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Math Help - Multiplication and division by 0

  1. #1
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    Question Multiplication and division by 0

    I have already looked for many answers but none of them appears to satisfy my question.

    I can imagine why everything divided by 0 makes no sense and has no solution.
    x divided by 0 means that you divide by nothing and therefore do not divide anything, so there is no answer available as you do not do anything. In other words, you are still left with your x / 0 waiting, stuck in time for infinity.

    Now I want to understand why x times 0 is possible and equals 0.

    I mean 1 times x = x (How many times do I add x? once, therefore x); 2 * x (How many times do I add x? Twice. therefore x + x). Now what about 0 * x? How many times? Not any time at all. Shouldn't this also be regarded as an impossible operation? You still don't do anything and are stuck in time for infinity, so why do we have an answer here? Please don't say "because this is how it is defined". I don't like that type of answers.

    If the answer is "how many times? 0 times, therefore 0, then this can also be applied to division and then say how many apples to you give John? none, so 0. However, it is impossible to divide by 0.
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  2. #2
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    Re: Multiplication and division by 0

    You can't split 10 apples into groups of zero. That's why \frac{x}{0} is undefined (unless x = 0, that's a different story). However, as the divisor approaches zero, the quotient approaches positive or negative infinity, for example,

    \frac{1}{.1} = 10

    \frac{1}{.001} = 1000

    \frac{1}{.00001} = 100000


    Also, anything times zero is zero and that makes sense. You have zero groups of 10 apples, so you have zero apples. Another way to think of it is, 0 \times x = (1 - 1) \times x = x - x = 0.
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    Re: Multiplication and division by 0

    OK...so the Math Bosses "decide" that n/0 = 0.
    What will that change? Reduce the price of groceries?

    Btw, hope they do change it: tired of message "you goofed, division by zero at ......"
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  5. #5
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    Re: Multiplication and division by 0

    Quote Originally Posted by Wilmer View Post
    so the Math Bosses "decide" that n/0 = 0.
    What will that change? Reduce the price of groceries?
    Actually, we can define n / 0 to be 0. This does not lead to contradiction if we keep the proviso that n ≠ 0 for the property that n / n = 1.

    Often the impossibility of division by zero is justified by the fact that, say, if x = 10 / 0, then x * 0 must equal 10, which is impossible. But how do we conclude that x = 10 / 0 imply x * 0 = 10? We multiply both sides of x = 10 / 0 by 0 and get x * 0 = (10 / 0) * 0 = 10 * (0 / 0) = 10 * 1 = 10. This reasoning would be impossible if we are not allowed to conclude that 0 / 0 = 1.
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    Re: Multiplication and division by 0

    IF a/0 = 0, for some non-zero a, then for any non-zero b:

    0 = b*0 = (b/1)*(a/0) = ab/0. taking b = 1/a, leads to 1/0 = 0, and thus b/0 = (b/1)*(1/0) = b*0 = 0, for EVERY non-zero b.

    furthermore, a/0 + c/d = (a*d + 0*c)/(0*d) = ad/0 = 0 (see above).

    but how can this be? surely a/0 + c/d = 0 + c/d = c/d. are we therefore to conclude that for all c/d, c/d = 0?

    unless you are willing to completely ruin the algebraic structure of the rational numbers (which seems a rather steep price to pay), i think it far preferable to leave a/0 undefined.
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    Re: Multiplication and division by 0

    Congratulations with the 2000th post!

    Quote Originally Posted by Deveno View Post
    furthermore, a/0 + c/d = (a*d + 0*c)/(0*d)
    This is not derivable. To factor out 1 / (0 * d) (i.e., to bring the two terms to the least common denominator), you need to multiply c / d by 0 / 0, but I made a disclaimer that we are allowed to conclude x / x = 1 only if x ≠ 0.

    The convention 0⁻ = 0 is used in the proof assistant I am using because it allows only total functions.
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    Re: Multiplication and division by 0

    the rule: a/b + c/d = (ad + bc)/(bd) isn't "derived" it's the *definition* of addition of two rational numbers. there's no factoring going on.

    someone pointed out a pertinant observation to me:

    rational numbers are actually equivalence classes on Z x Z*: namely a/b = cd iff ad = bc. what you propose is essentially "extending" this equivalence relation to Z x Z, in which case we have:

    a/0 = c/d iff a*d = c*0 = 0. if we agree that a is non-zero, this means d = 0 (which is a more elegant argument than i gave above), no matter *what* c is.

    that said, someone mentioned to me the concept of "Wheels", in which division by 0 (or rather, a similar involution) *is* defined, and extends the usual ring. there is a wikipedia article here:

    Wheel theory - Wikipedia, the free encyclopedia

    although it has much less information than i would like.
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    Re: Multiplication and division by 0

    Quote Originally Posted by Deveno View Post
    the rule: a/b + c/d = (ad + bc)/(bd) isn't "derived" it's the *definition* of addition of two rational numbers...

    rational numbers are actually equivalence classes on Z x Z*: namely a/b = cd iff ad = bc. what you propose is essentially "extending" this equivalence relation to Z x Z
    I agree that extending equality from Z x Z* to Z x Z using the same formula, i.e., (a, b) = (c, d) iff ad = bc, does not work because this relation is not transitive: (a, 0) = (0, 0) = (c, d) for any c and d, but (a, 0) ≠ (c, d) if a ≠ 0 and d ≠ 0. My idea had nothing to do with rational numbers per se. I was saying that if we add the axiom \forall x.\,x\ne 0\to x^{-1}=0 (edit: should be 0-1 = 0) to field axioms, we will get a consistent theory. So, given any field, including the field of rational numbers with the standard carrier and definition of equality, we can in addition define x / 0 to be 0 for x ≠ 0 (edit: for all x) and not get a contradiction. In particular, one cannot derive ∀a,b,c,d (a / b + c / d = (ad + bc) / (bd)) from field axioms even with this additional axiom; one needs the proviso b ≠ 0 ⋀ d ≠ 0.

    Quote Originally Posted by Deveno View Post
    that said, someone mentioned to me the concept of "Wheels", in which division by 0 (or rather, a similar involution) *is* defined, and extends the usual ring. there is a wikipedia article here:

    Wheel theory - Wikipedia, the free encyclopedia

    although it has much less information than i would like.
    I did not know about wheels, but what I am saying seems to be much less interesting than wheel theory. Wikipedia says that in wheels 0 * x is not necessarily 0 and x - x is not necessarily 0, so this is a substantially different mathematical object from a field.
    Last edited by emakarov; August 13th 2012 at 03:27 PM.
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    Re: Multiplication and division by 0

    Quote Originally Posted by Sserg View Post
    Now I want to understand why x times 0 is possible and equals 0.
    So you want to understand it, eh?
    Since multiplication is shorthand for consecutive addition (n*x=x+x+...+x) you can think of x times 0 like this:

    Consider the case of you having two piggy banks, a red and a blue one. Now suppose both piggy banks are empty and there's noone to put anything in them except for you, but you're broke, or your current balance is 0. Every day you walk up to the red piggy bank, collect the cash that is in it (remember it is empty) and add that cash to the amount inside the blue piggy bank (also empty). And you do that for n days in a row. Now on the (n+1)st day go inside a pub and take the blue piggy bank with you, and keep buying everyone rounds all night long. When the time comes for you to pay for all those rounds give the bartender your blue piggy bank and tell him he can keep the change. Then you'll understand...
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  11. #11
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    Re: Multiplication and division by 0

    Quote Originally Posted by emakarov View Post
    I agree that extending equality from Z x Z* to Z x Z using the same formula, i.e., (a, b) = (c, d) iff ad = bc, does not work because this relation is not transitive: (a, 0) = (0, 0) = (c, d) for any c and d, but (a, 0) ≠ (c, d) if a ≠ 0 and d ≠ 0. My idea had nothing to do with rational numbers per se. I was saying that if we add the axiom \forall x.\,x\ne 0\to x^{-1}=0 to field axioms, we will get a consistent theory. So, given any field, including the field of rational numbers with the standard carrier and definition of equality, we can in addition define x / 0 to be 0 for x ≠ 0 and not get a contradiction. In particular, one cannot derive ∀a,b,c,d (a / b + c / d = (ad + bc) / (bd)) from field axioms even with this additional axiom; one needs the proviso b ≠ 0 ⋀ d ≠ 0.

    I did not know about wheels, but what I am saying seems to be much less interesting than wheel theory. Wikipedia says that in wheels 0 * x is not necessarily 0 and x - x is not necessarily 0, so this is a substantially different mathematical object from a field.
    i am not sure which field axioms you are referring to. it seems to me that if x-1 = 0, for all x ≠ 0, then xx-1 = 1 implies 0 = 1, since in any ring (and fields are rings): 0x = 0 (as one can see by considering 0x + 0x = (0 + 0)x = 0x), and 0 and 1 must be distinct in any field.

    i spent a lot of time trying to consider how i could interpret your statement as somehow meaningful, but i have not been able to come up with a viable model that does not violate some field axiom.

    but...there's many things i *don't* know, and if you can exhibit a system in which x/0 makes sense, and x/y corresponds to our usual notion of "x divided by y" (at least for non-zero integers), i would very much be interested in seeing it.

    (P.S.: i don't know what you mean by "the standard carrier and definition of equality", so i can't comment on that. i always thought a/b = c/d meant ad = bc, but perhaps i am mistaken (this goes back to Eudoxos' theory of ratios, if i recall correctly))
    Last edited by Deveno; August 13th 2012 at 02:33 PM.
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    Re: Multiplication and division by 0

    Quote Originally Posted by Deveno View Post
    i am not sure which field axioms you are referring to. it seems to me that if x-1 = 0, for all x ≠ 0, then xx-1 = 1 implies 0 = 1
    Oops, I meant adding the axiom 0-1 = 0 and keeping \forall x.\,x\ne 0\to xx^{-1}=1. I am sorry about that. I wrote it correctly in posts #5 and #7. If 0-1 = 0, then x / 0 = (by definition) x * 0-1 = 0 for all x.

    My explanations may not be clear, but here is the precise claim: Field axioms together with 0-1 = 0 form a consistent theory, i.e., one cannot derive a contradiction from it.

    Quote Originally Posted by Deveno View Post
    i don't know what you mean by "the standard carrier and definition of equality"
    By rational numbers with the standard carrier and definition of equality I mean Z x Z* with (a, b) = (c, d) iff ad = bc. Note that I am suggesting we could define [(a, b)] / [(0, d)] = [(0, d)] where a ∈ Z and b, d ∈ Z* and [...] denotes the equivalence class. I am not suggesting defining [(a, 0)] = [(0, d)] (which could be written as a / 0 = 0) where a ∈ Z, d ∈ Z*. This would be redefining equality on rational numbers. We take existing set of rational numbers with existing equality and add an extra clause to the definition of division on these rational numbers.
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  13. #13
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    Re: Multiplication and division by 0

    if i understand you correctly, you are talking about replacing any expression of the form x/0 (for non-zero x) with 0. but so doing does not make (F,*) into a group, in other words 0-1 no longer means: "0 inverse", it means: 0-1 ordinarily means nothing, so in this one case, we replace it by 0.

    as i observed before, we can no longer *define* a/b + c/d as (ad + bc)/(ad) (or more properly, the equivalence class of said expression). this "new structure" has several exceptions to the rules for the rational numbers.

    in other words, there is no logical difference between what you propose and a structure (W,+,*,@) where:

    1) (W,+,*) is a field
    2) @:W-{0} x{0}→{0}, defined in the only way possible.

    in other words, you are essentially creating a "new operation" : @, with a@0 = 0.

    i agree that this does not violate the field axioms. in fact, it has nothing to do with them. in fact: @ is only right-distributive over addition:

    (w+w')@0 = w@0 + w'@0 unless you want left-distributivity just to mean:

    w@(0 + 0) = w@0 + w@0 (since 0 is the only "right-argument" for @).

    in fact, i suspect that we can just as easily define a/0 = Bob, and leave any other expression involving "Bob" undefined.
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