168(ii) d(a, b) = d(b, a);(iii) d(a, c) 2 d(a, b) + d(b, c).- Approximation and Location of Zeros
The third property is the triangle inequality; roughly speaking, it says that
two points which are close to a given point cannot be too far from each
other.
There is an analogue of this idea of distance and closeness that we can
formulate for integers. Let p be a fixed prime. Beginning with the idea that
0 is the only integer which is divisible by every prime power pk, we regard
an integer as being “close” to zero if it is divisible by a high power of p.
Accordingly, we define, for the integer n,
InIp = P-’
if p” is the highest power of p which divides n.
Verify that, in the case p = 3, 1171s = 1, 11813 = l/9, 19721s = l/243.
What is 1125012? 112501s? 1125Olr?
We now define the p-adic distance between the two integers m and n by
dp(m, n) = Im - nip.Verify that dP satisfies the properties (i), (ii), (iii) listed above.
Suppose we seek an integer root of the polynomial equation f(x) = 0
over 2. Then we want a succession of approximations u which will make
If(~ smaller and smaller, so that the p-adic distance from u to the desired
solution will tend to 0. Following Newton’s lead, we use the approximation
u to obtain the approximation v = u - f(~)/f’(~).
However, there is a problem with this. All the quantities we deal with
are supposed to be integers, and v will not in general be an integer. The
problem is that the reciprocal of an integer is not an integer, so we need
some way of replacing the quantity l/f’(~) by something which makes sense
in this context. Review the Exploration on Hensel’s Lemma, and explain
how this difficulty is surmounted. Explain why the choice of v ensures that
If(v)lp will be smaller than If(u)
E.45. Continued fractions: Lagrange’s Method of Approximation.
Suppose we know that a polynomial p(t) vanishes for some value r of t
between the integers a and a + 1. Then r = a + (l/s) where s > 1. If in
the equation p(r) = 0 we make the substitution r = a + (l/s), we get a
polynomial equation in s: q(s) = 0. Let b be an integer for which the root
s lies between b and b + 1. Write s = b + (l/u), and continue the process,
solving a polynomial equation for U, placing ‘u between consecutive integers,
and so on. We finally end up with