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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
    http://www.math.su.se/~jesper/research/wheels/ [Broken]
     
    Last edited by a moderator: May 1, 2017
  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.
     
  22. Aug 24, 2004 #21

    Hurkyl

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    There are quite a few problems: I'll demonstrate the problem in three common perspectives:

    Perspective 1: We live in the real numbers.

    Statement 1 and 2 are both meaningless, since there are no "infinitely large" numbers. 90 is simply not in the domain of tan, and similarly (1, 0) is not in the domain of division. (IOW, tan 90 and 1/0 are simply undefined)
    Statement 3 is similarly meaningless, because the LHS and RHS are both undefined.
    Statement 4 is similarly meaningless; 1/0 is undefined, thus so should arctan (1/0).


    Perspective 2: We live in real projective space.
    Real projective space is the topological space of the real numbers plus one extra point called [itex]\infty[/itex].

    It does make sense1 in real projective space to extend the definitions of tangent and division so that [itex]\tan 90 = \infty[/itex] and [itex]1/0 = \infty[/itex]. Indeed, tan 90 = 1/0.

    However, there is no consistent way to specify [itex]\arctan \infty[/itex]. As we let x increase without bound, we see that arctan x approaches 90... but as we let x decrease without bound, we see that arctan x approaches -90. So, we have two conflicting extensions: going one way suggests that [itex]\arctan \infty = 90[/itex], but the other way suggests that [itex]\arctan \infty = -90[/itex]. Since 90 is obviously not -90, we cannot continuously extend arctan to be defined at [itex]\infty[/itex].


    Perspective 3: We live in the extended real numbers.
    The extended real numbers is a topological space that consists of the real numbers adjoined with two additional points on either end: [itex]+\infty[/itex] and [itex]-\infty[/itex].

    Statements 1 and 2 (and thus 3) are meaningless, because as we approach 90 from the left, we find [itex]\tan x \rightarrow +\infty[/itex], but from the right we have [itex]\tan x \rightarrow -\infty[/itex].

    However, unlike the real projective space, we can extend arctan as [itex]\arctan +\infty = 90[/itex] and [itex]\arctan -\infty = -90[/itex].



    And incidentally, my calculator (a TI-89) is capable of working in the extended real numbers. If I ask it to compute [itex]\tan^{-1} \infty[/itex], it gives me [itex]\pi/2[/itex]. (it works in radians by default, not degrees) If I ask it to compute [itex]\tan^{-1} tan (\pi / 2)[/itex], it gives me [itex]+/- \pi/2[/itex].



    Footnotes

    1: In particular, tan x has a removable discontinuity at 90. In other words, the limit as x approaches 90 of tan x is [itex]\infty[/itex]. (in real projective space, that is) The extension of 1/0 is similarly justified, but can also be justified in a purely algebraic way.
     
    Last edited: Aug 24, 2004
  23. Aug 24, 2004 #22
    Perspective 4: Uncommon sense. You can't argue with this logic: if something is, it is. Nothing happens until it happens. E=E=E. As 1/0 is undefined, and tan 90 is also undefined, are they both undefined? If they weren't would they be defined?, and as being defined, would they not be undefined? Yes! As 1/0 is undefined and tan 90 is undefined, and as undefined is undefined, they are both the same "value," even though they are not a real value, they are both the same value in that they fill the same space in the same manner because they both are, and are both equal, judging by the fact that they are equal "values" even if they are not values. Thus, if something is not a value, it is still a "value" judging by the fact that all "values" are the same value. Therefore, all things of the same value might not have a value, but might be a "value" (imaginary). Logically, all things that do not have a position in the series of real numbers must belong if a real number does not belong. However, not all unbelongsts have the same nature and must be classified as such (not classified by categories consisting of unbelongts of different natures, but classified in categores of unbelongsts of the same nature).

    Thus, 1/0 and tan 90 only belong in a position only if a real number does not belong in that same position. And seeing that they are both unbelongsts of the same nature, they can both be classified in the same category, and are thus interchangeable. My calculator says that if a machine processes white phosphorous (tan 90, our "dangerous" "value") into water, it will still ignite in the presence of oxygen once it leaves the machine. Uncommon sense tells you that your undoing (getting burned) will be undone by the machine that contains the raw materials of your undoing. How can you be undone if the object you would have been a victim of is undone? So, in conclusion, how can 1/0 or tan 90 be considered as values, and not placeholders, if they are undone into harmless objects (real numbers). Tan 90 need not be calculated if it is never exposed to air. Just as white phosphorous needs not to be acknowledged if it is never exposed to air. The machine would break down if white phosphorous was acknowledged, but it turns out it would never have to be acknowledged anyway if you let the expression of the machine operate with no interruption. So, finally, do not acknowledge tan 90 or 1/0 if the product is a real number-let the "machine" of the expression operate with no interruption.

    <<edit>> "As the biologist exclaimed when she held up the petrie dish in front of the city of Athens, 'Now that is culture!'"
     
    Last edited by a moderator: Aug 24, 2004
  24. Aug 24, 2004 #23
    90 is in fact in the domain of tan and (1,0) is in the domain of division. Take, for example, the fact that (5+2) is in the domain of addition just as much as 7 is in the domain of division. You can express the solution of a problem as the problem. 0=0, 1+1=1+1. Are those not true? 1/0 and tan 90, along with "0" are unique in the fact that their solution is the problem. 1/0 is expressed as 1/0, just as tan 90 is expressed as tan 90. I challenge you to write 1/0 in a different manner, there is no better way. So why do we not say that 1/0=1/0 and write the answer to the problem as the problem itself? If 1/0 is not in the domain of 1/0 then 5, or any other number is not in the domain of addition.

    Until expressing 5 as a different expression of numericality is standardized in the mathematical community, I shall not be obliged to scribe "1/0" in any different manner other than that which is the currently accepted expression of this expression.
     
  25. Aug 25, 2004 #24

    Hurkyl

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    Well, there are two immediate problems:
    (a) In mathematics, and really in all sciences, intuition and (un)common sense must yield to deductive logic.

    (b) (un)common sense and logic are two very different things.


    Undefined is not a "value" in mathematics. Deductive logic has a very rigid structure. To borrow from ordinary grammar to help explain, "(un)defined" is an adjective. Relations like "is equal to" are only allowed to work on nouns.

    Furthermore, you may not realize it, but the statement "tan 90 is undefined" is quite a bit of shorthand: what it really means is "90 is not in the domain of tan". Similarly, "1/0 is undefined" really means "the pair (1, 0) is not in the domain of division".


    In order to speak mathematically about such things, you have to give precise meaning to what you say. Mathematically, we would say that:

    The function tan x has a pole of degree 1 at x = 90.
    and
    The function 1/x has a pole of degree 1 at x = 0.

    Being a pole is a description of the behavior of the function near (but not at) the given point. In particular, for a function f(x) to have a pole of degree k at x = a, that means that the function [itex](x-a)^k f(x)[/itex] is bounded near a. (In the usual contexts, it must actually have a removable discontinuity at a)

    Because we have this precise description, we can use deductive logic to prove things about poles, and mathematical progress continues. We can discover what really happens in analysis instead of what our intuition says.


    I can't write 1/0 in any manner. While "1/0" is certainly a string of symbols, it does not represent anything mathematical. To again draw analogy with natural language, writing "1/0" is akin to writing "zqurvip": while they are both strings of symbols, neither have any meaning.
     
  26. Aug 25, 2004 #25
    Circular reasoning at its finest. Mathematics doesn't work by the rule "if something is, it is" . One starts with a set of axioms and proves things exist from there. Until something is axiomatically assumed or proven to exist, statements about it are devoid of meaning.
    Let's work with an analogy. Compare the following:

    a) red is red (red = red)
    b) Red is in English what rouge is in French.
    c) This color is red.
    d) Light is red iff its frequency lies between 625nm and 740nm.

    Statement (a) in both its forms carries no information. One cannot even conclude that 'red' exists (replace red by 'the fountain of youth'), or determine what 'red' is. It is trivially true by the meaning of 'is' or '='.
    Statement (b) can be used to make a mapping between English and French. It allows the inference 'red' is a meaningful word of the English language.
    Statement (c) makes an empirical definition of 'red' by relying on the ability of the human brain to perceive the text in red as different than the rest of the page.
    Statement (d) defines 'red' as an objective, measurable, physical property.

    Now, which one do you think "tan 90 = tan 90" resembles the most?

    In fact, why not define tan 90 = 1? It can be done; you are simply left with an ugly discontinuity of tan (x) at x=90. Similarly one can extend tan (x) to all of its poles - but it serves no purpose.

    As for 1/0, here's an answer to the challenge.
    For any nonzero k, a*k / b*k = a / b . This can be derived from the axioms that define the real field. Then 1/0=(1*2)/(0*2). Since 1*2=2 and 0*2=0, 1/0=2/0. Since I like 2 better than 1, 2/0 is a better way to write it. Prove me wrong.
     
    Last edited: Aug 25, 2004
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