Lehman PTS-3 Summer 2020

1. 𝐏𝐓𝐒 𝟑
Bridge to Calculus Workshop
Summer 2020
Lesson 18
Rational Exponents
"God made the integers; all the rest
is the work of Man." – Kronecker -

2. Lehman College, Department of Mathematics
Integer Exponents (1 of 5)
A product of identical numbers is usually written in
exponential notation. For example:
DEFINITION. Let 𝑎 be a real number and let 𝑛 ≥ 2 be a
positive integer, then the 𝑛-th power of 𝑎 is defined as:
The number 𝑎 is called the base, and 𝑛 is the exponent.
Example 1. Determine the following:
𝟓 ⋅ 𝟓 ⋅ 𝟓 = 𝟓 𝟑
𝒂 𝒏
= 𝒂 ⋅ 𝒂 ⋅ ⋯ ⋅ 𝒂
𝒏 𝐟𝐚𝐜𝐭𝐨𝐫𝒔
1
2
4
(a) (b) −2 3
(c) −23

3. Lehman College, Department of Mathematics
Integer Exponents (2 of 5)
Solution: Note the difference between (b) and (c):
Let us discover some of the rules for working with
exponential notation:
It appears that when we multiply two powers of the
same base, we add their exponents.
1
4
3
=(a)
(b) −2 4
=
(c) −24
=
1
4
1
4
1
4
=
1
64
−2 ⋅ −2 ⋅ −2 ⋅ −2 = 16
(−1) 2 ⋅ 2 ⋅ 2 ⋅ 2 = −16
53 ⋅ 54 = 5 ⋅ 5 ⋅ 5 ⋅
3 factors
5 ⋅ 5 ⋅ 5 ⋅ 5 =
4 factors
57
(−1)24
=

4. Lehman College, Department of Mathematics
Integer Exponents (3 of 5)
In general, for any real number 𝑎 and positive integers 𝑛
and 𝑚 with 𝑛, 𝑚 ≥ 2:
Let us look at division of powers. Consider:
It appears that when we find the quotient of two powers
of the same base, we subtract their exponents.
Then, for any real number 𝑎 (𝑎 ≠ 0) and any positive
integers 𝑛 and 𝑚 with 𝑛, 𝑚 ≥ 2 and 𝑛 − 𝑚 ≥ 2, then:
𝒂 𝒎
𝒂 𝒏
= 𝒂 𝒎+𝒏
55
53
=
5 ⋅ 5 ⋅ 5 ⋅ 5 ⋅ 5
5 ⋅ 5 ⋅ 5
= 5 ⋅ 5 = 52
𝒂 𝒎
𝒂 𝒏
= 𝒂 𝒎−𝒏

5. Lehman College, Department of Mathematics
Integer Exponents (4 of 5)
The formulas we have discovered so far are true for
exponents 𝑛 with 𝑛 ≥ 2. Consider the following example
If we want the formula to still work, then we require that:
Similarly, consider the following example:
If we want the formula to still work, then we require that:
5
53−2 =
53
52
=
5 ⋅ 5 ⋅ 5
5 ⋅ 5
=
515 =
1
53
53
=
5 ⋅ 5 ⋅ 5
5 ⋅ 5 ⋅ 5
=
53−3 = 501 =

6. Lehman College, Department of Mathematics
Integer Exponents (5 of 5)
In general, for any nonzero real number 𝑎:
Let us look at negative exponents. Consider:
If we want the formula to still work, then we require that:
In general, for any nonzero real number 𝑎, and for any
integer 𝑛, we have:
𝒂 𝟏 = 𝒂
53
55
=
5 ⋅ 5 ⋅ 5
5 ⋅ 5 ⋅ 5 ⋅ 5 ⋅ 5
=
1
5 ⋅ 5
=
1
52
𝒂−𝒏 =
𝟏
𝒂 𝒏
𝒂 𝟎
= 𝟏and
1
52
= 53−5
= 5−2
𝒂 𝒏
=
𝟏
𝒂−𝒏
and

7. Lehman College, Department of Mathematics
Tables of Powers of Integers (1 of 2)
Students must know the following tables for calculus:
1
𝒙 𝒙 𝟐
1
2 4
3 9
4 16
5 25
6 36
7 49
8 64
𝒙 𝒙 𝟐 𝒙 𝒙 𝟑
9 81
10 100
11 121
12 144
13 169
14 196
15 225
16 256
1 1
2 8
3 27
4 64
5 125

8. Lehman College, Department of Mathematics
Tables of Powers of Integers (2 of 2)
Students must know the following tables for calculus:
1
𝒙 𝒙 𝟒
1
2 16
3 81
4 256
𝒙 𝒙 𝟓
1 1
2 32
21 = 2
22 = 4
23
= 8
24
=
25 =
16
32
26
= 64
27
= 128
28
= 256
29
= 512
210
= 1024

9. Lehman College, Department of Mathematics
Leibniz and Newton (1 of 1)
Isaac Newton (1642-1727)
- English Mathematician
Gottfried Leibniz (1646-1716)
- German Mathematician

10. Lehman College, Department of Mathematics
Simplifying Radicals (1 of 3)
Example 2. Write the following in simplified radical form
Solution. Find the highest perfect square integer factor
of 8. Next, write 8 as a product of two factors.
Example 3. Write the following in simplified radical form
Solution. Find the highest perfect fifth power integer
factor of 128. Next, write 128 as a product of factors.
8
8 = 4 ⋅ 2
= 4 ⋅ 2
= 2 2
5
128

11. Lehman College, Department of Mathematics
Simplifying Radicals (2 of 3)
Solution. Find the highest perfect fifth power integer
factor of 128. Next, write 128 as a product of factors.
Example 4. Write the following in simplified radical form
Solution. Find the highest perfect cube integer factor of
81. Next, write 81 as a product of two factors.
5
32 ⋅ 4
=
5
32 ⋅
5
4
= 2
5
4
5
128 =
3
81
3
81 =
3
27 ⋅ 3
=
3
27 ⋅
3
3 = 3
3
3

12. Lehman College, Department of Mathematics
Simplifying Radicals (3 of 3)
Example 4. Write the following in simplified radical form
Solution. Find the highest perfect fourth power integer
factor of 162. Next, write 162 as a product of factors.
4
162
4
162 =
4
81 ⋅ 2
=
4
81 ⋅
4
2
= 3
4
2

13. Lehman College, Department of Mathematics
Rational Exponents (1 of 4)
We know that:
It follows that:
And that:
Therefore:
Suppose we wish to write 2 as a power of 2, so that
the rule about addition of powers holds:
That is:
It follows that:
So. For all real numbers 𝑎 ≥ 0:
4 =
4
2
4 ⋅ 4 = 2 ⋅ 2 =
9 ⋅ 9 = 9
2 ⋅ 2 = 2
2 ⋅ 22 =
2 𝑟
⋅ 2 𝑟
= 21
𝑟 + 𝑟 = 2𝑟 = 1 Therefore: 𝑟 =
1
2
𝒂 = 𝒂
𝟏
𝟐
therefore22
= 4

14. Lehman College, Department of Mathematics
Rational Exponents (2 of 4)
We know that:
It follows that:
And that:
Therefore:
Suppose we wish to write
3
2 as a power of 2, so that
the rule about addition of powers holds:
That is:
It follows that:
So. For all real numbers 𝑎:
3
8 =
8
2
3
8 ⋅
3
8 ⋅
3
8 = 2 ⋅ 2 ⋅ 2 =
3
27 ⋅
3
27 ⋅
3
27 = 27
3
2 ⋅
3
2 ⋅
3
2 = 2
3
2 ⋅ 23
2 =
2 𝑞
⋅ 2 𝑞
= 21
𝑞 + 𝑞 + 𝑞 = 3𝑞 = 1 Therefore: 𝑞 =
1
3
𝟑
𝒂 = 𝒂
𝟏
𝟑
3
2 ⋅
2 𝑞
⋅
therefore23 = 8

15. Lehman College, Department of Mathematics
Rational Exponents (3 of 4)
In general, for all real numbers 𝑎:
if this 𝑛-th root makes sense as a real number.
For nonzero integers 𝑚, 𝑛, and real number 𝑎 ≥ 0, we
define rational powers of 𝑎 as:
Example 5. Evaluate 253.
Solution.
𝒏
𝒂 = 𝒂
𝟏
𝒏
𝒂
𝒎
𝒏 = 𝒂 𝒎
𝟏
𝒏 =
𝒏
𝒂 𝒎 = 𝒂
𝟏
𝒏
𝒎
= 𝒏
𝒂
𝒎
253 = 25
3
= 53 = 125

16. Lehman College, Department of Mathematics
Rational Exponents (4 of 4)
Example 6. Express the following using rational powers
Solution. These expressions are used often in calculus.
1
𝑥
(a)
1
𝑥2 𝑥
(b)𝑥 𝑥 (c)
(a) 𝑥 𝑥 = 𝑥1
⋅ 𝑥
1
2 = 𝑥
3
2
1
𝑥
=(b)
1
𝑥
1
2
= 𝑥−
1
2
1
𝑥2 𝑥
=(c)
1
𝑥
5
2
=
1
𝑥2 ⋅ 𝑥
1
2
= 𝑥−
5
2

17. Lehman College, Department of Mathematics
Rational Exponents (4 of 4)
Example 7. Simplify the following expression:
Solution.
32
3
5
1
32
−
3
5
=
= 32
1
5
3
=
5
32
3
= 23
1
32
−
3
5
Negative exponent
Property of exponents
Exponent to root rule
Simplify
= 8 Evaluate