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Apostol's Analysis 1.11- Irrational Numbers Proof

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Given any real x > 0, prove that there is an irrational number between 0 and x.

    2. Relevant equations
    I'm not sure if the concepts of supremums or upper bounds can used.

    3. The attempt at a solution
    Take an irrational number say Pi. We can always choose a number n such that n*Pi is between 0 and x. Thus for every real x > 0, there is at least one irrational number between 0 and x.
    Is this correct? If so, what other methods are there?
     
  2. jcsd
  3. Apr 5, 2009 #2

    lanedance

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    hi cordyceps

    I'm fairly new to this suff, but I'm not 100% that argument is bulletproof yet, you might just need to elaborate on what n is and how to choose it... ie are there conditions on n such that n*Pi is irrantional? n being rational springs to mind...

    another way to do it could be to show the set (0,x) is uncountable, so must contain more numbers than just the rationals contained in (0,x)... or start the other way and assume (0,x) ains only rationals and look for a contradiction...
     
  4. Apr 5, 2009 #3
    Great idea! I just remembered a little of dedekind on irrational numbers, but his proof that the set of rational numbers is uncountable is pretty long. Do you know a shorter method?
    And where would you start for the contradiction method?

    Thanks for the reply.
     
  5. Apr 5, 2009 #4
    I looked in Apostol, and he has covered supremums prior to the statement of this problem. So, if the ideas are applicable, you should be allowed to use them.

    Like lanedance said, you need to be careful what your variables are. Is n an integer, rational, or irrational number? These problems require you to be very explicit about where your variables are coming from.

    The rational numbers are countable! I think you are referring to Dedekind cuts, but these are used to construct the irrational numbers from the rational numbers. The proof that irrational numbers are uncountable uses another method.

    For a good reference problem, check out http://books.google.com/books?id=7J...euIAP&sa=X&oi=book_result&ct=result&resnum=6" from Calculus by Michael Spivak.
     
    Last edited by a moderator: Apr 24, 2017
  6. Apr 5, 2009 #5
    Assuming you already know that the rationals are dense in the reals, let x be any positive real number and q any positive rational number such that 4q < x. Now we have 0 < pi < 4 and multiplying both sides by q, you have 0 < pi*q < 4q < x, implying that 0 < pi*q < x.

    Now you can prove that any rational multiplied by any irrational is an irrational number, and this gives you the result you want.
     
  7. Apr 5, 2009 #6

    epenguin

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    Not my field but please tell me if anything is wrong with this because it looks to me some of the above is unnecessarily complicated.

    If x is irrational it is easy to prove that x/2 is irrational, and it is between 0 and x. I'm sure you can think of many other numbers that would do just as well.

    If x is rational it is easy to prove [tex]x.\sqrt2/2[/tex] is irrational (assuming already proved that [tex]\sqrt2[/tex] is) and this is between 0 and x.

    Edit. JG89's argument is the same idea. If you have a proof that pi is irrational.
     
    Last edited: Apr 6, 2009
  8. Apr 5, 2009 #7
    @ epenguin, I like your solution better as it doesn't assume the OP knows the rationals are dense.
     
  9. Apr 5, 2009 #8
    If you do know that the rationals are dense, then you know that there is a rational r between any two real numbers, say a and b, so a<r<b. Consider a number i such that
    a+[tex]\sqrt{2}[/tex]<i<b+[tex]\sqrt{2}[/tex]... you see where this is going, right?
     
  10. Apr 6, 2009 #9

    lanedance

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    would another way be to jut set up one to one relation between (0,1) or R, and (0,x)?

    Then if you know (0,1) or R, contains uncountably many irrationals, so must (0,x)?
     
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