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Small understanding

  1. Mar 18, 2007 #1

    This is not a homework question since I am out of college for a long time.

    I was trying to understand the following that [0,1] /\ Q is not closed in R.
    My understanding is that u must take a sequence (since this is a metric space) of the form m/n s.t m < n and create a sequence.

    So I was trying to construct sequences like 1/2, 2/3, 3/4 but they seemed to be all ending within [0,1] /\ Q. I am not sure but do I have to take a sum or sth but I am not sure how to prove it.
  2. jcsd
  3. Mar 18, 2007 #2


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    (I am assuming that Q means rational nos. and R means real nos.). Take any irrational number between 0 and 1 and let the rational number sequence be the the sequence where the nth term is the truncation of the decimal expansion of the irrational after n decimal places.
  4. Mar 18, 2007 #3


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    Just giving an example of what mathman said: [itex]\sqrt{2}{2}[/itex] is irrational and is 0.70710678118654752440084436210485...
    Each number in the sequence 0.7, 0.70, 0.707, 0.7071, 0.70710, 0.707106,... is a rational number because it is a terminating decimal; but the sequence as a whole converges to the irrational number [itex]\frac{\sqrt{2}}{{2}}[/itex]
    Last edited: Mar 20, 2007
  5. Mar 18, 2007 #4


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    You're working from the wrong direction. Rather than trying to come up with a sequence of rationals and hope that its limit is irrational... you should pick the irrational, and then try and find a sequence of rationals that converges to it.
  6. Mar 19, 2007 #5
    Thanks a lot for clearing the lacunae in my understanding. I think the basic idea is that u need to show that if u wiggle the rational number a bit u will fall into the set of irrational numbers, which is clearly proved by the sqrt(22) example.

    Thank u a lot for the same.
  7. Mar 19, 2007 #6
    Actually that was sqrt(2)/2 (how could the square root of 22 be less than 1?). You should read what Hurkyl said. The simplest irrational you can think of in the interval is sqrt(2)/2, and constructing a sequence that converges to it is easy, as shown by HallsofIvy.
  8. Mar 19, 2007 #7


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    'Cause I'm nothing if not simple!
  9. Mar 19, 2007 #8
    Do you understand why this argument suffices?
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