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

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 quasi Posts: 9,516 Registered: 7/15/05
Re: mediant of two integers
Posted: Aug 18, 2013 5:20 PM

Joshua Lindman wrote:
>quasi wrote:
>

>> 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

>
>I believe this should work when product(y_i) > 0
>(i.e., all pos. or all neg.)

For n > 2, prod(y_i) > 0 does not force all pos or all neg.

Thus, it's possible to have prod(y_i) > 0 (and no denominators
equal to zero) but for the Baseball Inequality to fail.

For example:

x_1, x_2, x_3 = 1, 1, 1

y_1, y_2, y_3 = -1, - 1, 1

Then prod(y_i) > 0 and

x_1/y_1 <= x_2/y_2 <= x_3/y_3

but

x_1/y_1 > sum(x)/sum(y)

violating the conclusion of the Baseball Inequality.

Of course, in the case where the y_i _are_ all negative, the
Baseball Inequality does hold since, in that case, the sequences

x_1, x_2, ..., x_n
y_1, y_2, ..., y_n

can be replaced by the sequences

-x_1, -x_2, ..., -x_m
-y_1, -y_2, ..., -y_n

The fractions are the same for both sequences, and for the new
sequences, all y-values are positive.

>> such that
>>
>> x_1/y_1 <= x_2/y_2 <= ... <= x_n/y_n
>>
>> then
>>
>> x_1/y_1 <= sum(x)/sum(y) <= x_n/y_n
>>
>> where sum(x) and sum(y) denote x_1 + x_2 + ... + x_n
>>
>> and y_1 + y_2 + ... + y_n, respectively.
>>
>> Moreover, the inequality is strict on both sides unless
>>
>> x_1/y_1 = x_2/y_2 = ... = x_n/y_n
>>
>> in which case the inequality becomes an equality.

quasi

Date Subject Author
8/17/13 i.love.jeevitha@gmail.com
8/17/13 Virgil
8/17/13 William Elliot
8/17/13 i.love.jeevitha@gmail.com
8/18/13 William Elliot
8/18/13 i.love.jeevitha@gmail.com
8/18/13 William Elliot
8/23/13 i.love.jeevitha@gmail.com
8/18/13 quasi
8/18/13 i.love.jeevitha@gmail.com
8/18/13 quasi
8/18/13 Virgil
8/23/13 Karl-Olav Nyberg