8 : Negative Eponents and the Laws of Eponents Student Outcomes Students know the definition of a number raised to a negative eponent. Students simplify and write equivalent epressions that contain negative eponents. Classwork Discussion 0 minutes) This lesson, and the net, refers to several of the equations used in the previous lessons. It may be helpful if students have some way of referencing these equations quickly e.g., a poster in the classroom or handout). For your convenience an equation reference sheet has been provided on page 54. and identities 6) 8): y be positive numbers throughout this lesson. Recall that we have the following three For all whole numbers m and n : m n = m+n 6) m ) n = mn 7) y ) n = n y n 8) m Make clear that we want 6) 8) to remain true even when and n are integers. Before we can say that, we have to first decide what something like 3 5 should mean. Date: 4/3/5 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
8 Allow time for the class to discuss the question, What should 3 5 mean? As in Lesson 4, where we introduced the concept of the zeroth power of a number, the overriding idea here is that the negative power of a number should be defined in a way to ensure that 6) 8) continue to hold when m and n are integers and not just whole numbers. Students will likely say that it should mean 3 5. Tell students that if that is what it meant, that is what we would write. When they get stuck, ask students this question, Using equation 6), what should 3 5 3 5 equal? Students should respond that they want to believe that equation 6) is still correct even when m and n are integers, and therefore, they should have 3 5 3 5 =3 5 + 5) =3 0 =. Scaffolding: Ask students, if is a number, then what value of would make the What does this say about the value 3 5?. The value 3 5 must be a fraction because 3 5 3 5 =, specifically the reciprocal of 3 5. Then, would it not be reasonable to define 3 n, in general, as 3 n? MP. Definition: For any positive number and for any positive integer n, we define n = n. Note that this definition of negative eponents says is just the reciprocal of. In particular, juncture. would make no sense if The definition has the following consequence: For a positive 9) =0. This eplains why we must restrict, b = b for all integers to be nonzero at this b. Note that 9) contains more information than the definition of negative eponent. For eample, it implies that with b= 3 in 9)) 5 3 = 5 3. Date: 4/3/5 2
8 Proof of 9): There are three possibilities for b : b>0, b=0, and b<0. If the b in 9) is positive, then 9) is just the definition of b, and there is nothing to prove. If b=0, then both sides of 9) are seen to be equal to and are, therefore, equal to each other. Again, 9) is correct. Finally, in general, let b be negative. Then b= n for some positive integer n. The left side of 9) is b = n ). The right side of 9) is equal to: = = n n =n n where we have made use of invert and multiply to simplify the comple fraction. Hence, the left side of 9) is again equal to the right side. The proof of 9) is complete. Definition: For any positive number and for any positive integer n, we define n = n. Allow time to discuss why we need to understand negative eponents. Answer: As we have indicated in Lesson, 4, the basic impetus for the consideration of negative and, in fact, arbitrary) eponents is the fascination with identities ) 3) Lesson 4), which are valid only for positive integer eponents. Such nice looking identities should be valid for all eponents. These identities are the starting point for the consideration of all other eponents beyond the positive integers. Even without knowing this aspect of identities ) 3), one can see the benefit of having negative eponents by looking at the complete epanded form of a decimal. For eample, the complete epanded form of 328.5403 is: 3 0 2 )+2 0 )+8 0 0 )+5 0 )+4 0 2 )+0 0 3 )+3 0 4 ). By writing the place value of the decimal digits in negative powers of 0, one gets a sense of the naturalness of the complete epanded form as the sum of whole number multiples of descending powers of 0. Date: 4/3/5 3
8 Eercises 0 0 minutes) Students complete Eercise independently or in pairs. Provide the correct solution. Net, have students complete Eercises 2 0 independently. Eercise Verify the general statement b = b for =3 and b= 5. If b were a positive integer, then we have what the definition states. However, b is a negative integer, specifically b= 5, so the general statement in this case reads 3 5 ) = 3 5. The right side of this equation is 3 = = 35 5 =35. 3 5 Since the left side is also 3 5, both sides are equal. 3 5 ) = 3 5 =35. Eercise 2 What is the value of 3 0 2 )? 3 0 2 )=3 0 2 = 3 0 2 =0.03 Eercise 3 What is the value of 3 0 5 )? 3 0 5 )=3 0 5 = 3 0 5 =0.00003 Date: 4/3/5 4
8 Eercise 4 Write the complete epanded form of the decimal 4. 728 in eponential notation. 4.728=4 0 0 )+7 0 )+ 2 0 2 )+ 8 0 3 ) For Eercises 5 0, write an equivalent epression, in eponential notation, to the one given and simplify as much as possible. Eercise 5 5 3 = 5 3 Eercise 6 8 9 =8 9 Eercise 7 3 2 4 =3 2 4 = 3 2 4 Eercise 8 be a nonzero number. 3 = 3 Eercise 9 be a nonzero number. 9 9= Eercise 0, y be two nonzero numbers. y 4 = y 4 = y 4 Discussion 5 minutes) We now state our main objective: For any positive number, y and for all integers a and b, a b = a+b 0) b ) a = ab ) Date: 4/3/5 5
8 y ) a = a y a. 2) We accept that for positive numbers, y a b and all integers and, a b = a+b b ) a = ab y ) a = a y a. We claim a Identities 0) 2) are called the laws of eponents for integer eponents. They clearly generalize 6) 8). The laws of eponents will be proved in the net lesson. For now, we want to use them effectively. In the process, we will get a glimpse of why they are worth learning. We will show that knowing 0) 2) means also knowing 4) and 5) automatically. Thus, MP. it is enough to know only three facts, 0) 2), rather than five facts, 0) 2) and 4) and 5). Incidentally, the preceding sentence demonstrates why it is essential to learn how to use symbols because if 0) 2) were stated in terms of eplicit numbers, the preceding sentence would not even make sense. Note to Teacher: You could mention that 0) 2) are valid even when a and b are rational numbers make sure they know rational numbers refer to positive and negative fractions). The fact that they are true also for all real We reiterate the following: The discussion below assumes the validity of 0) 2) for the time being. We claim a b = a b for all integers 3) y ) a = a y a for any integer a. 4) a, b Note to Teacher: The need for formulas about comple fractions will be obvious in subsequent lessons and will not be consistently pointed out. This fact should be brought to the attention of students. Ask students why these must be considered comple fractions. Date: 4/3/5 6
8 Note that identity 3) says much more than 4): Here, a and b can be integers, rather than positive integers and moreover there is no requirement that a>b. Similarly unlike 5), the a in 4) is an integer rather than just a positive integer. Eercises and 2 5 minutes) Students complete Eercises and 2 independently or in pairs in preparation of the proof of 3) in general. Eercise Eercise 2 9 2 9 5=92 5 7 6 7 3 =76 7 3=76 7 3 =7 6+3 Proof of 3): a b a b By the product formula for comple fractions a b By b = b 9) a + b ) By a b = a+b 0) a b Eercises 3 and 4 0 minutes) Students complete Eercise 3 in preparation for the proof of 4). Check before continuing to the general proof of 4). Eercise 3 If we let b= in ), a be any integer, and y be any positive number, what do we get? y ) a = y a Date: 4/3/5 7
8 Eercise 4 Show directly that 7 5) 4 = 7 4 5 4. 7 5) 4 7 5) 4 By the product formula 7 5 ) 4 By definition 7 4 5 ) 4 By y ) a = a y a 2) 7 4 5 4 By b ) a = ab ) 7 4 5 4 By b = b 9) 7 4 5 4 By product formula Proof of 4): y ) a y ) a By the product formula for comple fractions y ) a By definition a y ) a By y ) a = a y a 2) a y a By b ) a = ab ), also see Eercise 3 a y a By b = b 9) a y a Date: 4/3/5 8
8 Students complete Eercise 4 independently. Provide the solution when they are finished. Closing 5 minutes) Summarize, or have students summarize, the lesson. By assuming 0) 2) were true for integer eponents, we see that 4) and 5) would also be true. 0) 2) are worth remembering because they are so useful and allow us to limit what we need to memorize. Eit Ticket 5 minutes) Date: 4/3/5 9
8 Name Date : Negative Eponents and the Laws of Eponents Eit Ticket Write each answer as a simplified epression that is equivalent to the given one. 7654 3 4 = f be a nonzero number. f 4 = 67 28796 = a, b be numbers b 0). a b = Date: 4/3/5
8 g be a nonzero number. g = Eit Ticket Sample Solutions 7654 3 4 = Write each answer as a simplified epression that is equivalent to the given one. 76543 4 f be a nonzero number. f 4 = f 4 67 28796 =67 28796 = 67 28796 a, b be numbers b 0). a b =a b = a b g be a nonzero number. g = g Problem Set Sample Solutions. Compute: 3 3 3 2 3 3 0 3 3 2 =3 3 =27 Compute: 5 2 5 0 5 8 5 0 5 0 5 8 =5 2 =25 Compute for a nonzero number, a : a m a n a l a n a m a l a 0 =a 0 = Date: 4/3/5
8. Without using 0), show directly that 7.6 ) 8 =7.6 8. 7.6 ) 8 7.6 ) 8 By definition 8 7.6 8 By y ) n n y n 5) 7.6 8 7.6 8 By definition 2. Without using 0), show prove) that for any whole number n and any positive number y, y ) n = y n. y ) n y ) n By definition n y n By y ) n n y n 5) y n y n By definition 3. Show directly without using 3) that 2.8 5 2.8 7 2.8 2. 2.8 5 2.8 7 2.8 5 2.8 7 By the product formula for comple fractions Date: 4/3/5
8 2.8 5 2.8 7 By definition 2.8 5 2.8 7 By the product formula for comple fractions 2.8 5+7 By a b = a+b 0) 2.8 2 2.8 2 By definition Date: 4/3/5
8 Equation Reference Sheet For any numbers, y 0 in 4 ) and y 0 in 5 )) and any positive integers m, n, the following holds: m n = m+n ) m ) n = mn 2 ) y ) n = n y n 3 ) m n =m n 4 ) y ) n = n y n 5 ) For any numbers, y and for all whole numbers m, n, the following holds: m n = m+n 6 ) m ) n = mn 7 ) y ) n = n y n 8 ) For any positive number and all integers b, the following holds: b = b 9 ) For any numbers, y and all integers a, b, the following holds: Date: 4/3/5
8 a b = a+b 0 ) b ) a = ab ) y ) a = a y a 2 ) a b =a b 3 ) y ) a = a y a 4 ) Date: 4/3/5