Chapter 1: Fundamentals

Transcription

1 Chapter 1: Fudametals 1.1 Real Numbers Irratial umbers are real umbers that cat be expressed as ratis f itegers. That such umbers exist was a prfud embarrassmet t the Pythagrea brtherhd, ad they are said t have drwed e f their members (Hippasus, ca. 470 BC) fr revealig the secret. T prve that (fr example) is irratial, we will use a idirect prf (r reducti ad absurdum, t mix ur classical refereces). Suppse were ratial. The we culd write = a, which we take t be i lwest terms b a (a ad b have cmm factr). But squarig reveals = b, r b = a. We realize b is a eve umber, ad s a must als be eve. This meas a must als be eve. Let a = k. The b = (k) = 4k, s b = k, ad b, hece b, is eve. We have fud a ctradicti! We assumed a ad b have cmm terms, but if they are bth eve, they have a cmm factr f. S ur assumpti that = a b is t ratial. must be false, ad We will later deal with a mre geeral type f umber (cmplex). All the real umbers, which ca be put it a e-t-e crrespdece with the pits a lie, are but a small part f the set f cmplex umbers. Recall the abslute value f x, writte x, is defied t be x, if x < 0, but x, if x 0. Hw d we simplify the expressi x 10? If x 10, the expressi iside the abslute value sigs is egative, we write simply x 10. But if we kw x < 10, the expressi is egative ad we write its ppsite: (x 10) r 10 x. We will fte have t deal with sets f umbers that may be expressed i set tati, iterval tati, r graphically. Fr example, the set f real umbers betwee 1 ad + (iclusive) may be writte: {x 1 x }, read the set f all x such that 1 x ad x [ 1, ] 1, where we draw slid dts at each ed f the segmet. Fr < r >, use pe dts ad paretheses. Nte fr edpits f ±, always use paretheses. Fr example, (, 3] crrespds t x 3 r 1

2 1. Expets ad Radicals 1 1 Recall a 3 = a a a ad defie a 3 = =. All the familiar prperties f expets wrk 3 a a a a with egative as well as psitive expets. I particular, a m a = a m +. Thus a a a = a 3 a 3 = a = a 0 = 1. (This shws why we defie ay a (but zer) raised t the a a a pwer f zer t be e.) We may exted the laws f expets ad radicals t ratial expets by defiig ( ) m m m a a a = =. Here the demiatr f the fracti is the idex f the radical Fr example, 7 = 7 = ( 7 ) = 3 = 9. Later we will exted the laws f expets ad radicals t all real umbers. A summary f the laws f expets ad radicals (r rts) is give belw. Laws f Expets m m a a a + m a m = a a m ( a ) Laws r Radicals ab = a b a a = b b = = a ( ) ab a b a a = b b m m m = a a b = b a m a b = m b a a a = a = a, if is dd = a, if is eve a = b b = a (if is eve, require a 0 ad b 0) Nte carefully the defiiti f the pricipal th rt. Eve thugh 4 = 16 ad ( ) 4 = 16, s that 16 has tw real furth rts ( ad ), the tati 4 16 is reserved fr the psitive rt ale (). If we wat t idicate the egative rt, we write 4 16 =. This leads us t defie a = a whe is eve. Thus ( ) = 16 = =. Similarly we write 4 x 4 x ad 4 x 1 x 3 but 4 8 = = x = x (sice x is always 0). We will later allw eve rts f egative umbers as a ew kid f umber (cmplex).

3 The laws f expets ad rts are applied i sciece whe usig expetial (r scietific) tati. Oe writes x = a 10, where 1 a < 10 ad is a iteger. O ur TI-86 calculatr, eter umbers i this frmat by usig the key marked EE. Algebra teachers ffer guidelies fr whe expressis are i simplest frm t assure that ly e frm f a aswer is "crrect". This is the mtivati behid "ratializig demiatrs", i which e seeks t leave radical i the demiatr a a a Fr example, a = a a = a = a. It is fte easier t d this by first chagig frm radical frm t ratial expet frm. (Teachers fte request all aswers be expressed i terms f psitive expets ly.) a a Thus a = = =. Nte we did t have t write a i the demiatr f these a a a a examples whe takig the rt. I ur class, the rules fr "simplest frm" will evlve. Fr w, they are: Always cmbie like terms. Cmbie fractis. Cacel cmm factrs frm umeratr ad demiatr f fractis. Leave cmpud fractis. Leave radicals i demiatrs. Express all aswers i terms f psitive expets. Leave square rt with square factrs i the radicad, cube rt with cubic factrs, ad s frth. Write yur aswers i simplest frm uless directed therwise by yur teacher r directis fr specific prblems. A web site that ffers examples with slutis fr prblems frm this ad future sectis f ur class is TheMathPage ( Secti 7 ( has prblems ivlvig radicals. 3

4 1.3 Algebraic Expressis Use the distributive prperty t expad algebraic expressis. (x + 4)(3x ) = x 3x + x + 4 3x + 4 = 6x 4x + 1x 8 = 6x + 8x 8. Remember t cmbie all like terms (simplify). Memrize special prduct frmulas t speed this prcess alg. (a b)(a + b) = a b (a + b) = a + ab + b (a b) = a ab + b (a + b) 3 = a 3 + 3a b + 3ab + b 3 (a b) 3 = a 3 3a b + 3ab b 3 Mre prducts f the frm (a ± b) ca be expaded usig Pascal's triagle (belw) r the bimial therem (mre this later). Pascal's triagle was kw i may cultures fr may years befre Pascal, but i this curse we ctiue the DWEM traditi f amig therems ad laws after dead white Eurpea males Each rw starts with 1 ad the fllwig etries are the sums f etries i the previus rw t the immediate left ad right. Thus the 10 i ur last rw is the sum f 4 ad 6. Abve are the cefficiets i the expasi f (a + b) 0, (a + b) 1, (a + b), (a + b) 3, (a + b) 4, ad (a + b) 5. Fr the expasi f (a b), the sigs alterate. Example: the result f expadig (3x + y ) 3 = (3x) 3 + 3(3x) (y ) + 3(3x)(y ) + (y ) 3 r 7x x y + 36xy 4 + 8y 6 i simplest frm. Smetimes we wat t reverse the abve prcess ad factr a algebraic expressi it a prduct f simpler expressis. Sme useful frmulas t memrize are listed belw. a b = (a b)(a + b) a + ab + b = (a + b) a ab + b = (a b) a 3 b 3 = (a b)(a + ab + b ) *ew this year - memrize a 3 + b 3 = (a + b)(a ab + b ) *ew this year - memrize Try "takig ut" cmm factrs first. If there are tw terms, try the differece f squares r sum/differece f cubes. If there are three terms, use trial ad errr t factr it a prduct f bimials. If there are mre tha three terms, use grupig. T factr expressis with fractial r egative expets, always take ut the factr with the smallest expet. Example: Factr the expressi belw ( + ) + ( + ) = ( + ) + + = ( + ) [ + ] x x 1 x x 1 x x 1 (x 1) x x x 1 x 1 4

5 1.4 Fractial Expressis Simplify algebraic fractis by cacellig cmm factrs i umeratr ad demiatr. Nte that which is cacelled must be a factr f the etire umeratr r demiatr. x + x 3 ( x+ 3)( x 1) x 1 = = x + 6x+ 9 ( x+ 3) x+ 3. D t try t cacel the rigial x r x terms. They are t factrs. Multiply algebraic fractis by multiplyig umeratr times umeratr ad demiatr times demiatr. Cacel befre r after multiplyig t leave aswer i simplest frm. T divide algebraic fractis, ivert the divisr ad multiply. x + 3x + 1 x + 6x + 5 ( x+ 1)( x+ 1) ( x 1)( x 3) x+ 1 x 1 = = x + x 15 x 7x+ 3 ( x+ 5)( x 3) ( x+ 5)( x+ 1) x+ 5 ( )( ) ( ) Add r subtract algebraic fractis by first fidig a cmm demiatr ad the addig r subtractig the umeratrs ad writig ver the cmm demiatr. The simplify if pssible = + = x+ 3 x 9 x+ 3 ( x+ 3)( x 3) ( x 3) 1 1 x 3+ 1 x + = = ( x 3) ( x+ 3) ( x+ 3)( x 3) ( x+ 3)( x 3) ( x+ 3)( x 3) The best cmm demiatr t use is the LCD. Factr each demiatr ad take the prduct f distict factrs, each factr t its highest pwer i the factred demiatr. Cmpud fractis, where umeratr r demiatr r bth are fractis, are hadled by cmbiig terms t make e fracti i the umeratr, e i the demiatr, ad the ivertig ad multiplyig i the usual way. 1 1 y x x y x y x y = = = 1 1 x + y 1 1 y+ x + x y xy y x xy ( y x)( y+ x) xy y x = = xy y+ x xy y+ x xy Is it pssible t slve the last example i a simpler maer? Ratializig the umeratr r demiatr is easy if a factr f the frm a+ b c appears. Simply multiply by the cjugate a b c ad the radical will vaish frm part f the fracti. 1 1 x+ 1+ = x+ 1 x x+ 1 x x+ 1+ x = x x+ 1+ x = x+ 1 x x+ 1+ x 5

6 1.5 Equatis Recall i slvig equatis e may add equal quatities (+ r ) t bth sides. Oe may als multiply bth sides by ay zer quatity. Recall a prduct is zer if ad ly if e r bth f its factrs are zer. Simple quadratic equatis may be slved by takig rts r factrig. (x 4) = 1 has slutis x 4 = 1 ad x 4 = 1, r x = {5, 3}. I geeral, if x = c, the x = c ad x = c. We recall that i takig a eve rt we shuld write: ( ) x 4 = 1 x 4 = 1 x 4= 1rx 4= 1 x 3x + = 0 factrs: (x )(x 1) = 0 ad s x = {, 1}. Cmpletig the square will always slve a quadratic. x 8x + 15 = 0 becmes x 8x + = 15 + where we add a quatity t make the lhs a perfect square. If we have a 1 as x cefficiet, the umber t add is half the x cefficiet, squared. Thus we add 16. x 8x + 16 = r (x 4) = 1. This has slutis x = {5, 3} The mai use f cmpletig the square is t prve the quadratic frmula. Yu are respsible fr kwig this prf, credited t Brahmagupta ad al-khwarizmi (see the Histry f Math archive). Give ax + bx + c = 0, with a 0. b c b b c b x + x + = x + x + = + a a a a a a b c b 4ac b b 4ac x + = + = + = a a 4a a 4a 4a b b 4ac b b 4ac ± x+ =± x = a a a The radicad, b 4ac is called the discrimiat. If b 4ac > 0, the the equati has tw distict, real rts. If b 4ac = 0, the the equati has e real sluti (called a duble rt). If b 4ac < 0, the the equati has real sluti (but it des have cmplex rts). Csider the tw slutis x 1 = It is easily see that x 1 + x = this asserti! b a + b b 4ac a ad x1 x = c a ad x = b b 4ac a. The reader shuld supply the steps t justify. 6

7 Special Equatis Prjectile mti is eatly slved ear the surface f the earth usig British Imperial uits sice the accelerati due t gravity is a cveiet 3 ft/s dwward. Let v 0 be the startig velcity f a prjectile ad h 0 its iitial height at time t = 0. The at ay later time t the height f the prjectile is: h = h 0 + v 0 t + (1/)at. Let up be + ad dw, ad take a = 3 ft/s s h = h 0 + v 0 t 16t where h ad h 0 are i ft, ad v 0 i ft/s. If a ball is thrw straight up at 16 ft/s frm a iitial height we take t be 0m, hw much time passes befre the ball returs t its startig pit? Let h = h 0 = 0, ad v 0 = 16. Slve 0 = t 16t t fid t = 0 r 1s as the times the ball was at height 0m. Thus the ball is i the air fr 1s. Equatis ivlvig fractis: multiply each side by the lcd t clear fractis. Check fr extraeus slutis! = 4( x 1)( x+ ) + = 4( x 1)( x+ ) x 1 x+ 4 x 1 x+ 4 4 x x 1 = 5x + 5x 10 0 = 5x 3x 14 0 = 5x + 7 x 7 x = r 5 Bth f these slutis check whe substituted i the rigial equati. Equatis with radicals: islate the radical, take bth sides t a pwer t elimiate the radical. Repeat as eeded. Check fr extraeus slutis! ( ) ( ) ( )( ) x = x x + 1 = x 1 ( x + 1) = ( x 1) x + 1= x x = x 4x x = 0r 4 Check usig the rigial equati! Oly x = 4 checks; 0 is extraeus. Furth-degree equatis: if f the right frm, let y = x ad cvert t quadratic. 4 x 13x + 40 = 0 y 13y + 40 = 0 where y = x y 5 y 8 = 0 y= 5,8 x = ± 5, ± ( )( ) { } { } All slutis check. Equatis with fractial pwers: if f the right frm, let y = quadratic frm x 5x + 6= 0 y 5y+ 6= 0 where y= x 3 m x ad cvert t 3 ( y 3)( y ) = 0 y= { 3,} x = y = { ± 3 3, ± } Abslute value equatis: use the defiiti f abslute value. x 4 = 1 x 4 = 1 r (x 4) = 1 x 4 = 1 r x 4 = 1 Thus x = 5 r 3. 7

8 1.6 Mdelig with Equatis There is substitute fr thikig whe slvig wrd prblems, but smetimes rgaizig wrk i tabular frm is helpful. The basic wrd prblems f algebra iclude rate, time, distace prblems, mixture prblems, ad wrk prblems. After escapig jail i Crw Pit, Jh Dilliger drives ff at 14 mph. Te miutes later the plice leave Crw Pit i ht pursuit at 16 mph. Hw lg des it take the plice t catch the crimial? Recall distace = rate time. rate time distace Dilliger 14 miles/hur t hurs 14t miles Plice 16 miles/hur (t 1/6) hurs 16(t 1/6) miles The plice catch the crimial whe bth have travelled the same distace, s we slve 14t = 16(t 1/6) t fid t = 4/3 hurs. A bttle ctais 750mL f fruit puch with a ccetrati f 50% pure juice. Jill driks 100mL f the puch ad the refills the bttle with a equal amut f a cheaper brad f puch, If the ccetrati f juice i the bttle is the 48%, what was the ccetrati f the cheaper brad added by Jill? amut puch (ml) % ccetrati amut juice (ml) rigial (0.50) remved (0.50) added 100 x 100(x / 100) fial (0.48) Here we have ted the amut f puch times the ccetrati as a decimal gives the amut f pure juice. We slve the equati 750(0.50) 100(0.50) + 100(x / 100) = 750(0.48) t fid x = 35% Dimedes ad Eluria have bee set t milk the family gat herd. It takes Eluria 70 miutes t d the jb ale, whereas Dimedes ale takes 80 miutes. Hw lg des it take if they wrk tgether? rate (per miute) time (miutes) fracti f jb Dimedes 1/80 t t/80 Eluria 1/70 t t/70 Here we have ted fr the rate that Dimedes, fr example, wh takes 80 miutes t milk the herd ale, des 1/80 f the whle jb each miute. We see that the rate multiplied by the time gives the fracti f the jb each cmpletes. Of curse, the whle jb must be de s we slve t/80 + t/70 = 1. This gives t = 36 miutes. Please te the use f wrd prblems thrughut the ages pages 76 ad 77 f ur text. 8

9 1.7 Iequalities Multiplyig (r dividig) each side f a iequality by the same egative quatity reverses the directi f the iequality. Takig reciprcals f each side f a iequality ivlvig psitive quatities reverses the directi f the iequality. Addig (r subtractig) the same psitive r egative quatities, ad multiplyig (r dividig) the same psitive quatities, t bth sides f a iequality leaves the directi f the iequality uchaged. Cmbiig iequalities with the lgical r results i the ui f sluti sets. Cmbiig with ad results i the itersecti f sluti sets. Slve 3 3x 1 < 1. This is equivalet t 3 3x 1 ad 3x 1 < 1. Add 1 ad divide by 3: 4/3 x ad x < 13/3. The itersecti f these tw sets is: [4/3, 13/3). Nte e ca achieve this result by treatig the rigial prblem as a exteded iequality ad wrkig left, middle, ad right. Nliear Iequalities Here we mve all terms t e side. If fractis are ivlved, make the zer side f the iequality a sigle fracti. Factr t fid key umbers that divide the real umber lie it itervals. With fractis, ay umber that makes the demiatr zer is als a key umber. The iequality ca chage frm psitive t egative ly at these key umbers. Use cveiet test umbers i the itervals t fid if all the umbers i the iterval d r d t satisfy the iequality. Carefully check the iterval edpits. Slve x 3x 18 0 (x 6)(x + 3) 0 3 ad 6 are key umbers dividig the real umber lie it three itervals. Nte x = 3 ad 6 bth make (x 6)(x + 3) = 0 ad s satisfy the iequality. (, 3) Let x = 4: (x 6)(x + 3) ( )( ) = + Nt 0 ( 3, 6) Let x = 0: (x 6)(x + 3) ( )(+) = Yes, 0 (6, ) Let x = 10: (x 6)(x + 3) (+)(+) = + Nt 0 Nte x = 3 ad 6 bth make (x 6)(x + 3) = 0 ad s satisfy the iequality. Thus the sluti iterval is [ 3, 6]. Slve the iequality belw that ivlves fractis x 6( x 1) x( x 1) x 1 x x 1 x x( x 1) x + x+ 6 ( x+ 3)( x+ ) 0 0 x x 1 x x 1 ( ) ( ) Thus the key umbers are, 0, 1, ad 3 which divide the umber lie it 5 itervals. The reader shuld prceed as i the example abve t shw the sluti is the ui f the itervals [, 0) (1, 3]. 9

10 T slve abslute value iequalities, islate the abslute value e side f the iequality. x < c is equivalet t x < c ad x < c (the latter becmig x > c). x > c is equivalet t x > c r x > c (the latter becmig x < c). Slve, fr example, x + 5 <. This is equivalet t slvig the tw iequalities x + 5 < ad x + 5 > ad takig the itersecti f the sluti sets as the sluti t the abslute value prblem. We fid ( 7, 3) t be the sluti. This is the iterval f all pits less tha tw uits distat frm 5. 10

11 1.8 Crdiate Gemetry Distace frm (x 1, y 1 ) t (x, y ): d = ( x x ) + ( y y ) 1 1 (te always -egative). Midpit f segmet jiig (x 1, y 1 ) t (x, y ): x + x y + y, 1 1. Pit f divisi frmula: the pit D that divide the segmet frm P 1 (x 1, y 1 ) t P (x, y ) PD 1 such that r DP = has x = x 1 + r (x x 1 ) ad y = y 1 + r (y y 1 ). If r = 1/, this reduces t the midpit frmula. T fid the x-itercept(s) f a graph, set y = 0 i the equati f the graph. T fid the y-itercept(s) f a graph, set x = 0 i the equati f the graph. The circle may be defied as the set f pits i the plae equidistat frm a give pit. The ceter f the circle we idicate as (h, k) ad its radius as r. The usig the distace frmula abve, fr ay pit (x, y) the circle we have: r = ( x h) ( y k) ( x h) + ( y k) + r = r as the stadard (r ceter-radius) frm fr the equati f a circle. Write the circle equati belw i stadard frm. Fid the ceter ad radius. x + y x + 4y +1 = 0 x x + + y + 4y + = 1 Cmpletig the squares, we fid: x x y + 4y + 4 = (x 1) + (y + ) = which gives ceter (1, ) ad radius =. 11

12 1.9 Slvig Equatis ad Iequalities Graphically The TI-86 Guidebk may be dwladed frm the fllwig site: ( The Virtual-TI calculatr prgram is highly recmmeded. It may be fud at: ( See the web page usig Mathematica fr graphig ad algebra at: Sme wrds f advice: put i lts f paretheses, pay atteti t the viewig rectagle, ad thik abut what the calculatr displays. Belw is a example f slvig x 7x + 1 = 0 graphically with the TI-86. Use the TRACE cmmad r MATH ROOT t fid where the graph crsses the x-axis (where y1 = 0). As expected, sice this equati ca be slved algebraically, we fid x = 3 ad x = 4 t be the slutis. Slvig the iequality x 3 5x 8 graphically with Mathematica, Reduce[x^3-5 x^+8 0,x]//N x x 4.66 The sluti is [ 1.14, 1.5] [4.63, ) Plt[x 3-5 x +8,{x,-, 6}]

13 1.10 Lies Slpe: the slpe f a -vertical lie betwee pits A(x 1, y 1 ) ad B(x, y ) is y y1 rise m = =. x x 1 ru The slpe f a hriztal lie is zer. The slpe f a vertical lie is udefied. The equati f a hriztal lie is f the frm y = cstat. The equati f a vertical lie is f the frm x = cstat. Tw lies are parallel if their slpes are equal (if slpe is defied). Tw lies are perpedicular if their slpes are egative reciprcals (if slpe is defied). The equati f a lie may (assumig the existece f the slpe, m) be expressed as: Slpe-Itercept Frm Pit-Slpe Frm Geeral Frm y = mx + b, where b is the y-itercept. y y 1 = m(x x 1 ), where (x 1, y 1 ) is a pit the lie. Ax + By + C = 0, where A, B, C are itegers with cmm factr, A ad B are t bth zer, A 0. Stadard Frm Ax + By = C, ther restrictis as i Geeral Frm. x y Tw-Itercept Frm + = 1, where a is the x-itercept ad b is the y-itercept a b The stadard frm will t be used i this curse, but may be see stadardized tests. Nte the Geeral Frm may be cverted t the ther frms ad vice-versa. A C Ax + By + C = 0 y= x+ B B. A C C Thus the slpe m =, the y-itercept is b =, ad the x-itercept a =. B B A The restrictis A, B, ad C i the geeral frm are fr the cveiece f algebra teachers wh fte require all liear equatis t be writte i this frm. Strict adherece t the restrictis assures ly e frm fr the crrect aswer. Oe may write liear equatis t mdel supply ad demad fr a cmmdity. Fr wheat, fllwig the textbk, e may have: Supply: y = 8.33p y i millis f bushels, p i dllars per bushel Demad: y = 1.39p I the supply equati, y is the amut prduced, ad p is the price. I the demad equati, y is the amut sld at price p. The equilibrium pit is where these lies itersect; the amut prduced equals the amut sld, ad the price at which this ccurs wuld be a stable, sustaiable price fr the cmmdity. N wheat is sld ( y = 0 i demad equati) whe p = $16.80 N wheat is prduced (y = 0 i supply equati) whe p = $1.75 These lies itersect at ($3.90, 17.9), the equilibrium pit. 13

14 1.11 Mdelig Variati Direct Variati: Iverse Variati: If y = k x fr sme cstat k 0, we say y varies directly as x r y is directly prprtial t x. The cstat k is the prprtiality cstat. The graph is a lie with slpe k ad y-itercept 0. y1 y Give tw pits the lie (x 1, y 1 ) ad (x, y ), k = =. x x 1 If y = k / x, fr sme cstat k 0, we say y varies iversely as x r y is iversely prprtial t x. The graph is that f a reciprcal fucti. Give tw pits the lie (x, y ) ad (x, y ), k = x y = x y Jit Variati: If z = k x y, where k 0, we say z varies jitly as x ad y, r z is jitly prprtial t x ad y. May quatities i sciece are related as abve. I chemistry, we te the gas laws belw. Byle's Law: Pressure is iversely prprtial t vlume: P = k / V Charles' Law: Vlume is directly prprtial t temperature: V = k T Gay-Lussac's Law: Pressure is directly related t temperature: P = k T These laws are all special cases f the ideal gas law, but histrically they were discvered piecemeal, ad chemists are great believers f the DWEM amig rule. Example: A gas is fud t ccupy.4l at 0 C. What vlume will it ccupy at 100 C? Nte e must use abslute temperature whe applyig Charles' Law. Thus we write: V V1 V.4L = = V = 30.6L T T 373K 73K 1 14