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Homework Help: Irrational Number Proof

  1. Oct 14, 2008 #1
    Using the fact that [tex]\sqrt{2}[/tex] is irrational, we can actually come up with some interesting facts about other numbers. Consider the number t=1/[tex]\sqrt{2}[/tex], which is also irrational. Let a and b be positive integers, and a<b. We will prove that any rational approximation a/b of t will differ by at least 1/(4b^2), that is, for all integers 0<a<b.
    Notice that a<b because (1/[tex]\sqrt{2}[/tex])<1.
    (a) We prove the statement by contradiction. Assume the conclusion is false. Show it must follow that
    (b) Show that, as a result,

    The Attempt:
    (a) |(1/[tex]\sqrt{2}[/tex])-(a/b)<1/(4b^2)
    Combining the fractions on the left side, we get
    Multiplying the denominator of the left side to both sides gives
    |b-a[tex]\sqrt{2}[/tex]| < [tex]\sqrt{2}[/tex]/4b

    (b) I started part b by squaring both sides of the inequality I found in part a.
    |b-a[tex]\sqrt{2}[/tex]|^2 < ([tex]\sqrt{2}[/tex]/4b)^2
    |b^2-2(a[tex]\sqrt{2}[/tex])+2a^2| < 2/(16b^2)

    And then I'm not sure about where to go from here to get the result indicated in part b. Any tips?
  2. jcsd
  3. Oct 14, 2008 #2


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    Multiply both sides of the result of a) by b+a*sqrt(2).
  4. Oct 15, 2008 #3
    thanks =) it worked out

    However, I have an alternate part to this question.

    For any integers 0<a<b, prove that |b^2-2a^2|>1

    My Attempt:
    I used the axioms of ordering to find that a^2<b^2.
    Rearranging the inequality gave me b^2-a^2>0.
    Subtracting a^2 from both sides, b^2-2a^2>-a^2

    I'm not sure where to go from here or if this is the right way to go. Any hints?
  5. Oct 16, 2008 #4


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    [itex]b^2- 2a^2\ge -a^2[/itex] means that [itex]b^2- 2a^2|\ge a^2[/itex] doesn't it? And how small can a2 be if a is an integer?
  6. Oct 16, 2008 #5


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    No, I don't think that follows. It's even simpler than that. |b^2-2a^2| is also a nonnegative integer. If it's not greater than or equal to one, then it's zero.
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