Results 1 to 2 of 2

Math Help - Proving inequality For every integer n > 0 there is some real number x > 0 s.t. x < 1

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    40

    Proving inequality For every integer n > 0 there is some real number x > 0 s.t. x < 1

    For every integer n > 0 there is some real number x > 0 s.t. x < 1/n..

    I need a proof. Contrapositive? I don't know what to do.

    Thanks.





    Another question, too:

    A fraternity has a rule for new members: each must always tell the truth or always lie. They know
    which does which. If I meet three of them on the street and they make the statements below,
    which ones (if any) should I believe?
    A says: \All three of us are liars."
    B says: \Exactly two of us are liars."
    C says: \The other two are liars."
    Last edited by qtpipi; February 12th 2009 at 11:03 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Maybe you want to say an irrational number x>0 because \frac{1}{n+1} is a real number such that 0<\frac{1}{n+1}<\frac{1}{n}

    Let y>0 be an irrational number. Then the sequence \left(\frac{y}{k}\right)_{k>0} decreases and has 0 for limit. So there is a m\in\mathbb{N} such that \frac{y}{m}<\frac{1}{n}.
    We just have to prove that \frac{y}{m} is irrational: if that wasn't the case, since m\mathbb{Q}=\mathbb{Q}, then m\times\frac{y}{m} would be a rational number, contradiction. Thus \frac{y}{m} is an irrational number, and since there exist positive irrational numbers, that's ok.

    An example: \forall n\in\mathbb{N}-\{0\},\ 0<\frac{\sqrt{2}}{2n}<\frac{1}{n}


    At first sight, I would believe B

    If A tells the truth, then he lies... Impossible
    If C tells the truth (which implies that B lies), then B is right... Impossible
    If B lies, then A tells the truth... Impossible
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: February 8th 2011, 12:04 PM
  2. Replies: 9
    Last Post: March 16th 2010, 09:38 PM
  3. Replies: 4
    Last Post: February 2nd 2010, 05:09 PM
  4. Replies: 8
    Last Post: April 7th 2009, 01:15 PM
  5. Replies: 2
    Last Post: August 6th 2008, 03:58 AM

Search Tags


/mathhelpforum @mathhelpforum