Numer systems: the Rel Numer System syllusref eferenceence Core topic: Rel nd complex numer systems In this ch chpter A Clssifiction of numers B Recurring decimls C Rel nd complex numers D Surds: suset of irrtionl numers E Simplifying surds F Addition nd sutrction of surds G Multipliction of surds H Division of surds I The Distriutive Lw J Rtionlising denomintors K Rtionlising denomintors using conjugte surds L Further properties of rel numers modulus M Solving equtions using solute vlues N Solving inequtions
Mths Quest Mths C Yer for Queenslnd Introduction The numer systems used tody evolved from sic nd prcticl need of primitive people to count nd mesure mgnitudes nd quntities such s livestock, people, possessions, time nd so on. Erly cultures nd societies used their ody prts, such s fingers nd toes, s sis for their numertion systems. As the need for lrger numers grew, symols were developed to represent them. Ancient Egyptins, for exmple, used the symol of the lotus flower to represent the numer 000, nd Romns used the letter M to represent 000. Romn numerls cn e seen tody on some clock nd wtch fces. At the end of movie credits Romn numerls re often used to indicte the yer in which the movie ws mde. For exmple, MCMXCIX represents the yer. As societies grew nd rchitecture nd engineering developed, numer systems ecme more sophisticted. Numer use developed from solely whole numers to frctions, decimls nd irrtionl numers. We shll explore these different types of numers nd clssify them into their specific groups. Consider solutions to equtions such s: x = 0, x =, 0x = 00 Wht do they hve in common? Ech of the sttements is true for wholenumer vlue of x. This type of eqution represents mny rel-life situtions; for exmple, how mny people will I need to collect $ from to cover the cost of hiring $0 gme? As you work through this chpter on the Rel Numer System nd chpter you will e introduced to types of numers tht evolved to fill need. The first types of numers to evolve were the whole numers; this then is extended into the set of integers. The Rel Numer System The Rel Numer System contins the set of rtionl nd irrtionl numers. It is denoted y the symol R. Rel numers R Irrtionl numers I (surds, non-terminting nd non-recurring decimls, π,e) Integers Z Rtionl numers Q Non-integer rtionls (terminting nd recurring decimls) Negtive Z Zero (neither positive nor negtive) Positive Z + (Nturl numers N) The set of rel numers contins numer of susets which cn e clssified s shown in the chrt ove.
Chpter Numer systems: the Rel Numer System Reltionship etween susets The reltionship which exists etween the susets of the Rel Numer System cn e illustrted in Venn digrm s shown on the right. We cn sy N Z, Z Q, nd so on, where mens is suset of. Q (Rtionl numers) Z (Integers) N (Nturl numers) ε = R I (Irrtionl numers) Integers (Z) The set of integers consists of positive nd negtive whole numers, nd 0 (which is neither positive nor negtive). They re denoted y the letter Z nd cn e further divided into susets. Tht is: Z = {...,,, 0,,,,...} Z + = {,,,,,,...} Z = {,,,,,...} Positive integers re lso known s nturl numers (or counting numers) nd re denoted y the letter N. Tht is: N = {,,,,,,...} Integers my e represented on the numer line s illustrted elow. 0 Z N Z The set of integers The set of positive integers The set of negtive integers or nturl numers Note: Integers on the numer line re mrked with solid dot to indicte tht they re the only points in which we re interested. Another type of eqution is similr to x = 0 x = x =, nd so on. In these equtions the solution is not whole numer ut rtionl numer. Hence the need to further extend the numer system. Clssifiction of numers: rtionl nd irrtionl Rtionl numers (Q) A rtionl numer (rtio-nl) is numer which cn e expressed s rtio of two integers in the form where 0 nd nd hve no common fctors. Rtionl numers re given the symol Q. Exmples re:,, -,, Rtionl numers my e expressed s terminting decimls. Exmples re: - 0 = 0., = 0., = 0., =. These deciml numers terminte fter specific numer of digits. 0
Mths Quest Mths C Yer for Queenslnd Rtionl numers my e expressed s recurring decimls (non-terminting or periodic decimls). For exmple: = 0.... or 0.. - - = 0.... or 0... = 0.... or 0.. = 0.0 0... or 0.. 0. These decimls do not terminte, nd the specific. digit (or numer of digits) is repeted in pttern. Recurring decimls re represented y plcing dot.. ove the repeting digit or pttern. 0 Q Rtionl numers re defined in set nottion s: Q = rtionl numers Q = {,, Z, 0, g.c.d (, ) = } where mens n element of nd g.c.d. (, ) = mens gretest common divisor of (, ) =. Rtionl numers my e represented on the numer line (s illustrted ove) nd include terminting nd recurring decimls. Rel numers cn e thought of s numers tht cn e grphed on numer line. Consider equtions of the type x = x = x = where the solution to the first eqution is ±. The exct vlue of cn e plotted on numer line y geometric construction ut results in deciml equivlent tht is non-terminting. Irrtionl numers (I) Numers tht cnnot e expressed s rtio etween two integers re clled irrtionl numers. Irrtionl numers re denoted y the letter I. Numers such s surds (for exmple, 0 ), decimls tht neither terminte nor recur, nd π nd e re exmples of irrtionl numers. The numers π nd e re exmples of trnscendentl numers; these will e discussed riefly lter in this chpter. Irrtionl numers my lso e represented on the numer line with the id of ruler nd compss. An irrtionl numer (ir-rtio-nl) is numer which cnnot e expressed s rtio of two integers in the form where 0. Irrtionl numers re given the symol I. Exmples re:,,,, π, e Irrtionl numers my e expressed s decimls. For exmple: =. 0... 0.0 = 0. 0 00... =. 0... =. 0... π =.... e =....
Chpter Numer systems: the Rel Numer System These decimls do not terminte, nd the digits do not repet themselves in ny prticulr pttern or order (tht is, they re non-terminting nd non-recurring). Rtionl nd irrtionl numers elong to the set of rel numers (denoted y the symol R). They cn e positive, negtive or 0. The rel numers my e represented on numer line s shown t right (irrtionl numers ove the line; rtionl numers elow it). To clssify numer s either rtionl or irrtionl:. Determine whether it cn e expressed s n integer, frction, or terminting or recurring deciml.. If the nswer is yes, the numer is rtionl; if the nswer is no, the numer is irrtionl. Consider n isosceles right-ngled tringle of side length unit. By Pythgors Theorem, (OB) = (OA) + (AB) ; therefore the length of the hypotenuse is units. By using compss, we cn trnsfer the length of the hypotenuse OB to the numer line (lelled C). This distnce cn now e mesured using ruler. Although this distnce will e inccurte due to the equipment used, there is n exct point on the numer line for ech irrtionl numer. This geometric model cn e extended to ny irrtionl numer in surd form. 0 R π (pi) The symol π (pi) is used for prticulr numer; tht is, the circumference of circle whose dimeter length is unit. It cn e pproximted s deciml which is nonterminting nd non-recurring. Therefore, π is clssified s n irrtionl numer. (It is lso clled trnscendentl numer nd cnnot e expressed s surd.) In deciml form, π =.... It hs een clculted to 000 000 ( million) deciml plces with the id of computer. WORKED Exmple Specify whether the following numers re rtionl or irrtionl. π c d π e 0. f g h is lredy rtio. is rtionl. units unit O 0 B π A C unit R Evlute. = The nswer is n integer, so clssify is rtionl.. Continued over pge
Mths Quest Mths C Yer for Queenslnd c Evlute. c =. 0... The nswer is non-terminting nd is irrtionl. non-recurring deciml; clssify. d Use your clcultor to find the vlue of π. The nswer is non-terminting nd non-recurring deciml; clssify π. d π =. 0... π is irrtionl. e 0. is terminting deciml; clssify it ccordingly. e 0. is rtionl. f Evlute. f = The nswer is whole numer, so clssify. is rtionl. g Evlute. g =.0 0 0... The result is non-terminting nd non-recurring deciml; clssify. is irrtionl. h Evlute. h = The result is numer in rtionl form. is rtionl. Grphics Clcultor tip! Cue nd nth roots To find cue root of numer (e.g. ), press MATH nd select : followed y the numer, in this exmple,. To clculte higher root of numer (e.g. ), first enter the type of root (e.g. ), then press MATH x nd select : followed y the numer.
rememer rememer Chpter Numer systems: the Rel Numer System. Rtionl numers cn e expressed in the form, where nd re integers with no common fctors nd 0. They include whole numers nd terminting nd recurring decimls.. Rtionl numers re denoted y the symol Q.. Irrtionl numers cnnot e expressed in the form, where nd re integers nd 0. They my e expressed only s non-terminting nd nonrecurring decimls.. Irrtionl numers re denoted y the symol I.. Rtionl nd irrtionl numers together constitute set of rel numers, denoted y the symol R. A Clssifiction of numers WORKED Exmple Specify whether the following numers re rtionl (Q) or irrtionl (I). c d e f 0.0 g h i j 0. k. l 00 m. n. o π p - q. r s 000 t.... u v π w x - y 0.000 z Specify whether the following numers re rtionl (Q), irrtionl (I) or neither. c - d e f g h - i π j k l m ( ) n - o p - q - r s t u v. w x y z - π π ( ) 0. 0 00 0 0 Which of the following est represents rtionl numer? A π B C - D E none of these
Mths Quest Mths C Yer for Queenslnd Which of the following est represents n irrtionl numer? A B C D 0.0 E Which of the following sttements regrding the numers 0.,,, is correct? A is the only rtionl numer. B nd re oth irrtionl numers. C 0. nd re the only rtionl numers. D, nd re ll irrtionl numers. E π π 0. is the only rtionl numer. Which of the following sttements regrding the numers, -,, is correct? A - nd re oth irrtionl numers. B is n irrtionl numer nd is rtionl numer. C is the only rtionl numer. D nd re oth irrtionl numers. E is rtionl numer nd - is n irrtionl numer. π Summry of set nottion The following symols re used to descrie reltionships in sets. Consider group of numers from to (i.e.,,,,,,,, ). These numers cn e referred to s set nd denoted y A such tht A = {,,,,,,,, }. We cn sy tht is n element of set A nd write this s A. Similrly 0 is not n element of set A nd this is written s 0 A. The elements nd oth elong to set A nd this cn e written s {, } A, where nd re suset of A. Recurring decimls A rtionl numer my e converted to deciml y dividing the numertor y the denomintor. The resulting deciml my e terminting deciml contining specific numer of digits, tht is: =. or = 0. or it my e recurring deciml contining repeting digit or pttern, tht is: = 0.... or - = 0....
Chpter Numer systems: the Rel Numer System For convenience, recurring decimls re represented y plcing dot over the repeting digit, for exmple:. 0.... cn e written s 0.... 0.... cn e written s 0... If two or more digits repet the sme pttern, then dots or the overscore used s shown:. 0.... cn e written s... or lterntively 0.. ( ) re. 0.... cn e written s 0... or lterntively 0.. Note: When using the overscore, plce it over the whole pttern. The dots, however, re plced over the first nd the lst digits only of the repeting pttern. WORKED Exmple Stte which of the following rtionl numers cn e expressed s recurring decimls. - To convert - to deciml, divide y. - = 0.0 0 0... Use the overscore to indicte the = 0.0 repeting pttern. Write your conclusion. - cn e written s recurring deciml. Convert to deciml (divide y ). = 0. The resulting deciml termintes, so stte your conclusion. cnnot e written s recurring deciml. Whole numers nd terminting decimls such s, 0. nd. cn esily e expressed s rtionl numers. For exmple:. we my write s. we my write 0. s =. we my write. s - or -. 00 000 000 In ech of these cses, the whole numer nd decimls re expressed in the form. Recurring decimls re rtionl numers. Therefore they cn e converted to the form.
0 Mths Quest Mths C Yer for Queenslnd WORKED Exmple Express the following recurring decimls s rtionl numers in their simplest form. 0.. 0... c. Let x represent the recurring x = 0.... [] deciml. This is eqution []. We need to multiply oth sides of the 0x =.... [] eqution y power of 0. The numer of zeros in the power of 0 should e equl to the numer of repeted digits. Since digit is repeted, multiply oth sides of eqution [] y 0. Lel the new eqution []. Sutrct eqution [] from eqution []. This removes ll the repeting digits from ehind the deciml point. [] []: 0x x =.... 0.... x = Divide oth sides of the eqution y. x - = x = Verify the nswer using clcultor nd you will otin the originl vlue, 0... Let x represent the recurring x = 0.... [] deciml. This is eqution []. Since digits re repeted, multiply oth sides of eqution [] y 00 nd lel the new eqution []. 00x =.... [] Sutrct eqution [] from eqution []. This removes ll the repeting digits from ehind the deciml point. Divide oth sides of the eqution y. Cncel down to the simplest form; tht is, divide through y. Verify the nswer using clcultor. [] []: 00x x =.... 0.... x = x = - x = - x = - c Let x represent the recurring deciml. This is eqution []. Since digits re repeted, multiply oth sides of eqution [] y 000 nd lel the new eqution []. c x =.... [] 000x =.... []
Chpter Numer systems: the Rel Numer System Sutrct eqution [] from eqution []. This removes ll the repeting digits from ehind the deciml point. Divide oth sides of the eqution y. Cncel down to the simplest form; tht is, divide through y. Verify the nswer using clcultor. [] []: 000x x =........ x = x - = - x = - x = The Mths Quest CD-ROM contins file which llows you to convert numers from one form to nother s percentge, frction or deciml. Mthcd Decimls nd frctions rememer rememer. Rtionl numers cn e converted to decimls y dividing the numertor y the denomintor. The resulting deciml cn e either terminting or recurring.. Terminting decimls contin specific numer of digits.. Recurring decimls contin repeting digit or repeting pttern of digits.. Recurring decimls re represented y plcing dots over the first nd the lst digits of the repeting pttern. Alterntively, n overscore cn e plced over the whole pttern tht repets.. Recurring decimls re rtionl numers nd my e expressed s rtio of two integers.
Mths Quest Mths C Yer for Queenslnd B Recurring decimls Mthcd Decimls nd frctions WORKED Exmple WORKED Exmple Stte which of the following rtionl numers cn e expressed s recurring decimls. c d - e - f - g h - i - j k l m n - o p - q r s t u - v - w - x - y z - Express the following recurring decimls s rtionl numers in their simplest form. 0.. 0.. c 0.. d 0.. e 0.. f 0.. g 0.. h 0.. i 0.. j 0.. k.. l 0... m 0... n... o... p 0... q 0... r 0... s 0... t... u 0.. v 0.. w 0... x 0.. y... z 0... - - - The recurring deciml 0... cn e expressed s: A - B - C - D - E 0 0 The recurring deciml 0... cn e expressed s: A B C D E 00 0-0 Which sttement regrding the frctions,, -,, is correct? A, nd - re the only frctions which represent terminting decimls. B nd re the only frctions which represent terminting decimls. C, - nd re frctions which represent recurring decimls. D - nd re the only frctions which represent recurring decimls. E nd re frctions which represent recurring decimls. The recurring deciml 0... cn e expressed in its simplest form s the following frction: 0 A B - C D - E 00 Irene nd Bell re rguing out the correct wy of writing the recurring deciml 0.00 0.... Irene sys it should e written s 0.00, while Bell thinks it is 0.00. Which of the girls is right? 00
Chpter Numer systems: the Rel Numer System Rel nd complex numers WORKED Exmple Clssify ech of the following elements of the set - {,,., -,, p } into the smllest suset in which they elong, using Q, I, Z, Z + nd Z. so π is n irrtionl numer. Numer is positive whole numer; clssify it ccordingly. Z + () Chnge - into deciml. - =. () The frction - cn e expressed s - Q terminting deciml; therefore it cn e clssified s rtionl numer. The frction is in the form, 0, so it is rtionl. The numer. is terminting deciml, so clssify it ccordingly.. Q () Simplify -. - = () The result is negtive whole numer, - Z so clssify - ccordingly. () Use your clcultor to find the vlue of. =.... () The result is non-terminting nd I non-recurring deciml, so cn e clssified s n irrtionl numer. () Chnge π into deciml. π = 0. 0... () The resulting deciml is neither π I terminting nor recurring, So fr we hve considered the set of rel numers nd its susets. However, there is nother set of numers clled complex numers. Consider the exmple of the eqution x + = 0. There re no solutions for x in the Rel Numer System s the eqution ecomes x = ± nd there is no rel numer which, when squred, gives s result. The concept of n imginry numer, denoted y i where i =, ws introduced to overcome this prolem. This mens tht x = ± ecomes x = ± i = ±i. So x + = 0 hs no solutions in the Rel Numer System ut hs the solutions x = i nd x = i in the Complex Numer System. Some other exmples of complex numers re i, i, + i, i. More work involving complex numers will e undertken in chpter.
Mths Quest Mths C Yer for Queenslnd Rel numer investigtions A rel numer cn e defined s numer tht cn e plotted on numer line. Even if the position of the numer on the line is only n pproximte vlue, s long s numer cn e represented y one point on line it cn e regrded s rel. This is not so with the numers you will del with in chpter. The following steps will enle you to plot irrtionl numers such s surds (for exmple nd ) on numer line. Mterils needed: ruler, set of compsses, set squre. Step Drw numer line pproximtely 0 cm long, with unit divisions of cm. How cn we drw line segment exctly units long? Using Pythgors Theorem we cn otin the tringle shown t right which shows us tht = +. Step Use set squre to construct right-ngled tringle s shown elow: 0 Step Use set of compsses to trnsfer the length of the hypotenuse to the numer line. 0 Step If second right-ngled tringle (of height cm) were constructed on this hypotenuse, wht would e the length of its hypotenuse? 0 Step Continue constructing in this wy to plot on the originl numer line. Use your numer line to give n pproximte vlue for.
Chpter Numer systems: the Rel Numer System Other numer systems Introduction Throughout erly civilistions, numers hve een represented nd recorded in vriety of wys. Our numertion system uses the 0 digits 0,,,,,,,,, nd comintions of these. It is clled the deciml or se 0 system (possily influenced y the fct tht we hve 0 fingers). Pst civilistions hve used se nd se 0 systems (gin influenced y the fingers on one hnd nd the totl numer of digits). Mesopotmins used se 0 system, which is still used tody for units of time (0 seconds in minute nd 0 minutes in hour). Numertion systems tht re used tody include inry or se system nd modulr or se system. Plce vlue The plce vlue system ws introduced s mens of recording numers. Look t the numer. In our numertion system the se 0 (deciml) system we interpret the numer (se 0) or 0 s: plus lots of 00 or 0 plus lots of 0 or 0 plus lots of or 0 0 Using the se system Numers in the se system use the digits 0,,, nd only. The numer (se 0) cn e written s se numer in the following wy: lots of or plus lot of or plus lots of or plus 0 lots of or 0 0 So 0 = 0 Using the se system Numers in the se system use the digits 0 nd only. The numer (se 0) cn e written s se numer in the following wy: lot of or plus 0 lots of or 0 plus 0 lots of or 0 plus 0 lots of or 0 plus lot of or plus lot of or plus lot of or plus 0 lots of or 0 plus lot of or 0 So 0 = 00 0 0 Notice how we need to use zeros to hold ech plce vlue. Investigte the following points relting to non-se 0 numers, giving exmples in ech cse: How could numers of different ses e compred to ech other? How re numers of the sme (non-se 0) system dded nd sutrcted? c How re numers of the sme (non-se 0) system multiplied nd divided? d How re frctions nd decimls of non-se 0 system represented?
Mths Quest Mths C Yer for Queenslnd Mthcd Modulo Binry systems As the nme suggests, this numertion system is sed on. In this system, 0 nd re the only two digits used. The inry system is used in electronic computers. Investigte how the inry system is used in electronic computers, circuits or compct discs. Devise sitution which clls for the use of inry system. Modulr rithmetic Modulr rithmetic involves clock rithmetic where, insted of sying tht the time is o clock, we sy it is o clock. This is clled modulr (mod ) rithmetic. Any integer cn e converted to modulr (mod ) rithmetic y sutrcting or ny multiple of from the integer. The reminder is clled the residue. For exmple: = + = + = + (mod ) (mod ) (mod ) The reminders or residues in this cse re, nd respectively. Investigte the purpose, usefulness nd limittions of modulr rithmetic. Include illustrtions of how numers of modulr rithmetic re represented vi clock pttern. rememer rememer. The Rel Numer System contins the set of rtionl numers (Q) nd the set of irrtionl numers (I).. Rtionl numers cn e written s rtio of two integers.. The set of rtionl numers includes the set of integers (Z).. The set of integers consists of positive whole numers (Z + ), negtive whole numers (Z ) nd 0.. Positive integers re lso clled nturl numers (N).. Irrtionl numers include surds, non-terminting nd non-recurring decimls, numers such s π nd e. (There is no seprte set for ny of those.). The Complex Numer System llows us to tke the squre root of negtive numer. It uses the concept of n imginry numer, i, where i =.
Chpter Numer systems: the Rel Numer System C Rel nd complex numers WORKED Exmple Clssify ech of the following into the smllest suset in which they elong, using Q, Z + nd Z. 0. c d 0.. - e + f 0.... g h - i j - 0 k l m ( ) n o p 0... q r s ( ) t u v w ( ) 00 x y z 00 Clssify ech of the following into the smllest suset in which they elong using Q, I, Z + nd Z. 0.... c d - e f g h i 0.... j 0. k l m - n o p π q r ( ) s t u - v. w x (π) y z π 0 - Clssify ech of the following into the smllest suset in which they elong, using Q, I, Z, Z + nd Z. 0... c 0.... d ( ) e ( ) f g 0. 00 h.... i - j k - l 0 m + n - o p - q + r - s 0.000 t. u - v - w. 0 - x - y z π π 0. 0. 0 π - The smllest suset in which + elongs is: A Q B I C Z + D Z E Z
Mths Quest Mths C Yer for Queenslnd The smllest suset in which elongs is: A Q B I C Z + D Z E Z Which of the following sttements regrding numers {,, 0, π, } is correct? A nd 0 re the only rtionl numers. B, 0 nd C nd 0 re positive integers. my e expressed s rtionl numers. D is the only irrtionl numer. E π is the only irrtionl numer. WorkSHEET. Which of the following sttements regrding the given set of numers + {,,,, } is correct? A All of the ove numers in the set re irrtionl. B + nd re the only irrtionl numers of the set. C + is rtionl numer of the set. D is the only rtionl numer of the set. E nd re the only irrtionl numers of the set. Represent the numer on the numer line. Use the method outlined in this section for representing on the numer line s guide. Use the Complex Numer System to simplify ech of the following. c d 00 Surds: suset of irrtionl numers We hve clssified prticulr group of numers s irrtionl nd will now further exmine surds one suset of irrtionl numers nd some of their ssocited properties. A surd is n irrtionl numer which is represented y root sign or rdicl, for exmple:,, Exmples of surds include:,,, Exmples tht re not surds include:,,, Numers tht re not surds cn e simplified to rtionl numers, tht is: =, =, =, =
Chpter Numer systems: the Rel Numer System WORKED Exmple Which of the following numers re surds? 0 c d e f Evlute. The nswer is rtionl (since it is whole numer), so stte your conclusion. = is not surd. Evlute 0. The nswer is irrtionl (since it is non-recurring nd non-terminting deciml), so stte your conclusion. c Evlute. The nswer is rtionl ( frction); stte your conclusion. d Evlute. The nswer is irrtionl ( nonterminting nd non-recurring deciml), so stte your conclusion. e Evlute. The nswer is irrtionl, so clssify ccordingly. f Evlute. The nswer is rtionl; stte your conclusion. 0 =. 0... 0 is surd. c = is not surd. d =. 0 00... is surd. e =. 00... is surd. f = is not surd. So, d nd e re surds. Proof tht numer is irrtionl As prt of your Mthemtics C course you re required to study vriety of types of proofs. One such method is clled Proof y contrdiction. This method is so nmed ecuse the logicl rgument of the proof is sed on n ssumption tht leds to contrdiction within the proof. Therefore the originl ssumption must e flse. An irrtionl numer is one tht cnnot e expressed in the form (where nd re integers). The next worked exmple sets out to prove tht is irrtionl.
0 Mths Quest Mths C Yer for Queenslnd WORKED Exmple Prove tht is irrtionl. Assume tht is rtionl; tht is, it cn e written s in simplest form. We need to show tht nd hve no common fctors. Squre oth sides of the eqution. = Rerrnge the eqution to mke the suject of the formul. If x is n even numer then x = n. = where 0 = [] is n even numer nd must lso e even; tht is, hs fctor of. = r Since is even it cn e written s = r. Squre oth sides. = r [] But = from [] Equting [] nd [] = r Repet the steps for s previously done for. r = - = r is n even numer nd must lso e even; tht is, hs fctor of. Both nd hve common fctor of. This contrdicts the originl ssumption tht = where nd hve no common fctor. is not rtionl. It must e irrtionl. The dilogue included in the worked exmple should e present in ll proofs nd is n essentil prt of the communiction tht is needed in ll your solutions. Note: An irrtionl numer written in surd form gives n exct vlue of the numer; wheres the sme numer written in deciml form (for exmple, to deciml plces) gives n pproximte vlue. rememer rememer. A numer is surd if: () it is n irrtionl numer (equls non-terminting, non-recurring deciml) () it cn e written with rdicl (or root sign) in its exct form.. A non-zero numer hs two squre roots, one positive nd one negtive. For exmple, the squre root of is ± or ±. Note tht gives the position squre root only while gives the negtive squre root.
Chpter Numer systems: the Rel Numer System D Surds: suset of irrtionl numers WORKED Exmple WORKED Exmple Which of the numers elow re surds? c d. e 0. f g h - i 000 j. k 00 l + 0 m n o 00 p q + r π s t u v ( ) w x y The correct sttement regrding the set of numers {, 0,,, } is: A nd re the only rtionl numers of the set. B is the only surd of the set. C nd re the only surds of the set. D 0 nd re the only surds of the set. E z 0 0 All of the numers of the set re surds. Prove tht the following numers re irrtionl, using proof y contrdiction: c 0.000 Which of the numers of the set {, -,,, } re surds? A - B only C only D nd E nd only Which sttement regrding the set of numers {π, -,,, + } is not true? A is surd. B nd re surds. C π is irrtionl ut not surd. D nd + re not rtionl. E - when simplified is rtionl numer. Which sttement regrding the set of numers {,,,,, } is not true? A when simplified is n integer. B nd re not surds. C is smller thn. D is smller thn. E,, nd re surds.
Mths Quest Mths C Yer for Queenslnd Complete the following sttement y selecting pproprite words, suggested in rckets: is definitely not surd, if is... (ny multiple of ; perfect squre nd cue). Find the smllest vlue of m, where m is positive integer, so tht m is not surd. Simplifying surds To simplify surd mens to mke numer (or n expression) under the rdicl ( ) s smll s possile. To simplify surd (if it is possile), it should e rewritten s product of two fctors, one of which is perfect squre, tht is,,,,,,,,, 00 nd so on. We must lwys im to otin the lrgest perfect squre when simplifying surds so tht there re fewer steps involved in otining the nswer. For exmple, could e written s = ; however, cn e further simplified to, so = ; tht is =. If, however, the lrgest perfect squre hd een selected nd hd een written s = =, the sme nswer would e otined in fewer steps. Simplify the following surds. Assume tht x nd y re positive rel numers. 0 c d 0x y WORKED Exmple Express s product of two fctors where one fctor is the lrgest possile perfect squre. = Express s the product of = two surds. Simplify the squre root from the = perfect squre (tht is, = ). Express 0 s product of two fctors, one of which is the lrgest possile perfect squre. 0 = Express s product of two = surds. Simplify. = Multiply together the whole numers outside the root ( nd ). =
Chpter Numer systems: the Rel Numer System c d Express s product of two fctors where one fctor is the lrgest possile perfect squre. Express s product of surds. c = = Simplify. = Multiply together the numers outside the squre root sign. Express ech of 0, x nd y s product of two fctors where one fctor is the lrgest possile perfect squre. d 0x y = x x y y Seprte ll perfect squres into one surd nd ll other fctors into the other surd. = x y xy Simplify x y. = x y xy Multiply together numers nd the = 0xy xy pronumerls outside the squre root sign. = The Mthcd file titled Surds cn e used to simplify surds in different forms. The nswer to question in worked exmple cn e seen in the screen elow. Surds Mthcd
Mths Quest Mths C Yer for Queenslnd rememer rememer. To simplify surd mens to mke numer (or n expression) under the rdicl s smll s possile. For exmple, is equl to, ut simpler thn, 0.. To simplify surd, write it s product of two fctors, one of which is the lrgest possile perfect squre. E Simplifying surds GC progrm Mthcd Surds Surds WORKED Exmple Simplify the following surds. c d e f g h i j 0 k l m n 0 o 0 p q r s t 00 u v 0 w x y 0 z Simplify the following surds., c 0 c 0 d WORKED Exmple WORKED Exmple e 0 f g h 0 i j k 0 l 0 m n 0 o p q r s t u v 0 w - x y 0 z 0 Simplify the following surds. Assume tht,, c, d, e, f, x nd y re positive rel numers. d c d e 0 f g h 0 i j k l 0x y 0
Chpter Numer systems: the Rel Numer System m x y n x y o 0x y p x y q c d r c d s 0c d t c 0 d 0 u ef v 0e f w e f x - e f y - x y z - 0x 0 y 0 When expressed in its simplest form, is equl to: A B C D E When expressed in its simplest form, is equl to: A B C D E When expressed in its simplest form, is equl to: A B C D E Assuming tht x nd y re positive rel numers, simplest form is equl to: - x y A xy x y B -x y y C when expressed in its x y y D x y y E xy x y Addition nd sutrction of surds Surds my e dded or sutrcted only if they re like. Exmples of like surds include, nd. Exmples of unlike surds include,, nd. In some cses surds will need to e simplified efore you decide whether they re like or unlike, nd then ddition nd sutrction cn tke plce. The concept of dding nd sutrcting surds is similr to dding nd sutrcting like terms in lger.
Mths Quest Mths C Yer for Queenslnd WORKED Exmple Simplify ech of the following expressions contining surds. Assume tht nd re positive rel numers. + + + c 00 + All terms re like, since they contin the sme surd ( ), so group like terms together nd simplify. + = ( + ) = Simplify surds where possile. + + = + + = + + = + + Add like terms to otin the simplified nswer. = + c Simplify surds where possile. c 00 + = 0 + = 0 + = + 0 Add like terms to otin the simplified nswer. = 0 WORKED Exmple Determine the perimeter of rectngle whose length is ( + ) m. ( 0) m nd width is Write down the rule for the perimeter of rectngle where l is the length nd w is the width. P = l + w Sustitute the vlues of l nd w into the P = ( 0) + ( + ) rule.
Chpter Numer systems: the Rel Numer System Expnd nd simplify where possile. P = 0 + 0 + Simplify surds where possile. = + 0 + Collect like terms. = = = + 0 + 0 + 0 + Stte the nswer, including the P = ( ) pproprite unit. m The Mths Quest CD-ROM contins Mthcd file tht cn e used to dd or sutrct surds. The screen elow shows n exmple. Addition nd sutrction of surds Mthcd rememer rememer. Only like surds my e dded nd sutrcted. Exmples of like surds:, nd. Exmples of unlike surds:, nd.. Surds my need to e simplified efore dding nd sutrcting.
Mths Quest Mths C Yer for Queenslnd F Addition nd sutrction of surds Mthcd WORKED Exmple Addition nd sutrction of surds Simplify the following expressions contining surds. Assume tht x nd y re positive rel numers. c + + + + d + + e g i k m o + + + f + + + h + j + + l + + 0 n xy + xy xy x + y + x y p x + y + xy x y WORKED Exmple Simplify the following expressions contining surds. Assume tht nd re positive rel numers. 00 00 + 0 c e g i k m o q 0 + 00 d + + f + + 0 0 + h + 0 + + j + 0 + + 0 l 0 + 0 0+ 0 + 00 n + + p + + 0 + 0 + 0 r 0 0 0 0 + 0 s + + t 0 + u + + v - + w + x + - -
Chpter Numer systems: the Rel Numer System WORKED Exmple c Simplify the following expressions contining surds. Assume tht nd re positive rel numers. + 0 + + c 0 + d + + e + f + g i k + h + + + j + + + l + When expressed in its simplest surd form, is equl to: A B C D E none of these When expressed in its simplest surd form, 0 is equl to: A B C 0 D E 0 When expressed in its simplest surd form, - 00 0 + is equl to: A + B + C D + E + When expressed in its simplest surd form, is equl to: A B C D E ( ) When expressed in its simplest surd form, 0c d cd c d is equl to: A cd B cd C cd D cd E cd
0 Mths Quest Mths C Yer for Queenslnd SkillSHEET. WORKED Exmple Find the perimeter of the following shpes, giving nswers in the simplest surd form. Specify the units. cm c + cm cm cm + cm + cm cm WorkSHEET. d e m f m + m m + m m Multipliction of surds To multiply surds, multiply together the expressions under the rdicls. For exmple, =, where nd re positive rel numers. When multiplying surds it is est to first simplify them (if possile). Once this hs een done nd mixed surd hs een otined, the coefficients re multiplied with ech other nd then the surds re multiplied together. For exmple, WORKED Exmple 0 m n = mn Multiply the following surds, expressing nswers in the simplest form. Assume tht x nd y re positive rel numers. c 0 0 0 d e f x y x y Multiply surds together, using = (tht is, multiply expressions under the roots). Note: This expression cnnot e simplified ny further. = = Multiply coefficients with ech other nd then multiply surds together. = = 0 = 0
Chpter Numer systems: the Rel Numer System c Multiply surds together. c 0 = = 0 0 Simplify the product surd if possile. = = d Simplify. d = = = Multiply coefficients with ech other = nd multiply surds together. Simplify product surd. = = = e Multiply coefficients with ech other nd multiply surds together. e 0 0 = Simplify product surd. = Simplify y dividing oth 0 nd 0 y 0 (cross-cncel). = = 0 0-00 0-00 0-0 0 = or - f Simplify ech of the surds. f x y x y = x x y x y = x y x x y = x y x x y Multiply coefficients with ech other = x y x x y nd surds together. = x y xy Simplify product surd. = x y xy = x y xy = x y xy When working with surds, we sometimes need to multiply surds y themselves; tht is, squre them. Consider the following exmples: ( ) = = = ( ) = = =
Mths Quest Mths C Yer for Queenslnd We oserve tht squring surd produces the numer under the rdicl. This is not surprising, since squring nd tking the squre root re inverse opertions nd, when pplied together, leve the originl unchnged. When surd is squred, the result is the numer (or expression) under the rdicl; tht is,, where is positive rel numer. ( ) = Evlute the re of squre of length simplest form. Write the rule for the re of squre. A = l Sustitute the vlue for l into the rule. A = ( xy) Simplify, using. = = xy = xy Write the nswer, including n pproprite unit. A = xy m ( xy) ( ) = ( ) WORKED Exmple rememer rememer m, expressing the nswer in the ( xy). When multiplying surds, simplify the surd if possile, then pply the following rules: () = () m n = mn, where nd re positive rel numers.. When surd is squred, the result is numer (or n expression) under the rdicl:, where is positive rel numer. ( ) = G Multipliction of surds Multiply the following surds, expressing nswers in the simplest form. c d e f g 0 0 h i j k l 0
Chpter Numer systems: the Rel Numer System m n o 0 p q 0 r s 0 t 0 u v w x y - 0 0 0 z 0 0 WORKED Exmple 0f Simplify the following expressions with surds. Assume tht,, x nd y re positive rel numers. xy x y x y x y c e g i d 0c f x y x y h 0x y x y j WORKED Exmple Find the re of the following shpes. Answers must e expressed in the simplest surd form nd the pproprite units specified. c SkillSHEET. cm cm m m d e f m m m m 0 m m 0 m m The product of 0 expressed in its simplest form is: A B 0 C 0 D 0 E 0 The product of x y x y expressed in its simplest form is: A x y 0xy B 0x y C x y 0xy D x y E x y xy
Mths Quest Mths C Yer for Queenslnd The product of x y x y expressed in its simplest form is: A - x y B - x y C -x y xy D -x y xy E -x y xy The re of the tringle expressed in its simplest form is: m m A 0 m B 0 m C m D m E 0 m The height of squre-sed pyrmid is 0 units nd the length of the side of its se is units. Find the volume of the pyrmid, expressing the nswer in the simplest surd form. (Volume = re of se height) Division of surds To divide surds, divide the expressions under the rdicls; tht is, =, where nd re whole numers. When dividing surds it is est to simplify them (if possile) first. Once this hs een done, the coefficients re divided next nd then the surds re divided. WORKED Exmple Divide the following surds, expressing nswers in the simplest form. Assume tht x nd y re positive rel numers. xy - - c d - x y Rewrite the frction, using =. - = - Divide numertor y the denomintor = (tht is, y ). Check if the surd cn e simplified ny further.
Chpter Numer systems: the Rel Numer System Rewrite the frction, using =. - = - Divide y. = Evlute. = c Rewrite surds, using =. c = Simplify the frction under the rdicl y dividing oth numertor nd denomintor y. = - Simplify surds. = - Multiply the whole numers in the = numertor together nd those in the denomintor together. Cncel down the common fctor of. = d xy Simplify ech surd. d = x y = Cncel down ny common fctors in = this cse xy. xy x x y 0 y xy - x y xy - x y WORKED Exmple Find the perpendiculr height of tringle, given tht its re is cm nd its se length is cm. The nswer must e expressed in the simplest surd form nd the pproprite unit specified. Write the rule for the re of tringle. A = h Sustitute the vlues for A nd into the rule. = h Cncel down the nd the. = h Continued over pge
Mths Quest Mths C Yer for Queenslnd Trnspose the eqution to mke h the h = - suject. Divide numertor nd denomintor y (cncel down). = - Simplify nd write the nswer, using the pproprite unit. h = cm rememer rememer When dividing surds, simplify the surd if possile, then pply the following rule: = = where nd re whole numers, nd 0. H Division of surds WORKED Exmple Simplify the following surds, expressing nswers in the simplest form. Assume tht x nd y re positive rel numers. - - c d e 0 - f 0 - g h 0 i - j k l m - n o - p 0 00 q r s - t 0 0 - - u x y v x y w xy - x x y x y x y x y xy
Chpter Numer systems: the Rel Numer System Simplify the following. Assume tht ll pronumerls re positive rel numers. xy x y x y x y - c x y x y x y xy d mn - e m - n - f m n mn 0 - - m n m n - m n mn Expressed in its simplest form, - is: A 0 B C D - E Expressed in its simplest form, is: A - B - C - D - E 0x Expressed in its simplest form, y is: 0x y xy A xy B 0 x - C y x D y 0 - E 0 xy 0x x Expressed in its simplest form, y x - y - is: xy x y y A y x B x - x C x x - D - E x y y y - x y x x y WORKED Exmple Find the length of the unknown side in ech of the following. Answers must e expressed in the simplest surd form nd the pproprite units specified. c A = m A = 0 cm A = m w cm h m m
Mths Quest Mths C Yer for Queenslnd d e f V = 0 m V = π cm V = 0π cm h h h m m cm Are of se = π cm E Velocity v of the oject cn e found using the formul v =, where E is the kinetic m energy of the oject nd m is the mss of the oject. Express v s the simplest surd, if: E = 0 J, m = kg E = 0 J, m = 0 kg c E = 0 J, m = 0 kg A rectngulr fish tnk hs se 0 cm y 0 cm nd the height h. When of the tnk is filled, the volume of wter is L. Find: the height of the tnk (give the nswer s the simplest surd) the full cpcity of the tnk in litres. (Rememer tht cuic centimetre holds ml of wter.) The Distriutive Lw The Distriutive Lw sttes tht ( + c) = + c. When multipliction of surds involving rckets is required, the Distriutive Lw is pplied s in the cse with lgeric terms. Tht is: ( c) + = + c If there is negtive numer outside the rcket, then every term inside the rcket will undergo sign chnge since it hs een multiplied y the negtive numer. WORKED Exmple Expnd nd simplify the following where possile. ( ) c + ( ) ( 0 ) Write down the expression. ( + ) Apply the Distriutive Lw: Multiply the term outside the rcket with the first term inside the rcket, then multiply the term outside the rcket with the second term inside the rcket. = + Simplify. = 0 +
Chpter Numer systems: the Rel Numer System Write down the expression. ( ) ( ) Simplify. = Apply the Distriutive Lw to remove the rckets. = + Simplify. = c Write down the expression. c () Expnd the rckets, using the = 0 Distriutive Lw. () Be sure to multiply through with the negtive. = 0+ 0 Simplify. = = 0 + 0 0+ 0 ( 0 ) Expnd ( When expnding two inomil rckets the FOIL method is pplied; tht is, pirs of terms must e multiplied in the order First, Outer, Inner nd Lst. WORKED Exmple + )( ) Write down the prolem. Apply FOIL. Multiply the first terms of ech rcket. Multiply the outer terms of ech rcket. Multiply the inner terms of ech rcket. Multiply the lst terms of ech rcket.. Write your nswer in its simplest form. F Simplify. = 0+ = L ( )( ) + I O + + = 0+ = 0+ + Recll the perfect squre identities: ( + ) = + + ( ) = +
0 Mths Quest Mths C Yer for Queenslnd Expnd The perfect squre identities cn e pplied to surds s follows: WORKED Exmple ( ) ( + ) = ( ) + + ( ) = + + ( ) = ( ) + ( ) = +. Write your nswer in its simplest form. Write the expression. ( ) Apply the perfect squre identity. = ( ) + ( ) Simplify. = + = Note tht the expnsion of ( ) in the previous exmple could e lso done y writing it s product of two repeted fctors, ( )( ), nd pplying FOIL. Nturlly, the result would e the sme, ut the solution would tke longer. Recll the difference of two squres (DOTS) identity: ( )( + ) = The DOTS identity cn e pplied to surds s follows: ( )( ) = = - WORKED Exmple Expnd ( y x)( y + x). + ( ) ( ) Write the expression. ( y x)( y + x) Use DOTS identity for expnsion. = ( y) ( x) Simplify. = y x = y x In the ove exmple the inomil fctors which were multiplied together re conjugte pir (tht is, one rcket contins sum nd the other difference of the sme terms). Although the terms of the fctors re irrtionl, the nswer is not surd, ut n expression with rtionl terms.
Chpter Numer systems: the Rel Numer System The product of conjugte pir of surds (irrtionl numers) yields rtionl numer. Note tht to find the product of conjugte pir (s in worked exmple ), FOIL could e used s n lterntive to the DOTS identity. The ltter, however, leds to the nswer much more quickly. rememer rememer. When expnding rckets, the Distriutive Lw is pplied: ( + c) = + c. When expnding inomil rckets, FOIL is pplied: ( + )( c+ d) = c + d + c + d. Perfect squre identities: ( + ) = ( ) + + ( ). DOTS identity: = + + ( ) = ( ) + ( ) ) =. The product of conjugte pir of surds is rtionl. = + ( + )( I The Distriutive Lw WORKED Exmple WORKED Exmple Expnd nd simplify the following, where possile. ( + ) ( ) c ( ) d ( + ) e ( ) f ( ) g ( + ) h ( ) i ( ) j ( + 0) Expnd nd simplify where possile. ( ) ( + ) ( + )( ) c ( + )( ) d ( )( 0) e ( + )( ) f ( )( ) g ( ( + ) h ( )( ) i ( x + y)( x + y) j ( x 0y)( x + 0y) Surds SkillSHEET Mthcd.
Mths Quest Mths C Yer for Queenslnd WORKED Exmple WORKED Exmple Expnd nd simplify where possile. ( + ) ( + 0) c ( + ) d ( + ) e ( + ) f ( + ) g ( + ) h ( ) i ( ) j ( ) Expnd nd simplify the following where possile. ( + )( ) ( + )( ) c ( + )( ) d ( + )( ) e ( + )( ) f ( 0 + )( 0 ) g ( )( + ) h ( )( + ) i ( )( + ) j ( + )( ) k ( 0 + )( 0 ) l ( )( + ) m ( )( + ) n ( + )( ) o ( )( + ) p ( )( + ) q ( )( + ) r ( + )( ) s ( + )( ) t ( )( + ) u ( x y)( x + y) v ( x y)( x + y) w ( x y)( x + y) x (x x + y)(x x y) y (x y y x)(x y+ y x) z ( x y xy )( x y + xy ) When expressed in its simplest form, ( ) is equl to: A B C D 0 E When expressed in its simplest form, A ( + )( ) is equl to: 0 0 + B 0 0 + C E 0 0 0 + D 0 0 0 + 0 0 0 + When expressed in its simplest form, is equl to: A B C + D E ( )
Chpter Numer systems: the Rel Numer System When expressed in its simplest form, ( x y + xy)( x y xy) is equl to: A x y 0 xy + xy B x y xy C x y x y D x y xy E x y 0xy x + xy. Given tht x =, find: x x + x + Rtionlising denomintors If the denomintor of frction is surd, it cn e chnged into rtionl numer. In other words, it cn e rtionlised. As we discussed erlier in this chpter, squring simple surd (tht is, multiplying it y itself) results in rtionl numer. This fct cn e used to rtionlise denomintors s follows. = -, (where = ) If oth numertor nd denomintor of frction re multiplied y the surd contined in the denomintor, the denomintor ecomes rtionl numer. The frction tkes on different ppernce, ut its numericl vlue is unchnged, ecuse multiplying the numertor nd denomintor y the sme numer is equivlent to multiplying y. WORKED Exmple Express the following in their simplest form with rtionl denomintor. - c Write down the frction. Multiply oth the numertor nd denomintor y the surd contined in the denomintor (in this cse ). This hs the sme effect s multiplying the frction y, since - =. - = - - = - Continued over pge
Mths Quest Mths C Yer for Queenslnd Write down the frction. Simplify the surds. (This voids deling with lrge numers.) Multiply oth the numertor nd denomintor y. (This hs the sme effect s multiplying the frction y since =.) Note: We need to multiply only y the surd prt of the denomintor (tht is, y, rther thn y ). = - = - = - = - = Simplify. = - = - = Divide numertor nd denomintor y = - (cncel down). c Write down the frction. c Multiply oth the numertor nd denomintor y. Use rckets so you relise the whole numertor must e multiplied y. Apply the Distriutive Lw in the numertor. ( + c) = + c ( ) = - = - = -
Chpter Numer systems: the Rel Numer System Simplify. = = = - - rememer rememer To rtionlise the surd denomintor, multiply the numertor nd denomintor y the surd contined in the denomintor. This hs the effect of multiplying the frction y nd thus the numericl vlue of the frction remins unchnged, while the denomintor ecomes rtionl: = = - J Rtionlising denomintors Express the following in their simplest form with rtionl denomintor., c - d e - WORKED E xmple WORKED Exmple f - g - h - i - j 0 k l m - n o Express the following in their simplest form with rtionl denomintor. c + - c - d 0 + + e f g - h i - j k 0 - + -
Mths Quest Mths C Yer for Queenslnd When expressed in its simplest form, is equl to: A B C - D E When expressed in its simplest form, is equl to: 0 A B - C - D E 0 0 0 When expressed in its simplest form, is equl to: A B C 0 D E none of these When expressed in its simplest form, is equl to: 0 A - B 0 0 C D 0 0 E 0 0 0 0 - Solve for x, giving the nswers s the simplest surds with rtionl denomintors: x = x = c x = Rtionlising denomintors using conjugte surds As shown erlier in the chpter, the product of pirs of conjugte surds results in rtionl numer. (Exmples of pirs of conjugte surds include + nd, + nd, nd +.) This fct is used to rtionlise denomintors contining sum or difference of surds. To rtionlise the denomintor which contins sum or difference of surds, we multiply oth numertor nd denomintor y the conjugte of the denomintor.
Chpter Numer systems: the Rel Numer System Two exmples re given elow:. To rtionlise the denomintor of the frction -, multiply it y -. + +. To rtionlise the denomintor of the frction -, multiply it y -. + A quick wy to simplify the denomintor is to use the DOTS identity: WORKED Exmple ( )( = = + ) ( ) ( ) Rtionlise the denomintor nd simplify the following. + - - + Write down the frction. - Multiply the numertor nd = ( + ) denomintor y the conjugte of the denomintor. ( ) ( + ) + (Note tht - + =.) + Apply the Distriutive Lw in the = numertor nd the DOTS identity ( ) ( ) in the denomintor. Simplify. + = - + = - + Write down the frction. - + ( + ) ( ) Multiply the numertor nd = denomintor y the conjugte of the denomintor. ( + ) ( ) (Note tht - =.) Continued over pge
Mths Quest Mths C Yer for Queenslnd Apply FOIL in the numertor nd DOTS in the denomintor. = + + + - ( ) ( ) + Simplify. = - + = + = - + = - = - = You might wish to use clcultor to check if the finl nswer is correct. To do tht, evlute the originl frction nd the finl one (the one with the rtionl denomintor) nd check whether they oth equl the sme numer. Rtionlise the denomintors nd simplify the following. + + We will rtionlise the denomintor of ech term nd then dd them. Write down the first frction. WORKED Exmple 0 Multiply the numertor nd denomintor y the conjugte of the denomintor. ( + ) = - ( ) ( + ) + Apply the Distriutive Lw in the = - numertor nd DOTS in the denomintor. ( ) (Note tht when squring, we need to squre oth the nd the.) Simplify the denomintor. + = - Write down the second frction. +
Chpter Numer systems: the Rel Numer System Multiply the numertor nd denomintor y the conjugte of the denomintor. = - ( + ) ( ) - ( ) 0 Apply the Distriutive Lw in the numertor nd DOTS in the denomintor. = Simplify the denomintor. = Add the two frctions together. Bring them to the lowest common denomintor first. Add the numertors. = Simplify where pproprite. = + - + + = - + + = + - The following worked exmple demonstrtes the rtionlistion of the denomintor when it is trinomil (hs three terms). WORKED Exmple Simplify: - ( + ) Use set of rckets to group the trinomil into inomil. Multiply the numertor nd denomintor y the conjugte of the denomintor; tht is, ( + ) +. Use rckets round oth fctors so tht you will recognise tht ll terms need to e multiplied. - ( + ) = - ( + ) [( + ) + ] - ( + ) + Continued over pge
0 Mths Quest Mths C Yer for Queenslnd ( + ) + Use FOIL to expnd the denomintor. = ( + ) ( + ) + Expnd the squred terms of the = - denomintor. + + + + Group nd simplify the denomintor. = - + ( + + ) ( ) Rtionlise the denomintor s shown = - previously. Use rckets s in step. ( + ) ( ) + + Expnd the numertor, mking sure tht = - every term in the first set of rckets is multiplied y every term in the second set. + + Group like terms nd simplify. = Multiply numertor nd denomintor y to eliminte the negtive denomintor. + = - + + = rememer rememer. To rtionlise the denomintor contining sum or difference of surds, multiply oth the numertor nd denomintor of the frction y the conjugte of the denomintor. This elimintes the middle terms nd leves rtionl numer.. To simplify the denomintor quickly, use the DOTS identity: ( )( + ) = ( ) ( ) =. To rtionlise the denomintor of the frction, multiply it y + -.. To rtionlise the denomintor of the frction -, multiply it y +. +
Chpter Numer systems: the Rel Numer System K Rtionlising denomintors using conjugte surds WORKED Exmple Rtionlise the denomintor nd simplify. - - c - d + e f - g h + + i - j - k l 0 + + + m n - o - p 0+ + + 0 + q - r - s t + 0 0 + u - v w x + + - + - - + + - y + z - + WORKED Exmple 0 Rtionlise the denomintor nd simplify. - + - + + c - + d + + e - + + f + + - - + g + + + h + - + + i + + j + WORKED Exmple k + + + l - + +
Mths Quest Mths C Yer for Queenslnd Given tht x = - find ech of the following, giving the nswer in surd form + with rtionl denomintor: x + x x x Given tht x = find ech of the following, giving the nswer in surd form with rtionl denomintor: x + x c x + x x x d x e x + x + f x x + g h Is x = + solution for the eqution x 0x + = 0? Show ll working. Solve for x giving nswers in surd form with rtionl denomintors. x x + x x x x x + x + = + x x 0 = x + When simplified with rtionl denomintor, - is equl to: + + A B C - D E + + When simplified with rtionl denomintor, is equl to: + + + A - B + C - D E When simplified with rtionl denomintor, is equl to: + 0 + 0 + 0 + 0 A - B - C - D - E + - + + - + 0 0 If x = +, then x + when simplified with rtionl denomintor is equl to: x + + + + + A - B - C - D - E -
Chpter Numer systems: the Rel Numer System Further properties of rel numers modulus The modulus or solute vlue of numer is the mgnitude of tht numer. It represents the distnce of the numer from the origin (tht is, 0 on numer line). The modulus of x is denoted y x nd is lwys positive. Note: Do not confuse the modulus of numer with modulr rithmetic (see pge ). For exmple, = = 0 = 0 WORKED Exmple Evlute the following. 0 c d cd cd Write down the expression. 0 The modulus sign indictes tht we wnt only = 0 mgnitude of numer nd not the sign of it. So the negtive in front of the numer should e omitted. Write down the expression. Evlute ech modulus seprtely nd then simplify. = = c Write down the expression. c Evlute ech modulus seprtely, then simplify. = = d Write down the expression. d cd cd Evlute ech modulus seprtely, then simplify. cd cd = - c = d - = c d rememer rememer. The modulus (or solute vlue) of numer is the mgnitude of tht numer. It tells us how fr the numer is from zero, nd is lwys positive.. The modulus of x is denoted y x.
Mths Quest Mths C Yer for Queenslnd L Further properties of rel numers modulus WORKED Exmple Evlute the following where,, c, d 0: c 0. d e f g h i. j 0 k l m + n o p q r s t u v w 0 x 0 y z cd cd cd When simplified, ecomes: A B C D E When simplified, ecomes: A B C D E When simplified, + ecomes: A B C D E When simplified, ecomes: A - B C - D E Fill in the tle elow for the function y = x. x 0 y y c d Use the tle to plot (on the sme set of xes) the grph of y = x nd y = x. Stte the rnge of ech of the two functions. Compre the rnges of the two functions nd their grphs. Explin the difference.
Chpter Numer systems: the Rel Numer System Solving equtions using solute vlues If x = then, y definition of solute vlues, there re vlues of x tht stisfy this eqution tht sttes tht x is three units from 0. Tht is, x = or x = units units 0 Therefore, two seprte cses need to e considered when solving equtions involving solute vlues. WORKED Exmple Solve: x = - x =. Write the eqution. x = Remove the solute rckets nd Cse : Cse : write the positive (+ve) nd negtive x = x = ( ve) cses to e considered. Work the two cses side y side. x = or x = x cn equl either or, written ±. x = ± Write the eqution. x = Remove the rckets nd write the Cse : Cse : +ve nd ve cses. x = or x = Solve for x in oth cses. = x + = x = x = x x = x = Verify your solution y sustituting into the originl expression. Strt = = with the left-hnd side nd ensure tht it equls the right-hnd side. = = = Solutions re correct for oth cses. WORKED Exmple Solve x - = x +. Write the eqution. x = x + Continued over pge
Mths Quest Mths C Yer for Queenslnd Remove the rckets nd write the +ve nd ve cses. Use rckets for the RHS of the ve cse. Cse : Cse : x = x + or x = (x + ) Solve for x. = x x x = x = x x + x = + x = x = x = Verify your solution for oth cses y sustituting into the originl eqution. As the RHS should lwys e +ve, this solution x = is not suitle nd should e ignored. Notice how importnt this verifiction step is. We hve followed ll the correct steps ut logiclly rrived t n nswer tht is not possile. Verify ll results for these questions. Using x = = = (LHS) + = + (RHS) = + = (Not the correct solution since LHS RHS) Using x = = = (LHS) - + = + (RHS) = + = (Correct solution since LHS = RHS) The solution is x =. Solve x = x +. WORKED Exmple Write the eqution. x = x + Remove ll solute rckets nd write the +ve nd ve cses. Ressure yourself tht there re only two possile cses. (x ) = (x + ) is the sme s (x ) = (x + ) nd (x ) = (x + ) is the sme s (x ) = (x + ) Cse : Cse : x = x + or x = (x + )
Chpter Numer systems: the Rel Numer System Solve for x for oth cses. = x x x = x = x x + x = + x = x = x = Verify the solutions with Using x = respect to the originl = (LHS) eqution. ( ) + = + (RHS) = (Correct solution since LHS = RHS) Using x = rememer rememer + = = + = (Correct solution since LHS = RHS) Therefore x = nd x = re oth suitle solutions. To solve equtions with solute vlues:. remove the solute vlue symols nd stte the eqution s positive nd negtive cses. verify your solutions y sustituting your nswer into the originl eqution. M Solving equtions using solute vlues WORKED E xmple Solve for x. x = 0 x + = c x = x d x + = e + x = 0 f = SkillSHEET. WORKED Exmple Solve for x. x + = x x + = x c x + = x d x = x WORKED Exmple Solve the following for x. x = x + x = x + c x = x + d x = x
Mths Quest Mths C Yer for Queenslnd Solving inequtions You hve grphed inequtions on numer line in your junior mthemtics studies. These exmples require more cre nd you will notice tht the verifiction step is essentil to test the vlues you otin. WORKED Exmple Solve nd grph (x - )(x + ) > 0. /DRAW Write the ineqution. (x )(x + ) > 0 If > 0 then either nd re oth positive (+ve) or nd re oth negtive ( ve). This gives rise to cses. Rewrite the terms of the ineqution. Note: > 0 mens +ve, nd Note: < 0 mens ve Solve ech ineqution. Grph oth these inequtions nd decide which prt of the grph stisfies oth inequtions. Note tht the region grphed in the lst grph (x > ) stisfies oth prts of cse. Repet steps nd for cse. Grph oth these inequtions nd decide which prt of the grph stisfies oth inequtions. Note tht the region grphed in the lst grph (x < ) stisfies oth prts of cse. Either cse is true or cse is true t the one time ut not oth, s they re contrdictory. Comine oth cses on one numer line so tht either x > (from cse ) or x < (from cse ). Cse : If nd > 0 (x ) > 0 nd (x + ) > 0 x > x > 0 0 Cse : If nd < 0 (x ) < 0 nd (x + ) < 0 x < x < 0 0 0 Either x > or x < 0 Grphed Not grphed Grphed
Chpter Numer systems: the Rel Numer System Use tulr form to verify this solution. Note how the numer line is divided into regions. When completing this tle choose numer tht flls in ech region nd work out the sign only of ech expression. Becuse the originl product ws greter thn 0 (or positive) the tle hs verified the results on the grph. We do not wnt to include those vlues etween nd. x x + (x )(x + ) x < (let x = ) + grphed OK < x < (let x = 0) + not grphed OK The solution is either x > or x <. x > (let x = ) + + + grphed OK WORKED Exmple Solve nd grph: x < where x 0 - < 0 where x. x x + /DRAW Write the ineqution. < x x cn e either +ve or ve. When x is Cse : Cse : ve nd multiplied cross the inequlity sign, the sign must e If x > 0, < If x < 0, < x x reversed. < x > x < x > x Drw seprte grphs for oth these inequlities, ut rememer tht in cse, x < 0 so the only prt tht should e grphed is where x < 0. Comine these grphs on the one numer line nd stte your nswer. Rememer tht n initil condition of the prolem ws tht x 0 so tht hs een stisfied lso. or x > or x < 0 0 0 x < 0 nd x > Continued over pge
0 Mths Quest Mths C Yer for Queenslnd /DRAW x Write the ineqution. - < 0 x + Rememer tht < 0 mens ve. If < 0 Cse : If x < 0 nd x + > 0 then either < 0 or < 0, ut not oth t x < x > the one time. Grph oth these inequlities. 0 0 Since oth results from cse occur t the one time comine the two different grphs tht stisfy oth prts. 0 Repet steps to for cse. Cse : If x > 0 nd x + < 0 x > x < Since oth results of cse occur t the one time think out how to comine the two different grphs tht stisfy oth results. Note tht x cn t e greter thn nd less thn t the sme time. Therefore this solution is impossile. Reject this solution. Grph nd stte the finl solution (tht ws otined in step ove). 0 Cse Cse 0 The solution is < x <. Solve nd grph x <. WORKED Exmple /DRAW Write the inequlity. x < The solute vlue of the inequlity is Cse : Cse : less thn (tht is, within plces x < or x > of 0). So removl of the rckets mens the inequlity sign will e reversed in the negtive cse. Note tht x >. Solve for x. x < or x > Grph oth solutions on seprte numer lines. x < or x > x < 0 Cse Cse
Chpter Numer systems: the Rel Numer System Grph this comined solution nd stte the solution. /DRAW The solution is < x <. Verify the regions of this solution. For x < (let x = 0) 0 < (Not vlid nd not grphed) For < x < (let x = ) < < (Vlid nd grphed) For x > (let x = ) < < (Not vlid nd not grphed) 0 Solve nd grph - <. x WORKED Exmple /DRAW Write the inequlity. < x As with worked exmple, stte the Cse : Cse : two cses tht re possile. Solve for cse. - < or - > x x For cse if x > 0 (x > ) < (x ) < x < x < x x > Since x > (initil condition) x > stisfies this condition. Drw the grph for this solution. Repet step for cse. Reverse the inequlity sign when you multiply y negtive. Reverse the inequlity sign when you divide y negtive. Rememer to lwys check with the initil condition. 0 Cse - x > If x < 0 (x < ) < (x ) < x + < x Continued over pge
Mths Quest Mths C Yer for Queenslnd Drw the grph for this solution. Drw the comined grph for these two solutions nd stte the nswer. Verify the results for the three regions on the grph. All solutions hve een verified. Even though this verifiction is firly lengthy step it gives you confidence tht your solutions re more thn likely going to e correct. /DRAW > x x < from the initil condition; therefore x < stisfies this condition. 0 0 The solution is x < nd x >. For x < (let x = 0) < 0 - < ( Vlid nd grphed) For < x < (let x = ; rememer x ) - - < < < (Not vlid nd not grphed) For x > (let x = ) < - < (Vlid nd grphed) x Solve nd grph - < where x. x + /DRAW WORKED Exmple 0 x Write the inequlity - < x +
Chpter Numer systems: the Rel Numer System Rememer x hs the sme vlue in the numertor nd denomintor t ny one time. Seprte the solution into the +ve nd ve cses when the solute rckets re removed. /DRAW Cse : Cse : x x - < - > x + x + If the denomintor is ve, then the sign must e reverse when it is multiplied cross. Therefore, we will hve to mke su-cses for ech of cse nd cse. Grph this first solution. Cse (+ve): If x + > 0 (tht is, x > ) x < (x + ) x < x + < x x > But the initil condition is tht x >, therefore x > is vlid solution. 0 Reverse the sign due to the ve denomintor. Grph this solution. Cse ( ve): If x + < 0 (tht is, x < ) x - < x + x > (x + ) x > x + > x x < But the initil condition is tht x <, therefore x < is vlid solution. Determine cse s for cse then grph this solution. x - > x + Cse (+ve): If x + > 0 (tht is, x > ) x > (x + ) x > x x > + x > - But x > from initil condition so x >. 0 Continued over pge
Mths Quest Mths C Yer for Queenslnd Determine cse for the negtive denomintor. Rememer to reverse the sign when you multiply y the ve denomintor. Grph this solution. /DRAW Cse ( ve): If x + < 0 (tht is, x < ) x - > x + x < (x + ) x < x x < + x < But x < from the initil condition, so x <. Since the denomintor must e either positive or negtive t ny one time it is cse (+ve) nd cse (+ve) tht we need to comine s well s cse ( ve) nd cse ( ve) to produce the finl grph. Comine the two solutions. Verify the solutions y testing within the regions, using the originl inequlity. 0 Comining cse (+ve) nd cse (+ve) gives: 0 Comining cse ( ve) nd cse ( ve) gives: 0 0 0 The solution is x < nd x > For x < (let x = ) < + < (Vlid nd grphed) For < x < (let x = ) < + - < (Not vlid nd not grphed) For x > (let x = 0) 0 - < 0+ - < (Vlid nd grphed) This type of prolem demonstrtes higher level resoning for this study of numers. All possiilities need to e crefully considered nd exmined in thoughtful, methodicl mnner.
rememer rememer Chpter Numer systems: the Rel Numer System. When solving inequtions reverse the inequlity sign when you multiply or divide y negtive expression.. In ny eqution if the product of nd (tht is ) is positive then nd re either oth positive or oth negtive.. If product is negtive then either or is negtive.. If x is positive then x > 0 nd x >.. If x is negtive, then x < 0 nd x <. N Solving inequtions WORKED Exmple Solve nd grph the following inequtions. (x )(x ) < 0 (x )(x + ) > 0 c (x )(x ) < 0 d x 0x + < x x WORKED Exmple Solve nd grph the following inequtions. < (where x 0)) - < (where x ) x x x c - < (where x ) d - < (where x ) x x WORKED Exmple Solve nd grph the following. x < x + < x c x < d < WORKED Exmple, 0 Solve nd grph the following. - < (where x ) < (where x ) x + x c < x (where x ) d - x x + < (where x ) e x + x - > (where x ) f - x x > (where x ) WorkSHEET.
Mths Quest Mths C Yer for Queenslnd Approximtions for p Reserch the following historicl pproximtions for π nd present your findings in concise form. 000 BC Egypt: The pyrmids re uilt. The sides nd heights of the pyrmids of Cheops nd Sneferu t Gizeh re constructed in the rtio of :. Hence the rtio of one perimeter to heights is :. The vlue of π is pproximtely -. 000 BC Egypt: The Rhind Ppyrus, the oldest mthemticl text in existence, gives the following rule for constructing squre hving the sme re s given circle: Cut one-ninth off the circle s dimeter nd construct squre on the reminder. Using this method, π is found to equl ( - ). 0 BC Greece: Archimedes, engineer, rchitect, physicist nd mthemticin, 0 constructs polygons of sides to show tht - < π <. 0 BC Itly: Vituvius, rchitect nd engineer, mesures distnces using wheel nd determines tht π is equl to. AD Greece: Ptolemy writes his fmous work on stronomy, Syntxis 0 Mthemtic. He finds tht π is equl to + - +. AD 0 Chin: Tsu Ch ung-chih, expert in mechnics nd interested in mchinery, gives the vlue of π s. AD 0 Indi: Bhskr writes on stronomy nd mthemtics nd gives severl vlues of π, the most ccurte is -. AD Frnce: Viet finds π correct to nine deciml plces y considering polygons of. = sides. He lso discovers tht In this nd the next two series, exmine how the pproximtion improves s the numer of terms is incresed. AD 0 Englnd: John Wlis uses complicted nd difficult method to otin π from = - π 0 AD Englnd: Shrp clcultes π to deciml plces y evluting the π series = + AD Indi: The mthemticin Rmnujn presents the following s n pproximtion for π: π = + + + = - π + 0 0 0
Chpter Numer systems: the Rel Numer System Rel numers ppliction nd modelling Solve the following inequlity nd grph the solution on numer line x + - <. x Rtionlise the denomintor for -. + x Solve the ineqution > 0. x + By noting the expnsion of ( + ) nd the fct tht + = + +, determine +. Hence, determine +. If the integer points, n, nd the points midwy etween them, n +, re mpped on numer line, how fr wy from the nerest of these points cn ny point on the numer line e? Find n integer m such tht m < nd n integer k such tht k <. Explin the significnce of these results with respect to the topic of pproximtion of irrtionls to rtionls. (Your response should not rely on clcultor computtions.) The most common wy students lern to find the gretest common divisor of two integers is to fctorise oth numers into their prime fctors nd tke the common prime fctors. For exmple, to find the gretest common divisor of 0 nd we cn write: 0 = = So the gretest common divisor is =. However, finding the prime fctors is not lwys tht stright forwrd nd Euclid developed n lgorithm tht produces the gretest common divisor. To pply it we divide the smller integer into the lrger integer. Consider the integers 0 nd gin. When 0 is divided y the result is with reminder of 00. Thus 0 = + 00 Now divide the reminder (00) into the divisor () = 00 + nd the new reminder into the previous reminder, nd so on: 00 = + = + 0 The lst non-zero reminder is the gretest common divisor, ecuse it divides oth 0 nd nd every divisor of oth 0 nd must lso divide it.
Mths Quest Mths C Yer for Queenslnd We cn rewrite the ove equtions s 0 00 - = + - = + - = + = 00 00 - - 00 0 which cn e comined s continued frction s - = + - nd cn + - + lso e written s,,, (whole numers otined t ech division step). Express the quotient 00 0 s continued frction nd hence stte the gretest common divisor of these two integers. Show tht + + < (Do not use your clcultor.) Show tht = + - nd hence deduce tht + = + - = + - + - + - + + - + Now write the next two terms in such sequence. Diophntus of Alexndri, orn in either AD or AD 0, ws the uthor of Arithmetic nd wrote mny ppers investigting only whole numers. Diophntine equtions re involved in mny rel-life situtions, due to their whole numer solutions. The solutions re often found y tril nd error. The following prolem, which you might like to try, is sed on Diophntine equtions. (The nswer is given t the end of the prolem.) Five men nd monkey were shipwrecked on n islnd nd they set out to collect s mny coconuts s they could on their first dy so tht they would hve provisions until they were rescued. After mny hours collecting coconuts they piled them ll together nd went to sleep. Lter tht night, while the others slept one mn woke up, nd strted to worry out whether the coconuts would e divided firly. He decided to tke his one-fifth shre of the nuts nd gve the one remining coconut to the monkey. Throughout the night ll of the men did likewise; they took their fifth shre of wht ws left of the nuts nd gve the one remining nut to the monkey. In the morning they met to divide the nuts into five equl shres. Ech mn knew tht there were some nuts missing ut none dmitted he hd some extrs, of his own, hidden wy. How mny nuts were there to strt with? (Solution) The Diophntine equtions tht led to the solution of this prolem simplify to 0x 0 = y where x is the originl mount collected nd y is the numer ech is given in the morning. The smllest possile vlue for x is with other vlues t intervls of.
Chpter Numer systems: the Rel Numer System summry The Rel Numer System The set of rel numers (R) is divided into two min sets: rtionl nd irrtionl numers. These sets my e further divided into smller susets s illustrted on the chrt nd Venn digrm elow. Rel numers R Irrtionl numers I (surds, non-terminting nd non-recurring decimls, π,e) Integers Z Rtionl numers Q Non-integer rtionls (terminting nd recurring decimls) Negtive Z Zero (neither positive nor negtive) Positive Z + (Nturl numers N) Rtionl numers (Q) cn e expressed in ε = R the form, where nd re integers Q (Rtionl numers) with no common divisors nd 0. They my lso e expressed s whole numers, Z (Integers) I terminting decimls nd recurring N (Irrtionl (Nturl decimls. numers) numers) Irrtionl numers (I) cnnot e expressed in the form. They cn e expressed only s non-terminting nd non-recurring decimls nd include surds nd numers such s π nd e. Recurring decimls re rtionl numers which my e expressed s rtio of two integers. Surds re irrtionl numers represented y root symol (or rdicl), tht is:,, nd so on. π nd e re exmples of irrtionl numers which my e converted to nonterminting nd non-recurring decimls; however, they re not surds. Set nottion Set nottion is used when defining the Rel Numer System. The following symols re useful when working with sets: { } set is n element of is not n element of is suset of
0 Mths Quest Mths C Yer for Queenslnd Working with surds To simplify surd, it should e written s product of two fctors, one of which is the lrgest perfect squre. Like surds my e dded nd sutrcted; surds my need to e simplified efore dding nd sutrcting. Surds my e multiplied ccording to the rules: m n = mn When surd is multiplied y itself (squred), the result is the numer under the rdicl: Multipliction involving rckets:. The Distriutive Lw: = ( ) = ( + c) = + c. FOIL: ( + )( c d + ) = c + d + c + d. Perfect squres:. Difference of two squres DOTS: The product of conjugte pir of surds is rtionl. Surds my e divided ccording to the rule: ( + ) = + + ( ) = + ( + )( ) = Rtionlising denomintors:. If the denomintor contins surd, multiply oth numertor nd denomintor y the surd prt of the denomintor:. If the denomintor is sum or difference of surds, multiply oth the numertor nd the denomintor y the conjugte of the denomintor: - = = = = - + = - - + + = -
Chpter Numer systems: the Rel Numer System Modulus The modulus (or solute vlue) of numer is the mgnitude of tht numer nd is lwys positive. The modulus of x is denoted y x. x = x if x < 0 = 0 if x = 0 = x if x > 0 Solving equtions using solute vlues First remove the solute vlue symols nd stte the eqution s positive nd negtive lterntive cses. Verify your solutions for ll these questions y sustituting your nswer into the originl eqution. Solving inequtions Rememer tht if x > 0 then x is positive, nd vice vers. If product of two fctors is greter thn 0 then oth fctors must e either positive or negtive. Likewise, if product of two fctors is less thn 0 then only one of the fctors must e positive nd the other must e negtive, Orgnise your solution into two cses tht will develop rguments for ll possile vlues. The two vlues tht result for ech cse re vlues tht should occur t the one time. The grph you drw must e comintion of these two solutions for ech cse. Verify your solutions y choosing vlues tht fll in ech of the regions of your grph. When you multiply or divide y negtive fctor cross n inequlity sign, rememer to reverse the sign.
Mths Quest Mths C Yer for Queenslnd CHAPTER review A A Which of the given numers,,,,., 0.. π - 0.,, - re rtionl? A,,., 0.. π 0. nd - B - nd For ech of the following, stte whether the numer is rtionl or irrtionl nd give the reson for your nswer: c d 0.. e 0.0 C, nd - D,. nd E B B - 0. Which of the following sttements is not correct? A - Q B Q C Z + D Z E ( ) Z + 0 Which of the following frctions, -, -, -,,, cnnot e expressed s recurring decimls? A -,, B -, -, C -, - D E -,, - - B B C C The recurring deciml 0... cn e expressed s frction in its simplest form s: A - B C - D E 00 Express the following recurring decimls s frctions in the simplest form. 0... 0.. c 0... Which of the following sttements regrding the given set of numers, {, 0..,,,, 0 }, is correct? A,, 0, Z + B, cnnot e expressed s rtionl numers. C, 0.. 0 nd re the only rtionl numers of the set. D, nd 0 cnnot e expressed s rtionl numers. E None of the ove. Clssify ech of the following into the smllest suset in which they elong using Q, I, Z, Z + nd Z. (Simplify first where possile.) 0. - c - - d + - 0.0 -