Results 1 to 6 of 6

Math Help - proving an inequality

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    34

    proving an inequality

    Ok at first glance this seemed pretty easy, but I'm having a lot trouble with it. Let a,b be positive integers. Show that \sqrt{2} always lies between the two fractions a/b and (a+2b)/(a+b). Which fraction is closer to \sqrt{2}. Any help or hint would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello bleys

    Welcome to Math Help Forum!
    Quote Originally Posted by bleys View Post
    Ok at first glance this seemed pretty easy, but I'm having a lot trouble with it. Let a,b be positive integers. Show that \sqrt{2} always lies between the two fractions a/b and (a+2b)/(a+b). Which fraction is closer to \sqrt{2}. Any help or hint would be appreciated.
    Here's a start. Suppose first that \frac{a}{b}<\sqrt2

    Then since b>0,\; a<\sqrt2b

    \Rightarrow a+b < (1+\sqrt2)b

    \Rightarrow \frac{a+2b}{a+b}=1+\frac{b}{a+b}>1+\frac{b}{(\sqrt  2+1)b}=\frac{\sqrt2+2}{\sqrt2+1}=\frac{\sqrt2(1+\s  qrt2)}{1+\sqrt2}=\sqrt2

    So \frac{a}{b}<\sqrt2 < \frac{a+2b}{a+b}

    Do you want to see if you can take it from here?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
    34
    Welcome to Math Help Forum!
    Thanks!

    Oh alright, I need to learn how to manipulate surds better...
    The case where \frac{a}{b}>\sqrt2 just requires the inequality signs switched.
    Now, recover the first case where \frac{a}{b}<\sqrt2 < \frac{a+2b}{a+b}
    Suppose if possible that  |\frac{a+2b}{a+b} - \sqrt2 | < \frac{a}{b} - \sqrt2
    Then  \frac{a+2b}{a+b} < \frac{a}{b} , a contradiction. Then  |\frac{a+2b}{a+b} - \sqrt2 | > |\frac{a}{b} - \sqrt2|
    A similar argument can be applied to the other case.
    Is there a direct way to show this? I find using contradiction to be a little like cheating in this problem . The reason I didn't was that I wasn't sure my inequalities were valid when trying to recover the absolute value sign.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello bleys

    For the last part, can't we consider the differences between \left(\frac{a}{b}\right)^2 and 2, and \left(\frac{a+2b}{a+b}\right)^2 and 2?

    Which are:
    \left|\left(\frac{a}{b}\right)^2 - 2\right|=\left|\frac{a^2-2b^2}{b^2}\right|
    and
    \left|\left(\frac{a+2b}{a+b}\right)^2-2\right| =\left|\frac{(a+2b)^2-2(a+b)^2}{(a+b)^2}\right|
    =\left|\frac{2b^2-a^2}{(a+b)^2}\right|
    Clearly the second of these expressions is less than the first. So \frac{a+2b}{a+b} is closer to \sqrt2. (Incidentally, doesn't the fact that a^2-2b^2 and 2b^2-a^2 have opposite signs prove the first part of the question as well?)

    Grandad
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2009
    Posts
    34
    Why are you allowed to use squares? How do you know whether the distances are still preserved?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,790
    Thanks
    1687
    Awards
    1
    Quote Originally Posted by bleys View Post
    Why are you allowed to use squares? How do you know whether the distances are still preserved?
    |a|\le|b|\text{ if and only if }a^2\le b^2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. proving an inequality
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 6th 2011, 02:44 PM
  2. Replies: 3
    Last Post: December 12th 2010, 01:16 PM
  3. Proving Inequality
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: June 3rd 2010, 04:03 AM
  4. Proving inequality with a, b
    Posted in the Algebra Forum
    Replies: 5
    Last Post: January 9th 2010, 11:28 AM
  5. Proving inequality
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 16th 2009, 08:03 AM

Search Tags


/mathhelpforum @mathhelpforum