PASCAL’S TRIANGLE 163

and the method of extracting roots depends upon the expres-
sing of a number in the above form.

An approximate value of the cube root of 998 may be
found by this means:

998 =1000—2
=1000(I —+002)

therefore V998 =V 1000(1 —002)
—(10) (V'1—-002)
=10(1 —-002)}

We now expand the part of the expression (1 —-002)* where
x=—-002; and n=3%. The expression (1—-002)} by the
modified expansion, is approximately equivalent to:

1 —3%(-002)
=1—"007
=:9993
therefore - v/998 =10 X 9993
=9'993 (approx.)
Numbers may, of course, be raised to higher powers by
the same method:
9982=1000%(1 —-002)?
=10002%(1 —2 X '002)
=10002(-996)
=096,000
The degree of error is remarkably small, for the actual value
of 9982 is 996,004.
The principle of ignoring terms of an expansion if they are
too small to be of importance may also be employed to
evaluate expressions of the form:

(1+#) (1+9)
such as (1-002) (1-003)

for where both x and y are very small, their product xy will
be very much smaller still, so that:

(1+x) (1+y)=1+x-+y approx.

pERe s