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The distributive law

  1. Jan 21, 2004 #1
    Is there any way to prove the distributive law for integers? I heard that there is yet I don't understand how being that the distributive law is an axiom and therefore what the understanding our number system is based on.
  2. jcsd
  3. Jan 21, 2004 #2
    When working with the integers, you usually just consider the distributive law to be an axiom. So you don't prove it, since you've assumed it to be true.

    However, in some cases you wish to prove that using axioms for some other objects you can construct the integers. For example, you might want to prove that you can construct the integers using just sets and natural numbers. In that case you would have to prove that your construction obeys the distributive law.

    This approach has some advantages. Instead of having to assume that sets exist and that natural numbers exist, that integers exist, that rationals exist, etc. we can just instead assume that one exists (such as sets) and use them to construct all the other objects.

    But, as I said, there really isn't much point to trying to prove the axioms of the integers when all you want to work with are the integers.
  4. Jan 21, 2004 #3
    So to just prove a(b+c)= ab + ac in general would not be possible?
  5. Jan 21, 2004 #4

    The best you can do is to assume something else which is equivalent to the distributive law, and prove it from that. But if you aren't satisified with assuming the distributive law is true, you probably won't be satisified with assuming something else equivalent.
  6. Jan 22, 2004 #5


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    On the contrary, it is possible to prove the distributive law starting from "Peano's axioms" for the natural numbers. That is basically equivalent to "induction" and all proofs of properties of natural numbers are inductive.

    Here's a link to a PDF file that contains such proofs:
    http://academic.gallaudet.edu/courses/MAT/MAT000Ivew.nsf/ID/918f9bc4dda7eb1c8525688700561c74/ [Broken]$file/NUMBERS.pdf

    Click on "Numbers".
    Last edited by a moderator: May 1, 2017
  7. Jan 22, 2004 #6
    Well, we were talking about the integers, not the natural numbers.

    But my point was that we choose these axioms because they produce distributivity, so we aren't really "proving" the distributive law in a meaningful sense.

    It's sort of like proving 1+1=2. Of course we can do it, but it isn't really that satisifying. The constructions and definitions we use to prove it were specifically chosen because they give that result.
    Last edited by a moderator: May 1, 2017
  8. Jan 23, 2004 #7


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    I see your point but I disagree. The axioms for the counting numbers were not specifically chosen so that they give the distributive law but so that they give "induction" (i.e. "counting").
    It happens that they also give the distributive law but that has to be proven by someone! I wouldn't expect every algebra student to do it but it's a nice "five finger exercise" for mathematicians.

    The properties of the integers are derived from those of the counting numbers specifically to give the group properties (existance of an additive identity, additive inverses) and, again, it happens that the distributive law is true and that has to be proven.

    By the way, one does not have to prove that "1+ 1= 2" because that is basically how "2" is define. That "2+ 2= 4" is a theorem and has to be proven (by someone). It a simple two or three line proof, of course.
  9. Jan 23, 2004 #8
    Most (minimal) constructions of N that I have ever seen define 2 as the successor of 1, not as 1+1. Of course, the successor of any number n is n+1, but strictly speaking, 1+1=2 is a theorem. But I guess that depends on the details of how you decide to construct N and define +.

    I guess you are right in that any construction that provides induction should provide distribution. It is a derivation from a simpler principle and not just a restatement. I would have realized that if I didn't have a temporary rectal-cranial inversion.
  10. Jan 23, 2004 #9


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    Hmm, Yes, 2 is defined as the sucessor of 1 and n+1 is defined as "the sucessor of n" therefore--

    Gosh, I just might be forced to agree with you!
  11. Jan 23, 2004 #10
    Actually, I think it's even worse. Addition is usually defined with something like [itex]n+0=n[/itex] and [itex]n+s(m)=s(n+m)[/itex]. So you have to use that definition along with [itex]1=s(0)[/itex] to prove that [itex]n+1=s(n)[/itex] first.

    It is kind of neat sometimes to see some horribly counterintuitive construction turn into the natural numbers we know and love.
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