Results 1 to 5 of 5

Math Help - Rational number

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    29

    Rational number

    I need help proving that between any two real numbers there exists a rational number.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by spikedpunch View Post
    I need help proving that between any two real numbers there exists a rational number.

    r,s \in \mathbb{Q} \Longrightarrow x:=\frac{r+s}{2} \in \mathbb{Q}\,,\,\,so...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,259
    Thanks
    1798
    Tonio's proof is that between any two rational numbers, there exist an irrational number. If x is NOT rational, It's a little more complicated.

    Suppose that r and s are the given real numbers, with r< s. Again, look at t= (r+s)/2. That is a real number between r and s but may not be rational. Let \epsilon= t-r= (r+s)/2- 2r/2= (s- r)/2. There exist an increasing sequence of rational number converging to any real number (The sequence got by truncating the decimal expansion of the number at the n^{th} decimal place is such a sequence.) so there exist an increasing sequence of rational numbers, \{a_n\} converging to t. Since it converges to t, there exist N such that if n> N |a_n- t|< \epsilon. a_n for n> N is then larger than t- (s- r)/2= (s+r)/2- (s- r)/2= r and is less than t< s because it is an increasing sequence.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by HallsofIvy View Post
    Tonio's proof is that between any two rational numbers, there exist an irrational number. If x is NOT rational, It's a little more complicated.

    \color{red}\mbox{No. Tonio's proof is that between the two rational numbers r,s there exists} \color{red}\mbox{ another RATIONAL number which I denoted by x}

    \color{red}\mbox{This in fact is the easy part. To prove there exists an irrational} \color{red}\mbox{between those two is harder}

    \color{red}Tonio


    Suppose that r and s are the given real numbers, with r< s. Again, look at t= (r+s)/2. That is a real number between r and s but may not be rational. Let \epsilon= t-r= (r+s)/2- 2r/2= (s- r)/2. There exist an increasing sequence of rational number converging to any real number (The sequence got by truncating the decimal expansion of the number at the n^{th} decimal place is such a sequence.) so there exist an increasing sequence of rational numbers, \{a_n\} converging to t. Since it converges to t, there exist N such that if n> N |a_n- t|< \epsilon. a_n for n> N is then larger than t- (s- r)/2= (s+r)/2- (s- r)/2= r and is less than t< s because it is an increasing sequence.
    .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    68
    All tonio has shown is that between two rational numbers there exists a rational number; this does not imply that between two real numbers there exists a rational number.

    More precisely, we want to show that between two distinct real numbers there exists a rational number. Let x,y\in\mathbb R with x<y. Then 0<y-x \implies \exists\,n\in\mathbb N with \frac1n<y-x by the Archimedean principle. \therefore\ 1<ny-nx. Let m=\lfloor nx\rfloor (so m is the largest integer such that m\leqslant nx). Then nx<m+1 and also m+1<ny (otherwise m+1\geqslant ny \implies 1\geqslant ny-m\geqslant ny-nx). Hence

    x\ <\ \frac{m+1}n\ <\ y

    and \frac{m+1}n\in\mathbb Q.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Is this a rational number?
    Posted in the Algebra Forum
    Replies: 7
    Last Post: October 6th 2011, 09:40 AM
  2. Replies: 4
    Last Post: April 28th 2011, 07:20 AM
  3. Is the following a rational number?
    Posted in the Advanced Algebra Forum
    Replies: 12
    Last Post: December 6th 2009, 02:38 PM
  4. No rational number
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 23rd 2009, 01:20 PM
  5. Replies: 5
    Last Post: October 7th 2008, 01:55 PM

Search Tags


/mathhelpforum @mathhelpforum