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Division by ZERO

  1. Aug 20, 2004 #1
    Division by ZERO!!!

    Hey guys!!!

    Was just pondering...

    ALL my mathematical subject lecturers go on and on and on about trying to avoid division by zero. Fine - after all, it makes logical nonsense to divide by zero but my pondering thoughts led me somewhere...

    Is it possible to do some maths with the division by zero?? You know - like a NEW number system. So far - I haven't found one. Can anyone help me here??
     
  2. jcsd
  3. Aug 20, 2004 #2
  4. Aug 20, 2004 #3
    it depends what you mean by division by zero. Sometimes [tex]\epsilon^2=0[/tex] is taken as [tex]\epsilon[/tex] divides zero. In that case, you can perform symbolic calculations with [tex]\epsilon[/tex] exactly as you do with imaginary numbers. Indeed, [tex]\epsilon[/tex] is part of the generalized imaginary numbers. It has applications in geometry.
     
    Last edited: Aug 20, 2004
  5. Aug 20, 2004 #4
    So does that mean 0/0 is a complex number???
     
  6. Aug 20, 2004 #5
    No no no ! Generalized complex numbers !

    See, "i" is just defined as the root of an equation which has none in R(eals). Namely [tex]x^2+1=0[/tex]. The idea is : I define the new "imaginary" number "j" as the root of a convenient equation. That could be [tex]x^2+x+1=0[/tex] for instance. Or [tex]x^2=0[/tex] as well ! Depending on how "relevant" your initial equation is, you get more or less interesting results.
     
  7. Aug 20, 2004 #6
    I don't remember quite well, because I never used them. I think the [tex]\epsilon[/tex] example has to do with a geometry in which lines can be parallel or anti-parallel. They have orientation. I can't elaborate on this.
     
    Last edited: Aug 20, 2004
  8. Aug 21, 2004 #7
    Im not sure what u mean by a new number system, but no, you cant use dividing by zero in math because it leads to ilogical anwers. Let me show you an example...
    a = b
    a^2 = b^2
    a^2 - b^2 = a^2 - ab
    (a+b)(a-b) = a(a-b)
    a + b = a
    a + a = a
    2a = a
    2 = 1

    Obviously this cant be right? 2 doesnt equal 1, so how can this be?
    Well, in the math you do a step where you divide the factor (a-b) from both sides. Since a=b, then (a-b) = 0, and dividing by zero isnt allowed.
     
  9. Aug 22, 2004 #8

    Alkatran

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    There would have to be a set of rules which stopped 0/0 from returning any result you want. For example, you could say that x*0 <> 0, 0/x <> 0, and 0/0 = 1. If we can't reduce numbers using 0 then we don't have any problems.

    Take ArmoSkater87's example:
    a = b
    a^2 = b^2
    a^2 - b^2 = a^2 - ab
    (a+b)(a-b) = a(a-b)
    a + b = a
    a + a = a
    2a = a
    2 = 1

    Step 4 to Step 5 would violate x*0 <> 0 (or in this case, (a+b)*0 <> 0), stopping the error from ever being reached.
     
  10. Aug 23, 2004 #9

    Alkatran

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    I just wanted to know if there was any flaws in my above post. I don't want to waste my time thinking about this. :confused:
     
  11. Aug 23, 2004 #10

    HallsofIvy

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    The only flaw is that any structure in which 0/0 is defined is either the trivial structure containging ONLY 0, or is not a field. Yes, you can define structures that are not fields and they may be interesting, but they are a little too general.
     
  12. Aug 23, 2004 #11

    Alkatran

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    Can you give an example, following the rules I gave (x*0 <> 0, 0/x <> 0, 0/0 = 1) that doesn't make sense?
     
  13. Aug 23, 2004 #12

    Hurkyl

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    Yes.

    0 = x*0 - x*0 = x * (0-0) = x * 0
    Which contradicts x*0 <> 0
    1 = 0 / 0 = (0+0) / 0 = 0/0 + 0/0 = 1 + 1 = 2


    The laws of arithmetic (specifically the ring axioms) prove that x*0 = 0. (My first example above is the proof) Division is defined as the inverse of multiplication.

    If you're working with some system where x*0 is not equal to zero, or where division is not the inverse of multiplication, then you are not doing arithmetic; you're working on some other, strage, system.
     
  14. Aug 24, 2004 #13

    mathwonk

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    the only sense in which division by zero works in my experience is when we consider a function like [(x-a)(f(x))]/(x-a), and plug in x=a, geting f(a).

    I.e. whenever we take the limit, of [g(x)-g(a)]/(x-a), in some sense we are dividing zero by zero.
     
  15. Aug 24, 2004 #14

    Alkatran

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    Nice post. I was aware that x*0 = 0 is not an error, I was just saying if you put some rules in, operations on 0 wouldn't result in asymptotes.

    Another good example would have been log[0](x) = ? (0 to the power of ? gives x) etc etc... You'd actually have to say any operation involving 0 is unsolvable unless it is cancelled by another 0.

    Question: Some of the equations involving 0 contradict 1 = 1, but they are assumed wrong because 1=1 is one of the basic laws of math. I would argue that x*0 - x*0 = x * (0-0) is false (IN THE RULED SYSTEM) because to get from x*0 - x*0 you had to do an operation involving 0.... but this is all nonsense anyways. There's no point in having none of the 0 operations solvable instead of half just to give operations involving 0 twice sensable.
     
  16. Aug 24, 2004 #15

    matt grime

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    "I would argue that x*0 - x*0 = x * (0-0) is false"

    eh? when did 0 not become equal to zero?
     
  17. Aug 24, 2004 #16

    Hurkyl

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    Well, your premise is inaccurate.

    These arguments aren't "assumed wrong": they are invalid. The reason they are invalid is not because they "contradict 1 = 1", but because there is a step that does not logically follow from previous steps.

    For example, in the argument in #7, "(a + b) = a" does not logically follow from "(a+b)(a-b) = a(a-b)", or from any combination of previous steps.

    This flaw is the sole reason the argument is deemed invalid.


    Incidentally, there is a "law" of arithmetic that says

    If uw = vw, then u = v or w = 0 (or both).


    And your argument is invalid. Not only is no law, theorem, axiom, or anything that says "An equation involving operations with 0 is false".

    There is a very specific collection of axioms that are true for arithmetic, period. These axioms vary slightly for different systems (e.g. integers, complexes), but the point is that arithmetic is very clear and precise.
     
  18. Aug 24, 2004 #17

    Alkatran

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    I never said any such axioms existed. I agreed with you! :bugeye:
    Sorry for the confusion.
     
  19. Aug 24, 2004 #18

    robphy

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    Do you have a reference?
    I've been playing around with so-called hypercomplex numbers (not of the quaternion, octonian type).
     
  20. Aug 24, 2004 #19

    Hurkyl

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    Just about anything on Abstract Algebra should talk about this sort of thing. Formally, what is going on is that you're taking a polynomial ring modulo some polynomial.

    For example, the complex numbers are simply R / (i^2 + 1): the quotient field of all real polynomials in a free variable i, modulo the polynomial i^2 + 1.

    For example, i^3 and -i are both complex numbers, since they're real polynomials in i.

    They are equal complex numbers because they are equal mod i^2 + 1:
    i^3 = (i^2) i = (i^2 + 1 - 1) i = (i^2 + 1) i - i = 0i - i = -i (mod i^2 + 1)

    And, of course, you can show any complex number is equal to some complex number of the form a + bi.
     
  21. Aug 24, 2004 #20
    1. tan 90=infinitely large number
    2. 1/0=infinitely large number
    3. tan 90=1/0
    4. arctan (1/0) should equal 90. A calculator will not do it, but the human mind can. A calculator won't even do arctan (tan 90), or at least mine won't.
     
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