The Improving Mthemtics Eduction in Schools (TIMES) Project FRACTIONS AND THE INDEX LAWS IN ALGEBRA A guide for techers - Yers 8 9 June 2011 NUMBER AND ALGEBRA Module 32 8YEARS 9
Frctions nd the Index Lws in Algebr (Number nd lgebr : Module 32) For techers of Primry nd Secondry Mthemtics 510 Cover design, Lyout design nd Typesetting by Clire Ho The Improving Mthemtics Eduction in Schools (TIMES) Project 2009 2011 ws funded by the Austrlin Government Deprtment of Eduction, Employment nd Workplce Reltions. The views expressed here re those of the uthor nd do not necessrily represent the views of the Austrlin Government Deprtment of Eduction, Employment nd Workplce Reltions. The University of Melbourne on behlf of the Interntionl Centre of Excellence for Eduction in Mthemtics (ICE EM), the eduction division of the Austrlin Mthemticl Sciences Institute (AMSI), 2010 (except where otherwise indicted). This work is licensed under the Cretive Commons Attribution- NonCommercil-NoDerivs 3.0 Unported License. 2011. http://cretivecommons.org/licenses/by-nc-nd/3.0/
The Improving Mthemtics Eduction in Schools (TIMES) Project FRACTIONS AND THE INDEX LAWS IN ALGEBRA A guide for techers - Yers 8 9 June 2011 NUMBER AND ALGEBRA Module 32 Peter Brown Michel Evns Dvid Hunt Jnine McIntosh Bill Pender Jcqui Rmgge YEARS 8 9
{4} A guide for techers FRACTIONS AND THE INDEX LAWS IN ALGEBRA ASSUMED KNOWLEDGE The rithmetic of integers nd of frctions. Bsic methods of mentl rithmetic. The HCF nd LCM in rithmetic. Algebric expressions with whole numbers nd negtives. The index lws in rithmetic with nonzero whole numbers indices. The three index lws involving products in lgebr. Liner equtions MOTIVATION The vrious uses of lgebr require systemtic skills in mnipulting lgebric expressions. This module is the third of four modules tht provide systemtic introduction to bsic lgebric skills. In the module, Algebric Expressions, we introduced lgebr using only whole numbers nd the occsionl frction for the pronumerls. Then in the module Negtives nd the Index Lws in Algebr we extended the methods of tht module to expressions involving negtives. The present module extends the methods of lgebr so tht ll negtive frctions nd decimls cn lso be substituted into lgebric expressions, nd cn pper s the solutions of lgebric equtions. Although erlier modules occsionlly used negtive frctions, this module provides the first systemtic ccount of them, nd begins by presenting the four opertions of rithmetic, nd powers, in the context of negtive frctions nd decimls. The resulting number system is clled the rtionl numbers. This system is sufficient for ll the norml clcultions of every life, becuse dding, subtrcting, multiplying nd dividing rtionl numbers (prt from division by zero), nd tking whole number powers of rtionl numbers, lwys produces nother rtionl numbers. In prticulr, scientific clcultions often involve mnipultion of equtions with decimls.
The Improving Mthemtics Eduction in Schools (TIMES) Project {5} The module lso begins the discussion of lgebric frctions, which routinely cuse problems for students. In this module, the discussion is quickly restricted by excluding numertors nd denomintors with more thn one term. These more complicted lgebric frctions re then covered in more detil in the fourth module, Specil Expnsions nd Algebric Frctions. Once frctions hve been introduced into lgebr, the two index lws deling with quotients cn be stted in more systemtic form, nd then integrted with lgebr. This completes the discussion of the index lws in rithmetic nd lgebr s fr s nonzero whole number indices re concerned. CONTENT THE RATIONAL NUMBERS Our rithmetic begn with the whole numbers 0, 1, 2, 3, 4, We then constructed extensions of the whole numbers in two different directions. First, in the module, Frctions we dded the positive frctions, such s 2 3 nd 9 2 = 4 1 2, which cn be written s the rtio of whole number nd non zero whole number, nd we showed how to dd, subtrct, multiply nd divide frctions. This gve system of numbers consisting of zero nd ll positive frctions. Within this system, every non zero number hs reciprocl, nd division without reminder is lwys possible (expect by zero). Only zero hs n opposite, however, nd subtrction b is only possible when b. Secondly, in the module, Integers we strted gin with the whole numbers nd dded the opposites of the whole numbers. This gve system of numbers clled the integers, consisting of ll whole numbers nd their opposites,, 3, 2, 1, 0, 1, 2, 3, nd we showed how to dd, subtrct, multiply nd divide integers. Within this system of integers, every number hs n opposite, nd subtrction is lwys possible. Only 1 nd 1 hve reciprocls, however, nd division without reminder is only possible in situtions such s 21 7 when the dividend is n integer multiple of the divisor.
{6} A guide for techers Arithmetic needs to be done in system without such restrictions we wnt both subtrction nd division to be possible ll the time (except for division by zero). To chieve this, we strt with the integers nd positive frctions tht we hve lredy. Then we dd the opposites of the positive frctions, so tht our number system now contin ll numbers such s 2 3 nd 9 2 = 4 1 2 All these numbers together re clled the rtionl numbers the word rtionl is the djective from rtio. We hve ctully been using negtive frctions in these notes for some time. These remrks re intended to formlise the sitution. The usul lws of signs pply to multiplying nd dividing in this lrger system, so tht frction with negtive integers in the numertor, or in the denomintor, or in both, cn be written s either positive or negtive frction, 2 3 nd 2 3 = 2 3 nd 2 3 = 2 3 Becuse of these identities, it is usul to define rtionl number s the rtio of n integer nd non zero integer: A rtionl number is number tht cn be written s frction, where is n integer nd b is non zero integer. Thus positive nd negtive mixed numerls, terminting decimls, nd recurring decimls re rtionl numbers becuse they cn be written s frctions of integers. For exmple, 9 1 5 = 46 5 23 917 12, 23.917 = 1000, 1.09 = 11 In lter modules on topics such s surds, circles, trigonometry nd logrithms, we will need to introduce further numbers, clled irrtionl numbers, tht cnnot be written s frctions. Some exmples of such numbers re 2, π, sin 22, log 2 5. The module, The Rel Numbers will discuss the resulting more generl system of rithmetic. Show tht ech number below is rtionl by writing it s frction b, where nd b re integers with b 0. 5 b 3 7 c 5 1 3 d 0.002 e 4.7 f 1.3 5 = 5 1 b 3 7 = 3 7 c 51 3 = 16 3 d 0.002 = 2 1000 e 4.7 = 47 10 f 1.3 = 4 3
The Improving Mthemtics Eduction in Schools (TIMES) Project {7} Constructing the rtionl numbers There re two wys to construct the rtionl numbers. The first wy is to construct the positive frctions first nd then the negtives, which is roughly wht hppened historiclly. This involves three step procedure: 1 First construct the whole numbers 0, 1, 2, 3, 4, 2 Then construct the non negtive frctions, such s 3 4 nd 9 4, s the rtio of two whole numbers, where the second is non zero. This process involves identifying frction such s 2 6, where the two whole numbers hve common fctor, with its simplest form 1 3, nd identifying frction such s 5 1, where the denomintor is 1, with the whole number 5. 3 Lstly construct the rtionl numbers by constructing the opposites of every positive frction nd whole number. The other method is to construct the integers first nd then the frctions. The three steps re now: 1 First construct the whole numbers 0, 1, 2, 3, 4,.. 2 Then construct the integers by constructing the opposites, 4, 3, 2, 1 of ll the nonzero whole numbers. 3 Lstly, construct the rtionl numbers by constructing ll the frctions such s 3 4 nd 9 4 the rtio of two integers, where the second is non zero. This process gin involves identifying equivlent frctions, nd identifying frctions with denomintor 1 s integers. These two procedures re eqully cceptble mthemticlly, nd their finl results re identicl in structure. Students lern bout rtionl numbers, however, by working with exmples of them in rithmetic nd lgebr, nd there is little to be gined by such explntions. ADDING AND SUBTRACTING NEGATIVE FRACTIONS AND DECIMALS When dding nd subtrcting frctions nd decimls tht my be positive or negtive, the rules re combintion of the rules for integers nd the rules for dding nd subtrcting positive frctions nd decimls: To subtrct negtive number, dd its opposite. Perform ddition nd subtrction from left to right. When deling with mixed numerls, it is usully esier to seprte the integer prt nd the frction.
{8} A guide for techers The following exmples demonstrte such clcultions first in rithmeticl exmples, then in substitutions nd equtions. Simplify: 5 8 + 3 4 1 4 b 32.6 + ( 23.6) ( 26.4) 5 8 + 3 4 1 4 = 5 8 6 8 + 2 8 b 32.6 + ( 23.6) ( 26.4) = 32.6 23.6 + 26.4 = 1 8. = 35.4. Evlute x y + z when x = 20 2 3, y = 25 1 2 nd z = 353 4. b x = 3.6, y = 5.6 nd z = 2.07. x y + z = 20 2 3 25 1 2 + 353 4 = 20 + 8 12 25 6 12 35 9 12 = 40 7 12 = 40 7 12 Bewre tht in nottion 40 7 7 12, the negtive sign pplies to the whole number 4012, so tht 40 7 7 12 mens 40 + 12 = 40 7 12. b x y + z = ( 3.6) 5.6 + ( 2.07) = 3.6 5.6 2.07 = 4.07 EXERCISE 1 Simplify 1 + 1 2 1 3 + 1 4 1 5 + 1 6.
The Improving Mthemtics Eduction in Schools (TIMES) Project {9} MULTIPLYING NEGATIVE FRACTIONS AND DECIMALS As with ddition nd subtrction, we need to combine the rules lredy discussed: First determine the sign of the product, then del with the numbers. The product of two positives or two negtives is positive. The product of positive nd negtive is negtive. In the bsence of brckets, the order of opertions is: 1 Powers, 2 Multipliction nd division from left to right, 3 Addition nd subtrction from left to right. When multiplying frctions, first convert them to improper frctions. When multiplying decimls, convert them to frctions whose denomintors re powers of 10. (Some people prefer to use the rule, Before discrding triling zeroes, the number of deciml plces in the nswer equls the totl number decimls plces in the question. ) The following exmples demonstrte these procedures first in rithmeticl exmples, then with substitutions nd equtions. Simplify: 2 1 2 3 1 2 3 3 5 3 b 1.2 ( 0.2) 3 The product is negtive becuse there re five minus signs in the product. 2 1 2 3 1 2 3 3 3 5 = 5 2 10 3 10 3 3 5 3 5 3 5 = 6 b The product is positive becuse there re four minus signs in the product. 1.2 ( 0.2) 3 = 12 10 2 3 103 = 96 104 = 0.0096
{10} A guide for techers Simplify 3x 2 yz when: x = 1 1 2, y = 2 1 2 nd z = 4 1 5 b x = 1.1, y = 0.4 nd z = 12.1 3x 2 yz = 3 3 2 3 2 + 5 2 21 5 (The 3 is not squred indices come first) = 27 4 + 21 2 = 6 3 4 + 10 1 2 = 6 3 4 + 10 + 2 4 (Cre with mixed numerls: 63 4 mens 6 3 4 ). = 4 1 4 b 3x 2 yz = = 3 3 4 3 112 100 4 10 121 10 = 363 100 484 100 = 1.21 Solve ech eqution: x 3 + 43 4 = 53 4 b x 1.2 + 2.01 = 2 x 3 + 43 4 = 53 4 x 3 = 10 1 2 x= 31 1 2 4 3 4 ( 3) In the lst line, it ws esier to clculte 3 10 1 2 = 30 + 1 1 2 thn to use improper frctions b x 1.2 + 2.01 = 2 x 1.2 = 0.01 2.01 x= 1 100 12 10 1.2 = 0.012 EXERCISE 2 How mny fctors must be tken in the product ( 0.8) 0.9 ( 1.0) for the product to exceed 1?
The Improving Mthemtics Eduction in Schools (TIMES) Project {11} DIVIDING NEGATIVE FRACTIONS AND DECIMALS We hve lredy reviewed the order of opertions. The other necessry rules re: First determine the sign of the quotient, then del with the numbers. The quotient of two positives or two negtives is positive. The quotient of positive nd negtive is negtive. When dividing by frction, multiply by its reciprocl. An improper frction must first be converted to n improper frction. To divide by deciml, write the clcultion s frction, then multiply top nd bottom by power of 10 lrge enough to mke the divisor whole number. (Some people prefer to chnge both decimls to frctions nd then use frction methods.) Simplify: 1 1 2 2 1 2 b 0.9 ( 0.08) ( 0.06)2 1 1 2 2 1 2 = 3 2 2 5 = 3 5. b 0.9 ( 0.08) ( 0.06) 2 = 0.0072 0.0036 = 720 36 (Multiply top nd bottom by 10000.) = 20. Simplify 3 b c when: = 8 9, b = 1 1 10 nd c = 2. b = 0.09, b = 0.027 nd c = 0.1. 3 b c = 3 1 8 9 10 11 + 2 b 3 b c = 3 0.09 0.027 ( 0.1) = 80 90 33 + 2 = 3 27 + 0.1 = 14 33 = 10 + 0.1 = 9.9
{12} A guide for techers The reciprocl We hve seen tht the reciprocl of positive frction such s 2 5 is the frction 5 2 formed by interchnging the numertor nd denomintor, becuse 2 5 5 2 = 1 Similrly, the reciprocl of negtive frction such s 2 5 is 5 2, becuse 2 5 5 2 = 1 To form the reciprocl, interchnge numertor nd denomintor, nd do not chnge the sign. Reciprocls of positive nd negtive frctions re importnt in vrious lter pplictions, prticulrly perpendiculr grdients. Find, s mixed numerl when pproprite, the reciprocls of: 4 13 b 20 107 c 5 1 6 d 7 7 10 The reciprocl of 4 13 is 13 4 = 3 1 6. b The reciprocl of 20 107 107 is 20 = 5 7 20. c The reciprocl of 5 1 6 = 31 6 is 6 31. d The reciprocl of 7 7 10 = 77 10 10 is 77. Compound frctions 2 3 + 3 4 There is very strightforwrd wy to simplify compound frction such s 2 1 2 + 1 6 To simplify compound frction, multiply top nd bottom by the lowest common denomintor of ll the frctions. In the exmple bove, the lowest common denomintor of ll the frctions is 12, so multiplying top nd bottom by 12 2 3 + 3 4 2 1 2 + 1 6 = 8 + 9 30 + 2 = 17 32 It is lso possible to rewrite the frction s division 2 3 + 3 4 2 1 2 + 1 6, but such n pproch requires more steps.
The Improving Mthemtics Eduction in Schools (TIMES) Project {13} Simplify 1 1 3 + 1 1 2 1 1 3 1 1 2 b Evlute x + 2 x when x = 3 1 3 nd = 3 5 Multiply top nd bottom by 6. 1 1 3 + 1 1 2 1 1 3 1 1 2 = 8 + 9 8 9 = 1 17 b Multiply top nd bottom by 15. 1 3 3 + 6 5 x + 2 x = 3 1 3 3 5 = 50 + 18 50 9 = 32 59 = 39 59 Solving equtions with frctions An eqution with severl frctions cn be solved by the stndrd method, Move every term in x to one side, nd ll the constnts to the other. The resulting frction clcultions, however, cn be quite elborte. There is fr quicker pproch tht gets rid of ll the frctions in one single step, If n eqution involves frctions, multiply by the lowest common denomintor. Similrly, if n eqution involves decimls, multiply through by suitble power of 10. The following exmple uses first this quicker pproch, then the stndrd method, to solve two equtions. Compre the pproches nd decide for yourself. Use the quicker pproch bove, nd then the stndrd pproch, to solve these equtions. Solve: 5 6 x + 1 2 = 3 1 3 b 0.024x + 0.1 = 0.04 5 6 x + 1 2 = 3 1 3 Multiply by 6 to remove ll frctions. 5x + 3 = 20 6 5x = 23 3 x = 4 3 5 ( 5) or
{14} A guide for techers 5 6 x + 1 2 = 3 1 3 5 6 x = 3 2 6 3 6 1 2 5 6 x = 35 6 x = 6 5 23 6 6 5 = 4 3 5 b 0.024x + 0.1 = 0.04 Multiply by 1000 to remove ll decimls. 24x + 100 = 40 1000 24x = 60 100 x = 2.5 ( 24) or 0.024x + 0.1 = 0.04 0.024x = 0.06 0.1 = 0.06 0.024 ( 0.024) = 60 24 = 2.5 EXERCISE 3 Use n efficient method to find the men of the numbers 1, 0.9, 0.8,, 2. The system of rtionl numbers We hve now introduced four number systems in turn. First, we introduced the whole numbers 0, 1, 2, 3, 4, The set of whole numbers is closed under ddition nd multipliction, which mens tht when we dd or multiply two whole numbers, we obtin nother whole number. The whole numbers re not closed under either subtrction or division, however, becuse, for exmple, 7 10 is not whole number nd 7 10 is not whole number. Next, we dded the positive frctions to the whole numbers, so tht our number system now contined ll non negtive rtionl numbers, including numbers such s 4, 2 3, nd 55 7. This system is closed under division (except by zero), but is still not closed under subtrction.
The Improving Mthemtics Eduction in Schools (TIMES) Project {15} Then we introduced the integers, 3, 2, 1, 0, 1, 2, 3, consisting of ll whole numbers nd their opposites. This system is closed under subtrction, but not under division. Now tht we re working in the rtionl numbers the integers together with ll positive nd negtive frctions we finlly hve system tht is closed under ll four opertions of ddition, subtrction, multipliction nd division (except by zero). They re lso closed under the opertion of tking whole number powers. This set of rtionl numbers is therefore very stisfctory system for doing rithmetic, nd is quite sufficient for ll everydy needs. When we move to the even lrger system of rel numbers, every rel number cn be pproximted s close s we like by rtionl numbers (indeed by terminting decimls), so tht the rtionl numbers will remin extremely importnt. CANCELLING, MULTIPLYING AND DIVIDING ALGEBRAIC FRACTIONS An lgebric frction is frction involving pronumerls, x 3, 5 x, + b c, 3r s t. The pronumerls in the frction represent numbers, so we del with lgebric frctions using exctly the sme procedures tht we use in rithmetic. Substitution into lgebric frctions We hve seen in rithmetic tht the denomintor in frction cnnot be zero. This mens tht: In the expression 5 x, we cnnot substitute x = 0. In the expression + b c, we cnnot substitute c = 0. In the expression 3r s t, we cnnot substitute the sme vlue for s nd t. Aprt from this qulifiction, substitution into lgebric frctions works exctly the sme s substitution into ny lgebric expression. For exmple, if r = 5, s = 7 nd t = 4, then 3r s t = 15 7 4 = 5 Substitution of negtive numbers requires the usul cre with signs. If r = 2, s = 7 nd t = 5, 3r s t = 6 7 5 = 6 12 = 1 2
{16} A guide for techers Wht vlue of cnnot be substituted into + 3 3? b Wht vlue of mkes + 3 3 zero? c Evlute + 3 3 when = 2. When = 3, the denomintor is zero, nd the expression cnnot be evluted. b When = 3, the numertor is zero but the denomintor is not zero, so the expression is zero. c When = 2, + 3 3 = 2 + 3 2 3 = 1 5 = 1 5 The rest of this module will exclude denomintors, such s s t nd 3, tht hve two or more terms. Such expressions will be considered in the lter module, Specil Expnsions nd Algebric Frctions. Cncelling lgebric frctions Becuse pronumerls re just numbers, we cn cncel both pronumerls nd numbers in n lgebric frction. For exmple, 3xy 6x = y 2 nd 2 b b =. We cnnot cncel zero, so strictly we should exclude zero vlues of the pronumerl when we cncel. This would men writing 3xy 6x = y 2 provided x 0 nd 2 b = provided 0 nd b 0. b Such qulifictions become importnt much lter in clculus. At this stge, however, it is not pproprite to bring the mtter up, unless students insist. Simplify ech lgebric frction by cncelling ny common fctors: 5 10b b 9b 12bc c 4b2 6b d 12xy2 z 12x2yz 5 10b = 1 2b b 9b 12bc = 3 4c c 4b2 6b = 2b 3 d 12xy2 z 12x2yz = y x
The Improving Mthemtics Eduction in Schools (TIMES) Project {17} Dividing lgebric expressions When we divide, the lws of signs re the sme s the lws for multiplying: The quotient of two negtives or two positives is positive. The quotient of positive nd negtive is negtive. The following exmples illustrte ll the possibilities: 6x2 2x = 3x 6x2 2x = 3x 6x2 2x = 3x 6x2 2x = 3x As with multipliction, there re three steps in hndling more complicted expressions: First del with the signs. Then del with the numerls. Then del with ech pronumerl in turn. 3xy 6x2 = y 2x nd (21bc 2 ) ( 7b 2 c) = 3c b The lst nswer cn lso be written s 3c b nswer 3c b, with the negtive sign out the front of the frction. 3c or even s b, but the best wy to write the Frction nottion is usully used for quotients, but the division sign cn lso be used, s in the second exmple bove. Simplify ech quotient: ( 12b) ( 3bc) b 8x2 4x c 3bc 2bc3 d 92 6b2 c ( 12b) ( 3bc) = 4 c 3bc 2bc3 = 2 2c2 b 8x2 4x = 2x d 92 6b2 = 32 2b2 Multiplying lgebric frctions We hve seen how to multiply frctions in rithmetic, 15 5 10 21 = 5 7 10 7 = 50 49 First del with the signs. Then cncel ny common fctors from numertor nd denomintor. Then multiply the numertors, nd multiply the denomintors.
{18} A guide for techers Exctly the sme method pplies to multiplying lgebric frctions. For exmple, 3 x x = 3 1 1 = 3 5x 6y 6xy 25 = x 1 x 5 = x 2 5 Simplify: 2x 3 6 x b 2b c bc 2x 3 6 x = 2 1 2 1 b 2b c bc = 2 c c = 4 = 3 c2 Dividing by n lgebric frction We know the lw of signs for division, nd we know how to divide by frction. To divide by n lgebric frction, we pply these sme lws in succession 1 First determine the sign of the quotient. 2 To divide by frction, multiply by its reciprocl For exmple, 4 x 6 x = 4 x x 6 = 2 1 1 3 32b c 3b2 c = 32b c c 3b2 = 1 1 b = 2 3 = b Simplify ech quotient: 12 x 3 b 6x 2 x y 4x y2 c 3b c ( 6bc) d 22 c 2 c2 12 x 3 x = 12 x x 3 b 6x2 y c 3b c = 4 1 1 1 4x y2 = 6x2 y y 2 4y = 4 = 3xy 2 3b ( 6bc) = c 1 6bc d 22 c 2 c2 = 6x2 y y 2 4y = 22 c c2 2 = 1 c 1 2c = 1 c 1 = 1 2c2 = c
The Improving Mthemtics Eduction in Schools (TIMES) Project {19} The reciprocl of n lgebric frction Reciprocls of lgebric frctions re formed in exctly the sme wy s ws done in rithmetic. The reciprocl of 2x y is y 2x 2x, becuse y y 2x = 1. The reciprocl of 3 is 1 3, becuse ( 3 ) 1 3 = 1. The reciprocl of 7xy 3t is 3t 7xy, becuse 7xy 3t 3t 7xy = 1. To form the reciprocl, interchnge numertor nd denomintor, nd do not chnge the sign. Write down the reciprocl of ech lgebric frction: 2 b b 2 b c 6x2 7yz d 6x2 7yz b 2 b 2 b c 7yz d 7yz 6x 2 6x 2 THE INDEX LAWS FOR FRACTIONS The previous lgebr module Negtives nd the Index Lws in Algebr delt with the index lws for product of powers of the sme bse, for power of power, nd for power of product: To multiply powers of the sme bse, dd the indices: m n = m+n. To rise power to power, dd the indices: ( m ) n = mn. The power of product os the product of the powers: (b) n = n b n. In these identities, nd b re ny numbers, nd m nd n re ny nonzero whole numbers. We now turn to the remining two index lws, which involve quotients. The index lw for the power of quotient To see wht hppens when we tke power of frction, we simply write out the power. 2 3 2 5 = 5 2 5 2 5 5 b = b b b b b = 23 5 3 = 5 b 5
{20} A guide for techers The generl sitution cn thus be expressed s: The power of quotient is the quotient of the powers: n b = n n where nd b 0 re numbers, nd n is nonzero whole number. b The following exmples illustrte the use of the lw in rithmetic nd lgebr. Simplify: 1 2 5 b 3 4 c 1 2 d 1 3 10 4 2 3 3 1 2 c 4 1 2 5 = 1 5 2 5 b 3 4 3 10 = 4 10 4 = 1 81 32 = 10000 2 9 2 = 2 d 3 1 3 3 = 10 3 = 81 4 = 100 27 = 20 1 4 = 37 1 27 When pplied to n lgebric frction, the lws previously introduced for the power of product nd for the power of power, re often required s well. For exmple, 3 2x 3 8x 3y = 3 27y 3 nd 1 2 1 5 = 3 25 6 Expnd the brckets: 2 2x 3y b 5 3 b c 2 2 b 3 d 3 4 2 4 2 2x 4x 3y = 2 b 5 3 125 9y2 b = 3 b 3 c 2 2 b = 4 3 b6 d 3 4 2 4 = 81 16 16 The index lw for the quotient of powers of the sme bse The remining index lw from rithmetic dels with dividing powers of the sme bse. The module Multiples, Fctors nd Powers developed the lw in the form, To divide powers of the sme bse, subtrct the indices. For exmple, 7 5 7 3 = (7 7 7 7 7) (7 7 7) = 7 2. Frctions, however, hd not been introduced t tht stge, so the divisor could not be
The Improving Mthemtics Eduction in Schools (TIMES) Project {21} higher power thn the dividend. Now tht we hve frctions, we cn express the lw in frctionl nottion nd do wy with the restriction: 7 5 7 = 7 7 7 7 7 3 7 7 7 7 4 7 = 7 7 7 7 7 3 7 7 7 4 7 7 7 7 = 7 5 7 7 7 7 7 = 7 2 = 1 = 1 7 2 The bse cn lso be pronumerl, nd the lw cn now be written s: To divide powers of the sme bse, subtrct the indices. For exmple, 5 3 = 2 nd 4 4 = 1 nd 3 5 = 1 2. Lter, when zero nd negtive indices hve been introduced in the module The Index Lws, we will be ble to stte this lw in its usul, more concise form: m n = m n. Simplify ech expression: 3x 6 x4 b 8 b 6 6 b 8 c 34 4b8 4b4 38 d 3 b 2 6 b 6 c 6 3x 6 = 3x2 x 4 b 8 b 6 = 2 6 b c 34 4b4 = 1 8 b2 4b8 38 4 b d 3 b 2 6 b 6 c 6 1 = 4 3 b 4 c 6 ADDING AND SUBTRACTING ALGEBRAIC FRACTIONS When set of lgebric frctions hve common denomintor, it is strightforwrd to dd them. As lwys, we collect like terms, 3 x + x 7 x + 4 x = 5 x 4 x (which is perhps written s 5 4 x ) Simplify ech expression: 5 x x 4 x b 7 2x 11 2x 20 2x 5 x x 4 x = 1 x x, or 1 x b 7 2x 11 2x 20 2x = 24 2x = 12 x
{22} A guide for techers Using common denomintor When the frctions do not hve common denomintor, we need to find the lowest common denomintor, just s we did for rithmetic frctions. We shll restrict the discussion to numericl denomintors t this stge, nd leve denomintors with pronumerls to the module Specil Expnsions nd Algebric Frctions. x 4 x 6 = 3x 12 2x 12 = x 12. Simplify ech expression: x 2 x 3 + x 4 + x 12 b 3x _ 5x 8 6 x 2 x 3 + x 4 + x 12 = 6x _ 4x 12 12 + 3x 12 + x 12 b 3x _ 5x 8 6 = 9x _ 20x 24 24 = 6x 12 = 11x 24 = x 2 LINKS FORWARD As mentioned in the Motivtion section, this module is the third of four modules deling with bsic mnipultion of lgebric expressions. The four modules re: Algebric Expressions Negtives nd the Index Lws in Algebr Frctions nd the Index Lws in Algebr (the present module) Specil Expnsions nd Algebric Frctions Algebric frctions hve been introduced in this module, but the lgebr of lgebric frctions cn cuse considerble difficulty if it is introduced in full strength before more fundmentl skills hve been ssimilted. Algebric frctions will therefore be considered in more detils in the fourth module Specil Expnsions nd Algebric Frctions. The index lws will be gretly expnded in scope once negtive nd frctionl indices re introduced in the module Indices. One piece of clumsiness in the present module concerned the quotient of the difference of powers, where we needed to write down three different exmples x5 = x 2 x 3 nd x4 x 4 = 1 nd x3 x 5 = 1 x 2. Once the zero index nd negtive indices re brought into lgebr, ll this cn be replced
The Improving Mthemtics Eduction in Schools (TIMES) Project {23} with the single lw xm xn = x m n where m nd n re integers. Squre roots nd cube roots cn lso be represented using frctionl indices for exmple, 5 is written s 5 1 2 nd 3 5 is written s 5 1 3 nd yet the five index lws still hold even when the index is ny rtionl number. Negtive nd frctionl indices re introduced in the module, Indices nd Logrithms. Lter in clculus, n index cn be ny rel number. ANSWERS TO EXERCISES EXERCISE 1 The lowest common denomintor is 60. Grouping the numbers in pirs before dding them ll reduces the size of the numertor. Sum = 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 = 1 2 1 12 1 30 = 30 60 5 60 2 60 = 37 60 EXERCISE 2 We cn ignore ll negtive products. Since ( 0.8) 0.9 ( 1.0) 1.1 is less thn 1, the first product greter thn 1 is ( 0.8) 0.9 ( 1.0) 1.1 ( 1.2) 1.3 ( 1.4). EXERCISE 3 One pproch is to notice tht when the 31 numbers re dded, the first 21 numbers cncel out. The remining 10 numbers cn then be dded by grouping them in pirs: 1 + 0.9 + 0.8 + + ( 2) = 1.1 1.2 1.3 2. = ( 1.1 2) +( 1.2 1.9) + ( 1.3 1.8) + ( 1.4 1.7) + ( 1.5 1.6) = 3.1 5 = 15.5. Hence men = 15.5 31 = 31 62 = 0.5. A better pproch is relise tht 0.5 is the middle number, nd tht the other 30 numbers fll into pirs eqully spced one to the left nd one to the right of 0.5. Hence the men is 0.5.
The im of the Interntionl Centre of Excellence for Eduction in Mthemtics (ICE-EM) is to strengthen eduction in the mthemticl sciences t ll levelsfrom school to dvnced reserch nd contemporry pplictions in industry nd commerce. ICE-EM is the eduction division of the Austrlin Mthemticl Sciences Institute, consortium of 27 university mthemtics deprtments, CSIRO Mthemticl nd Informtion Sciences, the Austrlin Bureu of Sttistics, the Austrlin Mthemticl Society nd the Austrlin Mthemtics Trust. The ICE-EM modules re prt of The Improving Mthemtics Eduction in Schools (TIMES) Project. The modules re orgnised under the strnd titles of the Austrlin Curriculum: Number nd Algebr Mesurement nd Geometry Sttistics nd Probbility The modules re written for techers. Ech module contins discussion of component of the mthemtics curriculum up to the end of Yer 10. www.msi.org.u