1) For the triangle in Fig. 2, use 7-adic absolute values applied to the sides of the triangle, thereby compute: A, B and C and show it is isosceles
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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 + ....
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S - 7S = 1 (All other top and bottom terms cancel)
Whence:
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.
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.
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