Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Should be pretty simple proof

  1. Sep 3, 2009 #1
    ...but I feel like I did something wrong. Also, this was a problem in my Analysis book, hence my posting it here, although it doesn't explicitly deal with analysis.

    Prove that root(ab)=(a+b)/2 implies a=b. Assume 0[tex]\leq[/tex]a[tex]\leq[/tex]b. To prove the converse is true was another problem and was easy but anyway here's my work:

    Proof.

    Assume the contrary; that given root(ab)=(a+b)/2, a[tex]\neq[/tex]b.

    By the first multiplicative identity, 2*root(ab)=(a+b).

    Squaring both sides: 4ab=a2+2ab+b2

    By the first multiplicative identity and algebra, a2-2ab+b2=0.

    Factor: (a-b)(a-b)=0.

    Since a[tex]\neq[/tex]b by assumptions, a-b[tex]\neq[/tex]0 thus we can divide both sides by a-b: (a-b)=0. This implies a=b, which contradicts the original assumptions. Thus root(ab)=(a+b)/2 implies a=b.

    qed

    So is this correct? If not, where did I go wrong? I just have this feeling that it's a little...off...somewhere but I don't know how or where. Thanks in advance for the help, this has been bothering me for a bit now.
     
  2. jcsd
  3. Sep 3, 2009 #2
    This looks fine and is the equality condition in the 2-variable AM-GM inequality. Anyways, it is an exercise in basic analysis to show that if a and b are real numbers and ab = 0, then either a or b is 0. This might be the reason why you feel the proof is a bit weird, but it looks correct to me.
     
  4. Sep 8, 2009 #3
    it doesn't look like you need to prove by contradiction. just following the steps shows explicitly that a must equal b. then shouldn't need to assume 0=<a=<b (which forces a and b to be positive, both can be negative as well) ...otherwise it looks good.
     
  5. Sep 9, 2009 #4
    Hello Thread
    This is the first time that I participate this site and I have the following proof for your problem:
    (a-b)^2=a^2-2ab+b^2=a^2+2ab+b^2-4ab=(a+b)^2-4ab
    =4[((a+b)/2)^2-ab]
    =4([(a+b)/2)-sqrt{ab})([(a+b)[(a+b)/2)+sqrt{ab})
    =0 since (a+b)/2)=sqrt{ab}
    therefore (a-b)^2=0 implies a-b=0 and a=b Q.E.D
    Best Wishes
    Riad Zaidan
     
  6. Sep 9, 2009 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Why is (a+b)/2 equal to sqrt{ab}?

    The geometric mean of two numbers are hardly ever equal to the arithmetic mean..
     
  7. Sep 9, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Arildno, read the original post! The problem was to prove "If (a+b)/2= sqrt(ab), then a= b".
     
  8. Sep 9, 2009 #7

    lurflurf

    User Avatar
    Homework Helper

    let h=(b-a)/2
    so 0<=h
    sqrt(ab)=(a+b)/2
    sqrt(a^2+2ah)=a+h
     
  9. Sep 10, 2009 #8
    Dear arildno
    the problem was to prove that
    "If (a+b)/2= sqrt(ab), then a= b".So this is given
    Best Regards
     
  10. Sep 13, 2009 #9

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Ooops! Sorry about that..
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Should be pretty simple proof
  1. Simple proof (Replies: 12)

Loading...