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Calculus theory proof- Suppose a is irrational, prove√(1+a) is irrational.

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose a is irrational, prove√(1+a) is irrational.


    2. Relevant equations

    A number is rational if it can be expressed as p/q, p,q integers with q≠0

    3. The attempt at a solution

    I can reason through it intuitively but not sure how to demonstrate it formally. Any help or advice would be greatly appreciated
     
  2. jcsd
  3. Feb 28, 2012 #2
    I think you can use a proof by contradiction here. Assume you can write sqrt(1 + a) as p/q with p and q integers and see what happens.
     
  4. Feb 28, 2012 #3
    that was sort of my attempt at a solution. This is what i've got now.
    Assume a irrational, but √(1+a) is rational.
    Then √(1+a) = p/q for p,q integers q≠0.
    1+a = p^2/q^2 → a = p^2/q^2 - 1 → (p^2- q^2)/q^2
    Since p,q integers, p^2-q^2 and q^2 must be integers.
    Thus a must also be rational by definition, a contradiction.
    Thus if a is rational, √(1+a) is rational. The contrapositive is equivalent.
    Therefore if a is irrational, √(1+a) is irrational.

    Is this all valid and a valid conclusion? Thanks again!
     
  5. Feb 28, 2012 #4

    jbunniii

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    Correct up to this point.

    No, it's the other way around. If [itex]\sqrt{1+a}[/itex] is rational, then [itex]a[/itex] is rational. This is exactly what you just proved.

    And therefore:
     
  6. Feb 28, 2012 #5

    Deveno

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    suppose that a was irrational, but √(1+a) = r was rational.

    what can you say, then, about r2?
     
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