# Lesson 6: Proofs of Laws of Exponents

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1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 Student Outcomes Students extend the previous laws of exponents to include all integer exponents. Students base symbolic proofs on concrete examples to show that (x b ) a x ab is valid for all integer exponents. Lesson Notes This lesson is not designed for all students, but rather 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, (0) (2), are correct for all integers a and b and for all x, y 0. We recall (0) (2): For all x, y 0 and for all integers a and b, we have x a x b x a+b (0) (x b ) a x ab () (xy) a x a y a. (2) Scaffolding: Ask advanced learners to consider why it is necessary to restrict the values of x and y to nonzero numbers. They should be able to respond that if a or b is a negative integer, the value of the expression could depend on division by zero, which is undefined. MP.7 & MP.8 This is a tedious process as the proofs for all three are somewhat similar. The proof of (0) is the most complicated of the three, but students who understand the proof of the easier identity () should get a good idea of how all three proofs go. Therefore, we will only prove () completely. We have to first decide on a strategy to prove (). Ask students what we already know about (). Elicit the following from students MP. Equation (7) of Lesson 5 says for any nonzero x, (x m ) n x mn for all whole numbers m and n. How does this help us? It tells us that: (A) () is already known to be true when the integers a and b, in addition, satisfy a 0, b 0. Scaffolding: Keep statements (A), (B), and (C) visible throughout the lesson for reference purposes. 62 This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

2 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 MP. Equation (9) of Lesson 5 says that the following holds: (B) x m for any whole number m. xm How does this help us? As we shall see from an exercise below, (B) is the statement that another special case of () is known. We also know that if x is nonzero, then (C) ( x )m x m for any whole number m. This is because if 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. How does this help us? We will see from another exercise below that (C) is in fact another special case of (), which is already known to be true. The Laws of Exponents For x, y 0, and all integers a, b, the following holds: x a x b x a+b (x b ) a x ab (xy) a x a y a. Facts we will use to prove (): (A) () is already known to be true when the integers a and b satisfy a 0, b 0. (B) x m xm for any whole number m. (C) ( x )m xm for any whole number m. Exercises 3 (6 minutes) Students complete Exercises 3 in small groups. Exercise Show that (C) is implied by equation (5) of Lesson 4 when m > 0, and explain why (C) continues to hold even when m 0. Equation (5) says for any numbers x, y, (y 0) and any positive integer n, the following holds: ( x y )n xn yn. So, ( x )m m x m x m By (x y )n xn yn for positive integer n and nonzero y (5) Because m 63 This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

3 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 If m 0, then the left side is and the right side is ( m x ) ( 0 x ) By definition of x 0, x m x 0 By definition of x 0. Exercise 2 Show that (B) is in fact a special case of () by rewriting it as (x m ) x ( )m for any whole number m, so that if b m (where m is a whole number) and a, () becomes (B). (B) says x m x m. The left side of (B), x m is equal to x ( )m. The right side of (B), x m, is equal to (xm ) by the definition of (x m ) in Lesson 5. Therefore, (B) says exactly that (x m ) x ( )m. Exercise 3 Show that (C) is a special case of () by rewriting (C) as (x ) m x m( ) for any whole number m. Thus, (C) is the special case of () when b and a m, where m is a whole number. (C) says ( x )m The left side of (C) is equal to xm for any whole number m. ( x )m (x ) m By definition of x, and the right side of (C) is equal to x m x m By definition of x m, and the latter is equal to x m( ). Therefore, (C) says (x ) m x m( ) for any whole number m. 64 This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

4 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 MP.3 Discussion (4 minutes) In view of the fact that the reasoning behind the proof of (A) (Lesson 4) clearly cannot be extended to a case in which a and/or b is negative, it may be time to consider proving () 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 (): (i) a, b 0 (ii) a 0, b < 0 (iii) a < 0, b 0 (iv) a, b < 0. Why are there are no other possibilities? Do we need to prove case (i)? No, because (A) corresponds to case (i) of (). We will prove the three remaining cases in succession. Scaffolding: Have students think about the four quadrants of the plane. Read the meaning of the four cases aloud as you write them symbolically. Discussion (0 minutes) Case (ii): We have to prove that for any nonzero x, (x b ) a x ab, when the integers a and b satisfy a 0, b < 0. For example, we have to show that (5 3 ) 4 5 ( 3)4, or (5 3 ) The following is the proof: (5 3 ) 4 ( 5 3 ) 4 By definition (5 3 4 By ( x )m for any whole number m (C) xm ) 5 2 By (xm ) n x mn for all whole numbers m and n (A) 5 2 By definition In general, we just imitate this argument. Let b c, where c is a positive integer. We now show that the left side and the right side of (x b ) a x ab are equal. The left side is (x b ) a (x c ) a ( x c)a By x m for any whole number m (B) m x (x c ) a By ( x )m for any whole number m (C) xm x ac By (xm ) n x mn for all whole numbers m and n (A) Scaffolding: Keep the example, done previously with concrete numbers, visible so students can relate the symbolic argument to the work just completed. Remind students that when using concrete numbers, we can push through computations and show that the left side and right side are the same. For symbolic arguments, we must look at each side separately and show that the two sides are equal. 65 This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

5 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 The right side is x ab x a( c) x (ac) x ac. By x m for any whole number m (B) xm The left and right sides are equal; thus, case (ii) is done. Case (iii): We have to prove that for any nonzero x, (x b ) a x ab, when the integers a and b satisfy a < 0 and b 0. This is very similar to case (ii), so it will be left as an exercise. Exercise 4 (4 minutes) Students complete Exercise 4 independently or in pairs. Exercise 4 Proof of Case (iii): Show that when a < 0 and b 0, (x b ) a x ab is still valid. Let a c for some positive integer c. Show that the left and right sides of (x b ) a x ab are equal. The left side is (x b ) a (x b ) c The right side is So, the two sides are equal. (x b ) c x cb. x ab x ( c)b x (cb) x cb. By x m By (xm ) n x mn for all whole numbers m and n (A) By x m Discussion (8 minutes) The only case remaining in the proof of () is case (iv). Thus, we have to prove that for any nonzero x, (x b ) a x ab when the integers a and b satisfy a < 0 and b < 0. For example, (7 5 ) because (7 5 ) 8 (7 5 ) 8 By x m for any whole number m (B) xm By case (ii) (5 8) By x m for any whole number m (B) xm 66 This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

6 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 In general, we can imitate this explicit argument with numbers as we did in case (ii). Let a c and b d, where c and d are positive integers. Then, the left side is The right side is (x b ) a (x c ) d (x c ) d By x m for any whole number m (B) xm x cd By case (ii) By x m for any whole number m (B) x cd xm x cd. By invert-and-multiply for division of complex fractions x ab x ( c)( d) x cd. Scaffolding: Students may ask why the exponent c remained negative in the second line while d became positive. Reconcile this through the use of a concrete example or by pointing to the previous problem (7 5 ) 8 (7 5 ) 8 first line. The left side is equal to the right side; thus, case (iv) is finished. Putting all of the cases together, the proof of () is complete. We now know that () is true for any nonzero integer x and any integers a, b. Closing (2 minutes) Summarize, or have students summarize, the lesson. We have proven the laws of exponents are valid for any integer exponent. Exit Ticket (3 minutes) 67 This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

7 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 Name Date Exit Ticket. Show directly that for any nonzero integer x, x 5 x 7 x Show directly that for any nonzero integer x, (x 2 ) 3 x This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

8 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 8 Exit Ticket Sample Solutions. Show directly that for any nonzero integer x, x 5 x 7 x 2. x 5 x 7 x 5 x 7 x 5 x 7 By x m By the product formula for complex fractions x 5+7 By xm x n x m+n for whole numbers m and n (6) x 2 x 2 By x m 2. Show directly that for any nonzero integer x, (x 2 ) 3 x 6. (x 2 ) 3 (x 2 ) 3 By x m By case (ii) of () (2 3) x x 6 x 6 By x m Problem Set Sample Solutions. You sent a photo of you and your family on vacation to seven Facebook friends. If each of them sends it to five of their friends, and each of those friends sends it to five of their friends, and those friends send it to five more, how many people (not counting yourself) will see your photo? No friend received the photo twice. Express your answer in exponential notation. # of New People to View Your Photo Total # of People to View Your Photo (5 7) 7 + (5 7) + (5 2 7) 7 + (5 7) + (5 2 7) + (5 3 7) The total number of people who viewed the photo is ( ) This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

9 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson Show directly, without using (), that ( ) ( ) (.27 36) By definition ( ) 85 By ( x )m xm for any whole number m (C) By (xm ) n x mn for whole numbers m and n (7) By x m 3. Show directly that 3 ) 27 3 ) 56 3 ) ) 27 3 ) 56 3 )27 3 )56 3 ) 27 ( 2 3 ) 56 By definition By the product formula for complex fractions 3 )27+56 By xm x n x m+n for whole numbers m and n (6) 3 )83 3 ) 83 By x m 4. Prove for any nonzero number x, x 27 x 56 x 83. x 27 x 56 x 27 x 56 x 27 x 56 By definition By the product formula for complex fractions x By xm x n x m+n for whole numbers m and n (6) x 83 x 83 By x m 5. Prove for any nonzero number x, x m x n x m n for positive integers m and n. x m x n x m x n x m x n By definition By the product formula for complex fractions x m+n By xm x n x m+n for whole numbers m and n (6) x (m+n) By x m x m n 70 This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

10 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson Which of the preceding four problems did you find easiest to do? Explain. Students will likely say that x m x n x m n (Problem 5) was the easiest problem to do. It requires the least amount of writing because the symbols are easier to write than decimal or fraction numbers. 7. Use the properties of exponents to write an equivalent expression that is a product of distinct primes, each raised to an integer power (2 5)5 (3 3) 2 (2 3) This work is derived from Eureka Math and licensed by Great Minds. 205 Great Minds. eureka-math.org This file derived from G8-M-TE

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