m

NOTE. It should be remarked that it is not strictly true that (a) = (a) except with a limitation corresponding to that of Art. 164, or unless by the nth root of a quantity is meant orly the arithmetical root. For example, (a) has two values, namely a2, whereas (a)* has only the value +a".

IV. To find the meaning of a-", where m has any positive value.

167. We have in the preceding Article found that in order that the fundamental index law, am × aˆ = am+n, may always be obeyed, a" must have a definite meaning when n has any given positive or negative value. We have now to shew that, with the meanings thus obtained,

m+n

axa" = a+", (am)n

mn

= am, and (ab)" = a"b",

are true for all values of m and n. When these have been proved, the final result of Art. 159 is easily seen to be true in all cases.

m

I. To prove that a xa" = am+", for all values of m
and n.

We already know that this is true when m and n are
positive integers. Let m and n be

any positive fractions

Ρ

am × a" = aa × a3 = a* × √a", by definition

Thus the proposition is true for all positive values of
m and n. To shew that it is true also for negative values,
it is necessary and sufficient to prove that

n

a-mxa-n = a ̄m-n, and am × a ̄” = am−n,

m-n

a” × a ̃" × a" = am; therefore aTM =am xa ̄".

Hence am xa" = am+", for all values of m and n.

COR.

Since am-nx an
it follows that am÷a" = am-n.

mn

II. To prove that (am)" = am", for all values of m and n.

First, let n be a positive integer, m having any value
whatever.

Then (am)" = a” × am × am ×.......... to n factors,

Ρ

q

Next, let n be a positive fraction 2, where p and q are

positive integers.

P

Then (a")" = (a) = }/{(aTM)"},= {/(aTM”), since p is an
integer,

mp

= α2 = amn.

Finally, let n be negative, and equal to — p.

Hence for all values of m and n we have

(am)" = amn

III. To prove that (ab)"a"b", for all values of n.
We have proved in Art. 159 that (ab)" = a"b", where
n is a positive integer.

And, whatever m may be, provided that q is a positive
integer, we have

(ab)" = (ab)2 = '/(ab)" = (a"b"), since p is a positive integer.

Also (a"b")2=a"b", since q is a positive integer.

Thus (ab)" = a"b", for all positive values of n.

Finally, if n be negative, and equal to -m, we have

Ex. (iii). Simplify (a-21)-.

(a− 2 b3) − ₺ = a (− 2) ( − §) b‡ (−1) = a3b − 2 .

Ex. (iv). Simplify √(a ̄%b3c ̄‡)÷3/(a*bac−1)

a3

=

b2.

√(a ̄§b3c ̃3) ÷ ¥(a3b1c−1) = a ̄‡b‡ c ̄§ ÷ að b‡ c¬§

=abc-xa-b-ca-1b.

168. Rationalizing Factors. It is sometimes required to find an expression which when multiplied by a given irrational expression will give a rational product. The following are examples of rationalizing factors.

Since (a+b) (a - √b) = a2 - b, it follows that a±b is made rational by multiplying by ab.

So also abcd is made rational by multiplying by a√b+c√d.
Again from the known identity

2b2c2+2c2a2+2a2b2 - a4 - b4 — c4

=(a+b+c) ( − a+b+c) (a−b+c) (a + b − c ),

S. A.

14

it follows that the rationalizing factor of

√p +√q+√r is (−√p+√q+√r) (Jp−√q+√r) (Jp+√q − √r). The rationalizing factor of √p+√q+√r may also be found as

follows,

and

(√p+Ng+√r) (√p+√√q−√√r)=p+q-r+2√pq,

(p+q-r+2√pq) (p + q − r - 2√pq) = (p+q-r)2 - 4pq.

Thus the required rationalizing factor is

the rationalizing factor of a+b3 is seen to be a2 - ab3+b3⁄4.

169. To find the rationalizing factor of any binomial.

Let the expression to be rationalized be ax + bys.

Put X = ax, and Y=bys, and let n be the L.C.M. of q and s.

Then it is easily seen that X" and Y" are both rational.

Hence, from the identities

-2

(X + Y) {X”-1— X”¬2 Y+....+(−1)"−1 Y^-1}=X"+(−1)”−1Y" and (XY) (X”−1 +X"−2 Y +......+ Y~~1) = X" – Y",

-1

the rationalizing factors of X + Y and X - Y are seen to be respectively

X"-1 - X-2Y+......+ (− 1)"1Y"-1,