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An integer between n and n+1 where n is an integer.

  1. Apr 4, 2007 #1
    hi guys..
    Does this statement require a proof? It seems pretty obvious to me.

    "prove that there is no integer between n and n+1 where n is an integer."

    Also if it does require a proof, what I need to show? Just few hints will suffice.

    thanks
    jitendra
     
  2. jcsd
  3. Apr 4, 2007 #2

    AlephZero

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    The proof depends what system of axioms you use to define the integers.

    Starting from Peano's axioms and defining addition using Peano's successor function, this is fairly simple.

    On the other hand, Russell and Whitehead's "Principia Mathematica" takes 379 pages to get to the statement

    "From this proposition it will follow, when arithmetical addition has been defined, that 1+1=2"

    and proof that 1+1=2 is finally done on page 86 of Volume II - with the comment "The above proposition is occasionally useful."
     
  4. Apr 4, 2007 #3
    your name is AlbertEinstein. YOu should be able to prove it. Its rather an easy proof, What is the definition of an integer and go from there.
     
  5. Apr 4, 2007 #4
    Without going into axiomatics, a simple proof could use the division algorith.
     
  6. Apr 4, 2007 #5
    hmm..well and how the integers are exactly defined? I mean they are just the set of all natural numbers , with their negatives and zero. Is there any abstract way of defining them? Please describe in bit more details.

    thanks.
     
  7. Apr 4, 2007 #6
    What are natural numbers? What is a "negative" of a natural number? You need to build them from scratch, or axioms, as mathematicians call them. As mentioned by AlephZero, one way of defining (our long-known and beloved) numbers is by using Peano's axioms.
     
  8. Apr 4, 2007 #7

    StatusX

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    Note (by subtracting) it's equivalent to show there's no integer between 0 and 1. Since 0<1, such an integer would need to be positive. All positive integers are sums of some number of 1's (it might seem circular to say 3 is the sum of three ones, but it's the best you can do), and adding one to a number makes it larger, so all positive integers besides one are greater than one.
     
  9. Apr 4, 2007 #8
    you can also do it by means of set theory, i.e constructing N and Z by defining an inductive set for which its existence is being given to us by an the infinite set axiom. (which states that there exists an inductive set).
    and then prove this statement through tools of set theory.
    the question is where did you get this question from?
     
  10. Apr 4, 2007 #9
    Einstein wasn't that much of a mathematician.
     
  11. Apr 5, 2007 #10

    Gib Z

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    Yes but I'm sure if he could master Riemann Geometry he can prove this lol.

    Hehe heres a proof :D :

    Proof By Intimidation - Trivial. Lol I love that.

    But seriously now, Try proof by contradiction/example.
     
  12. Apr 5, 2007 #11

    AlephZero

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    I don't really see how you can make a sensible proof of this without using axiomatics.

    I think the result is "obvious" to anybody who knows how to count, but has not had any formal mathematics training at all. So the way to "prove" it is to start from some axioms (e.g. Peano's) which define exactly what you mean by "counting".

    If you want to use other theorems (e.g. the division algorithm) then you have to be sure you can prove the division algorithm without assuming this result, otherwise you have a circular argument.
     
  13. Apr 5, 2007 #12
    Well, I have just passed my high school and will be going in undergraduate course this year. therefore i don't know Peano axiom in detail. I think i have to study abstract algebra.Will that help me? Also, can i have a suggestion of a good introductory book treating the subject in detail as well as quite rigourous in approach.

    thanks.
     
  14. Apr 5, 2007 #13
    Apostol, in his calculus text (Vol.1), recommends this analysis text by Edmund Landau.

    This is book description
     
  15. Apr 6, 2007 #14
    thanks buddies..I will surely try to get my hands on the book..
    by the way, why was the last post deleted that gave a proof to my question?? Was that proof okay?
     
  16. Apr 6, 2007 #15
    mathworld.com search peano arithmetic or axiom
     
  17. Apr 6, 2007 #16

    HallsofIvy

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    If you mean the one by fopc, he may have reconsidered. He was proving that there is no integer between 0 and 1 but I don't think that's necessary. It follows from the definition of "<" in the Peano's axioms definition of the natural numbers that 1 is the smallest number. Certainly if n< k< n+1, then subtracting n from each part gives k-n< 1 (we need n< k because, in the natural numbers, b- a is only defined if a< b).
     
  18. Apr 7, 2007 #17
    "Was that proof okay?"

    No.

    Here's a modified version.
    It's still not OK, but it might serve
    the purpose of stimulating some criticism
    (which of course is always good).
    By the way, I agree with the post about
    starting essentially from scratch (say Peano).
    Probably the only way to give a rigorous proof.
    But also probably not appropriate for someone
    just leaving high school.


    Preliminary assumptions:

    1. N is set of Natural numbers. The set is well ordered, i.e., the set comes equipped with a well order relation. Therefore, every nonempty subset of N owns a least element.

    2. 0 is in N. Clearly, it is the least element for N itself.

    3. Z is symbol for integers.


    First, prove there is no k in N satisfying, 0 < k < 1.
    Consider it a lemma. It's purpose is to provide the contradiction sought in the following theorem.

    Let A = {k | k in N and 0 < k < 1}. Assume A nonempty. Since A is a nonempty and apparently a subset of N, it contains a unique least element, say m. Now 0 < m < 1.
    But then 0 < m^2 < m < 1, which is a contradiction since m is the least element of A. So A must be empty, and therefore no such k exists.


    *Note*
    I think you or anyone else who bothers to read this stuff will see problems with 0 < m^2 < m < 1. Anyway, I'll finish it up.


    Thm.
    For any n in Z, there is no k (in Z) satisfying n < k < n+1.

    Assume such a k exists. Then p = k - n is in Z. But now p must satisfy 0 < p < 1, which contradicts the result given above. (Assumption here is that Z includes N.)



    Incidently, I see "Foundations of Analysis", by E.Landau mentioned in a post. I'd also recommend this book.
    Landau will take you from Peano, step by step, through to the real and complex numbers. But be forewarned, he'll do it in what he describes as a "merciless telegraph style".
    But it's only about 125 pages and it's not an expensive purchase.
     
  19. Apr 14, 2009 #18
    To build up the positive integers (or the nonnegative integers) from scratch is tricky. Here are some of the ordering axioms for the nonnegative integers (the natural numbers.)

    1) For any n, n < n+1.

    2) For any n, if n < m+1 then n < or = m.

    It is from the second that we get the following argument:

    Assume there is some integer k such that n < k < n+1.
    Then by (2), k < n+1 implies that k < n or k = n. But this contradicts that n < k.
    Hence, there is NO integer k such that n < k < n+1.

    :P
     
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