Teacher Version

M3
Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
ALGEBRA II
𝟏
𝟏
Lesson 3: Rational Exponentsβ€”What are 𝟐𝟐 and πŸπŸ‘ ?
Student Outcomes
ο‚§
Students will calculate quantities that involve positive and negative rational exponents.
Lesson Notes
Students extend their understanding of integer exponents to rational exponents by examining the graph of 𝑓(π‘₯) = 2π‘₯
1
1
and estimating the values of 22 and 23 . The lesson establishes the meaning of these numbers in terms of radical
1
expressions and these form the basis of how we define expressions of the form 𝑏 𝑛 before generalizing further to
π‘š
expressions of the form 𝑏 𝑛 , where 𝑏 is a positive real number and π‘š and 𝑛 are integers, with 𝑛 β‰  0 (N-RN.A.1). The
lesson and Problem Set provide fluency practice in applying the properties of exponents to expressions containing
rational exponents and radicals (N-RN.A.2). In the following lesson, students will verify that the definition of an
π‘š
𝑛
expression with rational exponents, 𝑏 𝑛 = βˆšπ‘ π‘š , is consistent with the remaining exponential properties. The lesson
begins with students creating a graph of a simple exponential function, 𝑓(π‘₯) = 2π‘₯ , which they studied in Module 3 of
Algebra I. In that module, students also learned about geometric sequences and their relationship to exponential
functions, which is a concept that we revisit at the end of this module. In Algebra I, students worked with exponential
functions with integer domains. This lesson, together with the subsequent Lessons 4 and 5, helps students understand
why the domain of an exponential function 𝑓(π‘₯) = 𝑏 π‘₯ , where 𝑏 is positive number and 𝑏 β‰  1, is the set of real numbers.
To do so we must establish what it means to raise 𝑏 to a rational power and, in Lesson 5, to any real number power.
Classwork
Opening (1 minutes)
In Algebra I, students worked with geometric sequences and simple exponential functions.
Remind them that in Lesson 1, we created formulas based on repeatedly doubling and
halving a number when we modeled folding a piece of paper. We reviewed how to use
the properties of exponents for expressions that had integer exponents.
Opening Exercise (5 minutes)
Have students graph 𝑓(π‘₯) = 2π‘₯ for each integer π‘₯ from π‘₯ = βˆ’2 to π‘₯ = 5 without using a
graphing utility or calculator on the axes provided. Discuss the pattern of points and ask
students to connect the points in a way that produces a smooth curve. Students should
work these two problems independently and check their solutions with a partner if time
permits before you lead a whole class discussion to review the solutions.
Lesson 3:
Date:
Scaffolding:
ο‚§ Encourage students to
create a table of values to
help them construct the
graph.
ο‚§ For advanced learners,
have them repeat these
exercises with the function
𝑓(π‘₯) = 3π‘₯ and ask them
1
1
to estimate 32 and 33 .
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
Opening Exercise
a.
MP.3
𝟏
What is the value of 𝟐𝟐 ? Justify your answer.
A possible student response follows: I think it will be around 𝟏. πŸ“ because 𝟐𝟎 = 𝟏 and 𝟐𝟏 = 𝟐.
b.
Graph 𝒇(𝒙) = πŸπ’™ for each integer 𝒙 from 𝒙 = βˆ’πŸ to 𝒙 = πŸ“. Connect the points on your graph with a smooth
curve.
Ask for a few volunteers to explain their reasoning for their answers to Opening Exercise part (a). Then, debrief these
two exercises by leading a short discussion.
ο‚§
The directions in the Opening Exercise said to connect the points with a smooth curve. What does it imply
about the domain of a function when we connect points that we have plotted?
οƒΊ
ο‚§
That the domain of the function includes the points between the points that we plotted.
How does your graph support or refute your answer to the first exercise? Do you need to modify your answer
1
to the question: What is the value of 22 ? Why or why not?
οƒΊ
1
If the domain is all real numbers, then the value of 22 will be the 𝑦-coordinate of the point on the graph
1
where π‘₯ = . From the graph it looks like the value is somewhere between 1 and 2. The scaling on this
2
graph is not detailed enough for me to accurately refine my answer yet.
1
Transition to the next set of exercises by telling the students that we can better estimate the value of 22 by looking at a
graph that is scaled more precisely. Have students proceed to the next questions. They should work in small groups or
with a partner on these exercises.
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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M3
Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
ALGEBRA II
The graph of 𝑓(π‘₯) = 2π‘₯ shown below for the next exercises will appear in the student materials but you could also
1
1
display the graph using technology. Have students estimate the value of 22 and 23 from the graph. Students should
work by themselves or in pairs to work through the following questions without using a calculator.
The graph at right shows a close-up view of 𝒇(𝒙) = πŸπ’™ for βˆ’πŸŽ. πŸ“ < 𝒙 < 𝟏. πŸ“.
c.
Find two consecutive integers that are over and
𝟏
underestimates of the value of 𝟐𝟐 .
𝟏
𝟐𝟎 < 𝟐𝟐 < 𝟐𝟏
𝟏<
d.
𝟏
𝟐𝟐
<𝟐
𝟏
Does it appear that 𝟐𝟐 is halfway between the
integers you specified in Exercise 1?
𝟏
No, it looks like 𝟐𝟐 is a little less than halfway
between 𝟏 and 𝟐.
e.
Use the graph of 𝒇(𝒙) = πŸπ’™ to estimate the value
𝟏
of 𝟐𝟐 .
𝟏
𝟐𝟐 β‰ˆ 𝟏. πŸ’
f.
𝟏
Use the graph of 𝒇(𝒙) = πŸπ’™ to estimate the value of πŸπŸ‘ .
𝟏
πŸπŸ‘ β‰ˆ 𝟏. πŸπŸ“
Discussion (9 minutes)
Before getting into the point of this lesson, which is to connect rational exponents to
radical expressions, revisit the initial question with students.
ο‚§
1
What is the value of 22 ? Does anyone want to adjust their initial guess?
οƒΊ
MP.6
ο‚§
Our initial guess was a little too big. It seems like 1.4 might be a better
answer.
How could we make a better guess?
οƒΊ
We could look at the graph with even smaller increments for the scale
using technology.
If time permits, you can zoom in further on the graph of 𝑓(π‘₯) = 2π‘₯ using a graphing
calculator or other technology either by examining a graph or a table of values of π‘₯ closer
Scaffolding:
ο‚§ If needed, you can
demonstrate the
argument using perfect
squares. For example, use
a base of 4 instead of a
base of 2.
1 2
Show that (42 ) = 4
2
and (√4) = 4.
1
and closer to .
2
1
Next, we will make the connection that 22 = √2. Walk students through the following questions, providing guidance as
needed. Students proved that there was only one positive number that squared to 2 in Geometry, Module 2. You may
need to remind them of this with a bit more detail if they are struggling to follow this argument.
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
ο‚§
1
Assume for the moment that whatever 22 means that it satisfies 𝑏 π‘š βˆ™ 𝑏 𝑛 = 𝑏 π‘š+𝑛 . Working with this
1
1
assumption what is the value of 22 βˆ™ 22 ?
οƒΊ
ο‚§
1
1 1
What unique positive number squares to 2? That is, what is the only positive number that when multiplied by
itself is equal to 2?
οƒΊ
MP.7
1
It would be 2 because 22 βˆ™ 22 = 22+2 = 21 = 2.
By definition, we call the unique positive number that squares to 2 the square root of 2 and we write
√2.
Write the following statements on the board and ask your students to compare them and think about what they must
1
tell us about the meaning of 22 .
1
1
22 βˆ™ 22 = 2 and √2 βˆ™ √2 = 2
ο‚§
1
What do these two statements tell us about the meaning of 22 .
οƒΊ
Since both statements involve multiplying a number by itself and getting 2, and we know that there is
1
only one number that does that, we can conclude that 22 = √2.
1
At this point, you can have students confirm these results by using a calculator to approximate both 22 and √2 to several
1
decimal places. In the opening, we approximated 22 graphically, and now we have shown it to be an irrational number.
1
Next ask students to think about the meaning of 23 using a similar line of reasoning.
ο‚§
Scaffolding:
1
Assume that whatever 23 means it will satisfy 𝑏 π‘š βˆ™ 𝑏 𝑛 = 𝑏 π‘š+𝑛 . What is the
1
1
ο‚§ If needed, you can
demonstrate the
argument using perfect
cubes. For example, use a
base of 8 instead of a base
of 2.
1
value of (23 ) (23 ) (23 )?
οƒΊ
MP.7
ο‚§
1
1
1 1 1
3
3
3
What is the value of √2 βˆ™ √2 βˆ™ √2?
οƒΊ
ο‚§
1
The value is 2 because (23 ) (23 ) (23 ) = 23+3+3 = 21 = 2.
3
3
3
3
3
The value is 2 because √2 βˆ™ √2 βˆ™ √2 = ( √2) = 2.
1 3
Show (83 ) = 8
1
3
What appears to be the meaning of 2 ?
οƒΊ
3
1
3
and ( √8) = 8.
Since both the exponent expression and the radical expression involve
multiplying a number by itself three times and the result is equal to 2, we
3
know that 23 = √2.
1
3
Students can also confirm using a calculator that the decimal approximations of 23 and √2 are the same. Next, we ask
them to generalize their findings.
ο‚§
1
οƒΊ
MP.8
1
4
10
1
Can we generalize this relationship? Does 24 = √2? Does 210 = √2? What is 2𝑛 , for any positive integer 𝑛?
Why?
1
𝑛
2𝑛 = √2 because
1 𝑛
(2𝑛 )
𝑛 times
=
1
1
1
(2𝑛 ) (2𝑛 ) β‹― (2𝑛 )
⏟
=
⏞
1 1
1
2𝑛+𝑛+β‹―+𝑛
= 21 = 2.
𝑛 times
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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M3
Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
ALGEBRA II
1
𝑛
Have students confirm these using a calculator as well checking to see if the decimal approximations of 2𝑛 and √2 are
MP.8 the same for different values of 𝑛 such as 4, 5, 6, 10, …. Be sure to share the generalization shown above on the board to
1
𝑛
help students understand why it makes sense to define 2𝑛 to be √2.
𝑛
However, be careful to not stop here; there is a problem with our reasoning if we do not define √2. In previous courses,
only square roots and cube roots were defined.
We first need to define the 𝑛th root of a number; there may be more than one, as in the case where 22 = 4 and
(βˆ’2)2 = 4. We say that both βˆ’2 and 2 are square roots of 4. However, we give priority to the positive-valued square
root, and we say that 2 is the principal square root of 4. We often just refer to the square root of 4 when we mean the
principal square root of 4. The definition of 𝑛th root presented below is consistent with allowing complex 𝑛th roots,
which students will encounter in college if they pursue engineering or higher mathematics. If we allow complex 𝑛th
3
3
1
roots, there are three cube roots of 2: √2, √2 (βˆ’ +
2
√3
2
3
1
𝑖), and √2 (βˆ’ βˆ’
2
√3
2
3
𝑖), and we refer to the real number √2 as
the principal cube root of 2. There is no need to discuss this with any but the most advanced students.
𝑛
The √2 is the positive real number π‘Ž such that π‘Žπ‘› = 2. In general, if π‘Ž is positive, then the 𝑛th root of π‘Ž exists for any
positive integer 𝑛, and if π‘Ž is negative, then the 𝑛th root of π‘Ž exists only for odd integers 𝑛. This even/odd condition is
handled subtly in the definition below; the 𝑛th root exists only if there is already an exponential relationship 𝑏 = π‘Žπ‘› .
Present the following definitions to students and have them record them in their notes.
𝒏TH ROOT OF A NUMBER: Let π‘Ž and 𝑏 be numbers, and let 𝑛 β‰₯ 2 be a positive integer. If 𝑏 = π‘Žπ‘› ,
then π‘Ž is an 𝑛th root of 𝑏. If 𝑛 = 2, then the root is a called a square root. If 𝑛 = 3, then the root is called a cube
root.
PRINCIPAL 𝒏TH ROOT OF A NUMBER: Let 𝑏 be a real number that has at least one real 𝑛th root. The principal 𝑛th root of 𝑏
𝑛
is the real 𝑛th root that has the same sign as 𝑏 and is denoted by a radical symbol: βˆšπ‘.
Every positive number has a unique principal 𝑛th root. We often refer to the principal 𝑛th root of 𝑏 as just the 𝑛th
root of 𝑏. The 𝑛th root of 0 is 0.
Students have already learned about square and cube roots in previous grades. In Module 1 and at the beginning of this
lesson, students worked with radical expressions involving cube and square roots. Explain that the 𝑛th roots of a number
satisfy the same properties of radicals learned previously. Have students record these properties in their notes.
If π‘Ž β‰₯ 0, 𝑏 β‰₯ 0 (𝑏 β‰  0 when 𝑏 is a denominator) and 𝑛 is a positive integer, then
𝑛
𝑛 π‘Ž
βˆšπ‘Ž
𝑛
𝑛
𝑛
√ =𝑛 .
βˆšπ‘Žπ‘ = βˆšπ‘Ž βˆ™ βˆšπ‘ and
𝑏
βˆšπ‘
Background information regarding 𝑛th roots and their uniqueness is provided below. You may wish to share this with
your advanced learners or the entire class if you extend this lesson to an additional day.
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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M3
Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
ALGEBRA II
The existence of the principal 𝑛th root of a real number 𝑏 is a consequence of the fundamental theorem of
algebra: Consider the polynomial function 𝑓(π‘₯) = π‘₯ 𝑛 βˆ’ 𝑏. When 𝑛 is odd, we know that 𝑓 has at least one real
zero because the graph of 𝑓 must cross the π‘₯-axis. That zero is a positive number which (after showing that it is
the ONLY real zero) is the 𝑛th root. The case for when 𝑛 is even follows a similar argument.
To show uniqueness of the 𝑛th root, suppose there are two 𝑛th roots of a number 𝑏, π‘₯, and 𝑦 such that π‘₯ > 0,
𝑦 > 0, π‘₯ 𝑛 = 𝑏, and 𝑦 𝑛 = 𝑏. Then π‘₯ 𝑛 βˆ’ 𝑦 𝑛 = 𝑏 βˆ’ 𝑏 = 0. The expression π‘₯ 𝑛 βˆ’ 𝑦 𝑛 factors (see Lesson 7 in
Module 1).
0 = π‘₯ 𝑛 βˆ’ 𝑦𝑛
= (π‘₯ βˆ’ 𝑦)(π‘₯ π‘›βˆ’1 + π‘₯ π‘›βˆ’2 𝑦 + π‘₯ π‘›βˆ’3 𝑦 2 + β‹― + π‘₯𝑦 π‘›βˆ’2 + 𝑦 π‘›βˆ’1 )
Since both π‘₯ and 𝑦 are positive, the second factor is never zero. Thus, for π‘₯ 𝑛 βˆ’ 𝑦 𝑛 = 0, we must have
π‘₯ βˆ’ 𝑦 = 0 and it follows that π‘₯ = 𝑦. Thus, there is only one 𝑛th root of 𝑏.
A proof of the first radical property is shown below for teacher background information. You may wish to share this
proof with your advanced learners or the entire class if you extend this lesson to an additional day.
𝑛
𝑛
𝑛
Prove that βˆšπ‘Žπ‘ = βˆšπ‘Ž βˆ™ βˆšπ‘.
𝑛
Let π‘₯ β‰₯ 0 be the number such that π‘₯ 𝑛 = π‘Ž and let 𝑦 β‰₯ 0 be the number such that 𝑦 𝑛 = 𝑏, so that π‘₯ = βˆšπ‘Ž and
𝑛
𝑦 = βˆšπ‘. Then, by a property of exponents, (π‘₯𝑦)𝑛 = π‘₯ 𝑛 𝑦 𝑛 = π‘Žπ‘. Thus, π‘₯𝑦 must be the 𝑛th root of π‘Žπ‘. Writing
this using our notation gives
𝑛
𝑛
𝑛
βˆšπ‘Žπ‘ = π‘₯𝑦 = βˆšπ‘Ž βˆ™ βˆšπ‘.
After students have recorded this information in their notes, you can proceed with Example 1 and Exercise 1.
Example 1 (3 minutes)
This example will familiarize students with the wording in the definition presented above.
Example 1
a.
What is the πŸ’th root of πŸπŸ”?
πŸ’
π’™πŸ’ = πŸπŸ” when 𝒙 = 𝟐 because πŸπŸ’ = πŸπŸ”. Thus, βˆšπŸπŸ” = 𝟐.
b.
What is the cube root of πŸπŸπŸ“?
πŸ‘
π’™πŸ‘ = πŸπŸπŸ“ when 𝒙 = πŸ“ because πŸ“πŸ‘ = πŸπŸπŸ“. Thus, βˆšπŸπŸπŸ“ = πŸ“.
c.
What is the πŸ“th root of 𝟏𝟎𝟎, 𝟎𝟎𝟎?
π’™πŸ“ = 𝟏𝟎𝟎, 𝟎𝟎𝟎 when 𝒙 = 𝟏𝟎 because πŸπŸŽπŸ“ = 𝟏𝟎𝟎, 𝟎𝟎𝟎. Thus, πŸ“βˆšπŸπŸŽπŸŽ, 𝟎𝟎𝟎 = 𝟏𝟎.
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
Exercise 1 (2 minutes)
In these brief exercises, students will work with definition of 𝑛th roots and the multiplication property presented above.
Have students check their work with a partner and briefly discuss as a whole class any questions that arise.
Exercise 1
1.
Evaluate each expression.
a.
πŸ’
βˆšπŸ–πŸ
πŸ‘
Scaffolding:
b.
πŸ“
βˆšπŸ‘πŸ
If needed, continue to support
students that struggle with
abstraction by including
additional numeric examples.
𝟐
c.
πŸ‘
πŸ‘
βˆšπŸ— βˆ™ βˆšπŸ‘
πŸ‘
βˆšπŸπŸ• = πŸ‘
d.
πŸ’
πŸ’
πŸ’
βˆšπŸπŸ“ βˆ™ √𝟏𝟎𝟎 βˆ™ βˆšπŸ’
πŸ’
√𝟏𝟎, 𝟎𝟎𝟎 = 𝟏𝟎
Discussion (8 minutes)
1
1
1
Return to the question posted in the title of this lesson: What are 22 and 23 ? Now we know the answer, 22 = √2 and
1
1
3
1
𝑛
23 = √2. So far, we have given meaning to 2𝑛 by equating 2𝑛 = √2. Ask students if they believe these results extend
to any base 𝑏 > 0.
ο‚§
We just did some exercises where the 𝑏-value was a number different from 2. In our earlier work, was there
something special about using 2 as the base of our exponential expression? Would these results generalize to
1
1
1
1
expressions of the form 3𝑛 ? 7𝑛 ? 10𝑛 ? 𝑏 𝑛 , for any positive real number 𝑏?
οƒΊ
There is nothing inherently special about the base 2 in the above discussion. These results should
1
generalize to expressions of the form 𝑏 𝑛 for any positive real number base 𝑏 because we have defined
an 𝑛th root for any positive base 𝑏.
1
𝑛
Now that we know what the 𝑛th root of a number 𝑏, βˆšπ‘, means, the work we did earlier with base 2 suggests that 𝑏 𝑛
should also be defined as the 𝑛th root of 𝑏. Discuss this definition with your class. If the class is unclear on the definition,
1
1
do some numerical examples: (βˆ’32)5 = βˆ’2 because (βˆ’2)5 = βˆ’32, but (βˆ’16)4 does not exist because there is no
principal 4th root of a negative number.
1
For a real number 𝑏 and a positive integer 𝑛, define 𝑏 𝑛 to be the principal 𝑛th root of 𝑏 when it
1
𝑛
exists. That is, 𝑏 𝑛 = βˆšπ‘.
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
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M3
Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
ALGEBRA II
Note that this definition holds for any real number 𝑏 if 𝑛 is an odd integer and for positive real numbers 𝑏 if 𝑛 is an even
integer. You may choose to emphazise this with your class. Thus, when 𝑏 is negative and 𝑛 is an odd integer, the
1
1
expression 𝑏 𝑛 will be negative. If 𝑛 is an even integer, then we must restrict 𝑏 to positive real numbers only, and 𝑏 𝑛 will
be positive.
1
In the next lesson, we see that with this definition 𝑏 𝑛 satisfies all the usual properties of exponents so it makes sense to
define it in this way.
At this point, you can also revisit our original question with the students one more time.
ο‚§
1
1
What is the value of 22 ? What does it mean? What is the value of 𝑏 2 for any positive number 𝑏? How are
radicals related to rational exponents?
1
οƒΊ
1
We now know that 22 is equal to √2. In general 𝑏 2 = βˆšπ‘ for any positive real number 𝑏.
A number that contains a radical can be expressed using rational exponents in place of the radical.
1
As you continue the discussion, we extend the definition of exponential expressions with exponents of the form to any
𝑛
3
positive rational exponent. We begin by considering an example: What do we mean by 24 ? Give students a few
minutes to respond individually in writing to this question on their student pages and then have them discuss their
reasoning with a partner. Make sure to correct any blatant misconceptions and to clarify incomplete thinking as you
lead the discussion that follows.
Discussion
𝟏
𝟏
πŸ‘
πŸ‘
If 𝟐𝟐 = √𝟐 and πŸπŸ‘ = √𝟐, what does πŸπŸ’ equal? Explain your reasoning.
πŸ‘
𝟏
πŸ‘
Student solutions and explanations will vary. One possible solution would be πŸπŸ’ = (πŸπŸ’ ) , so it must mean that
MP.3
πŸ‘
πŸ’
πŸ‘
πŸπŸ’ = (√𝟐) . Since the properties of exponents and the meaning of an exponent made sense with integers and now for
𝟏
rational numbers in the form , it would make sense that they would work all rational numbers too.
𝒏
1
π‘š
Now that we have a definition for exponential expressions of the form 𝑏 𝑛 , use the discussion below to define 𝑏 𝑛 , where
π‘š and 𝑛 are both integers, 𝑛 β‰  0, and 𝑏 is a positive real number. Make sure students understand that our
π‘š
interpretation of 𝑏 𝑛 must be consistent with the exponent properties (which hold for integer exponents) and the
1
definition of 𝑏 𝑛 .
ο‚§
3
𝑛
οƒΊ
ο‚§
1
How can we rewrite the exponent of 24 using integers and rational numbers in the form ?
1 3
3
3
1
We can write 24 = (24 ) or we can write 24 = (23 )4 .
1
Now apply our definition of 𝑏 𝑛 .
οƒΊ
3
1 3
4
3
3
1
4
4
24 = (24 ) = ( √2) or 24 = (23 )4 = √23 = √8
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
ο‚§
3
3
4
Does this make sense? If 24 = √8, then if we raise 24 to the fourth power, we should get 8. Does this
happen?
3 4
3
3
3
3
3
οƒΊ
(24 ) = (24 ) (24 ) (24 ) (24 ) = 2(4βˆ™4) = 23 = 8.
οƒΊ
So, 8 is the product of four equal factors, which we denote by 24 . Thus, 24 = √8.
3
3
4
π‘š
π‘š
Take a few minutes to allow students to think about generalizing their work above to 2 𝑛 and then to 𝑏 𝑛 . Have them
write a response to the following questions and share it with a partner before proceeding as a whole class.
ο‚§
π‘š
Can we generalize this result? How would you define 2 𝑛 , for positive integers π‘š and 𝑛?
π‘š
οƒΊ
ο‚§
π‘š
𝑛
𝑛
π‘š
Conjecture: 2 𝑛 = √2π‘š , or equivalently, 2 𝑛 = ( √2) .
π‘š
π‘š
π‘š
π‘š
Can we generalize this result to any positive real number base 𝑏? What is 3 𝑛 ? 7 𝑛 ? 10 𝑛 ? 𝑏 𝑛 ?
οƒΊ
There is nothing inherently special about the base 2 in the above Discussion. These results should
π‘š
generalize to expressions of the form 𝑏 𝑛 for any positive real number base 𝑏.
π‘š
οƒΊ
𝑛
Then we are ready to define 𝑏 𝑛 = βˆšπ‘ π‘š for positive integers π‘š and 𝑛 and positive real numbers 𝑏.
This result is summarized in the box below.
1
For any positive integers π‘š and 𝑛, and any real number 𝑏 for which 𝑏 𝑛 exists, we define
π‘š
π‘š
𝑛
𝑛
𝑏 𝑛 = βˆšπ‘ π‘š , which is equivalent to 𝑏 𝑛 = ( βˆšπ‘)
π‘š
Note that this property holds for any real number 𝑏 if 𝑛 is an odd integer. You may choose to emphasize this with your
π‘š
class. When 𝑏 is negative and 𝑛 is an odd integer, the expression 𝑏 𝑛 will be negative. If 𝑛 is an even integer then we
must restrict 𝑏 to positive real numbers only.
Exercises 2–8 (4 minutes)
In these exercises, students use the definitions above to rewrite and evaluate expressions. Have students check their
work with a partner and briefly discuss as a whole class any questions that arise.
Exercises 2–12
Rewrite each exponential expression as an 𝒏th root.
2.
𝟏
πŸ‘πŸ
𝟏
πŸ‘πŸ = βˆšπŸ‘
3.
𝟏
πŸπŸπŸ“
𝟏
πŸ“
πŸπŸπŸ“ = √𝟏𝟏
Lesson 3:
Date:
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
4.
𝟏
πŸ’
( )
𝟏
πŸ“
𝟏
𝟏 πŸ“ πŸ“ 𝟏
( ) =√
πŸ’
πŸ’
5.
𝟏
πŸ”πŸπŸŽ
𝟏
𝟏𝟎
πŸ”πŸπŸŽ = βˆšπŸ”
Rewrite the following exponential expressions as equivalent radical expressions. If the number is rational, write it without
radicals or exponents.
6.
πŸ‘
𝟐𝟐
πŸ‘
𝟐𝟐 = βˆšπŸπŸ‘ = 𝟐√𝟐
7.
πŸ“
πŸ’πŸ
πŸ“
πŸ“
πŸ’πŸ = βˆšπŸ’πŸ“ = (βˆšπŸ’) = πŸπŸ“ = πŸ‘πŸ
8.
𝟏
πŸ–
( )
πŸ“
πŸ‘
πŸ“
πŸ“
πŸ‘ 𝟏
𝟏 πŸ‘ πŸ‘ 𝟏 πŸ“
𝟏 πŸ“
𝟏
( ) = √( ) = ( √ ) = ( ) =
πŸ–
πŸ–
πŸ–
𝟐
πŸ‘πŸ
Exercise 9 (3 minutes)
In this exercise, we ask students to consider a negative rational exponent. Have students work directly with a partner
and ask them to use thinking similar to the preceding discussion. Correct and extend student thinking as you review the
solution.
9.
Show why the following statement is true:
𝟏
πŸβˆ’πŸ =
𝟏
𝟏
𝟐𝟐
Student solutions and explanations will vary. One possible solution would be
𝟏
𝟏 βˆ’πŸ
πŸβˆ’πŸ = (𝟐𝟐 )
βˆ’πŸ
= (√𝟐)
=
𝟏
√𝟐
=
𝟏
𝟏
𝟐𝟐
𝟏
Since √𝟐 is a real number the properties of exponents hold and we have defined 𝒃𝒏. We can show these two
expressions are the same.
Share the following property with your class and show how the work they did in the previous exercises supports this
conclusion. Should you choose to you can verify these properties using an argument similar to the ones presented
π‘š
earlier for the meaning of 𝑏 𝑛 .
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
1
For any positive integers π‘š and 𝑛, and any real number 𝑏 for which 𝑏 𝑛 exists, we define
π‘š
1
π‘βˆ’ 𝑛 = 𝑛
βˆšπ‘ π‘š
or, equivalently,
1
π‘š
π‘βˆ’ 𝑛 =
π‘š.
𝑛
( βˆšπ‘)
Exercises 10–12 (3 minutes)
Rewrite the following exponential expressions as equivalent radical expressions. If the number is rational, write it without
radicals or exponents.
πŸ‘
10. πŸ’βˆ’πŸ
𝟏
πŸ‘
πŸ’βˆ’πŸ =
βˆšπŸ’πŸ‘
𝟏
=
πŸ‘
=
(βˆšπŸ’)
𝟏
πŸ–
𝟐
11. πŸπŸ•βˆ’πŸ‘
𝟐
πŸπŸ•βˆ’πŸ‘ =
𝟏
𝟐
πŸπŸ•πŸ‘
=
𝟏
πŸ‘
𝟐
=
(βˆšπŸπŸ•)
𝟏
𝟏
=
πŸ‘πŸ πŸ—
𝟏
𝟏 βˆ’πŸ
πŸ’
12. ( )
𝟏
βˆ’πŸ
𝟏 βˆ’πŸ
𝟏
We have ( ) = (√ )
πŸ’
πŸ’
𝟏
𝟏
𝟏
𝟏 βˆ’πŸ
𝟏 βˆ’πŸ
𝟏 βˆ’πŸ 𝟐
= ( ) = 𝟐. Alternatively, ( ) = (( ) ) = (πŸ’)𝟐 = βˆšπŸ’ = 𝟐.
𝟐
πŸ’
πŸ’
Closing (3 minutes)
Have students summarize the key points of the lesson in writing. Circulate around the classroom to informally assess
understanding and provide assistance. Their work should reflect the summary provided below.
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
Lesson Summary
𝒏TH ROOT OF A NUMBER: Let 𝒂 and 𝒃 be numbers, and let 𝒏 β‰₯ 𝟐 be a positive integer. If 𝒃 = 𝒂𝒏, then 𝒂 is an 𝒏th root of
𝒃. If 𝒏 = 𝟐, then the root is a called a square root. If 𝒏 = πŸ‘, then the root is called a cube root.
PRINCIPAL 𝒏TH ROOT OF A NUMBER: Let 𝒃 be a real number that has at least one real 𝒏th root. The principal 𝒏th root of 𝒃
𝒏
is the real 𝒏th root that has the same sign as 𝒃, and is denoted by a radical symbol: βˆšπ’ƒ.
Every positive number has a unique principal 𝒏th root. We often refer to the principal 𝒏th root of 𝒃 as just the 𝒏th
root of 𝒃. The 𝒏th root of 𝟎 is 𝟎.
For any positive integers π’Ž and 𝒏, and any real number 𝒃 for which the principal 𝒏th root of 𝒃 exists, we have
𝟏
𝒏
𝒃𝒏 = βˆšπ’ƒ
π’Ž
𝒏
𝒏
π’Ž
𝒃 𝒏 = βˆšπ’ƒπ’Ž = ( βˆšπ’ƒ)
π’Ž
𝟏
π’ƒβˆ’ 𝒏 = 𝒏
.
βˆšπ’ƒπ’Ž
Exit Ticket (4 minutes)
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
Name
Date
𝟏
𝟏
Lesson 3: Rational Exponentsβ€”What Are 𝟐𝟐 and πŸπŸ‘ ?
Exit Ticket
1.
Write the following exponential expressions as equivalent radical expressions.
1
a.
22
b.
24
c.
3βˆ’ 3
3
2
2.
3.
Rewrite the following radical expressions as equivalent exponential expressions.
a.
√5
b.
2 √3
c.
3
4
1
√16
Provide a written explanation for each question below.
1 3
1
a.
Is it true that (42 ) = (43 )2 ? Explain how you know.
b.
Is it true that (10003 ) = (10003 )3 ? Explain how you know.
c.
Suppose that π‘š and 𝑛 are positive integers and 𝑏 is a real number so that the principal 𝑛th root of 𝑏 exists. In
1 3
1 π‘š
1
1
general does (𝑏 𝑛 ) = (𝑏 π‘š )𝑛 ? Provide at least one example to support your claim.
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
Exit Ticket Sample Solutions
1.
Rewrite the following exponential expressions as equivalent radical expressions.
a.
𝟏
𝟐𝟐
𝟏
𝟐𝟐 = √𝟐
b.
πŸ‘
πŸπŸ’
πŸ‘
πŸ’
πŸ’
πŸπŸ’ = βˆšπŸπŸ‘ = βˆšπŸ–
c.
𝟐
πŸ‘βˆ’πŸ‘
𝟏
𝟐
πŸ‘βˆ’πŸ‘ =
2.
πŸ‘
βˆšπŸ‘πŸ
=
𝟏
πŸ‘
βˆšπŸ—
Rewrite the following radical expressions as equivalent exponential expressions.
a.
βˆšπŸ“
𝟏
βˆšπŸ“ = πŸ“πŸ
b.
πŸ’
πŸβˆšπŸ‘
𝟏
πŸ’
πŸ’
πŸ’
πŸβˆšπŸ‘ = βˆšπŸπŸ’ β‹… πŸ‘ = βˆšπŸ’πŸ– = πŸ’πŸ–πŸ’
c.
𝟏
πŸ‘
βˆšπŸπŸ”
𝟏
𝟏
πŸ‘
βˆšπŸπŸ”
𝟏
πŸ‘
𝟏
βˆšπŸπŸ”
3.
πŸ’
= (πŸπŸ’ )βˆ’πŸ‘ = πŸβˆ’πŸ‘
= (πŸπŸ”)βˆ’πŸ‘
Provide a written explanation for each question below.
a.
𝟏 πŸ‘
𝟏
Is it true that (πŸ’πŸ ) = (πŸ’πŸ‘ )𝟐? Explain how you know.
𝟏 πŸ‘
πŸ‘
(πŸ’πŸ ) = (βˆšπŸ’) = πŸπŸ‘ = πŸ–
𝟏
𝟏
(πŸ’πŸ‘ )𝟐 = πŸ”πŸ’πŸ = βˆšπŸ”πŸ’ = πŸ–
MP.3
&
MP.7
So the first statement is true.
b.
𝟏 πŸ‘
𝟏
Is it true that (πŸπŸŽπŸŽπŸŽπŸ‘ ) = (πŸπŸŽπŸŽπŸŽπŸ‘ )πŸ‘? Explain how you know.
Similarly the left and right sides of the second statement are equal to one another.
𝟏 πŸ‘
πŸ‘
πŸ‘
(πŸπŸŽπŸŽπŸŽπŸ‘ ) = (√𝟏𝟎𝟎𝟎) = πŸπŸŽπŸ‘ = 𝟏𝟎𝟎𝟎
𝟏
𝟏
(πŸπŸŽπŸŽπŸŽπŸ‘ )πŸ‘ = (𝟏𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎)πŸ‘ = 𝟏𝟎𝟎𝟎
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Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
c.
Suppose that π’Ž and 𝒏 are positive integers and 𝒃 is a real number so that the principal 𝒏th root of 𝒃 exists. In
𝟏 π’Ž
𝟏
general does (𝒃𝒏 ) = (π’ƒπ’Ž )𝒏? Provide at least one example to support your claim.
Thus, it appears the statement is true and it also holds for when the exponents are integers. Based on our
other work in this lesson, this property should extend to rational exponents as well.
Here is another example,
𝟏 𝟐
(πŸ–πŸ‘ ) = (𝟐)𝟐 = πŸ’
and
𝟏
𝟏
(πŸ–πŸ )πŸ‘ = (πŸ”πŸ’)πŸ‘ = πŸ’.
Thus,
𝟏 𝟐
𝟏
(πŸ–πŸ‘ ) = (πŸ–πŸ )πŸ‘ .
Problem Set Sample Solutions
1.
Select the expression from (A), (B), and (C) that correctly completes the statement.
(A)
a.
𝟏
π’™πŸ‘ is equivalent to
.
𝟏
πŸ‘
(B)
(C)
πŸ‘
πŸ‘
𝒙
βˆšπ’™
𝒙
√ π’™πŸ
𝟏
πŸ’
πŸ’
𝒙
(B)
b.
𝟐
π’™πŸ‘ is equivalent to
.
𝟐
πŸ‘
πŸ‘
πŸ‘
(βˆšπ’™)
(B)
c.
𝟏
π’™βˆ’πŸ’ is equivalent to
.
βˆ’ 𝒙
𝒙
𝟏
πŸ’
βˆšπ’™
(C)
d.
πŸ’
𝒙
𝟏
𝟐
( ) is equivalent to
.
𝟐
πŸ’
𝟐
𝒙
π’™πŸ
βˆšπ’™
(C)
2.
Identify which of the expressions (A), (B), and (C) are equivalent to the given expression.
(A)
a.
𝟏
πŸπŸ”πŸ
(
𝟏
)
πŸπŸ”
(B)
𝟏
βˆ’πŸ
(C)
𝟐
πŸ‘
πŸ–πŸ‘
πŸ”πŸ’πŸ
(A) and (B)
b.
𝟐 βˆ’πŸ
(πŸ‘)
βˆ’
πŸ‘
𝟐
πŸ—
πŸ’
( )
𝟏
𝟐
𝟏
πŸπŸ•πŸ‘
πŸ”
(B) only
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Date:
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
3.
Rewrite in radical form. If the number is rational, write it without using radicals.
a.
πŸ‘
πŸ”πŸ
b.
𝟏
𝟐
( )
𝟏
πŸ’
βˆšπŸπŸπŸ”
πŸ’
√
c.
𝟏
𝟏
𝟐
𝟐
πŸ‘(πŸ–)πŸ‘
d.
(
πŸ‘
πŸ”πŸ’ βˆ’πŸ‘
)
πŸπŸπŸ“
πŸ‘βˆšπŸ– = πŸ”
𝟐
πŸπŸπŸ“
πŸπŸ“
𝒍
(√
) =
πŸπŸ”
πŸπŸ”
πŸ‘
e.
𝟏
πŸ–πŸβˆ’πŸ’
𝟏
πŸ’
βˆšπŸ–πŸ
4.
=
𝟏
πŸ‘
Rewrite the following expressions in exponent form.
a.
βˆšπŸ“
b.
𝟏
πŸ“πŸ‘
βˆšπŸ“πŸ‘
d.
πŸ‘
πŸ‘
𝟐
(βˆšπŸ“)
𝟐
πŸ“πŸ
5.
βˆšπŸ“πŸ
𝟐
πŸ“πŸ
c.
πŸ‘
πŸ“πŸ‘
Use the graph of 𝒇(𝒙) = πŸπ’™ shown to the right to estimate the following powers of 𝟐.
𝟏
a.
πŸπŸ’
β‰ˆ 𝟏. 𝟐
b.
𝟐
πŸπŸ‘
β‰ˆ 𝟏. πŸ”
c.
πŸ‘
πŸπŸ’
β‰ˆ 𝟏. πŸ•
𝟎.𝟐
d.
𝟐
e.
𝟐𝟏.𝟐
f.
βˆ’πŸ“
𝟐
𝟏
Lesson 3:
Date:
β‰ˆ 𝟏. πŸπŸ“
β‰ˆ 𝟐. πŸ‘
β‰ˆ 𝟎. πŸ–πŸ“
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
6.
Rewrite each expression in the form π’Œπ’™π’ , where π’Œ is a real number, 𝒙 is a positive real number, and 𝒏 is rational.
πŸ’
βˆšπŸπŸ”π’™πŸ‘
a.
b.
πŸ‘
𝟏
πŸπ’™πŸ’
βˆ’
πŸ“π’™ 𝟐
πŸ‘
√𝟏/π’™πŸ’
c.
d.
πŸ’
βˆ’
𝒙 πŸ‘
πŸ’
πŸ‘
βˆšπŸ–π’™πŸ‘
πŸπ’™βˆ’πŸ
𝟏
πŸπŸ•
e.
f.
βˆšπŸ—π’™πŸ’
πŸ—π’™βˆ’πŸ
7.
πŸ“
βˆšπ’™
(
πŸπŸπŸ“ βˆ’πŸ‘
)
π’™πŸ
𝟏 𝟐
π’™πŸ‘
πŸ“
𝟏
Find a value of 𝒙 for which πŸπ’™πŸ = πŸ‘πŸ.
πŸπŸ“πŸ”
8.
πŸ’
Find a value of 𝒙 for which π’™πŸ‘ = πŸ–πŸ.
πŸπŸ•
9.
πŸ‘
πŸ‘
If π’™πŸ = πŸ”πŸ’, find the value of πŸ’π’™βˆ’ πŸ’.
πŸ‘
𝒙 = πŸπŸ”, so πŸ’(πŸπŸ”)βˆ’πŸ’ = πŸ’(𝟐)βˆ’πŸ‘ =
πŸ’ 𝟏
= .
πŸ– 𝟐
𝟏
πŸ—
10. If 𝒃 = , evaluate the following expressions.
a.
𝟏
π’ƒβˆ’πŸ
𝟏
𝟏
𝟏 βˆ’πŸ
( ) = πŸ—πŸ = πŸ‘
πŸ—
b.
πŸ“
π’ƒπŸ
πŸ“
𝟏 𝟐
𝟏 πŸ“
𝟏
( ) =( ) =
πŸ—
πŸ‘
πŸπŸ’πŸ‘
c.
πŸ‘
βˆšπŸ‘π’ƒβˆ’πŸ
𝟏 βˆ’πŸ
βˆšπŸ‘ ( ) = πŸ‘βˆšπŸπŸ• = πŸ‘
πŸ—
πŸ‘
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
11. Show that each expression is equivalent to πŸπ’™. Assume 𝒙 is a positive real number.
a.
πŸ’
βˆšπŸπŸ”π’™πŸ’
πŸ’
πŸ’
βˆšπŸπŸ” βˆ™ βˆšπ’™πŸ’ = πŸπ’™
πŸ‘
( βˆšπŸ–π’™πŸ‘ )
b.
βˆšπŸ’π’™πŸ
(πŸπ’™)𝟐
=
𝟏
(πŸ’π’™πŸ )𝟐
c.
𝟐
πŸ’π’™πŸ
= πŸπ’™
πŸπ’™
πŸ”π’™πŸ‘
πŸ‘
βˆšπŸπŸ•π’™πŸ”
πŸ”π’™πŸ‘
πŸ”
πŸ‘π’™πŸ‘
=
πŸ”π’™πŸ‘
= πŸπ’™
πŸ‘π’™πŸ
𝟏
12. Yoshiko said that πŸπŸ”πŸ’ = πŸ’ because πŸ’ is one-fourth of πŸπŸ”. Use properties of exponents to explain why she is or is not
correct.
𝟏 πŸ’
𝟏 πŸ’
𝟏
πŸ’
( )β‹…πŸ’
Yoshiko’s reasoning is not correct. By our exponent properties, (πŸπŸ”πŸ’ ) = πŸπŸ”
= πŸπŸ”πŸ = πŸπŸ”, but πŸ’πŸ’ = πŸπŸ“πŸ”. Since
𝟏
πŸ’
(πŸπŸ”πŸ’ ) β‰  πŸ’πŸ’ , we know that πŸπŸ” β‰  πŸ’.
πŸ’
𝟏
13. Jefferson said that πŸ–πŸ‘ = πŸπŸ” because πŸ–πŸ‘ = 𝟐 and πŸπŸ’ = πŸπŸ”. Use properties of exponents to explain why he is or is
not correct.
𝟏 πŸ’
πŸ’
MP.3
πŸ’
πŸ’
Jefferson’s reasoning is correct. We know that πŸ–πŸ‘ = (πŸ–πŸ‘ ) , so πŸ–πŸ‘ = πŸπŸ’ , and thus πŸ–πŸ‘ = πŸπŸ”.
𝟐
𝟐
𝟏
𝟐
𝟐
14. Rita said that πŸ–πŸ‘ = πŸπŸπŸ– because πŸ–πŸ‘ = πŸ–πŸ β‹… πŸ–πŸ‘, so πŸ–πŸ‘ = πŸ”πŸ’ β‹… 𝟐, and then πŸ–πŸ‘ = πŸπŸπŸ–. Use properties of exponents to
explain why she is or is not correct.
Rita’s reasoning is not correct because she did not apply the properties of exponents correctly. She should also
realize that raising πŸ– to a positive power less than 𝟏 will produce a number less than πŸ–. The correct calculation is
below.
𝟏 𝟐
𝟐
πŸ–πŸ‘ = (πŸ–πŸ‘ )
= 𝟐𝟐
=πŸ’
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
© 2014 Common Core, Inc. Some rights reserved. commoncore.org
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Lesson 3
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
ALGEBRA II
𝟏
𝟏 𝟐
𝟏
15. Suppose for some positive real number 𝒂 that (π’‚πŸ’ β‹… π’‚πŸ β‹… π’‚πŸ’ ) = πŸ‘.
What is the value of 𝒂?
a.
𝟏
𝟏
𝟏 𝟐
(π’‚πŸ’ β‹… π’‚πŸ β‹… π’‚πŸ’ ) = πŸ‘
𝟏 𝟏 𝟏 𝟐
(π’‚πŸ’+𝟐+πŸ’ ) = πŸ‘
(π’‚πŸ )𝟐 = πŸ‘
π’‚πŸ = πŸ‘
𝒂 = βˆšπŸ‘
b.
Which exponent properties did you use to find your answer to part (a)?
We used the properties 𝒃𝒏 β‹… π’ƒπ’Ž = π’ƒπ’Ž+𝒏 and (π’ƒπ’Ž )𝒏 = π’ƒπ’Žπ’ .
16. In the lesson, you made the following argument:
𝟏 πŸ‘
𝟏
𝟏
𝟏
(πŸπŸ‘ ) = πŸπŸ‘ β‹… πŸπŸ‘ β‹… πŸπŸ‘
𝟏 𝟏 𝟏
= πŸπŸ‘+πŸ‘+πŸ‘
= 𝟐𝟏
= 𝟐.
πŸ‘
πŸ‘
πŸ‘
Since √𝟐 is a number so that (√𝟐) = 𝟐 and
𝟏
πŸπŸ‘
𝟏 πŸ‘
𝟏
πŸ‘
is a number so that (πŸπŸ‘ ) = 𝟐, you concluded that πŸπŸ‘ = √𝟐. Which
exponent property was used to make this argument?
We used the property 𝒃𝒏 β‹… π’ƒπ’Ž = π’ƒπ’Ž+𝒏 . (Students may also mention the uniqueness of 𝒏th roots.)
Lesson 3:
Date:
Rational Exponentsβ€”What are 212 and 213 ?
7/28/17
© 2014 Common Core, Inc. Some rights reserved. commoncore.org
61
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Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.