: Ratioal Expoets What are ad? Studet Outcomes Studets will calculate quatities that ivolve positive ad egative ratioal expoets. Lesso Notes Studets exted their uderstadig of iteger expoets to ratioal expoets by examiig the graph of f(x) = 2 x ad estimatig the values of 2 1 2 ad 2 1. The lesso establishes the meaig of these umbers i terms of radical expressios ad these form the basis of how we defie expressios of the form b 1 before geeralizig further to expressios of the form b m, where b is a positive real umber ad m ad are itegers, with 0 (N-RN.A.1). The lesso ad problem set provide fluecy practice i applyig the properties of expoets to expressios cotaiig ratioal expoets ad radicals (N-RN.A.2). I the followig lesso, studets will verify that the defiitio of a expressio with ratioal expoets, b m = b m, is cosistet with the remaiig expoetial properties. The lesso begis with studets creatig a graph of a simple expoetial fuctio, f(x) = 2 x, which they studied i Module of Algebra 1. I that module, studets also leared about geometric sequeces ad their relatioship to expoetial fuctios, a cocept that we revisit at the ed of this module. I Algebra 1, studets worked with expoetial fuctios with iteger domais. This lesso, together with the subsequet Lessos ad 5, helps studets uderstad why the domai of a expoetial fuctio f(x) = b x, where b is positive umber ad b 1, is the set of real umbers. To do so we must establish what it meas to raise b to a ratioal power ad, i Lesso 5, to ay real umber power. Classwork Opeig (1 miutes) I Algebra 1, studets worked with geometric sequeces ad simple expoetial fuctios. Remid them that i Lesso 1, we created formulas based o repeatedly doublig ad halvig a umber whe we modeled foldig a piece of paper. We reviewed how to use the properties of expoets for expressios that had iteger expoets. Opeig Exercise (5 miutes) Have studets graph f(x) = 2 x for each iteger x from x = 2 to x = 5 without usig a graphig utility or calculator o the axes provided. Discuss the patter of poits ad ask studets to coect the poits i a way that produces a smooth curve. Studets should work these two problems idepedetly ad check their solutios with a parter if time permits before you lead a whole class discussio to review the solutios. Scaffoldig: Ecourage studets to create a table of values to help them costruct the graph. For advaced learers, have them repeat these exercises with the fuctio f(x) = x ad ask them to estimate 1 2 ad 1. Opeig Exercise MP. a. What is the value of? Justify your aswer.
A possible studet respose follows: I thik it will be aroud. 5 because 00 = ad =. b. Graph f(x) = x for each iteger x from x = to x = 5. Coect the poits o your graph with a smooth curve. y 6 2 28 2 20 16 12 8-2 -1 1 2 5 6 - x Ask for a few voluteers to explai their reasoig for their aswers to Opeig Exercise part (a). The, debrief these two exercises by leadig a short discussio. The directios i the Opeig Exercise said to coect the poits with a smooth curve. What does it imply about the domai of a fuctio whe we coect poits that we have plotted? That the domai of the fuctio icludes the poits betwee the poits that we plotted. How does your graph support or refute your aswer to the first exercise? Do you eed to modify your aswer to the questio: What is the value of 2 1 2? Why or why ot? If the domai is all real umbers, the the value of 2 1 2 will be the y-coordiate of the poit o the graph where x = 1. From the graph it looks like the value is somewhere betwee 1 ad 2. The 2 scalig o this graph is ot detailed eough for me to accurately refie my aswer yet. Trasitio to the ext set of exercises by tellig the studets that we ca better estimate the value of 2 1 2 by lookig at a graph that is scaled more precisely. Have studets proceed to the ext questios. They should work i small groups or with a parter o these exercises. The graph of f(x) = 2 x show below for the ext exercises will appear i the studet materials but you could also display the graph usig techology. Have studets estimate the value of 2 1 2 ad 2 1 from the graph. Studets should work by themselves or i pairs to work through the followig questios without usig a calculator.
The graph at right shows a close-up view of f(x) = x for 00. 5 < x <. 5. c. Fid two cosecutive itegers that are over ad uder estimates of the value of. y 2 00 < < < < 1.5 d. Does it appear that is halfway betwee the itegers you specified i Exercise 1? 1 No, it looks like is a little less tha halfway betwee ad. 0.5 e. Use the graph of f(x) = x to estimate the value of. -0.5 0.5 1 1.5 x. f. Use the graph of f(x) = x to estimate the value of.. 5 MP.6 Discussio (9 miutes) Before gettig ito the poit of this lesso, which is to coect ratioal expoets to radical expressios, revisit the iitial questio with studets. What is the value of 2 1 2? Does ayoe wat to adjust their iitial guess? Our iitial guess was a little too big. It seems like 1. might be a better aswer. How could we make a better guess? We could look at the graph with eve smaller icremets for the scale usig techology. If time permits, you ca zoom i further o the graph of f(x) = 2 x usig a graphig calculator or other techology either by examiig a graph or a table of values of x closer ad closer to 1. 2 Scaffoldig: If eeded, you ca demostrate the argumet usig perfect squares. For example, use a base of istead of a base of 2. Show that 1 2 2 = ad 2 =. Next, we will make the coectio that 2 1 2 = 2. Walk studets through the followig questios, providig guidace as eeded. Studets proved that there was oly oe positive umber that squared to 2 i Geometry, Module 2. You may eed to remid them of this with a bit more detail if they are strugglig to follow this argumet. Assume for the momet that whatever 2 1 2 meas that it satisfies b m b = b m+. Workig with this assumptio what is the value of 2 1 2 2 1 2? It would be 2 because 2 1 2 2 1 2 = 2 1 2 +1 2 = 2 1 = 2. What uique positive umber squares to 2? That is what is the oly positive umber that whe multiplied by MP.7
itself is equal to 2? By defiitio, we call the uique positive umber that squares to 2 the square root of 2 ad we write 2. Write the followig statemets o the board ad ask your studets to compare them ad thik about what they must tell us about the meaig of 2 1 2. 2 1 2 2 1 2 = 2 ad 2 2 = 2 What do these two statemets tell us about the meaig of 2 1 2. Sice both statemets ivolve multiplyig a umber by itself ad gettig 2, ad we kow that there is oly oe umber that does that, we ca coclude that 2 1 2 = 2. MP.7 At this poit, you ca have studets cofirm these results by usig a calculator to approximate both 2 1 2 ad 2 to several decimal places. I the opeig, we approximated 2 1 2 graphically, ad ow we have show it to be a irratioal umber. Next ask studets to thik about the meaig of 2 1 usig a similar lie of reasoig. Assume that whatever 2 1 meas it will satisfy b m b = b m+. What is the value of 2 1 2 1 2 1? The value is 2 because 2 1 2 1 2 1 = 2 1 +1 +1 = 2 1 = 2. What is the value of 2 2 2? The value is 2 because 2 2 2 = 2 = 2. Scaffoldig: If eeded, you ca demostrate the argumet usig perfect cubes. For example, use a base of 8 istead of a base of 2. Show 8 1 = 8 ad 8 = 8. What appears to be the meaig of 2 1? Sice both the expoet expressio ad the radical expressio ivolve multiplyig a umber by itself three times ad the result is equal to 2, we kow that 2 1 = 2. Studets ca also cofirm usig a calculator that the decimal approximatios of 2 1 ad 2 are the same. Next, we ask them to geeralize their fidigs. Ca we geeralize this relatioship? Does 2 1 = 2? Does 2 1 10 10 = 2? What is 2, 1 for ay positive iteger? Why? MP.8 2 1 = 2 because times 2 1 = 2 1 2 1 2 1 = 2 1 +1 + +1 = 2 1 = 2. times MP.8 Have studets cofirm these usig a calculator as well checkig to see if the decimal approximatios of 2 1 ad 2 are the same for differet values of such as, 5, 6, 10,. Be sure to share the geeralizatio show above o the board to help studets uderstad why it makes sese to defie 2 1 to be 2. However, be careful to ot stop here; there is a problem with our reasoig if we do ot defie 2. I previous courses, oly square roots ad cube roots were defied. We first eed to defie the th root of a umber; there may be more tha oe, as i the case where 2 2 = ad
( 2) 2 =. We say that both 2 ad 2 are square roots of. However, we give priority to the positive-valued square root, ad we say that 2 is the pricipal square root of. We ofte just refer to the square root of whe we mea the pricipal square root of. The defiitio of Pth root preseted below is cosistet with allowig complex Pth roots, th which studets will ecouter i college if they pursue egieerig or higher mathematics. If we allow complex P, 2 1 + i, ad 2 1 2 2 2 2 the pricipal cube root of 2. There is o eed to discuss this with ay but the most advaced studets. roots, there are three cube roots of 2: 2 i, ad we refer to the real umber 2 as The 2 is the positive real umber a such that a = 2. I geeral, if a is positive, the the Pth root of a exists for ay positive iteger, ad if a is egative, the the Pth root of a exists oly for odd itegers. This eve/odd coditio is hadled subtly i the defiitio below; the Pth root exists oly if there is already a expoetial relatioship b = a. Preset the followig defiitios to studets ad have them record them i their otes. NN TH ROOT OF A NUMBER: Let aa ad bb be umbers, ad let be a positive iteger. If bb = aa, the aa is a th root of bb. If =, the the root is a called a square root. If =, the the root is called a cube root. PRINCIPAL NN TH ROOT OF A NUMBER: Let bb be a real umber that has at least oe real th root. The pricipal th root of bb is the real th root that has the same sig as bb, ad is deoted by a radical symbol: bb. Every positive umber has a uique pricipal th root. We ofte refer to the pricipal th root of b as just the th root of b. The th root of 0 is 0. Studets have already leared about square ad cube roots i previous grades. I Module 1 ad at the begiig of this lesso, studets worked with radical expressios ivolvig cube ad square roots. Explai that the Pth roots of a umber satisfy the same properties of radicals leared previously. Have studets record these properties i their otes. If a 0, b 0 (b 0 whe b is a deomiator) ad is a positive iteger, the ab = a b ad a b = a. b Backgroud iformatio regardig Pth roots ad their uiqueess is provided below. You may wish to share this with your advaced learers or the etire class if you exted this lesso to a additioal day. The existece of the pricipal Pth root of a real umber b is a cosequece of the fudametal theorem of algebra: Cosider the polyomial fuctio f(x) = x b. Whe is odd, we kow that f has at least oe real zero because the graph of f must cross the x-axis. That zero is a positive umber which (after showig that it is the ONLY real zero) is the Pth root. The case for whe is eve follows a similar argumet. To show uiqueess of the Pth root, suppose there are two Pth roots of a umber b, x ad y such that x > 0, y > 0, x = b, ad y = b. The x y = b b = 0. The expressio x y factors (see Lesso 7 i Module 1). 0 = x y = (x y)(x 1 + x 2 y + x y 2 + + xy 2 + y 1 )
Sice both x ad y are positive, the secod factor is ever zero. Thus, for x y = 0, we must have x y = 0 ad it follows that x = y. Thus, there is oly oe Pth root of b. A proof of the first radical property is show below for teacher backgroud iformatio. You may wish to share this proof with your advaced learers or the etire class if you exted this lesso to a additioal day. Prove that ab = a b. Let x 0 be the umber such that x = a ad let y 0 be the umber such that y = b, so that x = a ad y = b. The, by a property of expoets, (xy) = x y = ab. Thus, xy must be the Pth root of ab. Writig this usig our otatio gives ab = xy = a b. After studets have recorded this iformatio i their otes, you ca proceed with Example 1 ad Exercise 1. Example 1 ( miutes) This example will familiarize studets with the wordig i the defiitio preseted above. Scaffoldig: If eeded, cotiue to support studets that struggle with abstractio by icludig additioal umeric examples. Example 1 a. What is the Pth root of 6? x = 6 whe x = because = 6. Thus, 6 =. b. What is the cube root of 125? x = 5 whe x = 5 because 5 = 5. Thus, 5 = 5. c. What is the 5Pth root of 0000, 000000? x 5 = 0000, 000000 whe x = 00 because 00 5 5 = 0000, 000000. Thus, 0000, 000000 = 00. Exercise 1 (2 miutes) I these brief exercises, studets will work with defiitio of Pth roots ad the multiplicatio property preseted above. Have studets check their work with a parter ad briefly discuss as a whole class ay questios that arise. Exercise 1 1. Evaluate each expressio. a. 8 5 b. c. 9 7 =
d. 5 0000 00, 000000 = 00 Discussio (8 miutes) Retur to the questio posted i the title of this lesso: What are 2 1 2 ad 2 1? Now we kow the aswer, 2 1 2 = 2 ad 2 1 = 2. So far, we have give meaig to 2 1 by equatig 2 1 = 2. Ask studets if they believe these results exted to ay base b > 0. We just did some exercises where the b-value was a umber differet from 2. I our earlier work, was there somethig special about usig 2 as the base of our expoetial expressio? Would these results geeralize to expressios of the form? 1 7? 1 10? 1 b, 1 for ay positive real umber b? There is othig iheretly special about the base 2 i the above discussio. These results should geeralize to expressios of the form b 1 for ay positive real umber base b because we have defied a Pth root for ay positive base b. Now that we kow what the Pth root of a umber b, b, meas, the work we did earlier with base 2 suggests that b 1 should also be defied as the Pth root of b. Discuss this defiitio with your class. If the class is uclear o the defiitio, do some umerical examples: ( 2) 1 5 = 2 because ( 2) 5 = 2, but ( 16) 1 does ot exist because there is o pricipal Pth root of a egative umber. For a real umber b ad a positive iteger, defie b 1 to be the pricipal th root of b whe it exists. That is, b 1 = b. Note that this defiitio holds for ay real umber b if is a odd iteger ad for positive real umbers b if is a eve iteger. You may choose to emphazise this with your class. Thus, whe b is egative ad is a odd iteger, the expressio b 1 will be egative. If is a eve iteger, the we must restrict b to positive real umbers oly, ad b 1 will be positive. I the ext lesso, we see that with this defiitio b 1 satisfies all the usual properties of expoets so it makes sese to defie it i this way. At this poit, you ca also revisit our origial questio with the studets oe more time. What is the value of 2 1 2? What does it mea? What is the value of b 1 2 for ay positive umber b? How are radicals related to ratioal expoets? We ow kow that 2 1 2 is equal to 2. I geeral b 1 2 = b for ay positive real umber b. A umber that cotais a radical ca be expressed usig ratioal expoets i place of the radical. As you cotiue the discussio, we exted the defiitio of expoetial expressios with expoets of the form 1 to ay positive ratioal expoet. We begi by cosiderig a example: what do we mea by 2? Give studets a few miutes to respod idividually i writig to this questio o their studet pages ad the have them discuss their reasoig
with a parter. Make sure to correct ay blatat miscoceptios ad to clarify icomplete thikig as you lead the discussio that follows. Discussio If = ad =, what does equal? Explai your reasoig. MP. Studet solutios ad explaatios will vary. Oe possible solutio would be = so it must mea that =. Sice the properties of expoets ad the meaig of a expoet made sese with itegers ad ow for ratioal umbers i the form, it would make sese that they would work all ratioal umbers too. Now that we have a defiitio for expoetial expressios of the form b, 1 use the discussio below to defie b m, where m ad are both itegers, 0, ad b is a positive real umber. Make sure studets uderstad that our iterpretatio of b m must be cosistet with the expoet properties (which hold for iteger expoets) ad the defiitio of b 1. How ca we rewrite the expoet of 2 usig itegers ad ratioal umbers i the form 1? We ca write 2 = 2 1 or we ca write 2 = (2 ) 1. Now apply our defiitio of b 1. 2 = 2 1 = 2 or 2 = (2 ) 1 = 2 = 8 Does this make sese? If 2 = 8, the if we raise 2 to the fourth power, we should get 8. Does this happe? 2 = 2 2 2 2 = 2 = 2 = 8. So, 8 is the product of four equal factors, which we deote by 2. Thus, 2 = 8. Take a few miutes to allow studets to thik about geeralizig their work above to 2 m ad the to b m. Have them write a respose to the followig questios ad share it with a parter before proceedig as a whole class. Ca we geeralize this result? How would you defie 2 m, for positive itegers m ad? Cojecture: 2 m = 2 m, or equivaletly, 2 m = 2 m. Ca we geeralize this result to ay positive real umber base b? What is m? 7 m? 10 m? b m? There is othig iheretly special about the base 2 i the above discussio. These results should geeralize to expressios of the form b m for ay positive real umber base b. The we are ready to defie b m = b m This result is summarized i the box below. for positive itegers m ad ad positive real umbers b. For ay positive itegers m ad, ad ay real umber b for which b 1 exists, we defie b m = b m, which is equivalet to b m = b m
Note that this property holds for ay real umber b if is a odd iteger. You may choose to emphasize this with your class. Whe b is egative ad is a odd iteger, the expressio b m will be egative. If is a eve iteger the we must restrict b to positive real umbers oly. Exercises 2 8 ( miutes) I these exercises, studets use the defiitios above to rewrite ad evaluate expressios. Have studets check their work with a parter ad briefly discuss as a whole class ay questios that arise. Exercises 2 8 Rewrite each expoetial expressio as a th root. 2. =. 5 5 5 =. 5 5 5 = 5. 6 00 6 00 = 00 6 Rewrite the followig expoetial expressios as equivalet radical expressios. If the umber is ratioal, write it without radicals or expoets. 6. = = 7. 5 5 = 5 = 5 = 5 = 8. 8 5 8 5 = 5 8 5 = 8 = 5 =
Exercise 9 ( miutes) I this exercise, we ask studets to cosider a egative ratioal expoet. Have studets work directly with a parter ad ask them to use thikig similar to the precedig discussio. Correct ad exted studet thikig as you review the solutio. Exercise 9 9. Show why the followig statemet is true: = MP. Studet solutios ad explaatios will vary. Oe possible solutio would be = = = = Sice is a real umber the properties of expoets hold ad we have defied bb. We ca show these two expressios are the same. Share the followig property with your class ad show how the work they did i the previous exercises supports this coclusio. Should you choose to you ca verify these properties usig a argumet similar to the oes preseted earlier for the meaig of b m. For ay positive itegers m ad, ad ay real umber b for which b 1 exists, we defie b m = 1 b m or, equivaletly, b m 1 = b m. Exercises 10 12 ( miutes) Exercises 10 12 Rewrite the followig expoetial expressios as equivalet radical expressios. If the umber is ratioal, write it without radicals or expoets. 10. = = = 8. 7 7 = 7 = 7 = = 9
12. We have = = =. Alteratively, = = () = =. Closig ( miutes) Have studets summarize the key poits of the lesso i writig. Circulate aroud the classroom to iformally assess uderstadig ad provide assistace. Their work should reflect the summary provided below. Lesso Summary NN TH ROOT OF A NUMBER: Let aa ad bb be umbers, ad let be a positive iteger. If bb = aa, the aa is a th root of bb. If =, the the root is a called a square root. If =, the the root is called a cube root. PRINCIPAL NN TH ROOT OF A NUMBER: Let bb be a real umber that has at least oe real th root. The pricipal th root of bb is the real th root that has the same sig as bb, ad is deoted by a radical symbol: bb. Every positive umber has a uique pricipal th root. We ofte refer to the pricipal th root of bb as just the th root of bb. The th root of 00 is 00. For ay positive itegers mm ad, ad ay real umber bb for which the pricipal th root of bb exists, we have bb = bb bb mm = bb mm = bb bb mm = bb mm. mm Exit Ticket ( miutes)
Name Date : Ratioal Expoets What are ad? Exit Ticket 1. Write the followig expoetial expressios as equivalet radical expressios. a. 2 1 2 b. 2 c. 2 2. Rewrite the followig radical expressios as equivalet expoetial expressios. a. 5 b. 2 c. 1 16
. Provide a writte explaatio for each questio below. a. Is it true that 1 2 = ( ) 1 2? Explai how you kow. b. Is it true that 1000 1 = (1000 ) 1? Explai how you kow. c. Suppose that m ad are positive itegers ad b is a real umber so that the pricipal Pth root of b exists. I geeral does b 1 m = (b m ) 1? Provide at least oe example to support your claim. Exit Ticket Sample Solutios 1. Rewrite the followig expoetial expressios as equivalet radical expressios. a. = b. = = 8 c. = = 9 2. Rewrite the followig radical expressios as equivalet expoetial expressios. a. 5 5 = 5 b. = = 8 = 8 c. 6
= ( ) = 6 = (6) 6. Provide a writte explaatio for each questio below. a. Is it true that = ( )? Explai how you kow. = = = 8 MP. & MP.7 ( ) = 6 = 6 = 8 So the first statemet is true. b. Is it true that 000000 = (000000 )? Explai how you kow. Similarly the left ad right sides of the secod statemet are equal to oe aother. 000000 = 000000 = 00 = 000000 (000000 ) = (000000000000000000) = 000000 c. Suppose that mm ad are positive itegers ad bb is a real umber so that the pricipal Pth root of bb exists. I geeral does bb mm = (bb mm )? Provide at least oe example to support your claim. Thus it appears the statemet is true ad it also holds for whe the expoets are itegers. Based o our other work i this lesso, this property should exted to ratioal expoets as well. Here is aother example, 8 = () = Ad (8 ) = (6) = Thus, 8 = (8 ). Problem Set Sample Solutios 1. Select the expressio from (a), (b), ad (c) that correctly completes the statemet. (a) (b) (c) a. x is equivalet to x x x (b) b. x is equivalet to (b) x x x c. x is equivalet to x x (c) x
d. x is equivalet to x x x (c) 2. Idetify which of the expressios (a), (b), ad (c) are equivalet to the give expressio. (a) (b) (c) a. 6 (a) ad (b) 6 8 6 9 7 6 b. (b) oly. Rewrite i radical form. If the umber is ratioal, write it without usig radicals. a. 6 6 b. c. (8) 8 = 6 d. 6 5 5 6 = 5 6 e. 8 = 8. Rewrite the followig expressios i expoet form. a. 5 5 b. 5 5
c. 5 5 d. 5 5 5. Use the graph of f(x) = x show to the right to estimate the followig powers of. a.. b.. 6 c.. 7 d. 00.. 5 e... f. 5 00. 85 6. Rewrite each expressio i the form kx where k is a real umber, x is a positive real umber ad is ratioal. a. 6x b. 5 x c. /x d. 8x 7 e. 9x f. 5 x x 5x x x 9x 5 x 7. Fid a value of x for which x =. 56 8. Fid a value of x for which x = 8. 7 9. If x = 6, fid the value of x. x = 6, so (6) = () = 8 =.
10. If bb =, evaluate the followig expressios. 9 a. bb 9 = 9 = b. bb 5 9 5 = 5 = c. bb 9 = 7 =. Show that each expressio is equivalet to x. Assume x is a positive real umber. a. 6x 6 x = x b. 8x x (x) (x ) = x x = x c. 6x 7x 6 6x 6x = = x x6/ x 12. Yoshiko said that 6 = because is oe-fourth of 6. Use properties of expoets to explai why she is or is t correct. Yoshiko s reasoig is ot correct. By our expoet properties, 6 = 6 = 6 = 6, but = 56. Sice 6, we kow that 6. 1. Jefferso said that 8 = 6 because 8 = ad = 6. Use properties of expoets to explai why he is or is t correct. MP. Jefferso s reasoig is correct. We kow that 8 = 8, so 8 =, ad thus 8 = 6. 1. Rita said that 8 = 8 because 8 = 8 8, so 8 = 6, ad the 8 = 8. Use properties of expoets to explai why she is or is t correct. Rita s reasoig is ot correct because she did ot apply the properties of expoets correctly. She should also realize that raisig 8 to a positive power less tha will produce a umber less tha 8. The correct calculatio is
below. 8 = 8 = = 15. Suppose for some positive real umber aa that aa aa aa =. a. What is the value of aa? aa aa aa = aa + + = (aa ) = aa = aa = b. Which expoet properties did you use to fid your aswer to part (a)? We used the properties bb bb mm = bb mm+ ad (bb mm ) = bb mm. 16. I the lesso, you made the followig argumet: = Sice = + + = =. is a umber so that = ad is a umber so that Which expoet property was used to make this argumet? =, you cocluded that that =. We used the property bb bb mm = bb mm+. (Studets may also metio the uiqueess of Pth roots.)