Download V1. Radical Expressions 0.1 Whole Number

V1. Radical Expressions
0.1
Whole Number Exponents
Mathematicians have found ways to concisely write repeated operations. Familiar to most is repeated addition, or simply, multiplication. Multiplication
expresses addition of the same number many times. For example, adding five
two’s can be written,
2 + 2 + 2 + 2 + 2 = 2(5).
{z
}
|
5
The natural extension of this leads to writing repeated multiplication. For
example, multiplying five two’s can be written as 2·2·2·2·2, which is exhausting,
or it can be expressed as follows,
2| · 2 ·{z
2 · 2 · 2} = 25
5
In this example of 25 , the two is called the base, and five is called the exponent.
In general, a whole number exponent is defined as
n
a
| · a · a{z· . . . · a} = a
n
for some number a and some whole number n.
Example.
15 = 1 · 1 · 1 · 1 · 1
=1
(0.1)5 = (0.1)(0.1)(0.1)(0.1)(0.1)
= 0.00001
5
2
2 2 2 2 2
= · · · ·
3
3 3 3 3 3
25
= 5
3
32
=
243
1
Challenge. Simplify (1.1)5 without a calculator to test your understanding.
The answer is presented at the end of this section. You should try it first!
Zero can be an exponent. Any number raised to the power of zero is one. The
motivation is illustrated below. Read the example by beginning at the boxed
fraction of each line. First read to the right to see how the fraction is reduced.
Then return to the boxed fraction and read to the left. When reading to the
left, recall that when dividing numbers with the same base, the exponents can
be simplified by subtraction.
x3−1 =
x·x·x
x·x·x
x3
=
= = x2
x
x
x
x2
x·x
x·x
=x
=
= x
x
x
x
x
=
= =1
x
x
x2−1 =
x0 = x1−1
There is one situation that can bring some consternation. Consider the following
two problems.
−24 = −1 · 24 by the rule of −1
= −1 · 16
The Order of Operations dictate to first simplify exponents.
= −16
Compare with,
(−2)4 = −2 · −2 · −2 · −2
= 16
Challenge. Simplify (1.1)5 .
(1.1)5 = (1.1)(1.1)(1.1)(1.1)(1.1)
= (1.21)(1.1)(1.1)(1.1)
= (1.331)(1.1)(1.1)
= (1.4641)(1.1)
= 1.61051
2
0.2
Square Roots and Radicals
√
Let a be any real number. If a2 = b, then b = a. A number that can be written
as another number squared is called a perfect square. In this definition, since
b can be written as a2 , b is called a perfect square. The number 49 is a perfect
square because it can be written as 72 . The square root of 49 answers the
question, “What number squared is 49?” The answer is√that 72 = 49. Thus,
the square root of 49 is 7, or in mathematical notation, 49 = 7.
Consider the following examples.
√
Example. Simplify 25. This is read, “Simplify the square root of 25”.
√
√
Twenty five is the perfect square of 52 . Since 52 = 25, then 25 = 52 = 5.
Thus, we say, “The square root of 25 is 5.”
√
Example. Simplify 1.
√
√
Since 12 = 1, then 1 = 12 = 1, or, “The square root of one is one.”
q
Example. Simplify 14 .
If ( 21 )2 = 14 , then
is one-half.”
q
Example. Simplify
1
4
=
√
q
( 12 )2 =
1
2
This is read “The square root of one-fourth
2.
Try as you may, there is no number that can be squared to obtain 2. When
a
√number
√ is encountered that is not a perfect square, the exact answer is that
2 = 2.
The concept of a perfect square can be extended to numbers raised to the
√ third
power. If a3 = b, then b is a perfect cube. In general, if an = b, then n b = a.
The n is called the index. When taking the square root of a number the index
n = 2 is implied.
√
Example. Simplify 64.
√
Since 82 = 64, then 64 = 8.
√
Example. Simplify 3 64. This is read, “Simplify the cube root of 64.”
√
Since 43 = 64, then 3 64 = 4. This is read, “The cube root of 64 is four.”
√
Example. Simplify 6 64.
√
Since 26 = 64, then 6 64 = 2. Or, “The sixth root of 64 is two.”
3
The next section examines strategies for finding roots of numbers that may not
be as obvious.
0.3
Practice Problems
Find the value of the expression, or simplify the expression, without the aid of
a calculator. This practice will help one to see patterns of numbers.
1.
4.
7.
10.
13.
16.
19.
22.
25.
28.
31.
34.
37.
40.
33
46
1.23
032
3
− 13
(23 )4
√
81
√
− 3
√
3
19
√
4
81
√
5
100, 000
√
4
625
√
6
√ 4096
38
2.
5.
8.
11.
14.
17.
20.
23.
26.
29.
32.
35.
38.
41.
54
1 3
2
(8.2)3
(−2)5
1.022
2 3
(−4)
√
−
q 16
4
25
√
3
−8
√
4
7
√
−√3 729
10
10
√
7
1
p
3
(−1)4
3.
6.
9.
12.
15.
18.
21.
24.
27.
30.
33.
36.
39.
4
(−7)2
2 4
5
123
(−2.3)4
(−0.2)3
√
√25
7
√
3
64
q
1
3
− 64
√
5
32
√
5
200
√
5
0
√
5
−1
0.4
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37.
39.
41.
Solutions
33 = 3 · 3 · 3 = 27
(−7)2 = (−7)(−7) = 49
1 3
= 12 · 21 · 12 = 18
2
3
1.2 = 1.728
123 = 1
(−2)5 = −32
3
1
− 13 = − 13 · − 13 · − 31 = − 27
3
(−0.2) = −0.008
2 3
= 4096
√(−4)
81
=
9
√
√
7
=
7
q
2
4
25 = √
5
√
3
3
19
=
19
q
1
3
− 64 = − 14
√
√
4
7= 47
√
5
100, 000
= 10
√
√
5
5
200
=
√
√200
10
10
10
=
10
√
6
4096 = 4
√
5
−1 = −1 √
p
3
(−1)4 = 3 1 = 1
2.
4.
6.
8.
10.
12.
14.
16.
18.
20.
22.
24.
26.
28.
30.
32.
34.
36.
38.
40.
5
54 = 5 · 5 · 5 · 5 = 625
46 = 4 · 4 · 4 · 4 · 4 · 4 = 4096
16
2 4
= 625
5
3
(8.2) = 551.368
032 = 0
(−2.3)4 = 27.9841
1.022 = 1.0404
(23 )4 = (2 · 2 · 2)4 = 84 = 8 · 8 · 8 · 8 = 4096
√
25 = 5
√
−√16 = −4
√
− 3=− 3
√
3
64 = 4
√
3
−8 = −2
√
4
81 = 3
√
5
32 = 2
√
3
−
729 = −9
√
4
625
=5
√
5
0
=
0
√
7
√1 = 1
38 = 81