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1 divided by zero

1 divided by zero=?

Poll closed Apr 8, 2005.
  1. Undefined

    8 vote(s)
    61.5%
  2. Infinity

    2 vote(s)
    15.4%
  3. Other(explain):

    3 vote(s)
    23.1%
  1. Mar 14, 2005 #1
    OK recently in class some of my classmates were debating whethter 1 divided by zero is undefined or infinity. I saw both points but im wondering what everyone else out there thinks. So 1/0= undefined or infinity or does it depend on other stuff?
     
  2. jcsd
  3. Mar 14, 2005 #2

    matt grime

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    Oh dear God, not again......
     
  4. Mar 14, 2005 #3
    Dividing by zero is undefined. There are ways around that using limits; however, in those cases, you are not dividing by zero, per se.
     
  5. Mar 14, 2005 #4

    Alkatran

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    lim(x->0)(1/x) = undefined
    lim(x->0+)(1/x) = infinity
    lim(x->0-)(1/x) = -infinity

    Take your pick. It all depends on the question.
     
  6. Mar 14, 2005 #5

    dextercioby

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    The interesting part is that he posted a poll...He probably meant division.All operations involving infinities & dividing through 0 are undefined...

    Daniel.
     
  7. Mar 14, 2005 #6

    arildno

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    I voted other: 0/1=0.
     
  8. Mar 14, 2005 #7
    This reminds me; if you ever see a book called "Zero: The Biography of a Dangerous Idea" and want to read it, be careful. The author frequently uses expressions like "dividing by zero/infinity". One of the passages goes like this: "[...]dividing anything by infinity gives zero; dividing anything by zero gives infinity. We can see the relationship these two numbers have throughout[...]"

    http://www.chapters.indigo.ca/item....talog=Books&N=35&Lang=en&Section=books&zxac=1

    I am told, however, that the historical account of zero of the book is rather fascinating.
     
  9. Mar 14, 2005 #8

    dextercioby

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    Don't read the book.Division under the reals has the property

    [tex] a:b=c\Rightarrow a=b\cdot c [/tex]

    You'll see that in the case of 0 or infinity,it doesn't work...(multivaluedness).

    Daniel.
     
  10. Mar 14, 2005 #9

    HallsofIvy

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    Define your terms. If you are working in the standard real number system, division of any number is "undefined"-literally. However, some people use "infinity" as a shorthand for "undefined": saying [itex]\lim_{x->0} \frac{1}{x}= \infty[/itex] is precisely the same as saying "the limit does not exist".
     
  11. Mar 14, 2005 #10
    Yes, inevitably.
    We must make sense out of the expression EyEx(x/0 = y).
    Is it tautologous or is it contradictory?

    i.e. is (1/0) a number?, or, is (0/0) a number?, or, is (2/0) a number? etc..

    It seems contradictory to me.
     
  12. Mar 14, 2005 #11

    dextercioby

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    As Halls said,how do you define your concepts:the concept of "number"...?

    Daniel.
     
  13. Mar 14, 2005 #12

    arivero

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    When x=0, the expression 1/x="multiplicative inverse of x"="the number y such that x times y equals unity" is a lot less ambiguous that 0/x, "the number y such that x times y equals zero".
     
  14. Mar 14, 2005 #13

    dextercioby

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    Yes,but when u see the overall picture,they're both faulty.

    Daniel.
     
  15. Mar 14, 2005 #14
    i think, this question can take place on a math-FAQ.

    people who think about division with zero seem to forget the meaning of division.
    steps of division are, you subtract the divisior from the divdend as long as there's something remaining dividend, ie it hits 0. for each exact subtraction, add 1 to the counter (ie, division). (i'm not covering the details for fractional parts, since my purpose in telling this is to remind the definition)

    so, applying this definition to 1/0: subtract 0 from 1 as long as you have a non-0, and add 1 to the counter each time. now, if you do this for 1/0, what's your counter?

    similar idea can be applied to 0/0: subtract 0 from 0 until as long as you have a non-0. now, indefinitiny approcahes at this point. no matter how you repeat this (and remember that you're adding a 1 to the result in each step), it'll be valid against the definition of division. 0/0 = 5, 0/0 = 9, 0/0 = anything.

    an application can be: [tex]x.0 = 0[/tex], this equation true for every value of x, ie, it's independent from the contents of x, as long as it's a valid number. dividing two sides by 0, you get [tex]x.\frac{0}{0} = \frac{0}{0}[/tex]. an equation independent from the value of x. it's pretty weird at the first glance, but if you remember what division is, the explanation reveals itself.

    i'd like to emphasize that division by zero in not undefined, it's indefinite (or maybe uncertain is a better word), and there're problem-dependent methods for getting rid of the indefinition.
     
    Last edited: Mar 14, 2005
  16. Mar 14, 2005 #15

    arildno

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    fdarkangel:
    Euclid's algorithm has nothing to do with the fact that the multiplicative inverse of 0 is undefined.
     
  17. Mar 14, 2005 #16
    I saw an interesting "proof" that illustrates the problems of dividing by zero:

    a=b
    a²=ab
    a²-b²=ab-b²
    (a-b)(a+b)=b(a-b)
    (a-b)(a+b)/(a-b)=b(a-b)/(a-b)
    a+b=b
    2b=b
    2=1

    The reason it comes out wrong is because (a-b)=0

    Kinda Fun.

    :D

    The Rev
     
  18. Mar 14, 2005 #17

    jcsd

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    We've had this question many times before:

    It cna easily be shown in any ring that:

    a*0 = a(b - b) = a(b - b) + 0 = a(b - b) + ab - (ab) = a(b - b + b) - (ab) = ab - (ab) = 0

    i.e. a*0 = 0 for all 'a';

    define divison by:

    a = b/c iff a*c = b

    let a = 1/0 therefore a*0 = 1, but a*0 = 0 for all 'a' so there is no 'a' which cna be identifed with 1/0.
     
  19. Mar 14, 2005 #18
    Of course.
     
  20. Mar 14, 2005 #19

    jcsd

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    Anything else to add....?

    The thing to note about the proof is that it applies to all rings.
     
  21. Mar 14, 2005 #20
    are you sure you've read my post?
     
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