## Thursday, February 14, 2013

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

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....

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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 + .........