84. _Roots and Surds._--The pth root of a number (S 43) may, if the
number is an integer, be found by expressing it in terms of its prime
factors; or, if it is not an integer, by expressing it as a fraction in
its lowest terms, and finding the pth roots of the numerator and of the
denominator separately. Thus to find the cube root of 1728, we write it
in the form 2^6 . 3^3, and find that its cube root is 2^2 . 3 = 12; or,
to find the cube root of 1.728, we write it as 1728/1000 = 216/125 =
(2^3 . 3^3)/5^3, and find that the cube root is (2.3)/5 = 1.2. Similarly
the cube root of 2197 is 13. But we cannot find any number whose cube is
2000.
It is, however, possible to find a number whose cube shall approximate
as closely as we please to 2000. Thus the cubes of 12.5 and of 12.6 are
respectively 1953.125 and 2000.376, so that the number whose cube
differs as little as possible from 2000 is somewhere between 12.5 and
12.6. Again the cube of 12.59 is 1995.616979, so that the number lies
between 12.59 and 12.60. We may therefore consider that there is some
number x whose cube is 2000, and we can find this number to any degree
of accuracy that we please.
A number of this kind is called a _surd_; the surd which is the pth root
of N is written [root p]N, but if the index is 2 it is usually omitted,
so that the square root of N is written [root]N.
85. _Surd as a Power._--We have seen (SS 43, 44) that, if we take the
successive powers of a number N, commencing with 1, they may be written
N^0, N^1, N^2, N^3, ..., the series of indices being the standard
series; and we have also seen (S 44) that multiplication of any two of
these numbers corresponds to addition of their indices. Hence we may
insert in the power-series numbers with fractional indices, provided
that the multiplication of these numbers follows the same law. The
number denoted by N^(1/3) will therefore be such that N^(1/3) X N^(1/3)
X N^(1/3) = N^(1/3 + 1/3 + 1/3) = N; i.e. it will be the cube root of N.
By analogy with the notation of fractional numbers, N^(2/3) will be
N^(1/3 + 1/3) = N^(1/3) X N^(1/3); and, generally, N^(p/q) will mean the
product of p numbers, the product of q of which is equal to N. Thus
N^(2/6) will not mean the _same_ as N^(1/3), but will mean the square of
N^(1/6); but this will be _equal_ to N^(1/3), i.e. ([root 6]N)^2 = [root
3]N.
86. _Multiplication and Division of Surds._--To add or subtract
fractional numbers, we must reduce them to a common denominator; and
similarly, to multiply or divide surds, we must express them as
power-numbers with the same index. Thus ([root 3]2) X ([root]5) =
2^(1/3) X 5^(1/2) = 2^(2/6) X 5^(3/6) = 4^(1/6) X 125^(1/6) = 500^(1/6)
= [root 6]500.
87. _Antilogarithms._--If we take a fixed number, e.g. 2, as base, and
take as indices the successive decimal numbers to any particular number
of places of decimals, we get a series of _antilogarithms_ of the
indices to this base. Thus, if we go to two places of decimals, we have
as the integral series the numbers 1, 2, 4, 8, ... which are the values
of 2^0, 2^1, 2^2, ... and we insert within this series the successive
powers of x, where x is such that x^100 = 2. We thus get the numbers
2^.01, 2^.02, 2^.03, ..., which are the antilogarithms of .01, .02, .03,
... to base 2; the first antilogarithm being 2^.00 = 1, which is thus
the antilogarithm of 0 to this (or any other) base. The series is formed
by successive multiplication, and any antilogarithm to a larger number
of decimal places is formed from it in the same way by multiplication.
If, for instance, we have found 2^(.31), then the value of 2^(.316) is
found from it by multiplying by the 6th power of the 1000th root of 2.
For practical purposes the number taken as base is 10; the convenience
of this being that the increase of the index by an integer means
multiplication by the corresponding power of 10, i.e. it means a
shifting of the decimal point. In the same way, by dividing by powers of
10 we may get negative indices.
88. _Logarithms._--If N is the antilogarithm of p to the base a, i.e. if
N = a^p, then p is called the logarithm of N to the base a, and is
written log_a N. As the table of antilogarithms is formed by successive
multiplications, so the logarithm of any given number is in theory
found by successive divisions. Thus, to find the logarithm of a number
to base 2, the number being greater than 1, we first divide repeatedly
by 2 until we get a number between 1 and 2; then divide repeatedly by
[root 10]2 until we get a number between 1 and [root 10]2; then divide
repeatedly by [root 100]2; and so on. If, for instance, we find that the
number is approximately equal to 2^3 X ([root 10]2)^5 X ([root 100]2)^7
X ([root 1000]2)^4, it may be written 2^(3.574), and its logarithm to
base 2 is 3.574.
For a further explanation of logarithms, and for an explanation of the
treatment of cases in which an antilogarithm is less than 1, see
LOGARITHM.
For practical purposes logarithms are usually calculated to base 10, so
that log10 10 = 1, log10 100 = 2, &c.