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An analysis proof(?)

  1. Aug 9, 2012 #1
    I'm going to be a math major soon, so I'm trying to learn analysis, proofs and some set theory too. Recently I started to read Rudin's analysis book and there is a question that I tried to solve(prove?). I think my proof is wrong or not complete, could you correct me please? http://pdfcast.org/pdf/proof-5 [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 9, 2012 #2

    micromass

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    I think there's a type. You typed that if x and r are rational, then x+r is irrational. This is of course false. Did you mean to say that if r is rational (and nonzero) and x is irrational, then x+r is irrational??

    Let's assume you meant that.

    I'm not gonna say your proof is wrong, but it's very weird. First of all, you spend your entire time proving that x+r is in [itex]\mathbb{R}[/itex]. I really don't see the point of this.

    And in the last line, you say "But since [itex]x\notin \mathbb{Q}[/itex], [itex]x+r\notin\mathbb{Q}[/itex] also". But this sentence is exactly what they asked you to prove!!! You should elaborate a bit on why this is true.
     
  4. Aug 9, 2012 #3
    Ah, yes it's a typo. Thank you very much...
     
  5. Aug 9, 2012 #4

    jgens

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    One quick comment: For set minus, you should write either [itex]\mathbb{R} \setminus \mathbb{Q}[/itex] or [itex]\mathbb{R}-\mathbb{Q}[/itex] instead of [itex]\mathbb{R}/\mathbb{Q}[/itex].
     
  6. Aug 9, 2012 #5
    Yes I know but I'm new to latex and couldn't figure out how to use math tools yet
     
  7. Aug 9, 2012 #6

    jgens

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    I see. Just use the "\setminus" command to get the slash facing the proper way.
     
  8. Aug 9, 2012 #7
    You could use the contrapositive. The contrapositive of "A implies B" is "not B implies not A". The contrapositive is logically equivalent to the original statement.
    Statement A:
    R is rational AND X is irrational

    Statement B:
    RX is irrational


    Proof:
    Suppose not B. So RX is rational. Assume (towards contradiction) that A is true. R must be rational also.
    We can write:
    RX=N1/N2.
    R=N3/N4. (Notation: Let "N#" variables be integers.)

    So,
    X=RX/R
    X=(N1/N2)/(N3/N4)=(N1*N4)/(N2*N3).
    Thus, X is rational.


    This contradicts A. Thus, A is false. We have shown Not B => Not A. Therefore, A => B.

    The proof is very similar with R+X. The only real difference is in the italicized region.

    Hope this helps.
     
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