COMPETENCY 1. ALGEBRA SKILL 1.1 1.1a. ALGEBRAIC STRUCTURES Kow why the real ad complex umbers are each a field, ad that particular rigs are ot fields (e.g., itegers, polyomial rigs, matrix rigs) Algebra must be systematically orgaized i order to create the axioms for sets of umbers. The biary relatio symbols +,, ad are operators that require two iputs ad obey the algebraic axioms of field theory. Ay set that icludes at least two ozero elemets that satisfies the field axioms for additio ad multiplicatio is a field. The real umbers, R, as well as the complex umbers, C, are each a field, with the real umbers beig a subset of the complex umbers. The field axioms are summarized below. Additio: Commutativity a+ b = b+ a Associativity a+ ( b+ c) = ( a+ b) + c Idetity a+ = a a+ a = Iverse ( ) Multiplicatio: Commutativity ab = ba Associativity a( bc) = ( ab) c Idetity a 1= a 1 Iverse a = 1 a ( a ) Additio ad multiplicatio: Distributivity a( b + c) = ( b + c) a = ab + ac Note that both the real umbers ad the complex umbers satisfy the axioms summarized above. MATHEMATICS 11, 111, 11 1
A rig is a itegral domai with two biary operatios (additio ad multiplicatio) where, for every o-zero elemet a ad b i the domai, the product ab is o-zero. A field is a rig where multiplicatio is commutative, or a b = b a, ad all o-zero elemets have a multiplicative iverse. The set Z (itegers) is a rig that is ot a field i that it does ot have the multiplicative iverse; therefore, itegers are ot a field. A polyomial rig is also ot a field, as it also has o multiplicative iverse. Furthermore, matrix rigs do ot costitute fields because matrix multiplicatio is ot geerally commutative. Note: Multiplicatio is implied whe there is o symbol betwee two variables. Thus, a b ca be writte ab. Multiplicatio ca also be idicated by a raised dot ( ). MATHEMATICS 11, 111, 11
1.1b. Apply basic properties of real ad complex umbers i costructig mathematical argumets (e.g., if a < b ad c <, the ac > bc) Basic Properties of Real ad Imagiary/Complex Numbers Real Numbers Irratioal Numbers Natural Numbers Whole Numbers Itegers Ratioal Numbers Real umbers are deoted by R ad are umbers that ca be show by a ifiite decimal represetatio such as 3.8675347... Real umbers iclude ratioal umbers, such as 4 ad 3/19, ad irratioal umbers, such as the ad π, ca be represeted as poits alog a ifiite umber lie. Real umbers are also kow as the uique complete Archimedea ordered field. Real umbers are to be distiguished from imagiary umbers. Real umbers are classified as follows: A. Natural umbers, deoted by N: the coutig umbers, 1,, 3,. B. Whole umbers: the coutig umbers alog with zero,, 1,, 3,. C. Itegers, deoted by Z: the coutig umbers, their egatives, ad zero,,, 1,, 1,,. MATHEMATICS 11, 111, 11 3
D. Ratioals, deoted by Q: all of the fractios that ca be formed usig whole umbers. Zero caot be the deomiator. I decimal form, these umbers will either be termiatig or repeatig decimals. Simplify square roots to determie if the umber ca be writte as a fractio. E. Irratioals: Real umbers that caot be writte as a fractio. The decimal forms of these umbers are either termiatig or repeatig. Examples iclude π, e ad. Imagiary ad complex umbers are deoted by!.! The set! is defied as { a+ bi : a, b! } ( meas elemet of ). I other words, complex umbers are a extesio of real umbers made by attachig a imagiary umber i, which satisfies the equality i = 1. Complex umbers are of the form a + bi, where a ad b are real umbers ad i = 1. Thus, a is the real part of the umber ad b is the imagiary part of the umber. Whe i appears i a fractio, the fractio is usually simplified so that i is ot i the deomiator. The set of complex umbers icludes the set of real umbers, where ay real umber ca be writte i its equivalet complex form as + i. I other words, it ca be said that!! (or! is a subset of! ). Complex umbers Real umbers The umber 3i has a real part ad imagiary part 3; the umber 4 has a real part 4 ad a imagiary part. As aother way of writig complex umbers, we ca express them as ordered pairs: Complex umber Ordered pair 3+i (3, ) 3+ 3i ( 3, 3 ) 7i (, 7) 6+i 6, 7 7 7 MATHEMATICS 11, 111, 11 4
These properties of real ad complex umbers ca be applied to the costructio of various mathematical argumets. A mathematical argumet proves that a propositio is true. Example: Prove that for every iteger y, if y is a eve umber, the y is eve. The defiitio of eve implies that for each iteger y there is at least oe iteger x such that y = x. y = x y = 4x Sice 4x is always evely divisible by two ( x is a iteger), y is eve for all values of y. MATHEMATICS 11, 111, 11 5 Example: If a, b ad c are positive real umbers, prove that c a+ b = b+ a c. ( ) ( ) Use the properties of the set of real umbers. c( a+ b) = c( b+ a) Additive commutativity = cb + ca Distributivity = bc + ac Multiplicative commutativity = b+ a c Distributivity ( ) Example: Give real umbers a, b, c ad d where ad = bc, prove that (a + bi)(c + di) is real. Expad the product of the complex umbers. ( a + bi )( c + di ) = ac + bci + adi + bdi Use the defiitio of i. ( a + bi )( c + di ) = ac bd + bci + adi Apply the fact that ad = bc. ( a + bi )( c + di ) = ac bd + bci bci = ac bd Sice a, b, c ad d are all real, ac bd must also be real.
1.1c. Kow that the ratioal umbers ad real umbers ca be ordered ad that the complex umbers caot be ordered, but that ay polyomial equatio with real coefficiets ca be solved i the complex field The previous skill sectio reviews the properties of both real ad complex umbers. Based o these properties, it ca be show that real ad ratioal umbers ca be ordered but complex umbers caot be ordered. Real umbers are a ordered field ad ca be ordered. As such, a ordered field F must cotai a subset P (such as the positive umbers) such that if a ad b are elemets of P, the both a + b ad ab are also elemets of P. (I other words, the set P is closed uder additio ad multiplicatio.) Furthermore, it must be the case that for ay elemet c cotaied i F, exactly oe of the followig coditios is true: c is a elemet of P, c is a elemet of P or c =. Likewise, the ratioal umbers also costitute a ordered field. The set P ca be defied as the positive ratioal umbers. For each a ad b that are elemets of the set! (the ratioal umbers), a + b is also a elemet of P, as is ab. (The sum a + b ad the product ab are both ratioal if a ad b are ratioal.) Sice P is closed uder additio ad multiplicatio,! costitutes a ordered field. Complex umbers, ulike real umbers, caot be ordered. Cosider the umber i = 1 cotaied i the set! of complex umbers. Assume that! has a subset P (positive umbers) that is closed uder both additio ad multiplicatio. Assume that i >. A difficulty arises i that i = 1<, so i caot be icluded i the set P. Likewise, assume i <. The problem oce agai arises that 4 i = 1>, so i caot be icluded i P. It is clearly the case that i, so there is o place for i i a ordered field. Thus, the complex umbers caot be ordered. Polyomial equatios with real coefficiets caot, i geeral, be solved usig oly real umbers. For istace, cosider the quadratic fuctio give below: ( ) x f x = + 1 MATHEMATICS 11, 111, 11 6
There are o real roots for this equatio, sice ( ) f x = = x + 1 x = 1 The Fudametal Theorem of Algebra, however, idicates that there must be two (possibly o-distict) solutios to this equatio. (See Skill 1.c for more o the Fudametal Theorem of Algebra.) Not that if the complex umbers are permitted as solutios to this equatio, the x =± i Thus, geerally, solutios to ay polyomial equatio with real coefficiets exist i the set of complex umbers. MATHEMATICS 11, 111, 11 7
SKILL 1. 1.a. POLYNOMIAL EQUATIONS AND INEQUALITIES Kow why graphs of liear iequalities are half plaes ad be able to apply this fact (e.g., liear programmig) Liear Iequalities To graph a liear iequality expressed i terms of y, solve the iequality for y. This reders the iequality i slope-itercept form (for example: y < mx + b). The poit (, b) is the y-itercept ad m is the slope of the lie. If the iequality is expressed oly i terms of x, solve for x. Whe solvig the iequality, remember that dividig or multiplyig by a egative umber will reverse the directio of the iequality sig. A iequality that yields ay of the followig results i terms of y, where a is some real umber, the the solutio set of the iequality is bouded by a horizotal lie: y < a y a y > a y a If the iequality yields ay of the followig results i terms of x, the the solutio set of the iequality is bouded by a vertical lie: x < a x a x > a x a Whe graphig the solutio of a liear iequality, the boudary lie is draw i a dotted maer if the iequality sig is < or >. This idicates that poits o the lie do ot satisfy the iequality. If the iequality sig is either or, the the lie o the graph is draw as a solid lie to idicate that the poits o the lie satisfy the iequality. The lie draw as directed above is oly the boudary of the solutio set for a iequality. The solutios actually iclude the half plae bouded by the lie. Sice, for ay lie, half of the values i the full plae (for either x or y) are greater tha those defied by the lie ad half are less, the solutio of the iequality must be graphed as a half plae. (I other words, a lie divides the plae i half.) Which half plae satisfies the iequality ca be foud by testig a poit o either side of the lie. The solutio set ca be idicated o a graph by shadig the appropriate half plae. MATHEMATICS 11, 111, 11 8
For iequalities expressed as a fuctio of x, shade above the lie whe the iequality sig is or >. Shade below the lie whe the iequality sig is < or. For iequalities expressed as a fuctio of y, shade to the right for > or. Shade to the left for < or. The solutio to a system of liear iequalities cosists of the portio of the graph where the shaded half plaes for all the iequalities i the system overlap. For istace, if the graph of oe iequality was shaded with red, ad the graph of aother iequality was shaded with blue, the the overlappig area would be shaded purple. The poits i the purple area would be the solutio set of this system. Example: Solve by graphig: x+ y 6 x y 6 Solvig the iequalities for y, they become the followig: y x+ 6 (y-itercept of 6 ad slope of 1) 1 y x 3 (y-itercept of 3 ad slope of 1/) A graph with the appropriate shadig is show below: x + y 6 1 8 6 4-1 -8-6 -4 - - 4 6 8 1-4 x y 6-6 -8-1 MATHEMATICS 11, 111, 11 9
Liear programmig, or liear optimizatio, ivolves fidig a maximum or miimum value for a liear fuctio subject to certai costraits (such as other liear fuctios or restrictios o the variables). Liear programmig ca be used to solve various types of practical, real-world word problems. It is ofte used i various idustries, ecological scieces ad govermetal orgaizatios to determie or project, for istace, productio costs or the amout of pollutats dispersed ito the air. The key to most liear programmig problems is to orgaize the iformatio i the word problem ito a chart or graph of some type. By plottig the iequalities that defie the problem, for istace, the rage of possible solutios ca be show visually. Example: A pritig maufacturer makes two types of priters: a Pritmaster ad a Speedmaster priter. The Pritmaster requires 1 cubic feet of space, weighs 5, pouds ad the Speedmaster takes up 5 cubic feet of space ad weighs 6 pouds. The total available space for storage before shippig is, cubic feet ad the weight limit for the space is 3, pouds. The profit o the Pritmaster is $15, ad the profit o the Speedmaster is $3,. How may of each machie should be stored to maximize profitability ad what is the maximum possible profit? First, let x represet the umber of Pritmaster uits sold ad let y represet the umber of Speedmaster uits sold. The, the equatio for the space required to store the uits is the followig. 1x+ 5y x+ y 4 Sice the umber of uits for both models must be o less tha zero, also impose the restrictios that x ad y. The restrictio o the total weight ca be expressed as follows. 5x+ 6y 3 5x+ 3y 15 The expressio for the profit P from sales of the priter uits is the followig. P = $15, x+ $3,y MATHEMATICS 11, 111, 11 1
The solutio to this problem, the, is foud by maximizig P subject to the costraits give i the precedig iequalities, alog with the costraits that x ad y. The equatios are grouped below for clarity. x y x+ y 4 5x+ 3y 15 P = $15, x+ $3,y The two iequalities i two variables are plotted i the graph below. The shaded regio represets the set of solutios that obey both iequalities. (Note that the shaded regio i fact oly icludes poits where both x ad y are whole umbers.) 6 y 5 4 3 x + y 4 1 5x + 3y 15 x 4 6 Note that the border of the shaded regio that is formed by the two iequalities icludes the solutios that costitute the maximum value of y for a give value of x. Note also that x caot exceed 6 (sice it would violate the secod iequality). The solutio to the problem, the, must lie o the border of the shaded regio, sice the border spas all the possible solutios that maximize the use of space ad weight for a give umber x. To visualize the solutio, plot the profit as a fuctio of the solutios to the iequalities that lie alog the border of the shaded area. MATHEMATICS 11, 111, 11 11
6 5 Profit (x$5,) 4 3 1 4 6 The profit curve shows a maximum at about x = 16. Test several values usig a table to verify this result. x y P (x$5,) 15 16 17 37 366 358 519 519. 514.6 Also double check to be sure that the result obeys the two iequalities. ( 16) + ( 366) = 398 4 5( 16) + 3( 366) = 1498 15 Thus, the optimum result is storage of 16 Pritmaster ad 366 Speedmaster priter uits. Example: Sharo's Bike Shoppe ca assemble a 3-speed bike i 3 miutes ad a 1-speed bike i 6 miutes. The profit o each bike sold is $6 for a 3 speed or $75 for a 1-speed bike. How may of each type of bike should it assemble durig a 8-hour day (48 miutes) to maximize the possible profit? Total daily profit must be at least $3. Let x be the umber of 3-speed bikes ad y be the umber of 1- speed bikes. Sice there are oly 48 miutes to use each day, the first iequality is the followig. 3x+ 6y 48 x+ y 16 MATHEMATICS 11, 111, 11 1
Sice the total daily profit must be at least $3, the the secod iequality ca be writte as follows, where P is the profit for the day. P = $6 x+ $75 y $3 4x+ 5y To visualize the problem, plot the two iequalities ad show the potetial solutios as a shaded regio. 8 7 y 6 5 x+ y 16 4 3 1 4x+ 5y 5 1 15 x The solutio to the problem is the ordered pair of whole umbers i the shaded area that maximizes the daily profit. The profit curve is added as show below. 8 y 7 6 5 P 4 3 1 5 1 15 x MATHEMATICS 11, 111, 11 13
Based o the above plot, it is clear that the profit is maximized for the case where oly 3-speed bikes (correspodig to x) are maufactured. Thus, the correct solutio ca be foud by solvig the first iequality for y =. x + ( ) 16 x 16 The maufacture of 16 3-speed bikes (ad o 1-speed bikes) maximizes profit to $96 per day. MATHEMATICS 11, 111, 11 14
1.b. Prove ad use the followig: the Ratioal Root Theorem for polyomials with iteger coefficiets; the Factor Theorem; the Cojugate Roots Theorem for polyomial equatios with real coefficiets; the Quadratic Formula for real ad complex quadratic polyomials; the Biomial Theorem A polyomial is a sum of terms where each term is a costat multiplied by a variable raised to a positive iteger power. The geeral form of a polyomial P(x) is 1 ( ) K P x = a x + a x + + a x + a x+ a 1 1 Polyomials writte i stadard form have the terms writte i decreasig expoet value, as show above. The above example is a degree polyomial (assumig a ) because the expoet of the first term is. A particular case of a polyomial is the quadratic equatio, which is degree two. The Ratioal Root Theorem The Ratioal Root Theorem, also kow as the Ratioal Zero Theorem, allows determiatio of all possible ratioal roots (or zeroes) of a polyomial equatio with iteger coefficiets. (A root is a value of x such that P(x) =.) Every ratioal root of P(x) ca be p writte as x =, where p is a iteger factor of the costat term q a ad q is a iteger factor of the leadig coefficiet a. p To prove the Ratioal Root Theorem, first assume that x = is a q root of P(x), where p ad q are itegers with a greatest commo deomiator (GCD) of 1. The: 1 p p p p p = + 1 + K + + 1 + = P a a a a a q q q q q Multiply the etire expressio by q. 1 1 + 1 + K + + 1 + = ap a p q apq apq aq MATHEMATICS 11, 111, 11 15
Sice each coefficiet a i is a iteger, as are p ad q, each term i the above expressio must also be a iteger. Cosequetly, ay partial sum of the terms is a iteger as well. Rewrite the sum as follows. 1 1 = 1 K 1 a p a p q a p q a pq a q ( 1 3 1 1 K 1 ) 1 3 1 ( a 1p q K ap q a1pq aq ) a p = q a p q a p q a pq a q ap q = Sice p is ot divisible by q (the GCD of p ad q is 1), a must be divisible by q sice the sum i the paretheses is a iteger. Likewise, ( 1 1 1 K ) 1 1 1 ( ap a 1p q K apq aq 1 ) = 1 1 aq p ap a p q apq aq aq p = By the same reasoig, a must be divisible by p. The theorem has thus bee prove, sice p is a factor of a ad q is a factor of a. Example: Fid the ratioal roots of ( ) 3 P x = 3x 7x + 3x. By the Ratioal Root Theorem, the roots must be of the form 1, x = m 1, 3 The cadidates are the 1 x =± 1, ±, ±, ± 3 3 Test each possibility. The oly result that works is x =. (Note that the Ratioal Root Theorem does ot guaratee that each potetial ratioal umber that icludes factors of the leadig ad costat terms is a root. The theorem oly states that roots will iclude these factors.) MATHEMATICS 11, 111, 11 16