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Homework Help: Slight deviation of proof, would it be correct for integers? Review for exam!

  1. Oct 16, 2006 #1
    Hello everyone.

    He told us he could of course change the parameters which he will of the proofs we have been working on so I'm testing out some cases but I want to make sure i'm doing it right.

    Here is an example of a proof the boook had:
    http://suprfile.com/src/1/3qaagxh/Untitled-1%20copy.jpg [Broken]

    Now where he has the statement: "If a and b are rational numbers...."
    I'm changing that to:
    "If a and b are integers...."
    and now here is my proof, i think its correct but I have to make sure.

    If a and b are integers, b != 0, and r is an irrational number, then a+ br is irrational.

    Proof by Contradiction:

    Suppose not. Suppose that a and b are integers, b != 0, and r is an irrational number such that a+br is rational. We must obtain a contradiction.

    Since a, b are integers and a + br are rational, a+br = m/n for some integers m, n with n != 0.


    a + br = m/n

    br = m/n - a
    r = (m-an)/bn

    where (m-an) and (bn) are integers since m, a, n, and b are integers, and bn is nonzero since b is nonzero. Therefore, r is rational, contradicting that r is irrational.

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 16, 2006 #2
    looks good to me. are you using the book my solow? btw, you you go to UPark?
    Last edited: Oct 16, 2006
  4. Oct 16, 2006 #3
    I'm using Discrete Mathematics with appplications 3rd by Susanna S. Epp, and yeah i go to Penn State UPark. CSE260 is the class, hah how did ya know.
  5. Oct 16, 2006 #4
    just a guess/facebook/google(thinkquest contest). many of my friend go there.
  6. Oct 16, 2006 #5
    yeah this place is pretty massive, well thanks for the help! I seem to post all questions but don't answer any hah. I did do thinkquest along time ago... i'm supprised that site is still up! well ttyl
  7. Oct 16, 2006 #6


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    Since all integers are rational numbers, once you had proven it for a, b rational, exactly the same proof must hold for a, b integer. Actually, I wouldn't have writen the rational numbers as fractions, i/j, etc. It is sufficient to use the fact that the rational numbers are closed under addition and multiplication:
    Suppose a+ br, a, b rational b non-zero, is rational: a+ br= c where c is rational. Then br= c- a so r= (c-a)(1/b). Since b is a non-zero rational number 1/b is rational and so r= (c-a)(1/b) is rational- contradiction.
  8. Oct 16, 2006 #7
    That does seem alot nicer, thanks!
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