Solutions to P-Adics Problems

 The  problems again:

1) For the triangle in Fig. 2 (July 8 post), use 7-adic absolute values applied to the sides of the triangle, thereby compute: A, B and C and show it is isosceles

Solution:

[A ] ₇   =  [14 - 1/7] ₇   =  abs {[2 x 7 ] ₇   -  1/[1 x 7] ₇ }

=   abs [1/7 - 7]  =   6 6/7   =  48/7

[B ] ₇   =  [1/7 -  21] ₇   =   abs {1/ [1 x 7 ] ₇   -  [3 x 7] ₇ }

=   abs [1/7 - 7]  =   6 6/7   =  48/7

[C ] ₇   =  [21  -   14] ₇   =  abs {[3 x 7 ] ₇   -  [2 x 7] ₇ }

= abs[1/7 - 1/7]  ₇ =    [0 ] ₇   = 1

From these calculations of  (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 multiplying both sides by 7:

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

Subtract the last form from the previous one:

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

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

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

Whence:

-6S = 1 and therefore, S = -1/6

3)  Using (2) as written, "invent" a new irrational number based on the p-adic form.

The most direct way to invent a new p-adic is to employ the same 7-adic form and simply increase powers in a slightly different way..

Then let the new irrational be S'  where S' is a variation of S such that:

S'  =   (7) ³   + (7) ⁶  + (7) ⁹  +   + (7) ¹²  + (7) ¹⁵  +.....

Hence, the new irrational S'  is obtained merely by changing the sequence of the powers of 7.