ACTIVITY 4:
Rational Exponents and Radicals (Section P.4,
pp.31-35)
DEFINITION OF NTH ROOT:
If n is a positive integer, then the principal n-th root
of a is the number b that, when raised to the n-th
power, gives a; that is
n
a b
means
b a
n
If n is even we must have a, b > 0.
This last requirement seems a bit odd? So lets
suppose for the moment that n = 2 and a = -4.
We would like to consider  4
By the above
4 b
So this is why
“a” has to be
nonnegative
means that b2  4 , but this is impossible!!!
Now suppose that n = 2 and a = 4, then the definition above
says that 4  b if and only if b 2  4 .
Consequently, “b” could equal 2 or -2.
PROPERTIES OF NTH ROOTS:
1.n ab 
a
n
2.

b
n
n
a
b
n
3.m n a 
an b
m*n
4.n a n  a
5. a  a
n
n
a
If n is odd
If n is even
This is because, we always choose the
nonnegative answer, and in the number
5 one could choose a to be negative
EXAMPLE 1: SIMPLIFY THE EXPRESSIONS BELOW.
24 4 54  4 24 * 54
 4 1296
 4 2434
4
4
4
Property 1  2 4 3
Property 5  2 * 3
6
4
Property 1
1296
24
3
2
12
6
2
2
54
27 2
9
3
3
3
3
27


8
3
3

2
3
 3
 
 2
3
Property 4
a 2b 3 a 4b  3 a 2ba 4b
6 2
3
 ab
 3 a 6 3 b2
Property 1
Property 1
 aa b  a
 aa3 b2 Property 4
 a 2 3 b2
3
3 33
2
3
33
a
33
b
2 Property 1
2 3 2
24  8 
Property 1
 2
Distributive
law
 2
12
64x
3
2
Combination of
property 1 and 5
3
3
 64x
6 6 12
 2 x
2
 2x
6
12
3
3 2
3
 3  1
2  3  1
3
Property 3
4
x y z 
4
2 6
2 2
xz y z
DEFINITION OF RATIONAL EXPONENTS:
For any rational exponent m/n in lowest terms, where
m and n are integers and n > 0, we define
m
n
a  n am
If n is even we require that a > 0.
EXAMPLE 2:
Radical Expression
3
5
3
b 
4
Exponential Expression
2
4
3
5
1

3
b
1
b3
1
x
5
5

1
b3
2
3
b
x
3
5
3
5
2
2
EXAMPLE 3.
Simplify the expressions below and eliminate any
negative exponents. When needed assume that
all letters denote positive numbers
10000
3
2
 10
 10

10000
3
4  2
4 3
*
1 2
 10 6
1
 6
10
100
100
10
10
10
10
 2 x y

4
4
 8 y

3
5
2

2
3
 2 x   y

3
4 3
4
 8 y 2 2 3

3
5
3
3 2 3
3
3
2


5
5
 2 x   y  2 y   2 x   y  2   y 




4 3
3 2
2 2
12
4
* 
* 
* 




3
4*3
5
1
1
3
1
3
 2  y   23 x12  y 5 22  y 3 
 2 x  y

  





  
3
4 3
4
3
2
4 3
4
3
2
3
2
12 4
12 4
16
 
 




3 2 12
5 12
5 12
5
3
5
3
15
  2 x  y
  2 x  y 
 2 x  y







25 x12
y
16
15
 12 4  12 * 3  4 * 5  36  20
 16
 


5
3
5*3
15
15
a b 
 1 2 
x y 
2 3

3
 2 1 
2 3  3 3   2 1 
 x b  a  b   x b 
 3 1   1 3 2 3  3 1  
 a 2 y 3  x   y   a 2 y 3 




6 9
 2 1
ab x b
3
6
3
2
x ya y

9
2
1
3
a b 10 x 2 x 3
y
19
3

6  9  1  2
ab
3
x
x y

6
1
3
9
2
a
3
2
a b 10 x 23
y
19
3


3
2
a 6a b 10 x 2
3
x y

 2 1 
a b x b 
x 3 y 6  23 13 
a y 
19
3
9
2
a x
10
b y
19
3
6 1 6 * 3  1 * 1 18  1 19
 


1 3
1* 3
3
3
6 3 6 * 2  3 * 1 12  3
9
 


1 2
1* 2
2
2
6 9

a
6
3
2
b 10 x 2
3
x y
19
3
EXAMPLE 4: RATIONALIZE THE DENOMINATORS:
8
5
2
3

1
4
2
3x y
3
5
8
5
22
2
3 5
2
2
1

4
2
3x y

4
3 4
85 22
5
2
35
33 x 2 y
3
2
3 x y
2
2

85 22
5
25
4

4
2
85 22
2
5
4 2

2
33 x 2 y
3x y
34
3
2
3 x y

4
4
33 x 2 y
34 x 4 y 4
33 x 2 y

3xy
4