Need help for an algebra problem - not for school.

[shamanking]

a_1, a_2, a_3, ... , a_2n are whole numbers. (I am going to write a_i as a i.)
|a 1 - a 2| = |a 2 - a 3| = ... = |a 2n-1 - a 2n| = |a 2n - a 1| = 1
If a i > a i-1 and a i > a i+1, than a i is called a ''big number''. If a i is smaller than its neighbors, it is called a "small number".
Prove that the sum of the "big numbers" minus the sum of the "small numbers" equals to n.

 

[JeffM]

a_1, a_2, a_3, ... , a_2n are whole numbers. (I am going to write a_i as a i.)
|a 1 - a 2| = |a 2 - a 3| = ... = |a 2n-1 - a 2n| = |a 2n - a 1| = 1
If a i > a i-1 and a i > a i+1, than a i is called a ''big number''. If a i is smaller than its neighbors, it is called a "small number".
Prove that the sum of the "big numbers" minus the sum of the "small numbers" equals to n.

The simplest possible case is n = 2.

Consider

\(\displaystyle b > 0\ and\ a_1 + b = a_2\ and\ a_2 + b = a_3\ and\ a_3 + b = a_4\implies\)

\(\displaystyle |a_1 - a_2| = |- b| = b\ and\ |a_2 - a_3| = |-b|= b\ and\ |a_3 - a_4| =|- b| = b.\)

\(\displaystyle Also\ b > 0\ and\ a_1 + b = a_2\ and\ a_2 + b = a_3\ and\ a_3 + b = a_4\implies\)

\(\displaystyle a_1 < a_2 < a_3 < a_4 \implies\ sum\ of\ big\ numbers = 0 = sum\ of\ small\ numbers.\)

\(\displaystyle 0 - 0 = 0 \ne 2.\)

The theorem is false, which will make proving it rather difficult.