First uncovered by Kurt Densel in the late 1800s, the p-adics are a specialized class of number whose key aspect is their absolute value. This depends on the prime number for which any given 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 (êp ) is simply 1. If we are looking at a p-adic absolute value of zero, the result is always zero. (E.g. [0]p =ê0 êp = 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:
[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 , and [24/25]2
The first is pretty easy, since: [4]2 = [2 x 2]2 = 1/4
The next isn't too terribly difficult either:
The first is pretty easy, since: [4]2 = [2 x 2]2 = 1/4
The next isn't too terribly difficult either:
[1/6]2 = [1]2/ [3 x 2]2 = 1/(1/2) = 2
And:
And:
[1/8]3 = [1]3/ [8]3 = 1/1 = 1
In this case, since the denominator (8) has no 3-factors, it must follow 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)
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 (= ¼). 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 the 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, e.g.:
5S = 5 + (5)2 + (5)3 + (5)4 + (5)5 + (5)6 + .........
Subtract term by term to get: S - 5S = 1 (all other terms above and below cancel out!)
So: -4S = 1 and S = - ¼
In other words, the sum S is less than 1 in the p-adic venue, 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!
Suggested Problems:
1. For the triangle shown in Fig. 2 below:
In this case, since the denominator (8) has no 3-factors, it must follow 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 below:
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 obtain the end computations based on the previous examples):
for A: [0 - 4]2 = 1/4for 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 (= ¼). 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 the 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, e.g.:
S = 1 + 5 + (5)2 + (5)3 + (5)4 + (5)5 + (5)6 + .........
5S = 5 + (5)2 + (5)3 + (5)4 + (5)5 + (5)6 + .........
Subtract term by term to get: S - 5S = 1 (all other terms above and below cancel out!)
So: -4S = 1 and S = - ¼
In other words, the sum S is less than 1 in the p-adic venue, 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!
Suggested Problems:
1. For the triangle shown in Fig. 2 below:
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:
3. Using the series in (2) as written, "invent" a new irrational number based on the p-adic form.
2. Find the value of the sum S for:
S = 1 + 7 + (7)2 + (7)3 + (7)4 + (7)5 + ....
3. Using the series in (2) as written, "invent" a new irrational number based on the p-adic form.
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