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Prove that There exists a sequence of rationals approaching any real number

  1. Jun 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Let x be any real number. Prove that there exists a sequence {Rn} of rationals different from x such that {Rn} converges to x.

    Use the Archimedian property, the fact that the rationals are dense in the reals, and the squeeze principle.

    2. Relevant equations
    The Archimedian property states that given any real number x, we may find a natural number n such that n>x.

    The squeeze principle states that if the sequence {an} converges to L, the sequence {bn} converges to L, {an}<{bn} for all n, then, if {an}≤{cn}≤{bn}, then cn converges to L.

    3. The attempt at a solution
    Proof:
    If x is rational, then let {an}=x+1/n which is a sequence of rationals that converges to x.
    Assume x is irrational and without loss of generality assume x>0. Using the Archimedian property we can find a natural number n such that 0<x<n. We will now construct two sequences {an}, {bn} which approach x from below and above respectively. Take the interval [0,n] and divide it into two equal parts. Then look at the interval [a1,b1] containing x. We know that X cannot be halfway point because x is irrational. Divide this interval into two again and consider the interval [a2,b2] containing x. Continue in this manner. Take {an} to be the sequence of left endpoints:{a1,a2...} and {bn} to be the sequence of right endpoints: {b1,b2...}. Both of these sequences are monotone and both are bounded by x. Thus {an} and {bn} both converge to x. Note that {an}<{(an+bn)/2}<{bn} and by the squeeze principle, {(an+bn)/2} converges to x. #

    I feel that this is the correct way to go about it but it seems to me that it is not necessary to use the squeeze principle when I've already found two sequences {an},{bn} that converge to x. However, my instructor insists that I have to use the squeeze principle so I feel as if my proof can be greatly shortened somehow and I'm just not seeing how...
     
  2. jcsd
  3. Jun 14, 2012 #2

    SammyS

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    Each of the sequences, {an} and {bn}, do appear to converge to x.

    However, the reason you give,
    "Both of these sequences are monotone and both are bounded by x. Thus {an} and {bn} both converge to x."​
    does not guarantee that either converges to x.

    Added in Edit:

    Suggestions for constructing the sequence, {Rn}:
    The sequences, {an} and {bn}, do not need to be rational. As long as each converges to x and an < bn, for each n, then density of rationals in the reals tells you that between each pair, an and bn, there is some rational number --- call it Rn .

     
    Last edited: Jun 15, 2012
  4. Jun 15, 2012 #3
    Ah so...
    Given a real number x, I may use the sequences, for example, x+1/n and x-1/n both of which approach x. Then, I can use density to say that between these two possibly irrational sequences I can find a rational sequence {Rn} so by the squeeze principle Rn converges to x?
     
  5. Jun 15, 2012 #4

    SammyS

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    Looks good to me.
     
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