p-adic Expansions

In Q2, a 2-adic expansion of Q, we found that

-1 = 1 + 2 + 4 + ...

which exactly means that the series ∑2-k converges in Q2 to -1. It makes no sense in R. As with the ordinary decimal expansions, one ought to exercise caution describing the right-hand side as "the sum of an infinite number of terms." It is just a limit in a well defined sense, namely in the metric of Q2. This said, the identity is strange, to say the least. An 1938 article opened with a similar (p-adic) expansion

One cannot blame a respectable mathematician for looking twice at the equation

 -1= 4 + 4·5 + 4·5² + 4·5³ + ...

However, if we add 1 to both sides of this equation, we have

 0= 5 + 4·5 + 4·5² + 4·5³ + ...
 0= 0 + 5·5 + 4·5² + 4·5³ + ...
 0= 0 + 0 + 5·5² + 4·5³ + ...
 0= 0 + 0 + 0 + 5·5³ + ...
 0= 0 + 0 + 0 + 0 + ...

with 0's as far out as we care to carry it.

This expansion designates a limit in Q5. The article continues,

it is obvious that the statement

 -1= 4 + 4·5 + 4·5² + 4·5³ + ...

is absurd if ordinary convergence is intended. The whole point to Hensel's theory is that this is not ordinary convergence, but a new type of convergence which, from the point of view of abstract algebra, is equally worthy of the name.

40 years after Hensel's discovery, the p-adic numbers have been still considered a novelty! A vindication came half a century later. Hensel's p-adic numbers played an important role in the developments that culminated with Andrew Wiles' proof of Fermat's Last Theorem. As F. Gouvêa put it, These are not so much actors in the play as they are part of the stage set: tools to allow the actors to do their job. We'll have a relatively simple but, nonetheless, a very surprising application on a separate page.

A more general p-adic expansion is an expression in the form

a0 + a1p + a2p² + a3p³ + ...,

with the digits ak satisfying the common restriction imposed on the residues modulo p:

0 ≤ ak < p.

Such expansions are called canonical. Canonical expansions are easily shown to be convergent in Qp, making it meaningful to assign them a value:

(1) α = a0 + a1p + a2p² + a3p³ + ...,

As a matter of fact, any element α∈Qp is a limit of a little more general expansion that starts with a (finite) negative index:

(2) α = a-np-n + ... + a1p + a2p² + a3p³ + ...

As an aside, members of Qp whose p-adic expansion starts with a zero index are exactly p-adic integers. It is interesting to compare the common decimal expansion of real numbers with the expansion p-adic numbers. The decimal expansion has the form

α = an10n + ... + a1101 + a0 + a-110-1 + a-2p-2 + ...

where the "short" part an10n + ... + a1101 + a0 represents the integer part of α and the rest its fractional part. For the p-adic number, the situation is reversed: the short part a-np-n + ... + a-1p-1 represents the fractional part of &alpha, and the infinite part to the right its integer part in the "p-adic" sense. For this reason, p-adic expansions are sometimes written right to left and, when the prime p is fixed, are shortened, for example, to ... a2a1a0a-1.

Given an expansion (1), observe that for all k > 0,

α  ≡  a0 + a1p + a2p² + ... + ak-1pk-1 (mod pk)

because all the remaining terms are divisible by pk. In particular,

α  ≡  a0 (mod p),
α  ≡  a0 + a1p (mod p2),
etc.

Let's have an example, say α = 2/3 and p = 5. Note that |2/3|5 = 1 so that 2/3 is a 5-adic integer and, as such, has an expansion that starts with a0. 0 ≤ a0 < 5 satisfies a0 ≡ 2/3 (mod 5). Multiply by 3: 3a0 ≡ 2 (mod 5). The only residue of division by 5 that solves this is a0 = 4.

To continue, (2/3 - 4) ≡ 5a1 (mod 5²), or -10/3 ≡ 5a1 (mod 5²). This simplifies to -2/3 ≡ a1 (mod 5) and is solved by a1 = 1. For a2, we have the equation:

2/3 ≡ 4 + 1·5 + 25·a2 (mod 5sup3;) -1/3 ≡ a2 (mod 5),

so that a2 = 3. -1/3 - 3 = -10/3. Dividing by 5 gives

-2/3 ≡ a3 (mod 5)

implying a3 = 1, etc. We see that

(3) 2/3 = 4 + 1·5 + 3·52 + 1·53 + 3·54 + 1·55 + ...

with a periodic sequence of coefficients. To check that our efforts have paid off, let's multiply this by 3:

2 = 12 + 3·5 + 9·52 + 3·12 + 9·53 + 3·54 + ...

But 12 = 2 + 2·5, so that 12 + 3·5 = 2 + 5·5 = 2 + 1·5² giving a 1 to carry to the next power of 5, 5³. Since the coefficient by 5³ is 9, the term becomes 10·5³ = 2·54, carrying 2 to the next term 3·55, etc. We see that the carry causes all the terms vanish, except for the first one, 2. Thus we do have an identity

3·2/3 = 2

which validates (3). In particular, 2/3 is a 5-adic integer!

There is another way to verify (3). Move 4 to the left and combine pairs of successive terms so as to get a geometric series:

2/3 - 4 = -10/3 = 16·5·(1 + 52 + 54 + ...).

For a convergent geometric series 1 + q + q² + ... we have the formula

1 + q + q² + ... = 1 / (1 - q).

Since the series 1 + 52 + 54 + ... is convergent in Q5,

1 + 52 + 54 + ... = 1 / (1 - 25) = -1/24.

And, as expected -80/24 = -10/3.

References

  1. E. B. Berger, Exploring the Number Jungle, AMS, 2000
  2. J. R. Goldman, The Queen of Mathematics, A K Peters, 1998
  3. F. Q. Gouvêa, Local and Global in Number Theory, in The Princeton Companion to Mathematics T. Gowers (ed.), Princeton University Press, 2008
  4. F. Q. Gouvêa, A Marvelous Proof, The American Mathematical Monthly, Vol. 101, No. 3 (Mar., 1994), pp. 203-222
  5. C. C. MacDuffy, p-adic Numbers of Hensel, The American Mathematical Monthly, Vol. 45, No. 8 (Oct., 1938), pp. 500-508

|Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny

Strange as it may appear, we'll write

q = 1 + p + ... + pm + ...

As we just said, q is well defined as a limit of sequence {xk} and this is how the identity should be understood. All arithmetic operations are permitted to carry over the limits so that

 q= 1 + p + ... + pm + ...
 pq= p + p² + ... + pm+1 + ...
 pq + 1= 1 + p + ... + pm + ... = q

So that pq + 1 = q, implying q = 1 / (1 - p). Taking p = 2 we obtain:

-1 = 1 + 2 + 4 + ...

One certainly needs time and effort to get used to such weirdness.

|Contact| |Front page| |Contents| |Up| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62041612

Search by google: