FUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation


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1 FUNCTIONS AND EQUATIONS. SETS AND SUBSETS.. Definition of set. A set is ny collection of objects which re clled its elements. If x is n element of the set S, we sy tht x belongs to S nd write If y does not belong to S, we write xεs. y S. The simplest wy to represent set is by listing its members. We use the nottion A = {,, 4} to denote the set whose elements re the numbers, nd 4. We cn lso describe set by describing its elements insted of listing them. For exmple we use the nottion C = {x : x +x 3 = 0} to denote the set of ll solutions to the eqution x +x 3 = 0. The set is {, 3}... Subsets.... Definition of subset. If ll the elements of set X re lso elements of set Y, then X is subset of Y nd we write X Y where is the setinclusion reltion. Consider the following five sets. A = {,, 4} B = {,, 4, 5} C = {,, 4, 5} D = {,, 4, 5, 8, 0} The following reltions re ll true. E = {,, 3, 4, 5, 8, 0} Dte: Jnury 3, 005.
2 FUNCTIONS AND EQUATIONS A B, A C, A D, A E B C, B D, B E C B, C D, C E D E... Definition of proper subset. If ll the elements of set X re lso elements of set Y, but not ll elements of Y re in X, then X is proper subset of Y nd we write X Y where is the proper setinclusion reltion. The following reltions re ll true. A B, A C, A D, A E B D, C D, B E C E D E..3. Definition of equlity of sets. Two sets re equl if they contin exctly the sme elements, nd we write For exmple B = C. X = Y...4. Exmples of sets. : ll corn frmers in Iow b: ll firms producing steel c: the set of ll consumption bundles tht given consumer cn fford d: the set of ll combintions of outputs tht cn be produced by given set of inputs.3. Set opertions..3.. Intersections. The intersection, W, of two sets X nd Y is the set of elements tht re in both X nd Y. We write W = X Y = {x : x X nd x Y }. The definition is symmetric: tht is, A B=B A. Also, A B AndA B=A A B. We hve (A B) C=A (B C) nd write this s A B C..3.. Empty or null sets. The empty set or the null set is the set with no elements. The empty set is written. For exmple, if the sets A nd B contin no common elements we cn write nd these two sets re sid to be disjoint. A B =
3 FUNCTIONS AND EQUATIONS Unions. The union of two sets A nd B is the set of ll elements in one or the other of the sets. We write We hve V = A B = {x : x Aorx B}. A B = B A A (B C) = (A B) C = A B C A A B A = A B B A.3.4. Distributive lws. A (B C) = (A B) (A C) A (B C) = (A B) (A C) Consider the following three sets. A = {,, 4} B = {,, 3, 5} Then we hve C = {, 4, 6} A B = {,, 3, 4, 5} A C = {,, 4, 6} B C = {,, 3, 4, 5, 6} A B C = {,, 3, 4, 5, 6} A B = {, } A C = {, 4} B C = {} We lso hve A B C = {} A (B C) ={,, 4} (A B) (A C) ={,, 4} A (B C) ={,, 4} (A B) (A C) ={,, 4}
4 4 FUNCTIONS AND EQUATIONS.3.5. Complements. The complement of set X is the set of elements of the universl set U tht re not elements of X, nd is written X C. Thus We hve X C = {x : U : x X}. C U C = U = (A C ) C = A A A C A A C = U = A B B C A C.3.6. De Morgn s lws. Consider the following three sets. (A B) C = A C B C (A B) C = A C B C U = {,, 3, 4, 5} A = {,, 4} Then B = {,, 3, 5} A B = {, } A B = {,, 3, 4, 5} = U A C = {3, 5} B C = {4} (A B) C = {3, 4, 5} (A B) C =.3.7. Set difference or reltive complement. If A nd B re two subsets of X, then define the set difference B \ A, or the reltive complement of A in B, s the set of elements in B tht re not in A. Thus We hve B \ A=B A C. B \ A = {x : x B nd x/ A} Consider the following three sets.
5 FUNCTIONS AND EQUATIONS 5 X = {,, 3, 4, 5} A = {,, 4} Then B = {, 3, 5} A \ B = {, 4} B \ A = {3, 5}. NUMBERS.. Nturl numbers. The nturl numbers re the elements of the set Z + = {,, 3,...}. A group of three dots,..., clled n ellipsis, indictes tht the numbers continue in the indicted pttern. If the ellipsis contins nothing to the right, it is ssumed tht the pttern continues forever. The nturl numbers re closed under ddition mening tht for numbers nd b tht re elements of Z +, then + b ε Z +. Similrly, b ε Z + nd we sy tht the nturl numbers re closed under multipliction... Prime numbers. A prime number is nturl number greter thn one tht is divisible only by itself nd, i.e., {,3,5,7,,3,7,9,3,... }. The first 5 prime numbers re s follows. TABLE. Prime Numbers Composite numbers. A composite number is nturl number greter thn one tht is not prime. {4, 6, 8, 9, 0,, 4, 5, 6, 8, 0,,, 4, 5, 6, 7, 8, 30...}.4. Integers. The integers re the elements of the set Z = {, 3,,, 0,,, 3, }. The integers re closed under ddition, subtrction nd multipliction. The positive integers re the sme s the nturl numbers..5. Rtionl numbers. The rtionl numbers re the elements of the set { } Q = b : εz, bεz {0}. The expression b is clled frction in the sense tht represents portion of b. We cll the numertor of the frction nd b the denomintor of the frction. An integer n is lso rtionl number becuse n = n. The rtionl numbers re closed under ddition, subtrction, multipliction nd division.
6 6 FUNCTIONS AND EQUATIONS.6. Irrtionl numbers. Numbers tht cnnot be written s rtios or quotients of integers re clled irrtionl numbers. Exmples include, 5,, π.to see why is not rtionl number see Appendix A..7. Rel numbers..7.. Definition of the rel numbers. The union of the sets of rtionl nd irrtionl numbers is the set of rel numbers. We cn think of the set of rel numbers, R, s extending long line to infinity in ech direction hving no breks or gps..7.. Writing rel numbers s deciml frctions. We cn write ny rel number s n infinite deciml frction. For exmple we cn write s In generl rel number cn be written s x = ±m.β β β 3..., where m is n integer nd β n ( n =,,..., ) is n infinite series of digits, ech in the rnge 0 to 9. Rtionl numbers cn be written s finite or recurring deciml frctions, tht is, fter certin plce in the deciml representtion, it either stops or continues to repet finite sequence of digits. For exmple, 3 4 = = = = Irrtionl numbers re nonrepeting deciml frctions. For exmple, π = = e = /3 = Numbers of the form n /m re irrtionl unless n is the m th power of n integer. For exmple, 65 /4 = 5, i.e., 65 = Addition. One cn view ddition s process of counting on or counting up. Strt with one of the numbers sy m nd view it s set with m objects. Then strt counting the elements of the set contining n objects strting with m+ for the first element of the second set. Consider the following exmple where n = 3 nd m = 5. Lbel the first set A nd the second set A. A = {,, 3} A = {,, 3, 4, 5} Counting the objects in both sets we obtin 8 so tht 3 dded to 5 gives 8. We often use the word plus for ddition nd use the sign +. Then = 8.
7 FUNCTIONS AND EQUATIONS Multipliction. Multipliction is the process of dding numbers together multiple times. For exmple, if we dd 3 nd 3 (3 + 3) we hve two groups of 3, or 3 two times. If we dd four nd four nd four (4+4+4) we hve three groups of 4, or 4 3 times. We sometimes use the word times to represent multipliction. We use the symbol or the symbol * or the symbol to represent multipliction. So 3 times is written 3 or 3* or 3. When the context is cler, we cn represent multipliction by writing two numbers next to ech other with spce between them s in 3 = 6. In such cses we often enclose the numbers in prentheses s in (3)() = 6. It is useful to memorize or t lest be ble to construct tble listing the first 0 or numbers multiplied by ech other. TABLE. Multipliction Tble Notice tht cross the second two or down the second column is counting by twos nd tht cross the fifth row or down the fifth column is counting by fives. Also notice the following.. All multiples of re even numbers.. The sum of the digits in ll multiples of 3 is lesser multiple of All multiples of 5 end in 0 or All multiples of 6 re even numbers nd the sum of the digits is lesser multiple of The sum of the digits in ll multiples of 9 is 9 or multiple of All multiples of 0 end in The items in the digonl of the tble representing ech number multiplied by itself re clled squres. Note tht they differ by,5,7,9,,3,5,7,9,, 3, 5, 7, 9, 3, 33, 35, 37, nd 39.
8 8 FUNCTIONS AND EQUATIONS 8. The tble is symmetric Field properties of ddition nd multipliction of rel numbers. For ny rel numbers, b, c: TABLE 3. Field Properties of Rel Numbers # Addition A Multipliction M 0 Closure +bε R b ε R. Commuttive +b=b+ b = b Associtive +(b+c)=(+b)+c (bc) = (b)c 3 Identity There is number 0 with +0=0+=. There is number with ()() = ()() =. There is number  with 4 Inverse +=+=0. D Distributive (b + c) = b + c If 0, there is number with () ( ( ) = ) () =. is lso written s nd = Definition of subtrction. For ny rel numbers nd b b=+b Definition of division. For the rel numbers nd b with b 0, b=/b= b = ()( b) = ()(/b) Some uniqueness theorems for rel numbers derived from the field properties nd the definitions. TABLE 4. Uniqueness theorems for rel numbers Theorem # Description Condition Result Theorem Uniqueness of 0 If z nd re elements of R such tht z + = then z=0 Theorem b Uniqueness of If u nd b 0 re elements of R such tht ub = b then u= Theorem Uniqueness of  If nd b re elements of R such tht + b = 0 then b= Theorem b Uniqueness of If 0 nd b re elements of R such tht b = then b =
9 FUNCTIONS AND EQUATIONS 9 3. OPERATIONS WITH REAL NUMBERS 3.. Summry of results. Using the field properties in section.7.5, one cn stte the following properties of rel numbers. TABLE 5. Opertions with Rel Numbers # Property Exmple Notes  b = +(b) 6=6+()=4 Subtrction (b)=+b 4(3)=4+3=7 3  = () 4 = ()4 4 (b+c) = b + c 3 (+5)= (3)() + (3)(5) = Distributive 5 (bc) = b  c 3 (4)= (3)()  (3)(4) = (+b) = ()(+b) = b (6+3) = 63 = 9 Distributive 7 (b) = +b (43) = 4+3 =  8 () = () = 9 (0) = 0 3(0) = 0 0 ()(b) = (b) = (b) (3)(4) = (3 4) = (3)(4)= ()(b) = b (3)(5) = 3 5=5 = ) = ) 3 b =( b 4 b =( b 5 b = b 0 ) = b 5 =( 5 9 =( 9 7 = 7 ) = 9 6 = 0 when =0 7 = when 0 5 =, 5 = 8 ( ( b ) =b 7 ) =7 9 = when 0 = 0 b c d = c bd = = 8 5 b c = ( ) ( c b= b ) 7 c 3 = 3 7 = 7 3 bc = ( )( ( b c) = )( ) b c 3 7 = 3 7 = b = ( )( c ) ( b c = c ) bc when c 0 7 = ( 5 ) 7)( 5 = b( c) = ( b)(c) = bc = ( b)( c) = bc 3( 5) = ( b) c 5 6 c + b c = +b c 7 c  b c = b c 8 b + c d = d+bc bd 9 b  c d = d bc bd b c d b c = ( )b c = b c = ( )( b) c = b c d = b d c = d bc = b c = c b = c b b c = b c = b c = bc = b c ( 3)(5) = 5 = ( 3)( 5) = 3(5) = 5 ( 3) 5 = ( )(3) 5 = (3) 5 = ( )( 3) 5 = (3) 5 = = +3 = = 3 9 = = = = = = = = = = 3 5 = 5 3 = 5 3 = = 3 5 = 3 5 = 3 5 = Working with frctions Properties of frctions.
10 0 FUNCTIONS AND EQUATIONS. If, b, c, nd d re rel numbers nd b nd d re not zero then Equlity of Frctions implies tht b = c if nd only if d = bc d. If, b, nd x re rel numbers nd b nd x re not zero then the Fundmentl Property of Frctions implies tht x bx = b 3... Multiplying frctions. To multiply two frctions, multiply the numertors nd denomintors seprtely, i.e., x y b = x y b = x y b = x yb For exmple, = 6 = Dividing frctions. To divide two frctions, use the inverse multipliction rule, i.e., For exmple, 3 4 = x y b = x y ( )( ) 3 4 b = xb y = 6 4 = Simplifying frctions. The Fundmentl Property of Frctions llows us to write ny frction in n infinite number of wys by multiplying the numertor nd denomintor by the sme number nd then repeting this process for different number. For exmple = 4 = 0 40 = =.5 5 = = 4 6 = = = = A frction which hs the smllest possible integers in both the numertor nd denomintor is sid to be in reduced form. The process of eliminting these integer multipliers in the numertor nd denomintor of frction is clled simplifying the frction or writing the frction in reduced form. For exmple, the frction cn be written in different form by eliminting the multiplier 5 s follows = = = 4 8 = 4 8 The multiplier 5 is often clled fctor. The frction 4 8 cn then be written in different form by eliminting the multiplier 4 s follows 4 8 = 4 4 = 4 4 = = is 5 4 = 0. This implies we could in one step s follows Notice tht the product of the two fctors we eliminted from 0 40 find the reduced form of 0 40
11 FUNCTIONS AND EQUATIONS 0 40 = 0 0 = 0 0 = = We cn reduce frction by finding common multipliers or fctors in the numertor nd denomintor of frction. A fil sfe wy to do this is to fctor number into the primes of which it is mde. We cn fctor s 3. And nd 3 re both prime. Definition (Prime Fctoriztion). The prime fctoriztion of number is group of prime numbers tht multiplied together is equl to tht number. For exmple 357 = nd 79 = The prime fctors in number my well repet s in 9 = 3. To find the prime fctoriztion of number we cn use the following lgorithm. Definition (Algorithm for Prime Fctoriztion).. If the number is divisible by, then write down s fctor nd proceed with the prime fctoriztion of the quotient.. If the number is not divisible by, then see if the number is divisible by 3. If the number is divisible by 3, then write down 3 s fctor nd proceed with the prime fctoriztion of the quotient. 3. If the number is not divisible by or 3, then see if the number is divisible by 5. If the number is divisible by 5, then write down 5 s fctor nd proceed with the prime fctoriztion of the quotient. 4. If the number is not divisible by or 3 or 5, then see if the number is divisible by 7. If the number is divisible by 7, then write down 7 s fctor nd proceed with the prime fctoriztion of the quotient. 5. Continue in this fshion until the quotient is prime number or the prime fctor you re trying is the smllest prime less thn the squre root of the smllest perfect squre lrger thn the number being fctored. Consider the prime fctoriztion of is not divisible by, so proceed with 3.. The sum of the digits in 693 is 8 so 693 is divisible by 3. Forming the quotient we obtin = is not divisible by, so proceed with The sum of the digits in 693 is 6 so 3 is divisible by 3. Forming the quotient we obtin 3 3 = is not divisible by, so proceed with is not divisible by 3, so proceed with is not divisible by 5, so proceed with is divisible by 7. The quotient is so we re finished becuse is prime number. The prime fctoriztion of 693 is then Finding the prime fctoriztions of the numertor nd denomintor of frction nd then cnceling like terms will lwys reduce frction to its simplest form, i.e. this will lwys identify the fctors tht re common in the numertor nd denomintor of frction. In mny cses, however, it my be esier to look for lrger common fctors thn those in the prime fctoriztion. Consider the following exmple = = = = 0 9 We might do this fster by noticing tht the numertor nd denomintor re divisible by becuse they re even nd divisible by 3 becuse the sum of the digits is divisible by 3 (6 nd 8). This would then give
12 FUNCTIONS AND EQUATIONS = = = Remembering our seven times tble we cn simplify s = = = 0 9 When using this pproch the following fcts or rules re useful.. All even numbers hve s fctor.. If the lst two digits of number divide by 4, the whole number divides by 4. For exmple the lst two digits of 4 divide by 4 nd 4/4 = If the lst three digits of number divide by 8, the whole number divides by 8. For exmple the lst three digits of 38 divide by 8 nd 38/8 = 39. Similrly for 6, 3, If number ends in 5 or 0, it is divisible by If the lst two digits of number re divisible by 5, then the whole number is divisible by 5. For exmple, the lst two digits of 675 re divisible by 5 so 675 is divisible by 5, i.e., 675/5 = 7. Similrly for 65, 35, All numbers ending in zero re divisible by 0. Numbers ending in two zeroes re divisible by 00 nd so on. 7. If the sum of the digits in number is divisible by 3, the whole number is divisible by If the sum of the digits in number is divisible by 9, the whole number is divisible by If the sum of the digits in number is divisible by 3 nd the number is even, the whole number is divisible by For, dd lternte digits, if the sums re identicl, differ by, or by multiple of, then the number is divisible by. Consider where + = nd /=. Or 8,347 where = so tht 8347/ = 577. Or consider 869 where = 7 nd 76 =. Therefore 869/= 79. A useful fct to remember in fctoring number is tht if it is not prime number, it will hve t lest one fctor less thn the squre root of the number which is the smllest perfect squre greter thn the number in question. Consider, for exmple, 39. None of the rules we listed seem to help in fctoring 39. Becuse 44 is the smllest perfect squre greter thn 39, we know tht the only primes numbers we need to try re, 3, 5, 7, nd. We cn esily eliminte, 3, nd 5. We cn eliminte becuse 0 3 nd 03 is not divisible by. Or we cn just remember our eleven times tble. So we cn try ) We find tht 39 = 9 6. This does not divide evenly so we know tht 39 is prime number Adding nd subtrcting frctions. To dd or subtrct two frctions, the denomintors must be the sme (common denomintor). If the denomintors re the sme, then the ddition is completed by dding the numertors, i.e., x y + y = x + y
13 FUNCTIONS AND EQUATIONS 3 For exmple = 5 6 If the denomintors in the frctions we wnt to dd re not the sme, we must rewrite them so tht they hve the sme denomintor. For exmple or But the 48 is not in reduced form, i.e., + 3 = = = = = = 7 6 If we first reduce 3 4 to 8, we obtin = = 7 6 So mking sure ll frctions in product re in reduced from will help us obtin sum tht is in reduced form. But how do we find common denomintors. In the exmples so fr, resonble choice hs seemed obvious. Is there more systemtic wy? One wy tht will lwys work is to simply multiply the denomintors together to obtin the common denomintor nd then multiply the numertors by the denomintors of the other frctions. For exmple = = = 56 8 = 4 3 = 7 6 Of course it would hve been esier to convert 8 to 6 t the beginning nd obtin = = 7 6 Multiplying ll the denomintors together will often led to very lrge number for the denomintor. An lterntive pproch is to find the lest common denomintor mong the denomintors of the frctions being dded or subtrcted. Definition 3 (Lest Common Denomintor(LCD)). The lest common denomintor for group of frctions is the smllest number tht cn be divided by ll the denomintors. For exmple, 6 is the lest common denomintor for nd 3, nd 4 is the lest common denomintor of nd 4, while 48 is the lest common denomintor of nd 6. Notice tht 48 is much less thn 6 = 9. The most strightforwrd wy to find the lest common denomintor for group of frctions is to first find the prime fctoriztion of ech denomintor. We then crete the lest common denomintor by using ech fctor the gretest number of times tht it ppers in ny one denomintor. For exmple, consider the lest common denomintor for nd 6. First find the prime fctoriztion of ech.
14 4 FUNCTIONS AND EQUATIONS = 3 The lest common denomintor is then 6 = 3 = 48 Consider slightly more complicted exmple where we wnt to dd, 5 prime fctoriztion of ech number. = 3 6 = 0 = 5, 7 6, 3 0 nd First find the 35 = 5 7 The lest common denomintor is then (***)*3*5*7=680. Notice tht (***), (**3), (**5), nd (5*7) re ll contined in this expression. Now multiply ech frction by the pproprite fctor so tht the denomintors re ll 680. Now dd nd reduce = = = = = = = = = = = = 680 = = = = Seprting Frctions. Given the rule for dding nd subtrcting frctions we cn lwys write sum over denomintor with single term s sum of frctions with the numertor of ech coming from the respectively terms to dded in the numertor of the originl frction nd the denomintor being the common denomintor. Consider the following exmple.
15 FUNCTIONS AND EQUATIONS = = 6 40 The is no simple wy to reduce frctions with more thn one term in the denomintor is in 5. On the 3+x other hnd, 3+x 5 = x 5 = 5 + x Rules of exponents Product rule for exponents. If m nd n re nturl numbers then x m x n = x m+n Note tht the product rule pplies to exponentil expressions with the sme bse. A product of two powers with different bses such s x 3 y 4, cnnot be simplified. Consider the following exmples =(5 5 5) (5 5) = = 5 5 x p x p = x p x p x p = x 3p Power rules for exponents. If m nd n re nturl numbers then (x m ) n = x mn (xy) n = x n y n For exmple, ( ) n x = xn y y, n y 0 (4 ) 3 =6 3 = = 56 6 = 4096 = 4 6 = = = 4096 (4 5) 3 =0 3 = = = 8000 = = 64 5 = Zero rule for exponents. x 0 =, x Negtive exponents. If n is n integer nd x 0, then For exmple 3 = 9. x n = x n x = n xn
16 6 FUNCTIONS AND EQUATIONS Quotient rule for exponents. If m nd n re integers nd x 0, then For exmple, x m x n = xm n = = = 9 = 34 = 3 = 9 Notice tht when the numertor nd denomintor contin the sme number, but to different powers, we cn cncel terms just like we did in simplifying frctions. Or consider the following exmple. 3 4 x y 3 3 x y = 3 3 x y = 7xy A frction to negtive power. If n is nturl number, then For exmple ( ) 3 = = ( 3 ) Some more exmples.. x 3 yx = x y ( ) n x ( y ) n =, x 0nd y 0 y x = 9 4 = 3 = ( 3. ). (x) = x 3. x = x 4. (3x) 3 4 9x4x = x = = x 3 x/ x = xx/ = x x /
17 FUNCTIONS AND EQUATIONS 7 8. t p t q t r t s = t p+q r s 9. 4 Ax /4 x / x = 4 Ax 3/4 x / 0. α x α xα x = α x α xα 3.4. Order of opertions. If n expression does not hve prentheses or grouping symbols, follow these steps,. Find the vlues of ny exponentil expressions. Perform ll multiplictions nd/or divisions, working from left to right. 3. Perform ll dditions nd/or subtrctions, working from left to right. 4. EXPRESSIONS AND EQUATIONS 4.. Vribles. A vrible is symbol tht cn be replced by ny one of set of numbers or other objects. For exmple, we might pick vrible denoted by r. Becuse r is vrible, it cn then be replced by vrious members of given set such s the integers. For exmple, if r is vrible representing the ge of n rbitrry person in yers, where frctions of yers re rounded down to the next integer, r could be replced by ny of the numbers {0,,,3,...,,..., } where the upper bound on the set is bit uncertin. (If one likes Biblicl dting, 969 might be resonble number.) 4.. Combining vribles nd numbers to form expressions Definition of n expression. When numbers nd vribles re combined, the result is clled n lgebric expression, or simply n expression Some exmples. : less thn twice number is written s x  b: Five times the sme number is written s 5x c: The product of output (y) nd price (p) less the cost of producing tht output c(y) is written py  c(y) Evluting n expression. Substituting for the vribles nd clculting result is clled evluting n expression. For exmple, consider the expression bh. If b is 5 nd h is 6, then the expression is evluted s (5)(6) = Fctoring expressions. Just s we fctored numbers, we cn fctor lgebric expressions. For exmple 63= And 5x +5x = 5x(x +3). here re some more exmples.. x 4x 6 = (6x + )(x 3). x x + 4 = (x ) 3. 4u +8u +4=(u +) = 4(u +) 4. x y 6x +6xy 8 = (xy 3)(x +6)
18 8 FUNCTIONS AND EQUATIONS The verbs of lgebr. The most common verbs used in lgebr re = equl > greter thn < less thn greter thn or equl to less thn or equl to not equl to pproximtely equl to Combining expressions nd verbs to form sentences. () Definition of sentence. We cn combine expressions nd verbs to form sentences. () Sentences tht re ssumed to be true re either definitions of terms or postultes. (3) Sentences tht re either true or flse re clled sttements or propositions. A true proposition might be All individuls who cn brethe on their own re live. A flse proposition might be All individuls who cn brethe on their own re lert nd ble to spek. (4) An open proposition is one tht is true for some vlues of the vribles but not others. For exmple, the proposition x  = 0 is true for some vlues of x but not ll vlues of x. (5) Sttements tht re shown to be true using definitions nd postultes re clled theorems Equtions. An eqution is sentence stting tht two expressions re equl. For exmple we might hve the eqution b + c = c + d. An eqution is n open proposition in tht it my be true for some vlues of the vribles but not others Formuls. A formul is sentence stting tht single vrible is equl to n expression with one or more vribles on the other side. Thus c = d b is both n eqution nd formul. But + b = b + is n eqution but not formul. If we evlute the expression in formul nd substitute the result for the vrible on the other side, the eqution will be true for those vlues of the vribles Exmples of equtions. Here re some exmples of typicl equtions one might encounter. : x 0 = 3x : x 6 = x + y 3: x + y = 5x 3 x 4: y = 4 5: x +3y = 0x + 6: y 6x = 4x 3 x 4x 7: 3 +y = 6 + 8: y = 5x 4y +5 9: 4x +3y = y 5 8 0: 6x +5y = x y : x + y = 5 : 3 + 3x 4 = y
19 FUNCTIONS AND EQUATIONS Some theorems useful in nlyzing equtions derived from the field properties nd definitions. Some theorems tht re useful in nlyzing nd simplifying equtions re contined in tble 6. TABLE 6. Theorems Useful for Anlyzing Equtions Theorem # Description Condition Result Theorem 3 Unique solutions to + x = b Let, b be rbitrry elements of R, then + x = b hs the unique solution x = () + b Let, b be rbitrry if 0, x = b hs the Theorem 3b Unique solutions to x = b elements of R, then unique solution x = b Theorem 4 Multipliction property of 0 For ny ε R ()(0) = 0 Theorem 4b Multipliction property of  For ny ε R ()() =  Theorem 4c Opposite of n opposite For ny ε R () = Theorem 4d Product of () nd () For ny ε R ()() = Theorem 4e Opposite of sum For ll nd b ε R (+b) = +b Theorem 5 Cnceltion for ddition Let x, y, z, be rbitrry elements of R. If x + y = x + z then y=z Theorem 6 Product of two fctors = 0 Let nd b be elements of R. We hve b = 0 if nd only if = 0 or b = 0 or both Theorem 6b Product of two fctors = 0 Let nd b be elements of R. We hve 0ndb 0iff b 0 Let be n element of R. Theorem 7 Reciprocl of reciprocl If 0, then / 0 nd ( ) = Let, b, c, be elements of R. Theorem 7b Cnceltion for multipliction If b = c nd 0, then b=c Theorem 7c Reciprocl of product Let nd b be elements of R. If 0 nd b 0, b = ( ) b Theorem 8 Theorem 9 Theorem 9b Theorem 9c Theorem 9d Theorem 9e Multipliction of frctions Cnceltion for frctions Addition of frctions Signs for frctions Frctions divided by frctions Cross multipliction for frctions Let, b, c, d, be elements of R. If b 0 nd d 0, then Let, b, c, be elements of R. If b 0 nd c 0, then Let, b, c, be elements of R. If c 0, then Let nd b be elements of R. If b 0, then Let, b, c, d, be elements of R. If b 0, c 0, nd d 0, then Let, b, c, d, be elements of R. If b 0 nd d 0, then bd 0 nd b = ( ) b c = c bd ( b ) = c bc c + b c = +b c b 0 nd ) ( ) = ( b ( /b c/d b = ( b ) c/d 0 nd ) = d bc = ( ) ( d b c ( ) ( b = c ) d = d+bc bd ) 5. SOLVING EQUATIONS 5.. Simple equtions. Most simple equtions cn be solved using the theorems from tble 6. The theorems on cnceltion provide wy to obtin equtions equivlent to given eqution by using the following eqution solving principle.
20 0 FUNCTIONS AND EQUATIONS 5.. Bsic Principle for Solving Equtions. We obtin equivlent equtions if on both sides of the equlity sign in n eqution we do the following: () dd the sme number () subtrct the sme number (3) multiply by the sme number 0 (4) divide by the sme number Exmples of simple equtions. () 3x +0=x +4 () 6x (x 3) = 3( x) 7 6 (x +) (3) x + x 8 x x = x (4) x x = 5 5 x 5.4. Qudrtic equtions. An eqution in which one or more of the vribles is squred (or two vribles re multiplied together) nd there re no vribles rised to the third power is clled qudrtic eqution Some importnt identities. : ( + b) = +b + b b: ( b) = b + b c: ( + b)( b) = b Completing the squre. Consider the following two expressions: x + bx () x +hx + h = (x + h) (b) An interesting question is wht do we hve to do to expression so tht it is equivlent to expression b. To find the nswer write the expression s follows: x + bx +? = x +hx + h () The first terms re equl nd the second terms will be equl if b = h. This implies tht h = /b. It is then cler tht h = ( b ). So we cn sy: Proposition (completing the squre). To complete the squre in the expression on x + bx, dd (/b) to it. Here re some simple exmples of expressions for prctice in completing the squre : x +8x : x +6x 3: x 3x 4: x + b x 5: x x + x 6: x + y x +4y 7: x 3 x 8: x 0x 9: x 0x +0
21 FUNCTIONS AND EQUATIONS 0: x + x : x + x 4 : x x ( µ + σ t ) Generl form of qudrtic eqution nd some simple exmples. The generl form of qudrtic eqution is x + bx + c = 0, ( 0) (3) If the qudrtic polynomil x + bx + c cn be fctored, then the eqution x + bx + c = 0cn be solved by using Theorem 6: If the product b = 0, then = 0 or b = 0, or both dn b equl zero. Consider the eqution x 4 = 0. We cn fctor it s x 4 = (x )(x +) Rewriting the eqution using this identity we obtin (x )(x +)=0 (x ) = 0 or (x + ) = 0 x = or x = Now consider the eqution x 3x 0 = 0. We cn fctor it s x 3x 0 = (x 5)(x +) Rewriting the eqution using this identity we obtin (x 5)(x +)=0 (x 5) = 0 or (x + ) = 0 x =5 or x = Now consider the eqution x 3.5x = 0. We cn fctor it s x 3.5x = (x 4)(x + ) Rewriting the eqution using this identity we obtin (x 4)(x + ) = 0 (x 4) = 0 or (x + ) = 0 x =4 or x = Finlly consider the eqution x 5x 7 = 0. We cn fctor it s x 5x 7 = (4x + )(3x 7) Rewriting the eqution using this identity we obtin
22 FUNCTIONS AND EQUATIONS (4x + )(3x 7) = 0 (4x +)=0or (3x 7) = 0 x = 4 or x = The qudrtic formul. In mny cses simple solution s bove is not obvious. Thus we derive the qudrtic formul. We do so in 7 steps s follows. x + bx + c = 0( 0) () Divide both sides by 0. x + b x + c = 0 ( 0) () Add c to ech side. x + b x = c ( 0) (3) Complete the squre by dding ( ) b. x + b x + b (4) Simplify the lhs s perfect squre. ( ) x + b = b (5) Simplify the rhs by dding frctions. ( x + b (6) Tke the squre root of ech side. x + b 4 = b 4 c c ( 0) 4 ) = b 4c 4 ( 0) = ± b 4c ( 0) ( 0) (7) Add b to both sides. x = b ± b 4c ( 0) Notice tht if b  4c is negtive, we do not hve rel solution to the eqution Exmples For Prcticing Use of the Qudrtic Formul. : x 4x 7 = 0 : 3x +x 4 = 0 3: 4x +x = 0 4: 0x +3x +3=0 5: 4x 44 = 0 6: 0x + 00 = 0 7: 6x 5x 3 = 0 6. FUNCTIONS OF A REAL VARIABLE 6.. Definition. A function of rel vrible x with domin D is rule tht ssigns unique rel number to ech number x in D. Functions re often given letter nmes such s f, g, F, or φ. We often cll x the independent vrible or the rgument of f. If g is function nd x is number in D, then g(x) denotes the number tht g ssigns to x. We sometimes mke the ide tht F hs n rgument (we substitute number for the vrible in F) explicit by writing F( ). In the cse of two vribles we sometimes use y = f(x) for the vlue of f evluted t the number x. Note the difference between φ nd φ(x). 6.. The domin of function. The domin is the set of ll vlues tht cn be substituted for x in the function f( ). If function f is defined using n lgebric formul, we normlly dopt the convention tht the domin consists of ll vlues of the independent vrible for which the function gives meningful vlue (unless the domin is explicitly mentioned).
23 FUNCTIONS AND EQUATIONS The rnge of function. Let g be function with domin D. The set of ll vlues g(x) tht the function ssumes is clled the rnge of g. To show tht number, sy, is in the rnge of function f, we must find number x such tht f(x) =. Here re some exmple functions for which to find the domin nd the rnge. : f(x) = x, 0 x 60 : g(x) = x /0, 0 x 60 3: h(x) = /x 4: φ(x) = 3x 5: x = y,y 0 6: f(x) = 3x x 4 7: g(x) = 4 3x 6.4. The grph of function. When the rule tht defines function f is given by n eqution in y nd x, the grph of f is the grph of the eqution, tht is the set of points (x, y) in the xyplne tht stisfies the eqution. Another wy to sy this is tht the grph of the function g is the set of ll point (x, g(x)), where x belongs to the domin of g The verticl line test. As set of points in the xyplne is the grph of function if nd only if verticl line intersects the grph in t most one point Liner functions Definition. A liner function of rel vrible x is given by y = f(x) = x + b, nd b re constnts. The grph of liner eqution is stright line. The number is clled the slope of the function nd the number b is clled the yintercept The slope of stright line given two distinct points on the line. Consider two distinct points on nonverticl stright line in the plne denoted by P = (x,y ) nd Q = (x,y ). Becuse the line is not verticl nd P nd Q re distinct, x x. The slope of the line is given by = y y x x,x x. (4) The point slope formul for line. Consider point P = (x,y ) nd ny line with slope. The line with slope pssing through the point cn be determined s follows. Pick n rbitrry point on the line nd denote it s (x, y). Then use the formul for the slope of line s follows. = y y x x,x x (5) Now solve the eqution for either y or y  y s follows. = y y,x x x x y y = (x x ) y = (x x )+y = x +(y x ). (6)
24 4 FUNCTIONS AND EQUATIONS Pointpoint formul for line. If we re given two points on line we cn find the eqution by first finding the slope nd then using the point slope formul. Let the two points on the line be denoted (x,y ) nd (x,y ), x x. The slope is given by = y y,x x x x. (7) If we substitute in the pointslope formul we obtin y y = (x x ) ( ) y y = (x x ) x x ( ) y y y = (x x )+y x x ( ) ( ) y y x (y y ) = x + y x x x x ( ) ( ) y y x y + x y + x y x y = x + x x x x ( ) ( ) y y x y x y = x +. x x x x (8) 7. COMPLEX NUMBERS 7.. Definition of complex number. A complex number is n ordered pir of rel numbers denoted by (x, x ). The first member, x, is clled the rel prt of the complex number; the second member, x,is clled the imginry prt. We define equlity, ddition, subtrction, nd multipliction so s to preserve the fmilir rules of lgebr for rel numbers Equlity of complex numbers. Two complex numbers x = (x, x ) nd y = (y, y ) re clled equl iff x = y, nd x = y. (9) 7... Sum of complex numbers. The sum of two complex numbers x + y is defined s x + y =(x + y,x + y ) (0) Difference of complex numbers. To subtrct two complex numbers, the following rule pplies Product of complex numbers. The product xy is defined s x y =(x y,x y ) () xy =(x y x y,x y + x y ). () The properties of ddition nd multipliction defined stisfy the commuttive, ssocitive nd distributive lws. 7.. The imginry unit. The complex number (0,) is denoted by i nd is clled the imginry unit. We cn show tht i =  s follows. i =(0, )(0, ) = (0, 0+0)=(, 0) = (3)
25 FUNCTIONS AND EQUATIONS Representtion of complex number. A complex number x = (x, x ) cn be written in the form x = x + ix. (4) Alterntively complex number z = (x,y) is sometimes written z = x + iy. (5) 7.4. Modulus of complex number. The modulus or bsolute vlue of complex number x = (x, x )is the nonnegtive rel number x given by x = x + x (6) 7.5. Complex conjugte of complex number. For ech complex number z = x + iy, the number z =xiy is clled the complex conjugte of z. The product of complex number nd its conjugte is rel number. In prticulr, if z = x + iy then zz =(x, y)(x, y) =(x + y, xy + yx) =(x + y, 0) = x + y (7) Sometimes we will use the nottion z to represent the complex conjugte of complex number. So z = x  iy. We cn then write z z =(x, y)(x, y) =(x + y, xy + yx) =(x + y, 0) = x + y (8) 7.6. Grphicl representtion of complex number. Consider representing complex number in two dimensionl grph with the verticl xis representing the imginry prt. In this frmework the modulus of the complex number is the distnce from the origin to the point. This is seen clerly in figure. FIGURE. Grphicl Representtion of Complex Number Imginry xis x Sqrt x ^ x ^ x x ix x 0 x Rel xis 7.7. Polr form of complex number. We cn represent complex number by its ngle nd distnce from the origin. Consider complex number z = x +iy. Now consider the ngle θ which the ry from the origin to the point z mkes with the x xis. Let the modulus of z be denoted by r. Then cos θ = x /r nd sin θ = y /r. This then implies tht z = x + iy = r cos θ + ir sin θ = r (cos θ + i sin θ ) Figure shows how complex number is represented in polr coordintes. (9)
26 6 FUNCTIONS AND EQUATIONS FIGURE. Grphicl Representtion of Complex Number in Polr Coordintes Imginry xis z r cos Θ i sin Θ y Θ 0 x Rel xis 7.8. Complex Exponentils. The exponentil e x is rel number. We wnt to define e z when z is complex number in such wy tht the principle properties of the rel exponentil function will be preserved. The min properties of e x, for x rel, re the lw of exponents, e x e x =e x+x nd the eqution e 0 =. If we wnt the lw of exponents to hold for complex numbers, then it must be tht e z = e x+iy = e x e iy (0) We lredy know the mening of e x. We therefore need to define wht we men by e iy. Specificlly we define e iy in eqution Definition 4. e iy = cos y + i sin y () With this in mind we cn then define e z =e x+iy s follows Definition 5. e z = e x e iy = e x (cos y + i sin y) () Obviously if x = 0 so tht z is pure imginry number this yields e iy = (cos y + i sin y) (3) It is esy to show tht e 0 =. If z is rel then y = 0. Eqution then becomes So e 0 obviously is equl to. e z = e x e i 0 = e x (cos (0) + i sin (0) ) = e x e i 0 = e x (+0)=e x (4) To show tht e x e iy =e x + iy or e z e z =e z + z we will need to remember some trigonometric formuls.
27 FUNCTIONS AND EQUATIONS 7 Theorem. sin(φ + θ) = sin φ cos θ + cos φ sin θ sin(φ θ) = sin φ cos θ cos φ sin θ cos(φ + θ) = cos φ cos θ sin φ sin θ (5) (5b) (5c) cos(φ θ) = cos φ cos θ + sin φ sin θ Now to the theorem showing tht e z e z z + z =e Theorem. If z = x + iy t nd z = x + iy re two complex numbers, then we hve e z e z =e z+z. Proof. e z = e x (cos y + i sin y ),e z = e x (cos y + i sin y ), e z e z = e x e x [cos y cos y sin y sin y + i(cos y sin y + sin y cos y )]. Now e x e x = e x+x, since x nd x re both rel. Also, (5d) (6) nd nd hence cos y cos y sin y sin y = cos (y + y ) (7) cos y sin y + sin y cos y = sin (y + y ), (8) e z e z = e x+x [cos (y + y )+i sin (y + y ) = e z+z. (9) Now combine the results in equtions 9, nd 3 to obtin the polr representtion of complex number z = x + iy = r (cos θ + i sin θ ) = r e iθ, (by eqution ) The usul rules for multipliction nd division hold so tht (30) z z = r e iθ r e iθ i(θ+ θ) = (r r ) e z = r ( ) e iθ r (3) = e i(θ θ) r e iθ z r
28 8 FUNCTIONS AND EQUATIONS 8. APPENDIX A PROOF THAT IS NOT A RATIONAL NUMBER We cn show tht Q (rtionls) s follows. Proof. The proof is by contrdiction. Suppose tht Q. Then there exist integers p nd q such tht p/q =, where we choose the smllest such p nd q. Now p /q =,orp = q, so tht p must be n even number. Since the squre of n odd number is lwys odd, p is even so we my write p = r, where r is lso n integer. We then hve p = 4r = q nd so q =r. Thus q, nd hence q, must be even. But if both p nd q re even, they hve common fctor of. This result contrdicts the ssumption tht p nd q were the smllest numbers to give p/q =. Thus, the sttement Q must be flse. 9. APPENDIX B PROOFS OF THEOREMS ON PROPERTIES OF REAL NUMBERS AND EQUATIONS Theorem 3. : If z nd re elements of R such tht z + =, then z = 0. b: If u nd b 0 re elements of R such tht u b = b, then u =. Proof of. The hypothesis is tht z + =. We dd the element , whose existence is given in (A4), to both sides to get (z + ) +( ) = +( ). If we use (A), (A4), nd (A3) in succession on the left side, we obtin (z + ) +( ) = z +( +( )) = z +0=z; If we use (A4) on the right side, we obtin Hence, we conclude tht z = 0. +( ) = 0. The proof of (b) is left s n exercise. Note tht the hypothesis b 0 is crucil. Theorem 4. : If nd b re elements of R such tht + b = 0, then b = . b: If 0 nd b re elements of R such tht b =, then b = /. Proof of. If + b = 0, then we dd  to both sides to get ( ) +( + b) = ( ) +0. If we use (A), (A4), nd (A3) on the left side, we obtin ( ) +( + b) = (( ) +) +b = 0 + b = b; if we use (A3) on the right side, we obtin ( ) +0=.
29 FUNCTIONS AND EQUATIONS 9 Hence, we conclude tht b = . The proof of (b) is left s n exercise. Note tht the hypothesis 0 is used. If we view the preceding properties in terms of solving equtions, we note tht (A4) nd (M4) enble us to solve the equtions + x = 0 nd x = (when 0) for x, nd theorem 4 implies tht the solutions re unique. Theorem 5. Let, b be rbitrry elements of R. Then: : the eqution + x = b hs the unique solution x = () + b; b: if 0, the eqution x = b hs the unique solution x = (/) b. Proof of. Using properties (A), (A4), nd (A3), we obtin +[( ) +b] = [ +( )] + b = 0 + b = b. which implies tht x = () + b is solution of the eqution + x = b. To show tht it is the only solution, suppose tht x is ny solution of the eqution, then + x = b, nd if we dd  to both sides, we get ( ) +( + x ) = ( ) +b. If we now use (A), (A4), nd (A3) on the left side, we obtin ( ) +( + x ) = ( + ) +x = 0 + x = x. Hence we conclude tht x = () + b. The proof of prt (b) is left s n exercise. Theorem 6. If nd b re ny elements of R, then: : 0=0, b: () = , c: () =, d: () () =, e: (+b) = () + (b). Proof of theorem 6. From (M3) we know tht =. Then dding 0 nd using (D) nd (A3) we obtin + 0 = + 0 = ( + 0) = =. Thus, by theorem 3 we conclude tht 0=0. Proof of theorem 6b. We use (D), in conjunction with (M3), (A4), nd prt (), to obtin +( ) = +( ) =( + ( )) = 0 = 0 Thus, from theorem 4, we conclude tht () =.
30 30 FUNCTIONS AND EQUATIONS Proof of theorem 6c. By (A4) we hve () + = 0. Thus from theorem 4, it follows tht = (). Proof of theorem 6d. In prt (b), substitute = . Then ( ) ( ) = ( ). Hence, the ssertion follows from prt (c) with =. Proof of theorem 6e. To prove prt e we use the following two lemms. Lemm. If, b, nd c re ny numbers, then + b + c = + c + b = b + + c = b + c + = c + + b = c + b +. Proof left to the reder. Lemm. If, b, c, nd d re numbers, then Proof. Using lemm nd the xioms, we hve ( + c) +(b + d) = ( + b) +(c + d). ( + c) +(b + d) = [( + c) +b] +d =( + c + b) +d = ( + b + c) +d =[( + b) +c] +d = ( + b) +(c + d). Now to prove prt e of the theorem. Proof. We know from the definition of negtive tht ( + b) +[ ( + b)] = 0. Furthermore, using lemm we hve ( + b) +[( ) +( b)] = +( )] + [b +( b)] = = 0. The result follows from theorem 4. Theorem 7. Let x, y, z be rbitrry elements of R. If x + y = x + z then y = z. Proof. If x + y = x + z, the xioms of ddition give y = 0 + y =( x + x) +y = x +(x + y) = x +(x + z) = ( x + x) +z = 0 + z = z. Theorem 8. For ny nd b ε R, we hve : b = 0 if nd only if = 0 or b = 0 or both. b: 0 nd b 0 if nd only if b 0.
31 FUNCTIONS AND EQUATIONS 3 Proof. We must prove two sttements in ech of prts () nd (b). To prove (), observe tht if = 0 or b = 0, or both, then it follows from theorem 6 tht b = 0. Going the other wy in (), suppose tht b = 0. Then there re two cses, either = 0 or 0. If = 0, the result follows. If 0, then we see tht b = b = ( )b = (b) = 0 = 0. Hence b = 0 nd () is estblished. To prove (b), first suppose 0 nd b 0. Then b 0, becuse 0 nd b 0 is the negtion of the sttement = 0 or b = 0 or both. Thus () pplies. For the second prt of (b), suppose b 0. Then 0 nd b 0 for, if one of them were zero, theorem 6 would pply to give b = 0. Theorem 9. Let, b, c, be elements of R. : If 0, then / 0 nd /(/) =. b: If b= c nd 0, then b = c. c: If 0 nd b 0, then ( b) =( ) (b ). Proof of. We re given 0, so tht / exists. If / = 0, then = (/) = 0 = 0, contrry to (M3). Thus / 0 nd since (/) =, theorem 4b implies tht /(/) =. Proof of b. If we multiply both sides of the eqution b= c by / nd pply the ssocitive property (M), we get ((/) ) b = ((/) ) c. Thus, b= c which is the sme s b = c. Proof of c. The proof of this theorem is like the proof of theorem 6 with ddition replced by multipliction, 0 replced by, nd (), (b) replced by,b. The detils re left to the reder. Note tht if 0, then 0 becuse =nd 0. Then theorem 8b my be used with b =. Theorem 0. If b 0 nd d 0, then b d 0 nd ( ) ( b c ) d = c b d. In order to prove the theorem we will use the following lemms, similr to lemm nd lemm. They cn be proven by the reder. For the lemms, we define bc s the common vlue of (b)c nd (bc). Lemm 3. If, b, nd c re numbers, then bc = cb = bc = bc = cb = cb. Lemm 4. If, b, c nd d re numbers, then (c) (bd) = (b) (cd). Proof. Tht b d 0 follows from theorem 8. Using the nottion for frctions, lemm 4 nd theorem 9c, we find ( ) ( c =( b b d) ) (cd ) =( c) (b d ) = ( c)(bd) = c b d.
32 3 FUNCTIONS AND EQUATIONS Theorem. : If b 0 nd c 0, then b = c b c. b: If c 0, then c + b ( + b) c = c. c: If b 0, then b 0 nd ( ) b = b = ( ) b. d: If b 0, c 0, nd d 0, then (c/d) 0 nd (/b) (c/d) e: If b 0 nd d 0, then b + c d + bc =. d bd Proof. Left to the reder. = d b c = ( b ) ( d c ).
Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
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