is a huge negative
    number.

    More generally, if f(x) --> 1 and g(x) --> 0 as x --> c, the
    quotient f(x)/g(x) is going to 'blow up', and in particular
    it can't possibly have a limiting value. We know the result
    -- there is no limit -- the moment we know that f(x)
    approaches 1 and g(x) approaches 0.

    But if f(x) and g(x) *both* approach 0 as x approaches c,
    it's a different story. Depending on exactly what the
    functions f and g are, the quotient f(x)/g(x) might have a
    limit, or it might not; and if it has one, that limit could
    be any real number. For example, suppose that g(x) = x^2
    and c = 0. If f(x) = x, then f(x)/g(x) = x/x^2 = 1/x as
    long as x isn't 0; as x approaches 0, 1/x 'blows up' like
    the previous example, so, as before, f(x)/g(x) has no limit
    as x approaches 0. Now suppose that f(x) = kx^2 for some
    real number k. Then f(x)/g(x) = kx^2/x^2 = k as long as x
    isn't 0; in this case f(x)/g(x) is constantly k as x
    approaches 0, and the limit of f(x)/g(x) is therefore k,
    which could be any number at all. That is, knowing that
    f(x) and g(x) both approach 0 gives us *no* information
    about the limit of f(x)/g(x); in order to find it (or show
    that it doesn't exist), we must get our hands dirty working
    with the specific functions f and g.

    The limit of a quotient with a 0 limit on top and a 0 limit
    on the bottom is therefore _indeterminate_: its value cannot
    be inferred from the limits of the numerator and
    denominator, but must be worked out individually in each
    case. The limit of a quotient with a 1 limit on top and a 0
    limit on the bottom, however, is determined just by those
    two limits: it doesn't exist. Note, though, that although
    people often talk about these as '0/0 limits' and '1/0
    limits', those are just verbal shorthand: there's really no
    0/0 or 1/0 involved here at all.

    Brian
     
    Brian M. Scott, Jan 28, 2006
    #4
  5. Turco

    Owen Guest

    n/0 is an arithmetic concern and not a concern of 'limits'.

    How we define x/y determines what x/0 means.

    If we define x/y as (the z: x=y*z), then x/0 does not exist for all x,
    including x=0.
    (because (the z: x=y*z) is not unique)

    That is to say, 1/0 and 0/0 are defined but, they do not exist.

    If we say: ~(y=0) -> x/y = (the z: x=y*z), then
    x/0 is not defined for any x, including x=0.

    That is to say, 1/0 and 0/0 are not defined.

    Either way, 1/0 and 0/0 are not sensible terms!
     
    Owen, Feb 10, 2006
    #5
  6. Turco

    in2infinity

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    Yes the amount of number space that is traversed. 0/0 = 0 no number space traversed. whereas 1/0=0 means a unit of one is traverse in order to arrive at the result.
     
    in2infinity, Apr 21, 2022
    #6
  7. Turco

    nycmathguy

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    I thought 0/0 means indeterminate in Calculus. No?
     
    nycmathguy, Apr 28, 2022
    #7
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