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Topic: mediant of two integers
Replies: 12   Last Post: Aug 23, 2013 2:57 AM

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Virgil

Posts: 8,833
Registered: 1/6/11
Re: mediant of two integers
Posted: Aug 18, 2013 5:21 PM
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In article <a9288644-3414-4df1-9a6f-809c8eaa139d@googlegroups.com>,
Joshua Lindman <i.love.jeevitha@gmail.com> wrote:

> On Sunday, August 18, 2013 1:56:58 AM UTC-4, 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.
--





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