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Rational and irrational numbers. (semi- )

  1. Sep 21, 2009 #1
    Rational and irrational numbers. (semi-urgent)

    I need to figure this out by tomorrow =/

    1. The problem statement, all variables and given/known data

    a. If a is rational and b is irrational, is a+b necessarily irrational? What if a and b are both irrational?
    b. If a is rational and b is irrational, is ab necessarily irrational?
    c. Is there a number of a such that a^2 is irrational, but a^4 is rational?

    2. Relevant equations

    none.

    3. The attempt at a solution

    a. I think I have this first part. You can prove it by contradiction.

    R= some rational number

    a+b = R
    b = R-a
    A rational number minus a rational number is a rational number. This would mean b = rational, which is not true. therefore a+b is irrational.

    This second part, if both are irrational? I was thinking:

    a+b = R

    a = R-b, or b = R-a. I'm not sure how this helps me x(
    -----
    b. If a is rational and b is irrational, is ab necessarily irrational?

    No idea, but here's my attempt:

    Two cases. If a = 0, ab = 0, and 0 is rational so ab is rational.

    if a =/= 0...proof by contradiction maybe?

    a*b = rational
    a*b = a*b
    b = a*b*a^-1?

    But then b=b? And that doesn't help me.
    -----
    c Is there a number a such that a^2 is irrational, but a^4 is rational?

    Well again I have no idea but here's my attempt:

    a=b
    a^2 = ab
    ab = x
    b = x/a
    b = x * a^-1

    ....again no idea, please help. x(
     
  2. jcsd
  3. Sep 21, 2009 #2

    VietDao29

    User Avatar
    Homework Helper

    Re: Rational and irrational numbers. (semi-urgent)

    So far so good. :)

    When finding a proof seems hopeless, one should try to search for a counter-example.

    According to part a,

    [tex]1 + \sqrt{2}[/tex], and [tex]-\sqrt{2}[/tex] are both irrational, what if you take the sum of them?

    First case, okay. :)

    Second case:

    Well, just do as you did in part a. Like this:

    [tex]a \in \mathbb{Q} \backslash \{ 0 \} 0, b \notin \mathbb{Q}[/tex]

    Assume that
    [tex]ab = r[/tex], where r is a rational number.
    [tex]\Rightarrow b = ra ^ {-1}[/tex]

    What can you say about b in the above expression?

    -------------------------------------

    Additional Problem:

    If a, and b are both irrational numbers, is ab also irrational?

    Think about 4-th root. :)
     
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