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Sum of a rational number and an irrational number ...

  1. Jul 30, 2017 #1
    1. The problem statement, all variables and given/known data

    I am trying without success to provide a rigorous proof for the following exercise:

    Show that the sum of a rational number and an irrational number is irrational.

    2. Relevant equations

    I am working from the following books:

    Ethan D. Bloch: The Real Numbers and Real Analysis

    and

    Derek Goldrei: Classic Set Theory

    Both use a Dedekind Cut approach to the construction of the real numbers (but Goldrei also uses Cauchy Sequences ... )

    I am taking the definition of an irrational number as equivalent to an irrational cut as defined by Bloch as follows:

    ?temp_hash=cee404bddcc16a3ee55c431470ff5b05.png

    Bloch's definition of a Dedekind Cut plus a Lemma indicating the that there are at least as many of them as there are rational numbers are relevant ... and read as follows:

    ?temp_hash=cee404bddcc16a3ee55c431470ff5b05.png






    3. The attempt at a solution

    I have been unable to make a meaningful start on this problem ...


    Peter


    *** EDIT ***

    Reflecting on this problem ... it has become apparent to me that Bloch's definition of the addition of real numbers (in terms of Dedekind Cuts is relevant ... so I am providing the relevant definition ... as follows:

    ?temp_hash=5e153c4d3963047e0f6ee5c2aa26e760.png


    The note preceding the above definition mentions Lemma 1.6.8 which reads as follows:

    ?temp_hash=5e153c4d3963047e0f6ee5c2aa26e760.png
     

    Attached Files:

    Last edited: Jul 30, 2017
  2. jcsd
  3. Jul 30, 2017 #2
    After reflecting on this problem ... I now think that the answer to the exercise above may be disappointingly trivial ...

    Consider the following:

    Let ##a \in## ##\mathbb{Q}## and let ##b \in \mathbb{R}## \ ##\mathbb{Q}##

    Then suppose a + b = r

    Now ... assume r is rational

    Then b = r - a ...

    But since r and a are rational ... we have r - a is rational ..

    Then ... we have that an irrational number b is equal to a rational number ...

    Contradiction!

    So ... r is irrational ..



    Is that correct?

    Peter
     
    Last edited: Jul 30, 2017
  4. Jul 30, 2017 #3

    fresh_42

    Staff: Mentor

    Yes.
     
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