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Two questions

  1. Dec 2, 2004 #1
    Is there a proof of a*b=0 <--> (a=0 or b=0) that does NOT involve division? Also, my discrete math professor has claimed that mathematicians tend to avoid the concept of division. How accurate is this claim for discrete mathematicians? What is the reason for it?
     
  2. jcsd
  3. Dec 2, 2004 #2

    matt grime

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    exceptionally accurate. Divisibilty is a property of multiplication.

    x divides y if there is a z such that xz=y

    Avoid dividing numbers unnecessarily - it avoids supposing things ought to exist that may not exist.

    For instance the "correct" thing to say is something like 2x=3 has no solution in Z because 2 does not divide 3. DO NOT write... becuase 3/2 is not in Z. What is 3/2? What if the ring weren't Z? Your proof relies on the existence of some larger object (the raionals) which you've not proved even exist in a meaningful way, and it precludes your answer being applied in other rings where there is no obious embedding into something that behaves like the rationals.


    As for the proof ab=0 => a or b =0, in what space are you thinking? Z, R? C? In Z we may assume that if niether a nor b is zero they are both positive. But 0<a<=ab is one of the axioms of Z with ordering, which is a contradiction.
     
  4. Dec 2, 2004 #3

    mathwonk

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    no. this property is not always true for all numbers systems. it is true if and only if the number system can be extended to one allowing division. so this property is an axiom. after making the axiom, one can prove there is a larger number system in which division by non zero numbers is possible. conversely, if there is such an extension, then thios property must have held.

    we also tend to avoid driveby shootings. what is he talking about? maybe he means that we, like many other human beings, find fractions hard. hence the first step in any problem involving fractions is always to "multiply out the botoms".
     
  5. Dec 2, 2004 #4

    jcsd

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    In any ring, such as the real numbers, rational numebrs integers, etc,:

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

    add ab

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

    therfeore by identity 0b = 0.

    A simlair proof holds for left mulplication by zero.
     
  6. Dec 2, 2004 #5

    mathwonk

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    he asked the converse too, jcsd. i.e. to prove that ab cannot be zero unless a or b is zero. you did the easy direction.
     
  7. Dec 2, 2004 #6

    jcsd

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    OOps I didn't see that, in that case I should of added then that a*b = 0 implies that either a or b is zero is only true in rings that are integral domains (which obviously includes all the rings I mentioned but not rings in general).
     
  8. Dec 2, 2004 #7

    mathwonk

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    which was my point, i.e. a ring is an integral domain if and only if it can be embedded in a field, i.e. a ring where division is available. hence essentially the proof cannot be done without division.
     
  9. Dec 2, 2004 #8
    Thanks for your replies. I actually am only interested in the ab=0 --> (a=0 or b=0) direction. The problem came up as part of a proof about integers. Could you approach it by saying that if a and b are both not 0, then each must be positive or negative, and for each case use the multiplication rules that pos * pos = pos, neg * pos = neg, and so on? Or would that require a lot of implicit assumptions?

    But anyway, without actually having such a proof in hand... let me see if I understand the situation. The proof of ab=0 --> (a=0 or b=0) does NOT require division explicitly, but for it to work it must be part of a system where division is an option--so you might as well use division anyway since then the proof is easy. Is that right?

    Anyway, is it safe to say that mathematicians use division fairly freely when they are working with the REAL numbers? (To hedge my bet, include applied mathematicians too)
     
  10. Dec 2, 2004 #9

    mathwonk

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    a proof uses whatever is eprmitted. so to do this proof you use whatever you know about integers. so what are you given? are you given that the rpoduct of positives is positive?

    if so you can prove the product of negatives is positive and the product of one of each is negative. that would do it. on the other ahnd maybe you are explicitly given that the product of non zero factors is non zero. or maybe you are given that the rationals exist. but actulaly you cannot prove the rationals existm until you know that the rpooduct of non zero integers is non zero. so if you do not show us what axioms you have for the integers we cannot tell you what is mneeded for a proof.

    i.e. proof is a relaitve thing. it depends on what you already know. and we do not know that.

    for example you can deduce that the product of two positive integers is positive, from the analogous property for sums, and induction, as follows:

    let a be positive, then a(1) = a is also positive. if ad is positive for all positive integers d less thasn n then an = a(n-1+1) = a((n-1)+a, which is a positive integer a(n-1) plus another positive integer a. this is positive if i know that the sum of two positive integers is positive. so do you know that?

    you see I cannot do it without knowing what you are given.
     
    Last edited: Dec 2, 2004
  11. Dec 2, 2004 #10

    mathwonk

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    the point is one cannot prove anmything about the integers until one knows what definition o=f the integers you are using. so i mthoiught i would consult some of my books and give you a typicsal dfinition of the integers. to my surprize the only one i could find at hand thjat even defiens them is artin's algebra book, who gives p[eanos axioms.

    he says the positive integers are a set S equipped with a "successor' function f:S-->S (intuitively taking n to n+1), such that f is injective, there is a special integer 1 which is not the successor of any other, and if T is any subset of S such that it contains 1 and also contains the successor of every one of its elements then T = S.


    he thens proves a few of the various proeprties of integers and elaves to us to prove that if ab=ac then b=c, which is equivalent to what tyou want.


    the key to this is as i sketched above, using the injectivity of the successor function. i.e. you never get the same answer twice by adding more ones to a given integer.

    but to repeat: before you can do this proof, you must ask what is the definition of integers you are using.
     
  12. Dec 3, 2004 #11

    matt grime

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    ^^^ commutative^^^ ring
     
  13. Dec 7, 2004 #12
    I'm not actually _given_ anything--this isn't an assigned question and the course I am taking is sadly not that rigorous anyway. But the proof I am looking for, if it exists, must not use division or any assumption which would normally be derived through the use of division. The integers I'm working with would be ... -2, -1, 0, 1, 2, ... although I don't know just how a formal definition of that should go.
     
  14. Dec 7, 2004 #13

    matt grime

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    You have to use some of the formal properties of the integers, such as the ordering on them, since the result isn't true in Z mod 4, for instance. I gave you the answer: if a and b are positive, which we may assume, then ab>0
     
  15. Dec 7, 2004 #14
    Are you sure that you don't need division to show that a>0 and b>0 --> ab>0 from basic premises?
     
  16. Dec 8, 2004 #15

    matt grime

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    The ordering is part of the integers, it comes for free - there are other rings which have division but do not have an ordering (eg the complex numbers). Put it interms of peano axioms if you must, but all I'm doing is using the fact that if a and b are positive integers, then ab=>a. OK? The ordering is deducible from the basic axioms without need to define division at all.
     
  17. Dec 8, 2004 #16
    So where does the integers being an integral domain come in at all?
     
  18. Dec 9, 2004 #17

    matt grime

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    If you assume they are an integral domain then by definition there are no zero divisors, and the question is trivial, as you are asking us to prove it is an integral domain.
     
  19. Dec 9, 2004 #18
    Okay, I think I understand now. Thanks for your help.
     
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