Solutions to p-adic Problems

These are the solutions to the problems at the end of the blog on p-adic numbers, http://www.brane-space.blogspot.com/2013/02/what-are-p-adic-numbers.html


1) Take the The 7-adic value for each side;


[A]₇ = [14 - 1/7]₇   = ç[2 x 7]₇ - 1/[1 x 7]₇ ç

=  ç 1/7 - 7 ç= 6 6/7 = 48/7


 

[B]₇ = [1/7 - 21]₇ = ç1/[1 x 7]₇ - [3 x 7]₇

= ç7 - 1/7 ç =  6  6/7 = 48/7


[C]₇ = [21 - 14]₇ =  ç 3 x 7] ₇ - [2 x 7]₇ ç=  ç1/7 - 1/7ç₇ = [0]₇ = 1

From these calculations of p-adic (7-adic) absolute values we find:


side A = side B = 48/7,

Hence in the p-adic format the triangle is isosceles.

(2) Find the value of the sum S for:

S = 1 + 7 + (7)² + (7)³ + (7)⁴ + (7)⁵ + ....

Rewrite the sum (multiplying both sides by 7):

7S = 7 +  (7)² + (7)³ + (7)⁴ + (7)⁵ +  (7)⁶....

Subtract the lower form from the upper, viz.:

S = 1 + 7 + (7)² + (7)³ + (7)⁴ + (7)⁵ +  (7)⁶....

-7S = 7 + (7)² + (7)³ + (7)⁴ + (7)⁵ +  (7)⁶....

--------------------------------------------

S - 7S = 1 (all other top and bottom terms cancel)

whence:


-6S = 1 and therefore, S = - 1/6


  (3) This is straightforward. Let N be the new irrational and equal to S' where S' is a variant sum derived from S, i.e. by excluding all even-exponent terms in (7),  such that:

N= S’ =

7 + (7)³+ (7)⁶ + (7)⁹ + (7)¹² + (7)¹⁵ + .........

WHY does this answer qualify?