Virgil
Posts:
6,972
Registered:
1/6/11


Re: mediant of two integers
Posted:
Aug 18, 2013 5:21 PM


In article <a928864434144df19a6f809c8eaa139d@googlegroups.com>, Joshua Lindman <i.love.jeevitha@gmail.com> wrote:
> On Sunday, August 18, 2013 1:56:58 AM UTC4, quasi wrote: > > Joshua Lindman wrote: > > > > > > > > > >What is the most general form of the median inequality? > > > > > > > > > >Let a/b < c/d (rationals, where a,b,c,d integers) > > > > > > > > > >Then, > > > > > > > > > >a/b < (a+c)/(b+d) < c/d > > > > > > > > > > > > > > >I don't see any restrictions on a,b,c,d (except that b,d nonzero) > > > > >in Wikipedia: > > > > >http://en.wikipedia.org/wiki/Mediant_%28mathematics%29 > > > > > > > > No check the link again. > > > > > > > > Firstly, they choose the letters in a different order. > > > > > > > > They use the fractions a/c, b/d, and for the inequality they > > > > impose the restrictions c > 0 and d > 0. > > > > > > > > For your choice of fractions a/b, c/d, the corresponding > > > > restrictions are b > 0 and d > 0. > > > > > > > > Here's a more general version. > > > > > > > > I call it "The Baseball Inequality". > > > > > > > > The proof is straightforward. > > > > > > > > Given a sequence x_1, x_2, ..., x_n of real numbers and a > > > > sequence y_1, y_2, ..., y_n of positive real numbers such > > > > I believe this should work when product(y_i) > 0 (i.e., all pos. or all neg.) > That product will be positive if any even number of the y_i are negative so that does not require that they all be all positive or all negative for n > 2. 

