Stephen F Austin. The Real Numbers. chapter

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1 chpter 0 The Prthenon ws built in Athens over 400 yers go. The ncient Greeks developed nd used remrkbly sophisticted mthemtics. The Rel Numbers Success in this course nd in your future mthemtics courses will require good understnding of the bsic properties of the rel number system. Thus this book begins with review of the rel numbers. This chpter hs been lbeled Chpter 0 to emphsize its review nture. The first section of this chpter strts with the construction of the rel line. This section contins s n optionl highlight the ncient Greek proof tht no rtionl number hs squre equl to. This beutiful result ppers here not becuse you will need it for clculus, but becuse it should be seen by everyone t lest once. Although this chpter will be mostly review, thorough grounding in the rel number system will serve you well throughout this course nd then for the rest of your life. You will need good lgebric mnipultion skills; thus the second section of this chpter reviews the fundmentl lgebr of the rel numbers. You will lso need to feel comfortble working with inequlities nd bsolute vlues, which re reviewed in the lst section of this chpter. Even if your instructor decides to skip this chpter, you my wnt to red through it. Mke sure you cn do ll the exercises. 1

2 chpter 0 The Rel Numbers 0.1 The Rel Line By the end of this section you should section objectives understnd the correspondence between the system of rel numbers nd the rel line; pprecite the proof tht no rtionl number hs squre equl to. The integers re the numbers..., 3,, 1, 0, 1,, 3,..., The use of horizontl br to seprte the numertor nd denomintor of frction ws introduced by Arbic mthemticins bout 900 yers go. where the dots indicte tht the numbers continue without end in ech direction. The sum, difference, nd product of ny two integers re lso integers. The quotient of two integers is not necessrily n integer. Thus we extend rithmetic to the rtionl numbers, which re numbers of the form m n, where m nd n re integers nd n 0. Division is the inverse of multipliction, in the sense tht we wnt the eqution m n n = m to hold. In the eqution bove, if we tke n = 0 nd (for exmple) m = 1, we get the nonsensicl eqution = 1. This eqution is nonsensicl becuse multiplying nything by 0 should give 0, not 1. To get round this problem, we leve expressions such s 1 0 undefined. In other words, division by 0 is prohibited. The rtionl numbers form terrificlly useful system. We cn dd, multiply, subtrct, nd divide rtionl numbers (with the exception of division by 0) nd sty within the system of rtionl numbers. Rtionl numbers suffice for ll ctul physicl mesurements, such s length nd weight, of ny desired ccurcy. However, geometry, lgebr, nd clculus force us to consider n even richer system of numbers the rel numbers. To see why we need to go beyond the rtionl numbers, we will investigte the rel number line. Construction of the Rel Line Imgine horizontl line, extending without end in both directions. Pick n rbitrry point on this line nd lbel it 0. Pick nother rbitrry point to the right of 0 nd lbel it 1, s in the figure below.

3 section 0.1 The Rel Line Two key points on the rel line. Once the points 0 nd 1 hve been chosen on the line, everything else is determined by thinking of the distnce between 0 nd 1 s one unit of length. For exmple, is one unit to the right of 1, nd then 3 is one unit to the right of, nd so on. The negtive integers correspond to moving to the left of 0. Thus 1 is one unit to the left of 0, nd then is one unit to the left of 1, nd so on. The symbol for zero ws invented in Indi more thn 1100 yers go Integers on the rel line. If n is positive integer, then 1 n is to the right of 0 by the length obtined by dividing the segment from 0 to 1 into n segments of equl length. Then n is to the right of 1 n by the sme length, nd 3 n is to the right of n by the sme length gin, nd so on. The negtive rtionl numbers re plced on the line in similr fshion, but to the left of 0. In this fshion, we ssocite with every rtionl number point on the line. No figure cn show the lbels of ll the rtionl numbers, becuse we cn include only finitely mny lbels. The figure below shows the line with lbels ttched to few of the points corresponding to rtionl numbers Some rtionl numbers on the rel line. We will use the intuitive notion tht the line hs no gps nd tht every conceivble distnce cn be represented by point on the line. With these concepts in mind, we cll the line shown bove the rel line. We think of ech point on the rel line s corresponding to rel number. The undefined intuitive notions (such s no gps ) will become more precise when you rech more dvnced mthemtics courses. For now, we let our intuitive notions of the rel line serve to define the system of rel numbers. Is Every Rel Number Rtionl? We hve seen tht every rtionl number corresponds to some point on the rel line. Does every point on the rel line correspond to some rtionl number? In other words, is every rel number rtionl? If more nd more lbels of rtionl numbers were plced on the figure bove, the rel line would look incresingly cluttered. Probbly the first people to ponder these issues thought tht the rtionl numbers fill up the entire rel line. However, the ncient Greeks relized tht this is not true. To see how they cme to this conclusion, we mke brief detour into geometry.

4 chpter 0 The Rel Numbers This theorem is nmed in honor of the Greek mthemticin nd philosopher Pythgors who proved it over 500 yers go. The Bbylonins discovered this result thousnd yers erlier thn tht. Recll tht for right tringle, the sum of the squres of the lengths of the two sides tht form the right ngle equls the squre of the length of the hypotenuse. The figure below illustrtes this result, which is clled the Pythgoren Theorem. The Pythgoren Theorem for right tringles: c = + b. c b Now consider the specil cse where both sides tht form the right ngle hve length 1, s in the figure below. In this cse, the Pythgoren Theorem sttes tht the length c of the hypotenuse hs squre equl to. 1 c 1 An isosceles right tringle. The Pythgoren Theorem implies tht c =. Becuse we hve constructed line segment whose length c stisfies the eqution c =, point on the rel line corresponds to c. In other words, there is rel number whose squre equls. This rises the question of whether there exists rtionl number whose squre equls. We could try to find rtionl number whose squre equls by experimenttion. One striking exmple is ( 99 ) 9801 = ; here the numertor of the right side misses being twice the denomintor by only 1. Although ( 99 ) 70 is close to, it is not exctly equl to. Another exmple is The squre of this rtionl number is pproximtely , which is very close to but gin is not exctly wht we seek. Becuse we hve found rtionl numbers whose squres re very close to, you might suspect tht with further cleverness we could find rtionl number whose squre equls. However, the ncient Greeks proved this is impossible. This course does not focus much on proofs, nd probbly your clculus course will not be proof oriented either. The Greek proof tht there is no rtionl number whose squre equls could be skipped without endngering your future success. However, the Greek proof, s one of the

5 section 0.1 The Rel Line 5 gret intellectul chievements of humnity, should be experienced by every educted person. Thus it is presented below for your enrichment. Wht follows is proof by contrdiction. We will strt by ssuming tht the desired result is flse. Using tht ssumption, we will rrive t contrdiction. So our ssumption tht the desired result ws flse must hve been wrong. Thus the desired result is true. Understnding the logicl pttern of thinking tht goes into this proof will be vluble sset if you continue to other prts of mthemtics beyond clculus. No rtionl number hs squre equl to. Proof: Suppose there exist integers m nd n such tht ( m ) =. n By cnceling ny common fctors, we cn choose m nd n to hve no fctors in common. In other words, m n is reduced to lowest terms. The eqution bove is equivlent to the eqution m = n. This implies tht m is even; hence m is even. Thus m = k for some integer k. Substituting k for m in the eqution bove gives 4k = n, or equivlently k = n. This implies tht n is even; hence n is even. We hve now shown tht both m nd n re even, contrdicting our choice of m nd n s hving no fctors in common. This contrdiction mens our originl ssumption tht there is rtionl number whose squre equls must be flse. When you hve excluded the impossible, whtever remins, however improbble, must be the truth. Sherlock Holmes The result bove shows tht not every point on the rel line corresponds to rtionl number. In other words, not every rel number is rtionl. Thus the following definition is useful: Irrtionl numbers A rel number tht is not rtionl is clled n irrtionl number.

6 6 chpter 0 The Rel Numbers We hve just seen tht, which is the positive rel number whose squre equls, is n irrtionl number. The rel numbers π nd e, which we will encounter in lter chpters, re lso irrtionl numbers. Once we hve found one irrtionl number, finding others is much esier, s shown in the exmple below. exmple 1 The ttitude of the ncient Greeks towrd irrtionl numbers persists in our everydy use of irrtionl to men not bsed on reson. Show tht 3 + is n irrtionl number. Suppose 3 + is rtionl number. Becuse = (3 + ) 3, this implies tht is the difference of two rtionl numbers, which implies tht is rtionl number, which is not true. Thus our ssumption tht 3 + is rtionl number must be flse. In other words, 3 + is n irrtionl number. The next exmple provides nother illustrtion of how to use one irrtionl number to generte nother irrtionl number. exmple Show tht 8 is n irrtionl number. Suppose 8 is rtionl number. Becuse = 8 8, this implies tht is the quotient of two rtionl numbers, which implies tht is rtionl number, which is not true. Thus our ssumption tht 8 is rtionl number must be flse. In other words, 8 is n irrtionl number. problems The problems in this section my be hrder thn typicl problems found in the rest of this book. 1. Show tht is n irrtionl number.. Show tht 5 is n irrtionl number. 3. Show tht 3 is n irrtionl number. 4. Show tht 3 is n irrtionl number Show tht is n irrtionl number. 6. Explin why the sum of rtionl number nd n irrtionl number is n irrtionl number. 7. Explin why the product of nonzero rtionl number nd n irrtionl number is n irrtionl number. 8. Suppose t is n irrtionl number. Explin why 1 is lso n irrtionl number. t 9. Give n exmple of two irrtionl numbers whose sum is n irrtionl number. 10. Give n exmple of two irrtionl numbers whose sum is rtionl number. 11. Give n exmple of three irrtionl numbers whose sum is rtionl number. 1. Give n exmple of two irrtionl numbers whose product is n irrtionl number. 13. Give n exmple of two irrtionl numbers whose product is rtionl number.

7 section 0. Algebr of the Rel Numbers 7 0. Algebr of the Rel Numbers By the end of this section you should section objectives recll how to mnipulte lgebric expressions using the commuttive, ssocitive, nd distributive properties; understnd the order of lgebric opertions nd the role of prentheses; recll the crucil lgebric identities involving dditive inverses nd multiplictive inverses. The opertions of ddition, subtrction, multipliction, nd division extend from the rtionl numbers to the rel numbers. We cn dd, subtrct, multiply, nd divide ny two rel numbers nd sty within the system of rel numbers, gin with the exception tht division by 0 is prohibited. In this section we review the bsic lgebric properties of the rel numbers. Becuse this mteril should indeed be review, no effort hs been mde to show how some of these properties follow from others. Insted, this section focuses on highlighting key properties tht should become so fmilir to you tht you cn use them comfortbly nd without effort. Exercises woven throughout this book hve been designed to shrpen your lgebric mnipultion skills s we cover other topics. Commuttivity nd Associtivity Commuttivity is the forml nme for the property stting tht order does not mtter in ddition nd multipliction: Commuttivity + b = b + nd b = b Here (nd throughout this section), b, nd other vribles denote either rbitrry rel numbers or expressions tht tke on vlues tht re rel numbers. For exmple, the commuttivity of ddition implies tht x + x 5 = x 5 + x. Neither subtrction nor division is commuttive becuse order does mtter for those opertions. For exmple, , nd 6 6. Associtivity is the forml nme for the property stting tht grouping does not mtter in ddition nd multipliction: Associtivity ( + b) + c = + (b + c) nd (b)c = (bc) Expressions inside prentheses should be clculted before further computtion. For exmple, ( + b) + c should be clculted by first dding nd b, nd then dding tht sum to c. The ssocitive property of ddition

8 3 chpter 0 The Rel Numbers sserts tht this number will be the sme s + (b + c), which should be clculted by first dding b nd c, nd then dding tht sum to. Becuse of the ssocitive property of ddition, we cn dispense with prentheses when dding three or more numbers, writing expressions such s + b + c + d without worrying bout how the terms re grouped. Similrly, becuse of the ssocitive property of multipliction we do not need prentheses when multiplying together three or more numbers. Thus we cn write expressions such s bcd without specifying the order of multipliction or the grouping. Neither subtrction nor division is ssocitive becuse the grouping does mtter for those opertions. For exmple, (9 6) = 3 = 1, but 9 (6 ) = 9 4 = 5, which shows tht subtrction is not ssocitive. Becuse subtrction is not ssocitive, we need wy to evlute expressions tht re written without prentheses. The stndrd prctice is to evlute subtrctions from left to right unless prentheses indicte otherwise. For exmple, 9 6 should be interpreted to men (9 6), which equls 1. The Order of Algebric Opertions Consider the expression This expression contins no prentheses to guide us to which opertion should be performed first. Should we first dd nd 3, nd then multiply the result by 7? If so, we would interpret the expression bove s ( + 3) 7, which equls 35. Or to evlute Note tht ( + 3) 7 does not equl + (3 7). Thus the order of these opertions does mtter. should we first multiply together 3 nd 7, nd then dd to tht result. If so, we would interpret the expression bove s + (3 7), which equls 3. So does equl ( + 3) 7or+ (3 7)? The nswer to this question depends on custom rther thn nything inherent in the mthemticl sitution. Every mthemticlly literte person would interpret to men + (3 7). In other words, people in the modern er hve dopted the convention tht multiplictions should be performed before dditions unless prentheses dictte otherwise. You need to become ccustomed to this

9 section 0. Algebr of the Rel Numbers 9 convention, which will be used throughout this course nd ll your further courses tht use mthemtics. Multipliction nd division before ddition nd subtrction Unless prentheses indicte otherwise, products nd quotients re clculted before sums nd differences. Thus, for exmple, +bc is interpreted to men +(bc), lthough lmost lwys we dispense with the prentheses nd just write + bc. As nother illustrtion of the principle bove, consider the expression 4m + 3n + 11(p + q). The correct interprettion of this expression is tht 4 should be multiplied by m, 3 should be multiplied by n, 11 should be multiplied by p + q, nd then the three numbers 4m, 3n, nd 11(p + q) should be dded together. In other words, the expression bove equls (4m) + (3n) + ( 11(p + q) ). The three newly dded sets of prentheses in the expression bove re unnecessry, lthough it is not incorrect to include them. However, the version of the sme expression without the unnecessry prentheses is clener nd esier to red. When prentheses re enclosed within prentheses, expressions in the innermost prentheses re evluted first. Evlute inner prentheses first In n expression with prentheses inside prentheses, evlute the innermost prentheses first nd then work outwrd. The size of prentheses is sometimes used s n optionl visul id to indicte the order of opertions. Smller prentheses should be used for Stephen more inner prenthe- F Austin ses. Thus expressions enclosed in smller prentheses should usully be evluted before expressions enclosed in lrger prentheses. Evlute the expression ( 6 + 3(1 + 4) ). Here the innermost prentheses surround 1+4. Thus strt by evluting tht expression: ( ) (1 + 4) = ( ). }{{} 5 Now to evlute the expression , first evlute 3 5, getting 15, then dd tht to 6, getting 1. Multiplying by completes our evlution of this expression: ( ) (1 + 4) = 4. }{{} 5 }{{} 1 exmple 1

10 10 chpter 0 The Rel Numbers The Distributive Property The distributive property connects ddition nd multipliction, converting product with sum into sum of two products. Distributive property (b + c) = b + c Becuse multipliction is commuttive, the distributive property cn lso be written in the lterntive form ( + b)c = c + bc. The distributive property provides the justifiction for fctoring expressions. Sometimes you will need to use the distributive property to trnsform n expression of the form (b+c) into b+c, nd sometimes you will need to use the distributive property in the opposite direction, trnsforming n expression of the form b +c into (b +c). Becuse the distributive property is usully used to simplify n expression, the direction of the trnsformtion depends on the context. The next exmple shows the use of the distributive property in both directions. exmple Simplify the expression (3m + x) + 5x. First use the distributive property to trnsform (3m + x) into 6m + x: (3m + x) + 5x = 6m + x + 5x. Now use the distributive property gin, but in the other direction, to trnsform x + 5x to ( + 5)x: 6m + x + 5x = 6m + ( + 5)x = 6m + 7x. Putting ll this together, we hve used the distributive property (twice) to trnsform (3m + x) + 5x into the simpler expression 6m + 7x. One of the most common lgebric mnipultions involves expnding product of sums, s in the following exmple. exmple 3 Expnd ( + b)(c + d).

11 section 0. Algebr of the Rel Numbers 11 Think of (c + d) s single number nd then pply the distributive property to the expression bove, getting ( + b)(c + d) = (c + d) + b(c + d). Now pply the distributive property twice more, getting ( + b)(c + d) = c + d + bc + bd. If you re comfortble with the distributive property, there is no need to memorize the lst formul from the exmple bove, becuse you cn lwys derive it gin. Furthermore, by understnding how the identity bove ws obtined, you should hve no trouble finding formuls for more complicted expressions such s ( + b)(c + d + t). An importnt specil cse of the identity bove occurs when c = nd d = b. In tht cse we hve After you use this formul severl times, it will become so fmilir tht you cn use it routinely without needing to puse. Note tht every term in the first set of prentheses is multiplied by every term in the second set of prentheses. ( + b)( + b) = + b + b + b, which, with stndrd use of commuttivity, becomes the identity ( + b) = + b + b. Additive Inverses nd Subtrction The dditive inverse of rel number is the number such tht + ( ) = 0. The connection between subtrction nd dditive inverses is cptured by the identity b = + ( b). In fct, the eqution bove cn be tken s the definition of subtrction. You need to be comfortble using the following identities tht involve dditive inverses nd subtrction: Identities involving dditive inverses nd subtrction ( ) = ( + b) = b ( )( b) = b ( )b = ( b) = (b) ( b)c = c bc (b c) = b c

12 1 chpter 0 The Rel Numbers Be sure to distribute the minus signs correctly when using the distributive property, s in the exmple below. exmple 4 Expnd ( + b)( b). Strt by thinking of ( + b) s single number nd pplying the distributive property. Then pply the distributive property twice more, pying creful ttention to the minus signs: ( + b)( b) = ( + b) ( + b)b = + b b b = b You need to become sufficiently comfortble with the following identities so tht you cn use them with ese. Identities rising from the distributive property ( + b) = + b + b ( b) = b + b ( + b)( b) = b exmple 5 Without using clcultor, evlute = (40 + 3)(40 3) = 40 3 = = 1591 Multiplictive Inverses nd Division The multiplictive inverse of b is sometimes clled the reciprocl of b. The multiplictive inverse of rel number b 0 is the number 1 b such tht b 1 b = 1. The connection between division nd multiplictive inverses is cptured by the identity b = 1 b. In fct, the eqution bove cn be tken s the definition of division. You need to be comfortble using the following identities tht involve multiplictive inverses nd division:

13 section 0. Algebr of the Rel Numbers 13 Identities involving multiplictive inverses nd division b + c d + bc = d bd b c d = c bd c d = c d b c = c b b = b = b b = b Let s look t these identities bit more crefully. In ll the identities bove, we ssume tht none of the denomintors equls 0. The first identity bove gives formul for dding two frctions. The second identity bove sttes tht the product of two frctions cn be computed by multiplying together the numertors nd multiplying together the denomintors. Note tht the formul for dding frctions is more complicted thn the formul for multiplying frctions. The third identity bove, when used to trnsform c d into c d, is the usul simplifiction of cnceling common fctor from the numertor nd denomintor. When used in the other direction to trnsform c c d into d, the third identity bove becomes the fmilir procedure of multiplying the numertor nd denomintor by the sme fctor. In the fourth identity bove, the size of the frction brs re used to indicte tht b c should be interpreted to men /(b/c). This identity gives the key to unrveling frctions tht involve frctions, s shown in the following exmple. Never, ever, mke the mistke of thinking tht b + c +c equls d b+d. Simplify the expression b c d. exmple 6 The size of the frction brs indictes tht the expression to be simplified is (/b)/(c/d). Dividing by c d is the sme s multiplying by d. Thus we hve c b c d = b d c = d bc. When fced with complicted expressions involving frctions tht re themselves frctions, remember tht division by frction is the sme s multipliction by the frction flipped over.

14 14 chpter 0 The Rel Numbers exercises For Exercises 1 4, determine how mny different vlues cn rise by inserting one pir of prentheses into the given expression ( m + 4(n + 5p) ) + 6n m n For Exercises 5 18, expnd the given expression. 5. (x y)(z + w t) x y (x + y r )(z + w t) 7. (x + 3) x y + 8. (3b + 5) 9. (c 7) 10. (4 5) 11. (x + y + z) 1. (x 5y 3z) 13. (x + 1)(x )(x + 3) 14. (y )(y 3)(y + 5) 15. ( + )( )( + 4) 16. (b 3)(b + 3)(b + 9) 17. xy(x + y) ( 1 x 1 ) y 18. z(z ) ( 1 z + 1 ) m n m n ( x x y y y ) x 1 ( 1 y x y 1 ) x + y (x + ) x 1 x+ 1 x x y z x+ For Exercises 19 40, simplify the given expression s much s possible (m + 3n) + 7m x 4 y+3 y 3 x+4 problems Some problems require considerbly more thought thn the exercises. Unlike exercises, problems usully hve more thn one correct nswer. 41. Explin how you could show tht = 499 in your hed by using the identity ( + b)( b) = b. 4. Show tht 3 + b 3 + c 3 3bc = ( + b + c)( + b + c b bc c). 43. Give n exmple to show tht division does not stisfy the ssocitive property. 44. The sles tx in Sn Frncisco is 8.5%. Diners in Sn Frncisco often compute 17% tip on their before-tx resturnt bill by simply doubling the sles tx. For exmple, $64 dollr food nd drink bill would come with sles tx of $5.44; doubling tht mount would led to 17% tip of $10.88 (which might be rounded up to $11). Explin why this technique is n ppliction of the ssocitivity of multipliction.

15 section 0. Algebr of the Rel Numbers A quick wy to compute 15% tip on resturnt bill is first to compute 10% of the bill (by shifting the deciml point) nd then dd hlf of tht mount for the totl tip. For exmple, 15% of $43 resturnt bill is $ $.15, which equls $6.45. Explin why this technique is n ppliction of the distributive property. 46. The first letters of the phrse Plese excuse my der Aunt Slly re used by some people to remember the order of opertions: prentheses, exponentition (which we will discuss in lter chpter), multipliction, division, ddition, subtrction. Mke up ctchy phrse tht serves the sme purpose but with exponentition excluded. 47. () Verify tht (b) = From the exmple bove you my be tempted to think tht b c d = c b d provided none of the denomintors equls 0. Give n exmple to show tht this is not true. worked-out s to Odd-numbered Exercises Do not red these worked-out s before first struggling to do the exercises yourself. Otherwise you risk the dnger of mimicking the techniques shown here without understnding the ides. Best wy to lern: Crefully red the section of the textbook, then do ll the odd-numbered exercises (even if they hve not been ssigned) nd check your nswers here. If you get stuck on n exercise, rered the section of the textbook then try the exercise gin. If you re still stuck, then look t the workedout here. For Exercises 1 4, determine how mny different vlues cn rise by inserting one pir of prentheses into the given expression Here re the possibilities: 19( 1 8 ) = ( 1 8) = 38 19( 1) 8 = 38 (19 1) 8 = (8 ) = 1 19 (1 8) = (1 8 ) = ( ) = ( 8) = ( 8 ) = 139 Other possible wys to insert one pir of prentheses led to vlues lredy included in the list bove. For exmple, (19 1 8) = 3. Thus ten vlues re possible; they re 418, 38, 38, 3, 1, 13, 17, 3, 113, nd Here re the possibilities: ( ) = ( ) = 40 (6 + 3) = (4 + 5 ) = (4 + 5) = 60 Other possible wys to insert one pir of prentheses led to vlues lredy included in the list bove. For exmple, ( ) = 46.

16 4 chpter 0 The Rel Numbers Thus five vlues re possible; they re 8, 40, 46, 48, nd 60. For Exercises 5 18, expnd the given expression. 5. (x y)(z + w t) (x y)(z + w t) 7. (x + 3) 9. (c 7) 11. (x + y + z) = x(z + w t) y(z + w t) = xz + xw xt yz yw + yt (x + 3) = (x) + (x) (x + y + z) = 4x + 1x + 9 (c 7) = (c) (c) = 4c 8c + 49 = (x + y + z)(x + y + z) = x(x + y + z) + y(x + y + z) + z(x + y + z) = x + xy + xz + yx + y + yz + zx + zy + z = x + y + z + xy + xz + yz (x + 1)(x )(x + 3) 15. ( + )( )( + 4) = ( (x + 1)(x ) ) (x + 3) = (x x + x )(x + 3) = (x x )(x + 3) = x 3 + 3x x 3x x 6 = x 3 + x 5x 6 ( + )( )( + 4) = ( ( + )( ) ) ( + 4) 17. xy(x + y) ( 1 x 1 ) y ( 1 xy(x + y) x 1 ) y = ( 4)( + 4) = 4 16 ( y = xy(x + y) xy x ) xy ( Stephen y x ) F Austin = xy(x + y) xy = (x + y)(y x) = y x For Exercises 19 40, simplify the given expression s much s possible (m + 3n) + 7m 4(m + 3n) + 7m = 8m + 1n + 7m = 15m + 1n 13. (x + 1)(x )(x + 3) = = =

17 section 0. Algebr of the Rel Numbers = = 7 13 = m n = = = x y 4 = 5 x y 4 x x + 3 = = = 5 + (y 4)(x + 3) 5(x + 3) 10 + yx + 3y 4x 1 5(x + 3) xy 4x + 3y 5(x + 3) m n = m + 1 n n + 3 n ( x x y y y ) x = (m + 1)n + 3 n = mn + n + 6 n 1 ( x x y y y ) = 1 ( x x x y y x x y x y ) y = 1 ( x y ) x y xy = 1 ( (x + y)(x y) ) x y xy = = = (x + ) x = x + y xy = (x + ) x = x + x + x m = x + = x m = m x y 4 5 = m = = 4m m x y z x+ x y z x+ = x y x + z = x 4 yz

18 18 chpter 0 The Rel Numbers 0.3 Inequlities From now on, number mens rel number unless otherwise stted. section objectives By the end of this section you should recll the lgebric properties involving positive nd negtive numbers; understnd inequlities; be ble to use intervl nottion for the four types of intervls; be ble to use intervl nottion involving nd ; be ble to work with unions of intervls; be ble to mnipulte nd interpret expressions involving bsolute vlue. Positive nd Negtive Numbers The words positive nd negtive hve mny uses in English in ddition to their mthemticl mening. Some of these uses, such s in the phrse photogrphic negtive, re relted to the mthemticl mening. Positive nd negtive numbers A number is clled positive if it is right of 0 on the rel line. A number is clled negtive if it is left of 0 on the rel line. Every number is either right of 0, left of 0, or equls 0. Thus every number is either positive, negtive, or negtive numbers positive numbers All of the following properties should lredy be fmilir to you. Exmple: + 3 = 5 ( ) + ( 3) = 5 is negtive ( ) is positive 3 = 6 ( ) ( 3) = 6 ( 3) = 6 1 is positive 1 is negtive Algebric properties of positive nd negtive numbers The sum of two positive numbers is positive. The sum of two negtive numbers is negtive. The dditive inverse of positive number is negtive. The dditive inverse of negtive number is positive. The product of two positive numbers is positive. The product of two negtive numbers is positive. The product of positive number nd negtive number is negtive. The multiplictive inverse of positive number is positive. The multiplictive inverse of negtive number is negtive.

19 section 0.3 Inequlities 19 Lesser nd Greter We sy tht number is less thn number b, written <b,if is left of b on the rel line. Equivlently, <bif nd only if b is positive. In prticulr, b is positive if nd only if 0 <b. <b. b We sy tht is less thn or equl to b, written b, if<bor = b. Thus the sttement x<4 is true if x equls 3 but flse if x equls 4, wheres the sttement x 4 is true if x equls 3 nd lso true if x equls 4. We sy tht b is greter thn, written b>,ifb is right of on the rel line. Thus b>mens the sme s <b. Similrly, we sy tht b is greter thn or equl to, written b, ifb>or b =. Thus b mens the sme s b. We now begin discussion of series of simple but crucil properties of inequlities. The first property we will discuss is clled trnsitivity. Trnsitivity If <bnd b<c, then <c. To see why trnsitivity holds, suppose <bnd b<c. Then is left of b on the rel line nd b is left of c. This implies tht is left of c, which mens tht <c; see the figure below. For exmple, from the inequlities 15 < 4 nd 4 < 1 5 we cn conclude tht 15 < 1 5 Stephen. F Austin b c Trnsitivity: <bnd b<cimplies tht <c. Often multiple inequlities re written together s single string of inequlities. Thus <b<cmens the sme thing s <bnd b<c. Our next result shows tht we cn dd inequlities. Addition of inequlities If <bnd c<d, then + c<b+ d. To see why this is true, note tht if <bnd c<d, then b nd d c re positive numbers. Becuse the sum of two positive numbers is positive, this implies tht (b ) + (d c) is positive. In other words, (b + d) ( + c) is positive. This mens tht + c<b+ d, s desired. The next result sttes tht we cn multiply both sides of n inequlity by positive number nd preserve the inequlity. However, if both sides of n inequlity re multiplied by negtive number, then the direction of the inequlity must be reversed. For exmple, from the inequlities 8 < 3 nd 4 < 17 we cn conclude tht <

20 0 chpter 0 The Rel Numbers Multipliction of n inequlity For exmple, from the inequlity 7 < 8 we cn conclude tht 3 7 < 3 8 nd ( 3) 7 >( 3) 8. Suppose <b. If c>0, then c < bc. If c<0, then c > bc. To see why this is true, first suppose c>0. We re ssuming tht <b, which mens tht b is positive. Becuse the product of two positive numbers is positive, this implies tht (b )c is positive. In other words, bc c is positive, which mens tht c < bc, s desired. Now consider the cse where c < 0. We re still ssuming tht <b, which mens tht b is positive. Becuse the product of positive number nd negtive number is negtive, this implies tht (b )c is negtive. In other words, bc c is negtive, which mens tht c > bc, s desired. An importnt specil cse of the result bove is obtined by setting c = 1, which gives the following result: For exmple, from the inequlity < 3 we cn conclude tht > 3. Additive inverse nd inequlities If <b, then > b. In other words, the direction of n inequlity must be reversed when tking dditive inverses of both sides. The next result shows tht the direction of n inequlity must lso be reversed when tking multiplictive inverses of both sides, unless one side is negtive nd the other side is positive. For exmple, from the inequlity < 3 we cn conclude tht 1 > 1 3. Multiplictive inverse nd inequlities Suppose <b. If nd b re both positive or both negtive, then 1 > 1 b. If <0 <b, then 1 < 1 b. To see why this is true, first suppose nd b re both positive or both negtive. In either cse, b is positive. Thus 1 b > 0. Thus we cn multiply both sides of the inequlity <bby 1 b, preserving the direction of the inequlity. This gives 1 b <b 1 b, which is the sme s 1 b < 1, or equivlently 1 > 1 b, s desired. The cse where <0 <bis even esier. In this cse 1 is negtive nd 1 b is positive. Thus 1 < 1 b, s desired.

21 section 0.3 Inequlities 1 Intervls We begin this subsection with n imprecise definition. Set A set is collection of objects. This definition is imprecise becuse the words collection nd objects re vgue. The collection of positive numbers is n exmple of set, s is the collection of odd negtive integers. Most of the sets considered in this book re collections of rel numbers, which t lest removes some of the vgueness from the word objects. If set contins only finitely mny objects, then the objects in the set cn be explicitly displyed between the symbols {}. For exmple, the set consisting of the numbers 4, 17 7, nd cn be denoted by {4, 17 7, }. Sets cn lso be denoted by property tht chrcterizes objects of the set. For exmple, the set of rel numbers greter thn cn be denoted by {x : x>}. Here the nottion {x :...} should be red to men the set of rel numbers x such tht nd then whtever follows. There is no prticulr x here. The vrible is simply convenient device to describe property, nd the symbol used for the vrible does not mtter. Thus {x : x>} nd {y : y>} nd {t : t>} ll denote the sme set, which cn lso be described (without mentioning ny vribles) s the set of rel numbers greter thn. A specil type of set occurs so often in mthemtics tht it gets its own nme, which is given by the following definition. Intervl An intervl is set of rel numbers tht contins ll numbers between ny two numbers in the set. For exmple, the set of positive numbers is n intervl becuse ll numbers between ny two positive numbers re positive. As nonexmple, the set of integers is not n intervl becuse 0 nd 1 re in this set, but 1, which is between 0 nd 1, is not in this set. As nother nonexmple, the set of rtionl numbers is not n intervl, becuse 1 nd re in this set, but, which is between 1 nd, is not in this set. Intervls re so useful in mthemtics tht specil nottion hs been designed for them. Suppose nd b re numbers with <b. We define the following four intervls with endpoints nd b:

22 7π chpter 0 The Rel Numbers Intervls The open intervl (, b) with endpoints nd b is the set of numbers between nd b, not including either endpoint: (, b) ={x : <x<b}. The definition of [, b] lso mkes sense when = b; the intervl [, ] consists of the single number. The closed intervl [, b] with endpoints nd b is the set of numbers between nd b, including both endpoints: [, b] ={x : x b}. The hlf-open intervl [, b) with endpoints nd b is the set of numbers between nd b, including but not including b: [, b) ={x : x<b}. The term hlf-closed would mke s much sense s hlf-open. The hlf-open intervl (, b] with endpoints nd b is the set of numbers between nd b, including b but not including : (, b] ={x : <x b}. With this nottion, prenthesis indictes tht the corresponding endpoint is not included in the set, nd stright brcket indictes tht the corresponding endpoint is included in the set. Thus the intervl (3, 7] includes the numbers 4, 17, 5.49, nd the endpoint 7 (long with mny other numbers), but does not include the numbers or 9 or the endpoint 3. Sometimes we need to use intervls tht extend rbitrrily fr to the left or to the right on the rel number line. Suppose is rel number. We define the following four intervls with endpoint : Intervls Exmple: (0, ) denotes the set of positive numbers. The intervl (, ) is the set of numbers greter thn : (, ) ={x : x>}. The intervl [, ) is the set of numbers greter thn or equl to : [, ) ={x : x }. Exmple: (, 0) denotes the set of negtive numbers. The intervl (,) is the set of numbers less thn : (,)={x : x<}. The intervl (,] is the set of numbers less thn or equl to : (,]={x : x }.

23 section 0.3 Inequlities 3 Here the symbol, clled infinity, should be thought of simply s nottionl convenience. Neither nor is rel number; these symbols hve no mening in this context other thn s nottionl shorthnd. For exmple, the intervl (, ) is defined to be the set of rel numbers greter thn (note tht is not mentioned in this definition). The nottion (, ) is often used becuse writing (, ) is esier thn writing {x : x>}. As before, prenthesis indictes tht the corresponding endpoint is not included in the set, nd stright brcket indictes tht the corresponding endpoint is included in the set. Thus the intervl (, ) does not include the endpoint, but the intervl [, ) does include the endpoint. Both of the intervls (, ) nd [, ) include.5 nd (long with mny other numbers); neither of these intervls includes 1.5 or 857. There do not exist intervls with closed brcket djcent to or. For exmple, [, ] nd [, ] do not mke sense becuse the closed brckets indicte tht both endpoints should be included. The symbols nd cn never be included in set of rel numbers becuse these symbols do not denote rel numbers. In lter chpters we will occsionlly find it useful to work with the union of two intervls. Here is the definition of union: Some books use the nottion (, ) to denote the set of rel numbers. Union The union of two sets A nd B, denoted A B, is the set of objects tht re contined in t lest one of the sets A nd B. Thus A B consists of the objects (usully numbers) tht belong either to A or to B or to both A nd B. Similrly, the union of three or more sets is the collection of objects tht re con- tined in t lest one of the sets. Write (1, 5) (3, 7] s n intervl. exmple 1 As cn be seen from the figure here, every number in the intervl (1, 7] is either in (1, 5) or is in (3, 7] or is in both (1, 5) nd (3, 7]. The figure shows tht (1, 5) (3, 7] = (1, 7] The next exmple goes in the other direction, strting with set nd then writing it s union of intervls. Write the set of nonzero rel numbers s the union of two intervls. exmple The set of nonzero rel numbers is the union of the set of negtive numbers nd the set of positive numbers. In other words, the set of nonzero rel numbers equls (, 0) (0, ).

24 4 chpter 0 The Rel Numbers Absolute Vlue The bsolute vlue of number is its distnce from 0; here we re thinking of numbers s points on the rel line. For exmple, the bsolute vlue of 3 equls 3, s cn be seen in the figure below. More interestingly, the bsolute vlue of 3 equls The bsolute vlue of number is its distnce to 0. The bsolute vlue of number b is denoted by b. Thus 3 = 3 nd 3 = 3. Here is the forml definition of bsolute vlue: Absolute vlue The bsolute vlue of number b, denoted b, is defined by b if b 0 b = b if b<0. (x + y) = x + y regrdless of the vlue of x + y, s you re sked to explin in Problem 56. For exmple, 3 < 0, nd thus by the formul bove 3 equls ( 3 ), which equls 3. The concept of bsolute vlue is firly simple just strip wy the minus sign from ny number tht hppens to hve one. However, this rule cn be pplied only to numbers, not to expressions whose vlue is unknown. For exmple, if we encounter the expression (x + y), we cnnot simplify this expression to x + y unless we know tht x + y 0. If x + y hppens to be negtive, then (x + y) = (x + y); stripping wy the negtive sign would be incorrect in this cse. Inequlities involving bsolute vlues cn be written without using n bsolute vlue, s shown in the following exmple. exmple 3 () Write the inequlity x < without using n bsolute vlue. (b) Write the set {x : x < } s n intervl. () A number hs bsolute vlue less thn if only nd only if its distnce from 0 is less thn, nd this hppens if nd only if the number is between nd. Hence the inequlity x < could be written s <x<. (b) The inequlity bove implies tht the set {x : x < } equls the open intervl (, ).

25 section 0.3 Inequlities 5 In the next exmple, we end up with n intervl not centered t 0. () Write the inequlity x 5 < 1 without using n bsolute vlue. (b) Write the set {x : x 5 < 1} s n intervl. The set exmple 4 () The bsolute vlue of number is less thn 1 precisely when the number is between 1 nd 1. Thus the inequlity x 5 < 1 is equivlent to 1 <x 5 < 1. {x : x 5 < 1} is the set of points on the rel line whose distnce to 5 is less thn 1. Adding 5 to ll three prts of the inequlity bove trnsforms it to the inequlity 4 <x<6. (b) The inequlity bove implies tht the set {x : x 5 < 1} equls the open intervl (4, 6). In the next exmple, we del with slightly more bstrct sitution, using symbols rther thn specific numbers. You should begin to get comfortble working in such situtions. To get good understnding of n bstrct piece of mthemtics, strt by looking t n exmple using concrete numbers, s in Exmple 4, before going on to more bstrct setting, s in Exmple 5. Suppose b is rel number nd h>0. () Write the inequlity x b < h without using n bsolute vlue. (b) Write the set {x : x b <h} s n intervl. The set exmple 5 () The bsolute vlue of number is less thn h precisely when the number is between h nd h. Thus the inequlity x b <his equivlent to h <x b<h. {x : x b <h} is the set of points on the rel line whose distnce to b is less thn h. Adding b to ll three prts of the inequlity bove trnsforms it to the inequlity b h<x<b+ h. (b) The inequlity bove implies tht the set {x : x b <h} equls the open intervl (b h, b + h). bh b bh {x : x b <h} is the open intervl of length h centered t b.

26 6 chpter 0 The Rel Numbers Equtions involving bsolute vlues must often be solved by considering multiple possibilities. Here is simple exmple: exmple 6 Find ll numbers t such tht 3t 4 =10. The eqution 3t 4 =10 implies tht 3t 4 = 10 or 3t 4 = 10. Solving these equtions for t gives t = 14 or t =. Substituting these vlues for t 3 bck into the originl eqution shows tht both 14 nd re indeed s. 3 A more complicted exmple would sk for ll numbers x such tht The worked-out to Exercise 5 shows how to del with this sort of sitution. exercises x 3 + x 4 =9. To find the s to this eqution, think of the set of rel numbers s the union of the three intervls (, 3), [3, 4), nd [4, ) nd consider wht the eqution bove becomes for x in ech of those three intervls. In Exercises 1 6, find ll numbers x stisfying the given eqution. 1. x 6 = x+ = 5 x 4. 5x + 8 =19 5. x 3 + x 4 =9 3. x+1 = x 1 6. x x =7 In Exercises 7 16, write ech union s single intervl. 7. [, 7) [5, 0) 8. [ 8, 3) [ 6, 1) 9. [, 8] ( 1, 4) 10. ( 9, ) [ 7, 5] 11. (3, ) [, 8] 1. (, 4) (, 6] 13. (, 3) [ 5, ) 14. (, 6] ( 8, 1) 15. ( 3, ) [ 5, ) 16. (, 10] (, 8] 17. Give four exmples of pirs of rel numbers nd b such tht + b = nd + b = Give four exmples of pirs of rel numbers nd b such tht + b = 3 nd + b = 11. In Exercises 19 30, write ech set s n intervl or s union of two intervls. 19. {x : x 4 < 1 10 } 0. {x : x + < } 1. {x : x + 4 < ε }; here ε>0 [Mthemticins often use the Greek letter ε, which is clled epsilon, to denote smll positive number.]. {x : x < ε }; here ε> {y : y <ε}; here ε>0 4. {y : y + b <ε}; here ε>0 5. {x : 3x < 1 4 } 6. {x : 4x 3 < 1 5 } 7. {x : x > } 8. {x : x > 9} 9. {x : x 5 3} 30. {x : x + 6 } The intersection of two sets of numbers consists of ll numbers tht re in both sets. If A nd B re sets, then their intersection is denoted by A B. In Exercises 31 40, write ech intersection s single intervl. 31. [, 7) [5, 0) 3. [ 8, 3) [ 6, 1) 33. [, 8] ( 1, 4) 34. ( 9, ) [ 7, 5] 35. (3, ) [, 8] 36. (, 4) (, 6] 37. (, 3) [ 5, ) 38. (, 6] ( 8, 1) 39. ( 3, ) [ 5, ) 40. (, 10] (, 8]

27 section 0.3 Inequlities 7 problems 41. Suppose nd b re numbers. Explin why either <b, = b, or>b. 4. Show tht if <bnd c d, then +c <b+d. 43. Show tht if b is positive number nd <b, then b < + 1 b In contrst to Problem 47 in Section 0., show tht there do not exist positive numbers, b, c, nd d such tht b + c d = + c b + d. 45. () True or flse: If <bnd c<d, then c b<d. (b) Explin your nswer to prt (). This mens tht if the nswer to prt () is true, then you should explin why c b<d whenever <bnd c<d;if the nswer to prt () is flse, then you should give n exmple of numbers, b, c, nd d such tht <bnd c<dbut c b d. 46. () True or flse: If <bnd c<d, then c < bd. (b) Explin your nswer to prt (). This mens tht if the nswer to prt () is true, then you should explin why c < bd whenever <bnd c<d;if the nswer to prt () is flse, then you should give n exmple of numbers, b, c, nd d such tht <bnd c<dbut c bd. 47. () True or flse: (b) If 0 <<bnd 0 <c<d, then d < b c. Explin your nswer to prt (). This mens tht if the nswer to prt () is true, then you should explin why d < b whenever 0 <<bnd 0 <c<d; c if the nswer to prt () is flse, then you should give n exmple of numbers, b, c, nd d such tht 0 <<bnd 0 <c<dbut d b c. 48. Explin why every open intervl contining 0 contins n open intervl centered t Give n exmple of n open intervl nd closed intervl whose union equls the intervl (, 5). 50. Give n exmple of n open intervl nd closed intervl whose intersection equls the intervl (, 5). 51. Give n exmple of n open intervl nd closed intervl whose union equls the intervl [ 3, 7]. 5. Give n exmple of n open intervl nd closed intervl whose intersection equls the intervl [ 3, 7]. 53. Explin why the eqution 8x 3 = hs no s. 54. Explin why = for every rel number. 55. Explin why b = b for ll rel numbers nd b. 56. Explin why = for ll rel numbers. 57. Explin why = b b for ll rel numbers nd b (with b 0). 58. () Show tht if 0 nd b 0, then + b = + b. (b) (c) (d) (e) Show tht if 0 nd b<0, then + b + b. Show tht if <0 nd b 0, then + b + b. Show tht if <0 nd b<0, then + b = + b. Explin why the previous four items imply tht + b + b for ll rel numbers nd b.

28 8 chpter 0 The Rel Numbers 59. Show tht if nd b re rel numbers such tht + b < + b, 60. Show tht b b then b < 0. for ll rel numbers nd b. worked-out s to Odd-numbered Exercises In Exercises 1 6, find ll numbers x stisfying the given eqution. 1. x 6 =11 The eqution x 6 = 11 implies tht x 6 = 11 or x 6 = 11. Solving these equtions for x gives x = 17 or x = x+1 = x 1 The eqution x+1 x 1 = implies tht x+1 x+1 = or =. Solving these equtions for x gives x = 3orx = 1 3 x 1 x x 3 + x 4 =9 First, consider numbers x such tht x 4. In this cse, we hve x 3 0 nd x 4 0, which implies tht x 3 =x 3 nd x 4 =x 4. Thus the originl eqution becomes x 3 + x 4 = 9, which cn be rewritten s x 7 = 9, which cn esily be solved to yield x = 8. Substituting 8 for x in the originl eqution shows tht x = 8 is indeed (mke sure you do this check). Second, consider numbers x such tht x < 3. In this cse, we hve x 3 < 0 nd x 4 < 0, which implies tht x 3 =3 x nd x 4 = 4 x. Thus the originl eqution becomes 3 x + 4 x = 9, which cn be rewritten s 7 x = 9, which cn esily be solved to yield x = 1. Substituting 1 for x in the originl eqution shows tht x = 1 is indeed (mke sure you do this check). Third, we need to consider the only remining possibility, which is tht 3 x<4. In this cse, we hve x 3 0 nd x 4 < 0, which implies tht x 3 =x 3 nd x 4 =4 x. Thus the originl eqution becomes x x = 9, which cn be rewritten s 1 = 9, which holds for no vlues of x. Thus we cn conclude tht 8 nd 1 re the only vlues of x tht stisfy the originl eqution. In Exercises 7 16, write ech union s single intervl. 7. [, 7) [5, 0) The first intervl is the set {x : x<7}, which includes the left endpoint but does not include the right endpoint 7. The second intervl is the set {x :5 x<0}, which includes the left endpoint 5 but does not include the right endpoint 0. The set of numbers tht re in t lest one of these sets equls {x : x<0}, s cn be seen in the figure below: 5 7 Thus [, 7) [5, 0) = [, 0). 9. [, 8] ( 1, 4) 0 The first intervl is the set {x : x 8}, which includes both endpoints. The second intervl is the set {x : 1 <x<4}, which does not include either endpoint. The set of numbers tht re in t lest one of these sets equls {x : x 8}, s cn be seen in the following figure:

29 section 0.3 Inequlities 9 8 t lest one of these sets equls {x : 5 x}, s cn be seen in the figure below: Thus [, 8] ( 1, 4) = [, 8]. 11. (3, ) [, 8] The first intervl is the set {x :3< x}, which does not include the left endpoint nd which hs no right endpoint. The second intervl is the set {x : x 8}, which includes both endpoints. The set of numbers tht re in t lest one of these sets equls {x : x}, s cn be seen in the figure below: 3 8 Thus (3, ) [, 8] = [, ). 13. (, 3) [ 5, ) The first intervl is the set {x : x< 3}, which hs no left endpoint nd which does not include the right endpoint. The second intervl is the set {x : 5 x}, which includes the left endpoint nd which hs no right endpoint. The set of numbers tht re in t lest one of these sets equls the entire rel line, s cn be seen in the figure below: 5 3 Thus (, 3) [ 5, ) = (, ). 15. ( 3, ) [ 5, ) The first intervl is the set {x : 3 <x}, which does not include the left endpoint nd which hs no right endpoint. The second intervl is the set {x : 5 x}, which includes the left endpoint nd which hs no right endpoint. The set of numbers tht re in 5 Thus ( 3, ) [ 5, ) = [ 5, ). 17. Give four exmples of pirs of rel numbers nd b such tht + b = nd + b = 8. First consider the cse where 0 nd b 0. In this cse, we hve + b 0. Thus the equtions bove become + b = nd + b = 8. There re no s to the simultneous equtions bove, becuse + b cnnot simultneously equl both nd 8. Next consider the cse where <0 nd b<0. In this cse, we hve + b<0. Thus the equtions bove become b = nd b = 8. There re no s to the simultneous equtions bove, becuse b cnnot simultneously equl both nd 8. Now consider the cse where 0, b<0, nd + b 0. In this cse the equtions bove become + b = nd b = 8. Solving these equtions for nd b, weget = 5 nd b = 3. Now consider the cse where 0, b<0, nd + b<0. In this cse the equtions bove become b = nd b = 8. Solving these equtions for nd b, weget = 3 nd b = 5. Now consider the cse where <0, b 0, nd + b 0. In this cse the equtions bove become + b = nd + b = 8.

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