Lesson 6: Proofs of Laws of Exponents
8•1
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 6: Proofs of Laws of Exponents Student Outcomes Students extend the previous laws of exponents to include all integer exponents. a
Students base symbolic proofs on concrete examples to show that
( x b ) = x ab
is valid for all integer
exponents.
Lesson Notes This lesson is not designed for all students, but for those who would benefit from a lesson that enriches their existing understanding of the laws of exponents. For that reason this is an optional lesson that can be used with students who have demonstrated mastery over concepts in Topic A.
Classwork Discussion (8 minutes) The goal of this lesson is to show why the laws of exponents, (10)–(12), are correct for all integers
b and for all For all
a
and
x , y >0 . We recall (10)–(12): x , y >0 and for all integers a a
b
x ∙x =x
and
b , we have
a+b
(10)
a
Lesson 6: Date:
( x b ) = x ab
(11)
( xy )a =x a y a .
(12)
Proofs of Laws of Exponents 4/3/15
1 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
MP. 7 &
8•1
This is a tedious process as the proofs for all three are somewhat similar. The proof of (10) is the most complicated of the three, but if one understands the proof of the easier identity (11), one will get a good idea of how all three proofs go. Therefore, we will only prove (11) completely. We have to first decide on a strategy to prove (11). Ask students what we already know about (11).
Elicit the following from the students Equation (7) of Lesson 5 says for any positive
x , ( x m ) n= x mn for
all whole numbers m and n .
MP.
How does this help us? It tells us that: (A) (11) is already known to be true when the integers satisfy
a
and
b , in addition,
Scaffolding: Keep statements (A), (B), and (C) visible throughout the lesson for reference purposes.
a ≥0 , b≥ 0 .
Equation (9) of Lesson 5 says that the following holds: −m
x
(B)
1 m x
=¿
for any whole number
m.
How does this help us? As we shall see from an exercise below, (B) is the statement that another special case of (11) is known. MP.
We also know that if
1 m 1 = m x x
()
(C) This is because if
x
is positive, then
for any whole number
m.
m is a positive integer, (C) is implied by equation (5) of Lesson 4, and if m=0 ,
then both sides of (C) are equal to
Lesson 6: Date:
1 .
Proofs of Laws of Exponents 4/3/15
2 This work is licensed under a
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
8•1
How does this help us? We will see from another exercise below that (C) is in fact another special case of (11), which is already known to be true.
The Laws of Exponents For
x
y >0
,
a
, and all integers
,
b
, the following holds:
x a ∙ x b= x a+b a
( x b ) = x ab ( xy )a =x a y a Facts we will use to prove (11): (A)
Exercises 1–3 (6 minutes)
(11) is already known to be true when the integers
a
and
b
satisfy
Students complete Exercises 1–3 in small groups. Exercise 1 Show that (C) is implied by equation (5) of Lesson 4 when even when
m=0
, and explain why (C) continues to hold
.
Equation (5) says for any numbers
holds:
m>0
x n xn = n y y
()
x
,
y
,
( y ≠ 0)
and any positive integer
n
, the following
. So,
1 m 1m = m x x
()
By
x n xn = n y y
()
for positive integer n and nonzero
y
(5)
¿
1 xm
Because
1m=1
If m = 0, then the left side is
Lesson 6: Date:
Proofs of Laws of Exponents 4/3/15
3 This work is licensed under a
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
m
8•1
0
1 1 = x x
() () ¿1
0
By definition of
x
By definition of
x0
and the right side is
1 1 = 0 m x x 1 1
¿ ¿ 1.
Exercise 2 −1
Show that (B) is in fact a special case of (11) by rewriting it as
m
, so that if
(B) says
b=m 1 m x
−m
x =¿
The left side of (B),
−m
x
The right side of (B),
(where
m
( x m ) =x (−1)m
is a whole number) and
a=−1
for any whole number
, (11) becomes (B).
.
is equal to
1 , xm
is equal to
x (−1) m
( xm)
.
−1
by the definition of
( xm)
−1
in Lesson 5.
−1
Therefore, (B) says exactly that
( x m ) =x (−1)m
.
Exercise 3 m
Show that (C) is a special case of (11) by rewriting (C) as
m
. Thus, (C) is the special case of (11) when
( x −1 ) =x m (−1 )
b=−1
and
a=m
for any whole number , where
m
is a whole
number.
Lesson 6: Date:
Proofs of Laws of Exponents 4/3/15
4 This work is licensed under a
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
8•1
m
1 1 = m x x
()
(C) says
for any whole number
m
.
The left side of (C) is equal to m
1 x
()
−1 m
¿(x )
−1
By definition of
x
By definition of
x
,
and the right side of (C) is equal to
1 xm and the latter is equal to
m
−m
¿x x m (−1 )
−m
,
m
. Therefore, (C) says
( x −1 ) =x m (−1 )
for any whole number
.
Discussion (4 minutes) In view of the fact that the reasoning behind the proof of (A) (Lesson 4) clearly cannot be extended to the MP.
case when
a
and/or
b is negative, it may be time to consider proving (11) in several separate
cases so that, at the end, these cases together cover all possibilities. (A) suggests that we consider the following four separate cases of identity (11): (i)
a , b≥ 0
(ii)
a ≥ 0 , b< 0
(iii)
a
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 6: Proofs of Laws of Exponents Student Outcomes Students extend the previous laws of exponents to include all integer exponents. a
Students base symbolic proofs on concrete examples to show that
( x b ) = x ab
is valid for all integer
exponents.
Lesson Notes This lesson is not designed for all students, but for those who would benefit from a lesson that enriches their existing understanding of the laws of exponents. For that reason this is an optional lesson that can be used with students who have demonstrated mastery over concepts in Topic A.
Classwork Discussion (8 minutes) The goal of this lesson is to show why the laws of exponents, (10)–(12), are correct for all integers
b and for all For all
a
and
x , y >0 . We recall (10)–(12): x , y >0 and for all integers a a
b
x ∙x =x
and
b , we have
a+b
(10)
a
Lesson 6: Date:
( x b ) = x ab
(11)
( xy )a =x a y a .
(12)
Proofs of Laws of Exponents 4/3/15
1 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
MP. 7 &
8•1
This is a tedious process as the proofs for all three are somewhat similar. The proof of (10) is the most complicated of the three, but if one understands the proof of the easier identity (11), one will get a good idea of how all three proofs go. Therefore, we will only prove (11) completely. We have to first decide on a strategy to prove (11). Ask students what we already know about (11).
Elicit the following from the students Equation (7) of Lesson 5 says for any positive
x , ( x m ) n= x mn for
all whole numbers m and n .
MP.
How does this help us? It tells us that: (A) (11) is already known to be true when the integers satisfy
a
and
b , in addition,
Scaffolding: Keep statements (A), (B), and (C) visible throughout the lesson for reference purposes.
a ≥0 , b≥ 0 .
Equation (9) of Lesson 5 says that the following holds: −m
x
(B)
1 m x
=¿
for any whole number
m.
How does this help us? As we shall see from an exercise below, (B) is the statement that another special case of (11) is known. MP.
We also know that if
1 m 1 = m x x
()
(C) This is because if
x
is positive, then
for any whole number
m.
m is a positive integer, (C) is implied by equation (5) of Lesson 4, and if m=0 ,
then both sides of (C) are equal to
Lesson 6: Date:
1 .
Proofs of Laws of Exponents 4/3/15
2 This work is licensed under a
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
8•1
How does this help us? We will see from another exercise below that (C) is in fact another special case of (11), which is already known to be true.
The Laws of Exponents For
x
y >0
,
a
, and all integers
,
b
, the following holds:
x a ∙ x b= x a+b a
( x b ) = x ab ( xy )a =x a y a Facts we will use to prove (11): (A)
Exercises 1–3 (6 minutes)
(11) is already known to be true when the integers
a
and
b
satisfy
Students complete Exercises 1–3 in small groups. Exercise 1 Show that (C) is implied by equation (5) of Lesson 4 when even when
m=0
, and explain why (C) continues to hold
.
Equation (5) says for any numbers
holds:
m>0
x n xn = n y y
()
x
,
y
,
( y ≠ 0)
and any positive integer
n
, the following
. So,
1 m 1m = m x x
()
By
x n xn = n y y
()
for positive integer n and nonzero
y
(5)
¿
1 xm
Because
1m=1
If m = 0, then the left side is
Lesson 6: Date:
Proofs of Laws of Exponents 4/3/15
3 This work is licensed under a
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
m
8•1
0
1 1 = x x
() () ¿1
0
By definition of
x
By definition of
x0
and the right side is
1 1 = 0 m x x 1 1
¿ ¿ 1.
Exercise 2 −1
Show that (B) is in fact a special case of (11) by rewriting it as
m
, so that if
(B) says
b=m 1 m x
−m
x =¿
The left side of (B),
−m
x
The right side of (B),
(where
m
( x m ) =x (−1)m
is a whole number) and
a=−1
for any whole number
, (11) becomes (B).
.
is equal to
1 , xm
is equal to
x (−1) m
( xm)
.
−1
by the definition of
( xm)
−1
in Lesson 5.
−1
Therefore, (B) says exactly that
( x m ) =x (−1)m
.
Exercise 3 m
Show that (C) is a special case of (11) by rewriting (C) as
m
. Thus, (C) is the special case of (11) when
( x −1 ) =x m (−1 )
b=−1
and
a=m
for any whole number , where
m
is a whole
number.
Lesson 6: Date:
Proofs of Laws of Exponents 4/3/15
4 This work is licensed under a
Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
8•1
m
1 1 = m x x
()
(C) says
for any whole number
m
.
The left side of (C) is equal to m
1 x
()
−1 m
¿(x )
−1
By definition of
x
By definition of
x
,
and the right side of (C) is equal to
1 xm and the latter is equal to
m
−m
¿x x m (−1 )
−m
,
m
. Therefore, (C) says
( x −1 ) =x m (−1 )
for any whole number
.
Discussion (4 minutes) In view of the fact that the reasoning behind the proof of (A) (Lesson 4) clearly cannot be extended to the MP.
case when
a
and/or
b is negative, it may be time to consider proving (11) in several separate
cases so that, at the end, these cases together cover all possibilities. (A) suggests that we consider the following four separate cases of identity (11): (i)
a , b≥ 0
(ii)
a ≥ 0 , b< 0
(iii)
a
