What are the p-adics? They're actually a specialized class of numbers first uncovered by Kurt Densel in the late 1800s. A key aspect is their absolute value, which depends on the prime number on which any p-adic is based. The p then means the particular prime. Given primes of 2, 3, 5, 7 - for example, one can find 2-adic, 3-adic, 5-adic and 7-adic absolute values which are always computed by taking the reciprocal of the highest multiple of p which divides into any given natural number, N.
If N has no multiples of p, then the absolute value is simply 1. If we are looking at a p-adic absolute value of zero, the result is always zero. (E.g. [0] p = 0)
If N has no multiples of p, then the absolute value is simply 1. If we are looking at a p-adic absolute value of zero, the result is always zero. (E.g. [0] p = 0)
Let's look at some examples before examining more elaborate applications. Consider the 3-adic versions of: 7, 5 and 1/3. What will the 3-adic absolute values be? Compute each in turn:
[7] 3 = 1 (since there are NO multiples of 3 to form the number N = 7)
[5] 3 = 1 (for the same reason, thus: [7] 3 = [5] 3 )
[1/3 ] 3 = [1] 3 / [3] 3 = 1/ (1/3) = 3
Since the reciprocal of 3 is 1/3 which we then divide into the numerator 1.
How about obtaining the p-adic absolute values for each of these?
[4] 2
[1/ 6 ] 2
[1/ 8 ] 3
[24 / 25 ] 2
The first is pretty easy, since: [4] 2 = [2 x 2] 2 = 1 /4
The next isn't terribly difficult either:
[1/ 6 ] 2 = [1 ] 2 / [3 x 2 ] 2 = 1/(1/2) = 2
and: [1/ 8 ] 3 = [1 ] 3 / [8 ] 3 = 1/ 1 = 1
Note in this case, since the denominator (8) has no 3-factors, it must be that [8 ] 3 = 1
Lastly:
[24 / 25 ] 2 = [3 x 2 x 2 x 2] 2 / [25] 2 = (1/8)/ 1 = 1/8
(Again, 25 has no multiples of 2 which can compose it, so [25 ] 2 = 1)
Even more intriguing are the spatial relations and differences, divergences between normal space and p-adic space. Consider the triangle (scalene) shown in Fig. 1 and the linear dimensions (absolute values) of its respective sides. We find: A = 4 ([4 - 0]); B = 6 ([10 - 4]); and C = 10 ([0 - 10]). Now compute the sides using 2-adic absolute values (I will assume the reader can perform the end computations based on the previous examples):
for A: [0 - 4] 2 = 1 / 4
for B: [4 - 10] 2 = [1/4 - 1/2] = 1/4
for C: [0 - 10] 2 = [10] 2 = 1/2
Amazingly, in the p-adic context we find the counter-intuitive result that side A equals side B. In other words, in this context, the triangle is found to be isosceles! A general rubric is that for any such computations of the p-adic absolute values contingent on a given triangle's sides - there will always be found an isosceles triangle - irrespective of how the triangle appears in normal space.
Even more bizarre results await when we examine apparently infinite series in the p-adic context. Thus, a series that first appears to go on to an infinitely large extent may be found much more different when p-adics enter. Consider the series given by a sum:
S = 1 + 5 + (5) 2 + (5) 3 + (5) 4 + (5) 5 + (5) 6 + .........
To treat S p-adically, multiply both sides by 5, then place the result under the original S and subtract:
->
S = 1 + 5 + (5) 2 + (5) 3 + (5) 4 + (5) 5 + (5) 6 + .+ .........
5 S = 5 + (5) 2 + (5) 3 + (5) 4 + (5) 5 + (5) 6 + .+ .........
______________________________________________
S - 5S = 1 (all other terms above and below cancel out!)
So: -4S = 1 and S = -1/4
In other words, the sum S is less than 1 in the p-adic form - totally counter-intuitive! We see that evidently the notion or concept of "closeness" emerges quite differently - certainly if we can turn an "infinite" (apparently) sum into one yielding a result less than one!
Problems for the Math whiz:
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.
2- Find the value of the sum S for: S = 1 + 7 + (7) 2 + (7) 3 + (7) 4 + (7) 5 + .......
S - 5S = 1 (all other terms above and below cancel out!)
So: -4S = 1 and S = -1/4
In other words, the sum S is less than 1 in the p-adic form - totally counter-intuitive! We see that evidently the notion or concept of "closeness" emerges quite differently - certainly if we can turn an "infinite" (apparently) sum into one yielding a result less than one!
Problems for the Math whiz:
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.
2- Find the value of the sum S for: S = 1 + 7 + (7) 2 + (7) 3 + (7) 4 + (7) 5 + .......
3- Using (2) as written, "invent" a new irrational number based on the p-adic form.
No comments:
Post a Comment