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Least Upper Bound and the Density of the Irrationals Theorem

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data
    For the following set if it has an upper bound, find two different upper bounds as well as the least upper bound (LUB), justifying your answer. If the set has no upper bound, state this and justify your answer.

    {x | 1 < x < √(7) and x is irrational}
    (a proof requires the density of the irrationals)

    2. Relevant equations
    density of the irrationals theorem


    3. The attempt at a solution
    I stated two upper bounds as √(10) and √(37)
    But where I get confused is with the LUB. Isn't it simply √(7)? Why does the question state that a proof using the density of the irrationals is necessary?

    Had x been declared as rational then this would be more clear to me, since there would be no LUB and I would need to prove this using the density of the rationals.

    Where am I going wrong?
     
  2. jcsd
  3. Mar 11, 2012 #2
    hi jr

    To prove that some given number is LUB, you need to prove two things. If k is the LUB First,you need to show that k is an upper bound and second, you need to prove that for any
    [itex]\epsilon > 0[/itex], there exists some [itex]x_1 [/itex] in the set such that [itex]k-\epsilon <x_1[/itex].

    Can you prove these two things for [itex]\sqrt{7}[/itex] ? Its here you need to use the
    density of irrationals...
     
  4. Mar 11, 2012 #3

    Dick

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    You aren't going wrong. The answer is sqrt(7) for the LUB. The answer to LUB {x | 1 < x < √(7) and x is a element of D} for any dense set D is sqrt(7). Maybe they are just asking you to prove this without proving sqrt(7) is irrational?
     
  5. Mar 11, 2012 #4
    I have the same puzzlement as you, so your confusion is reasonable. However, there's an important piece of information missing. When you wrote

    {x | 1 < x < √(7) and x is irrational}

    you did not specify the set that x is a member of. That's an important part of this problem. As you noted, if x is constrained to be rational, then we'll need to use the density of the rationals in the reals to solve this problem.

    There's another reason to always include the set that x is part of when you're using set specification notation. That's because if you don't constrain x, you can get famous set theory paradoxes. So it's not only a notational formality, it's a logical necessity. That's kind of pedantic point; but in general it's helpful to get in the habit of always writing

    {x in X | etc. }

    And in this case, it's an important clue to the problem.
     
  6. Mar 11, 2012 #5

    Dick

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    I would say {x | 1 < x < √(7) and x is irrational} is the same as saying {x is irrational | 1 < x < √(7)}. I think that is overly pedantic. I'll repeat that if you haven't shown that sqrt(7) is irrational, then the density of the irrationals may be helpful.
     
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