Chapter 2 The Number System (Integers and Rational Numbers)


 Antony Kelley
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1 Chpter 2 The Number System (Integers nd Rtionl Numbers) In this second chpter, students extend nd formlize their understnding of the number system, including negtive rtionl numbers. Students first develop nd explin rithmetic opertions with integers, using both tiles nd the number line s models. Then, they move, in both models, to the rtionl numbers by observing tht the unit represented by the tile, or the first hsh mrk on the number line, cn be chnged; for exmple it could be the qth prt of the unit originlly represented by the first tile, or the first hsh mrk on the line. As the lws of rithmetic remin the sme, they extend directly to the rithmetic for the qth prt of the originl unit. It ll holds together from the simple observtion tht q copies of the qth prt of the unit returns us to the originl unit. In this wy, the rithmetic of rtionl numbers is consequence of the properties of the four bsic opertions of ddition, subtrction, multipliction nd division. By pplying these properties, nd by viewing negtive numbers in terms of everydy contexts (i.e. money in n ccount or yrds gined or lost on footbll field), students explin nd interpret the rules for dding, subtrcting, multiplying, nd dividing with negtive numbers. Students reexmine equivlent forms of expressing rtionl numbers (frctions of integers, complex frctions, nd decimls) nd interpret deciml expnsions in terms of successive estimtes of the number representing point on the number line. They lso increse their proficiency with mentl rithmetic by rticulting strtegies bsed on properties of opertions. In Grde 6 students come to understnd tht positive nd negtive numbers re used together to describe quntities hving opposite directions or vlues, lerning how to plce nd compre integers on the number line. They use positive nd negtive numbers to represent quntities in relworld contexts, explining the mening of in ech sitution. Students recognize signs of numbers s indicting loctions on opposite sides of on the number line nd recognize tht the opposite of the opposite of number is the number itself. The development of rtionl numbers in Grde 7 is the beginning of development of the rel number system tht continues through Grde 8. In high school students will extend their understnding of number into the complex number system. Note tht in Grde 7 students ssocite points on the number line to rtionl numbers, while in Grde 8, the process reverses: they strt with lengths nd try to identify the number tht corresponds to the length. For exmple, students find tht the length of the hypotenuse of right tringle is p 5, they then try to identify the number tht corresponds to tht length. The min work of this chpter is in understnding how the rithmetic opertions extend from frctions to ll rtionl numbers. At first ddition is defined for the integers by djunction of lengths, nd subtrction s the ddition of the opposite; in symbols: b = + ( b). This ide is prelude to vector concepts to be developed in high school mthemtics. Students understnding of multipliction nd division is extended to integers so tht their properties continue to hold, using the understnding tht the number is the opposite of its opposite(e.g., ( 1) = 1). 7MF21
2 Section 2.1: Add nd Subtrct Integers; Number Line nd Chip/Tile Models Apply nd extend previous understndings of ddition nd subtrction to dd nd subtrct rtionl numbers; represent ddition nd subtrction on horizontl or verticl number line digrm. 7NS.1 Describe situtions in which opposite quntities combine to mke. For exmple, hydrogen tom hs chrge becuse its two constituents re oppositely chrged. 7.NS.1 Understnd p + q s the number locted distnce q from p, in the positive or negtive direction depending on whether q is positive or negtive. Show tht number nd its opposite hve sum of (re dditive inverses). Interpret sums of rtionl numbers by describing relworld contexts.7.ns.1b Understnd subtrction of rtionl numbers s dding the dditive inverse, p q = p + ( q). Show tht the distnce between two rtionl numbers on the number line is the bsolute vlue of their di erence, nd pply this principle in relworld contexts.7ns.1c We strt the study of the opertions of ddition nd multipliction for integers with review of 6th grde, where the integers re plced on the number line. The reson for this is tht, once students hve understood the rithmetic opertions with integers, in prticulr, the rules of signs, then the extension to ll rtionl numbers is nturl. This is becuse the rtionl numbers re plced on the line just by chnge of unit from the unit intervl to the 1/qth intervl for every positive integer q (see section 3). The only compliction is the ddition of rtionl numbers; this is discussed t the end of section 3. The gol in this section is to build intuition nd comfort with integer ddition nd subtrction so tht by the end of the section students cn reson through ddition nd subtrction of integers without model, nd then will be ble to extend those opertions to ll nturl numbers. Students strt by working with opposites (dditive inverses) to notice tht pirs of positives nd negtives result in zero pirs. They then move to dding integers. Students know from previous grdes tht the fundmentl ide of ddition is joining. After reviewing prior understnding of subtrction, we turn the number line model of subtrction (s dding the opposite). Students should notice tht joining positive nd negtive numbers often results in zero pirs nd tht which is left over is the finl sum. Students will develop this ide first with chip or tile model nd the rel line. Rules for operting with integers come from the permnence of the rules of rithmetic, by which we men tht extensions of the concept of number require tht the rules of rithmetic continue to hold. Students should understnd tht rithmetic with negtive numbers is consistent with rithmetic with positive numbers. When looking t number line, two numbers re opposites when they re the sme distnce wy from zero, but in opposite directions. Those numbers to the right of re the positive numbers, nd those on the left re the negtive numbers. For exmple, 3 represents the point tht is 3 units to the right of, nd 3 is its opposite, three units to the left. Two integers re opposites if they re ech the sme distnce wy from zero, but on opposite sides of the number line. One will hve positive sign, the other negtive sign. In the number line, +3 nd 3 re lbeled s opposites (it is customry to not write the plus sign in front of positive number, so +3 will be denoted by 3. Opposites Negtive Integers Positive Integers 7MF22
3 The full number line ppers nturlly in mesurements, such s the thermometer, elevtion, timelines, nd bnking. In ech of these pplictions, we re mking mesurements less thn zero of which the number line illustrtes ech of these concepts. Three observtions tht re importnt to notice: The integers re ordered: we sy is less thn b, nd write < b, whenever b is to the right of on the number line. The bsolute vlue of n integer is the distnce from the point on the line to. A number nd its opposite hve the sme bsolute vlue. Every nturl number hs n opposite, or dditive inverse. The negtive integers re opposites of the positive integers. For exmple, the opposite of 5 is 5, nd the bsolute vlue of both numbers is 5. Since the positive integers re the opposites of the negtive integers, we conclude tht ( 5) = 5, nd in generl ( ) = for ny integer.the opposite of is. Electron  Proton + Hydrogen Atom. Source: The ide of opposite occurs in the rel world in mny wys. The opposite of income is debt. If I find $5, I cn clim tht I hve +5 dollrs; if I lose $5, tht mounts to hving 5 dollrs. In chemistry hydrogen tom, like other toms, hs nucleus. The nucleus of hydrogen tom is mde of just one proton ( positive chrge). Around the nucleus, there is just one electron ( negtive chrge), which goes round nd round the nucleus. As result, the positive proton nd the negtive electron blnce the electricl chrge, so tht the hydrogen tom is electriclly neutrl (bsiclly zero). As with ny new topic, it is importnt to strt with fmilir contexts so tht students cn use prior knowledge to build mening. With integers, students often get confused bout which symbol is the opertion or in which direction they re moving when they compute, so hving context is prticulrly importnt. As students lern to compre nd compute, they cn use the contexts to ground their thinking nd justify their nswers. In this course we will understnd integers s points on the number line, nd will lso model the opertions on integers with the chip/tile model. We emphsize the number line model becuse it is key to understnding the extension of rithmetic, nd it provides the bridge between lgebr nd geometry tht hs been centrl to the development of mthemtics for millennium. The number line model mounts to working with definition of ddition nd subtrction for integers nd subsequently, rtionl numbers; nd the chip/tile model (quntity) mounts to working with theorem bout ddition nd subtrction ( = + ( ) =, represented by zero pirs ), s bsis for understnding how the properties of opertions should extend to integers. A number line model hs severl dvntges. In Grde 6 students lerned to locte nd dd whole numbers, frctions nd decimls on the number line. Additionlly, the number line is n importnt connection to the coordinte xis, which involves two perpendiculr numbers lines, explored in Grde 8. Previously ddition ws represented by linking the line segments together. Moving from whole numbers to integers, we recognize tht line segments hve beginning nd n end; tht is they re directed. So, now we hve to be more cler: which end of one intervl is linked to which end of the other? This clrifiction leds to n understnding of the reltion of subtrction to ddition nd to tking the opposite. In the number line model rrows re used to show distnce nd direction. The rrows help students think of integer quntities s directed distnces. An rrow pointing right is positive, nd negtive rrow points left. Ech rrow 7MF23
4 is quntity with both length (mgnitude) nd direction (sign). To put the integers on the line, we first select point to the right of nd designte it s 1. The point t the sme distnce from, but to the left, is 1, the opposite of 1. Now, for positive integer p, move to the right distnce of p units: the end point represents p. For negtive integer, do the sme thing, but on the left side of. Now, we dd integers on the line by djunction of the corresponding directed line segments. To dd the integers p nd q: begin t zero nd drw the line segment (rrow) to p. Strting t the endpoint p, drw the line segment representing q. Where it ends is the sum p + q. If p nd q re of the sme sign, the point p + q is frther from the origin thn q. If the second line segment is going in the opposite direction to the first, it cn bcktrck over the first, e ectively cnceling prt or ll of it out. The following figure depicts the sum p + q where p is negtive nd q is positive. p q p p + q q Exmple 1. Demonstrte ech of the following on number line b c ( 9) d. 2 + ( 3) Solution. To void confusion, students should strt by distinguishing between sign nd n instruction for n opertion. So, for exmple, in prt c., the symbol + is n opertion, wheres the  is sign. This will continue to be n issue, so it is vitl to strt ddressing this now. In the following we hve used colors nd rrowheds for stress, hoping tht students will incorporte tht in their understnding of vocbulry : this ddition cn be thought of s strting t nd counting 2 units to the right (in the positive direction on the number line) nd then counting on 3 more units to the right = 5 b : this ddition cn be thought of s strting t nd counting 4 units to the left (in the negtive direction on the number line) nd then counting on 3 more units to the right = 1 7MF24
5 c ( 9): strt t nd count 12 units to the right, then count on 9 more units to the left ( 9) = 3 d. 2 + ( 3): Here we count 2 units to the left, nd then 3 more units to the left = 5 Finlly, note tht if we dd the opposite of number to number, we return to : n + ( n) = = ( n) + n. for ny number n. This eqution sttes tht n nd n re dditive inverses, tht is, n nd n dd up to. Finlly, note tht prentheses cn be used to void strnge, or mbiguous expressions like 3 = 3 nd =. Insted, write ( 3) = 3 nd 3 + ( 3) =. Even hs n opposite, nmely, itself, since + = ; tht is, =. The second model mentioned utilizes colored chips/tile to represent integers, with one color representing positive integers nd nother color representing negtive integers; subject to the rule tht chips of di erent colors cncel ech other out. Historiclly, Chinese trders used system of colored rods to keep records of their trnsctions. A red colored rod denoted credits nd blck colored rod denoted debits (opposite of how we view debits/credits tody). Assign red chip to represent 1 (debit or negtive chrge) nd yellow chip to represent 1 (credit or positive chrge): = 1, = 1 A pir of red nd yellow chips is clled zero pir, becuse it represents the integer. We use the informl term zero pir to refer to two things tht re opposite or undo ech other. The sum of the two elements of zero pir is zero. We see zero pirs in mny contexts. For exmple, n tom with 5 protons nd 5 electrons hs neutrl chrge, or net chrge of zero. Exmple 2. I took 4 steps forwrd nd 3 steps bck. Wht is the result in words, nd how mny zero pirs re there? Solution. A visul model is shown nd depicts tht I m one step hed from where I stred nd tht there re 3 zero pirs. 7MF25
6 We emphsize tht =, expressing ( ) + =, the dditive inverse property of integer ddition. A red tile coupled with yellow tile is zero pir, tht is, they cncel ech other. Exmple 3. Demonstrte the ddition with the chip/tile model. Solution. We begin with 7 yellow chips nd 3 red chips in our first box. In the second box we recognize tht there re three sets of zero pirs circled with ech zero pir representing ; resulting in (in the lst box) four yellow chips or 4. It is importnt to understnd tht it is lwys possible to dd to or remove from collection of ny number of pirs consisting of one positive nd one negtive chip/tile without chnging the vlue (i.e. it is like dding equl quntities of debits nd credits). We lso recognize tht, = 3 + (3 + 4) = ( 3 + 3) + 4 = + 4 = 4. Although not explicitly stted, we re moving students towrds understnding, either with the chip/tile model or the number line model the properties of rithmetic. Implicit in the use of chips is tht the commuttive nd ssocite properties extend to integers, since combining chips cn be done in ny order. Integer chips re useful in understnding opertions with rtionl numbers, up to point. By chnging the unit concept from the intervl [, 1] to the 1/qth prt of tht intervl (for ny integer q) we cn illustrte, with chips, some of the rithmetic fcts for rtionl numbers. However, if we work with rtionl numbers with di erent denomintors, this model becomes unwieldy. For this reson, in section 3, where we turn to rithmetic with rtionl numbers, we will work exclusively with the line model. Next, students move to subtrction, first by reviewing, from previous grdes, tht there re two wys to think concretely bout subtrction. ) Tkewy: Ann Mri hs 5 gummy bers. José ets 3. How mny gummy bers does she now hve? 5 3, thus 2 gummy bers. b) Comprison: Ann Mri hs 5 gummy bers, José hs 3 gummy bers, how mny more gummy bers does Ann Mri hve thn José? Agin, the opertion is 5 3 resulting in 2 gummy bers. We build on the concept of comprison s concrete wy to think bout subtrction with integers on the rel line. We strt by locting integers on the line nd note tht when compring two integers, there is directionl or signed distnce between them e.g. when compring 5 nd 3 we cn think 5 is two units to the right of 3, (5 3 is 2), or 3 is two units to the left of 5, (3 5 is 2). In 2.1e students will exmine subtrction exercises nd notice tht b cn be written s + ( b) nd tht ( b) cn be written s + b: b = + ( b) ( b) = + b. As n exmple: if we subtrct 11 from 7, we move 11 units to the left from 7, to 11 to Tht is the sme s dding Exmple 4. Model 7 2 nd 2 7 on the number line. Solution. Think in terms of ddition: 7 2 = 7 + ( 2), nd 2 7 = 2 + ( 7). 7MF26
7 . Locte 7 nd 2 on the number line. Tke the directed line segment corresponding to 2 nd move it so tht its beginning point is t the point 7. Then the endpoint lnds t 5. b. Locte 2 nd 7 on the number line. Tke the directed line segment corresponding to 7 nd move it so tht its beginning point is t the point 2. Then the endpoint lnds t 5. Describing subtrction in terms of the number line model brings home the reliztion tht integers involve two concepts: mgnitude nd direction, sometimes designted s directionl distnce. We now see tht there is distinction between the distnce from to b nd how you get from to b, in other words, the directionl distnce. Adding nd subtrcting integers cn cuse students gret del of trouble prticulrly when they first confront exercises tht hve both ddition nd subtrction with integers. For exmple, in expressions like 5 7 students re often unsure if they should tret the 7 s negtive or positive integer nd if they should dd or subtrct. Put nother wy, students re not sure tht 5 7 = 5+( 7), nd worry tht it could men (5 7), or even 5 ( 7). For this reson, it is extremely helpful tht, from the beginning of this chpter, students rticulte wht they think the expression mens. We close section 2.1 with few words of cution. Subtrction in the set of integers is neither commuttive nor ssocitive: 5 3, 3 5 becuse 2, 32; (5 2) 1, 5 (2 1) becuse 2, 4. The expression is mbiguous becuse the order of opertions is not specified. If we operte from the right we red this s 5 (2 1) = 5 1 = 4, nd if we operte from the left, it is (5 2) 1 = 3 1 = 2. But sometimes you will see n expression like 521, in which cse it should be red s (5 2) 1. Tht is: when the expression is mbiguous, the defult is to operte from left to right. In ny cse, it is best to get in the hbit of using prentheses to eliminte mbiguities. Section 2.2: Multiply nd Divide Integers; Represented with Number Line Model Apply nd extend previous understndings of multipliction nd division nd of frctions to multiply nd divide rtionl numbers. 7.NS.2 Here we restrict ttention to integers; in the next section we shll move to rtionl numbers, nd del with the full stndrd in tht context.. The gol here for students in this section is twofold: 1) fluency with multipliction nd division of integers nd 2) understnding how multipliction nd division with integers is n extension of the rules of rithmetic s lerned in previous grdes. Multipliction of integers is n extension of multipliction of whole numbers, frctions nd decimls. Let us consider typicl elementry multipliction problem. Exmple 5. Slly bbyst for her mother for 3 hours. Her mother pid her $5 ech hour. How much money did Slly ern fter 3 hours? Solution. We note tht Slly erned $5 + $5 + $5 = $15, illustrted s which is the sum of 3 fives, written s 3 5 nd is red s 3 times 5, nd mens the totl number of objects in 3 groups if there re $5 in ech group. Additionl representtions include: 7MF27
8 Source: The bsic definition of multipliction of whole numbers is, if nd b re nonnegtive numbers, then b is red s times b, nd mens the totl number of objects in groups if there re b objects in ech group. Note: if we rotte the rectngle in the first representtion, we will hve 5 groups of 3 objects in ech group, representing 5 3. This is wy of seeing tht b = b for ny positive nd b. In elementry grde the dot symbol for multipliction is replced by cross, (not to be mistken for the letter x or x), or by n sterisk * (used in computer progrmming). Sometimes no symbol is used (s in b), or prentheses re plced to distinguish the fctors such s ()(b). In this text we will typiclly use the dot or prenthesis to express multipliction of expressions nd the cross for numbers, to be sure tht 3 5 is distinguished from 35. We note tht nd b re clled fctors, nd tht the first fctor stnds for the numbers of groups, nd the second fctor b stnds for the numbers of objects in ech group (lthough, s we hve pointed out, those roles re interchngeble). How cn you tell if the opertion needed to solve problem is multipliction? To nswer this question we consider the definition of multipliction once gin. Multipliction pplies to situtions tht involve equl groups. In the illustrtion for our previous problem, there were 3 distinct groups of $5 in ech group. Thus, ccording to the definition of multipliction, there re 3? 5 dollr bills in ll. Simply put, whenever collection of objects is rrnged into groups, nd there re b objects in ech group, then we know tht ccording to the definition of multipliction, there re b totl objects. Up to this point we hve exmined multipliction only for whole numbers, so, wht is the mening of multipliction of negtive numbers? Understnd tht multipliction is extended from frctions to rtionl numbers by requiring tht opertions continue to stisfy the properties of opertions, prticulrly the distributive property, leding to products such s (1)(1) = 1 nd the rules for multiplying signed numbers. Interpret products of rtionl numbers by describing relworld concepts. 7.NA.2 We will discuss this stndrd here in the context of integers, nd then in the next section extend tht logic to ll rtionl numbers. Exmple 6. Wht does 3 ( 5) men? Solution. Just s 3 5 cn be understood s (5) + (5) + (5) = 15, so 3 ( 5) cn be understood s ( 5) + ( 5) + ( 5) = 15. 7MF28
9 On the number line, we think of 3 5 s 3 jumps to the right (or up) on the number line, strting t. Similrly, 3 ( 5) is represented by 3 jumps of distnce 3 to the left strting t, s in the following figure: Using chip/tiles, 3 ( 5) mens 3 groups of 5 yellow tiles. And finlly in context: if I lose $5 on three consecutive dys, then I ve lost $15. Exmple 7. Wht does ( 3) 5 men? Solution. There re two wys student my ttck this using rules of rithmetic: one wy is to recognize tht ( 3) 5 is the sme s 5 ( 3) (e.g. the commuttive property) nd then pply the logic bove: = 15. A model would be illustrted s: Another is to turn to the understnding of integers from 6th grde. There the ide of 3 ws developed s the opposite of 3. Now, if the ssocite lw of rithmetic extends, (the opposite of 3) 5 is the sme s (the opposite of 3 5), or 15. On the number line, we record ( 3) 5 s the opposite of 3 5: Opposite Exmple Let us note tht this rgument gives us the eqution ( 1) ( 1) = 1. For, multipliction by 1 is reflection to the opposite side of, nd therefore lnds on the opposite point. So, this eqution is simply sying tht the opposite of 1 is 1. It is instructive to go through the resoning for this multipliction tble: In other words, the product of two numbers with the sme sign is positive, nd if the numbers hve opposite signs, it s negtive. Notice tht this is the sme rule s tht for the sum of even nd odd numbers: the sum of two numbers of the sme prity (both even or both odd) is even, if the prity is di erent, the sum is odd. 7MF29
10 Understnd tht integers cn be divided, provided tht the divisor is not zero, nd every quotient of integers (with nonzero divisor) is rtionl number. 7.NS.A.2b Just s the reltionship between ddition nd subtrction helps students understnd subtrction of rtionl numbers, so the reltionship between multipliction nd division helps them understnd division. To mke this more precise: for ddition, the opposite of number is, the solution of the problem + x =. On the number line, represents going out from to the point, nd dding brings us bck to the beginning. For this reson is clled the dditive inverse of. So, for multipliction, the opposite of is the solution of the eqution x = 1, denoted 1/. On the number line, multiplying 1 by mens dding 1 to itself b times. To invert this process (tht is, to get bck to 1), we hve to prtition the result into pieces of the sme length; tht is the prtitive definition of division. So, dividing by undoes multiplying by, so 1/ is the multiplictive inverse of ; tht is (1/) = 1, or 1/ is the solution to the eqution x = 1. It is this logic tht extends from frctions to ll rtionl numbers, s we shll see in the next section. One importnt point: since x = for every number x, there is no solution to the eqution x = 1; nd it is for this reson tht we cnnot divide by zero: it just plin does not mke ny sense. In short, in terms of the number line, the properties of the opertions of multipliction nd division crry over from frctions to ll numbers, positive nd negtive. Let us illustrte. Exmple 9.. Clculte ( 15) 5. b. Clculte 8 ( 4). c. Clculte ( 12) ( 3). Solution.. Since division by 5 is the inverse of multipliction by 5, the eqution tells us tht the solution is some number x, which when multiplied by 5 gives us 15. But we know tht ( 3) 5 = 15, so the nswer hs to be 3. b. If = 8/( 4), then is the solution of the problem 4x = 8. So, hs mgnitude 2, but is it negtive or positive? Let s check: ( 4)(2) = 8 nd ( 4)( 2) = 8. So = 2. c. ( 12) ( 3) is tht number which, when multiplied by 3 gives us 12. So, it hs to be negtive (for the product of two negtives is positive, so it cn t be positive) nd it hs to be of mgnitude 4. Therefor the nswer is 4. Section 2.3: Add, Subtrct, Multiply, Divide Positive nd Negtive Rtionl Numbers (ll forms). Apply nd extend previous understndings of ddition nd subtrction to dd nd subtrct rtionl numbers; represent ddition nd subtrction on horizontl or verticl number line digrm. 7.NS.1 Understnd p + q s the number locted distnce q from p, in the positive or negtive direction depending on whether q is positive or negtive. Show tht number nd its opposite hve sum of (re dditive inverses). Interpret sums of rtionl numbers by describing relworld contexts. 7.NS.1b Understnd subtrction of rtionl numbers s dding the dditive inverse, p q = p + ( q). Show tht the distnce between two rtionl numbers on the number line is the bsolute vlue of their di erence, nd pply this 7MF21
11 principle in relworld contexts. 7NS.1c Apply properties of opertions s strtegies to dd nd subtrct rtionl numbers. 7NS.1d Apply nd extend previous understndings of multipliction nd division nd of frctions to multiply nd divide rtionl numbers.7ns.2 Understnd tht multipliction is extended from frctions to rtionl numbers by requiring tht opertions continue to stisfy the properties of opertions. 7NS.2 Understnd tht integers cn be divided, provided tht the divisor is not zero, nd every quotient of integers (with nonzero divisor) is rtionl number. If p nd q re integers, then (p/q) = ( p)/q = p/( q ). Interpret quotients of rtionl numbers by describing rel world contexts. 7NS.2b Apply properties of opertions s strtegies to multiply nd divide rtionl numbers. 7.NS.2c Let s move on to rtionl numbers: plcing them s points on the number line, nd then extending the understnding of rithmetic opertions to ll nturl numbers. In the sixth grdes students lerned how to plce the rtionl numbers on the number line, nd we shll depend upon tht knowledge in this section, lthough we will strt new: by defining nd working with ddition nd subtrction of points on the line with specific point identified s, by considering the point, nd the line segment from to tht point s one nd the sme thing. The points on one side of re considered negtive nd those on the other side s positive. Ordinrily the line is drwn s horizontl line, nd the positive points re those on the right of. If the line is drwn verticlly, then the custom is to designte the segment bove s the positive segment. In some contexts the number line my be neither horizontl or verticl; it is importnt to relize tht its position in the plne (or, for tht mtter, in spce) is not relevnt; ll tht mtters is tht bse point hs been designted s, nd one side hs been designted s positive. Once tht hs been done, the concept of opposite (the opposite of is the symmetric point on the other side of ). We use the nottion to represent the opposite of. Without ctully using the term vector. it helps to use this ide to understnd opposite, nd the rithmetic opertions of ddition nd subtrction. For exmple, the ct of tking the opposite of is performed by reflecting in the origin, s illustrted in Figure 1. Figure 1. Tking the opposite:! Addition, denoted + b for two points nd b on the line, is defined (vectorilly) by djunction: the sum of nd b is the endpoint of the intervl formed by djoining the intervl to b to the intervl to (see figure 2). b + b Figure 2. Addition: (, b)! + b 7MF211
12 This brings the integers into the picture in this wy: for ny integer n, nd ny point on the line, n is the endpoint obtined by ppending the intervl [, ] to itself n times. For n >, this is cler, nd for n < we men the djunction on the opposite side of [, ]. In other words, for > nd n >, n is the endpoint of the intervl strting from nd moving to the right distnce of n lengths of size, nd ( n) is the endpoint of the intervl strting from nd moving to the left distnce of n lengths of size. This cn be summrized by the eqution (which works for either n or either positive or negtive: ( n) = n ( ) = (n ) Subtrction, denoted b, for two points nd b on the line, is defined s b = + ( b). Often, students will be confused by the symbol, since it represents both subtrction nd the opposite. A wy to del with this confusion is to understnd tht it mkes sense if we think of the opposite of s obtined by subtrcting from : =. In fct, in doing ddition nd subtrction, it is customry to delete the ddend if it ppers. For exmple, one might replce by nd then just omit the, getting. The bsic fct tht resolves the confusion is this: if b is subtrcted from, tht is the sme s dding b to ; tht is, b = +( b). Furthermore, ( ) =, nd ( b) = + b. Hving defined ddition nd subtrction for points ( s intervls) on the rel line, we cn pictorilly demonstrte tht the properties of ddition extend nturlly to the sme opertions on the line. Here re two exmples; students cn reinforce their understnding of those opertions by drwing digrms for other lws of rithmetic. Exmple 1. + b = b + b b + b b b + Figure 3. + b = b + 7MF212
13 Exmple 11. (b ) = b + b b b b + ( ) (b ) b b b b b + b b Figure 4. (b ) = b + Bringing this discussion together with tht of the previous section, we cn summrize in this wy: Tke line, nd point on the line. Let s ssume the line is horizontl. The point divides the line into two segments; the positive side on the right of the point, nd the negtive side on the left. Define opposite nd ddition s bove. There is reltion on this line: the expression < b mens tht the point lies to the left of b. This is the sme s sying b is positive. For ny two di erent points, b, either < b or b <. If < b nd b < c, then < c. For ny nd b, pple b the segment, or intervl, [, b] is the set of points between nd b, ordered from to b. In prticulr, [, ] is just the point.finlly we define ddition of points nd b by djunction of the intervls [, ] nd [, b]: put the endpoint of [, b] t the endpoint of [, ]: the combined intervl is the intervl [, + b]. 7MF213
14 Now select point on the right of, nd denote it s 1. Then, for ny positive integer n, n is represented by the sum of n copies of the intervl between nd 1 (where, for n negtive, we do the sme on the other side of the line strting with the opposite 1 of 1). C A M B D Figure 5 Finlly, we put rtionl numbers on the line. Given p/q, where p is n integer nd q is positive integer, first prtition the intervl [, 1] into q equl prts, nd denote the first intervl s [, 1/q]. If we djoin q copies of this intervl to ech other, we end up t the point 1, showing tht q(1/q) = 1. Now p/q is the endpoint of the intervl obtined by djoining p copies of [, 1/q] to ech other (on the left side of if p <. Incidentlly, the instruction prtition the intervl [, 1] into q equl prts cn ctully be done by construction with stright edge nd compss. For q = 2, students probbly lredy know how to bisect line segment (see figure 5 for reminder); di erent construction for ny positive integer q will be lerned in high school mthemtics. Multipliction nd division of rtionl numbers, s points on the line, re defined geometriclly just s they re on the integers, with the requirement tht the rules of rithmetic persist. Let us explin: First of ll, if p is positive integer nd is point on the line, Then p is defined s the djunction of p copies of the intervl [, ] from to. We cn extend the definition for p = : represents zero copies of ; in other words =. Now, if p is negtiive integer, p is defined s the djunction of p copies of the opposite of. Now, if q is positive integer /q is defined s the first piece of prtition of the intervl [, ] into q pieces of the sme length. This su ces to define multipliction of ny length by n integer, nd division of ny lengths by positive integer. But we wnt to show tht we cn multiply ny two rtionl numbers, nd divide ny rtionl number by nonzero rtionl number. To get there tkes few steps. We re used to the understnding (for positive integer p, nd n intervl [, ], tht p represents p copies of [, ] djoined together sequentil, nd (1/q) represents one qth prt of [.]. But tht is how we defined /q, so 1 q = q. Now, how do we multiply ny point on the line by rtionl number p/q? If the lws of rithmetic re to hold, we must hve p q = p (1 q ) = p q. In words, (p/q) is constructed in this wy: First divide into q equl prts nd then dd together p copies of tht length. Now, we might just hs well hve written 7MF214
15 p q = (1 )(p ) ; q in words::, (p/q) is constructed in this wy: First djoin together p copies of, nd then divide the result into q equl prts. One of the strengths of rithmetic is tht these two interprettions re correct, nd result from the commuttivity of multipliction. When we defined division, b, we sid tht the divisor b should be not zero. Why not? We know tht division is the inverse opertion of multipliction. For exmple, the unique (one nd only one) solution of 15/5 = 3, since 3 is the vlue for which the unknown quntity in (?)(5) = 15 is true. But the expression 15/ requires vlue to be found for (?) tht solves the eqution: (?)() = 15. But ny number multiplied by is nd so there is no number tht solves the eqution. Extension The bove is just one of mny explntions of the prohibition of division by. But students hve seen mny instnces in which n rithmetic gp is filled by inventing new numbers. Mybe we cn introduce new number, cll it, tht solves this eqution, tht is = 15. But if the lws of rithmetic re to continue to persist, then we d hve 15 = =, which is contrdiction: 15 is not the sme s. The persistent student my sk tht we might succeed by tking wy the rule x =.. This would force us to hve specific for every number, nd this shortly leds us into new contrdictions. Further explortion of this wy is doomed to filure: there is no wy to try to define division by unless we wnt to scrifice the rule tht division is the inverse of multipliction, in which cse, we re no longer tlking bout the rithmetic of the number line. The point here is this: if we wnt to extend the number field, it should be bsed on specific context; in our cse: the number line. Let s look t the number line context for our s. Locte 15/ with > on the number line. As gets smller nd smller, 15/ gets lrger nd lrger, nd if we slide to zero, 15/ slides o the number line on the right. So, 15/, if it cn mke ny sense t ll, is o the number line. Until we find new context where this mkes sense, we hve to bndon the hope to define division by zero. End Extension. But, now, wht sense re we to mke of division by frctions? Wht is the nswer to the question =? Let the lws of rithmetic be our guide: the question mrk s filled with number tht hs this property: when multiplied by 1/7, we get 1/5. Tht is the sme s sying: the question mrk s filled with number tht hs this property: when divided 7, tht is, when prtitioned into 7 equl prts, ech prt is copy of 1/5. The nswer then is 7 copies of 1/5, or 7/5: = 7 5. In conclusion, these constructions ssocite to every rtionl number (quotient of n integer by positive integer) point on the line. 7MF215
16 Extension Does this give us every point on the plne? This is question tht is posed nd discussed in Grde 8, but the persistent 7th grder my strt wondering bout this. In prctice, point is blot on the line, nd so hs positive length. So, we cn get within tht length with the deciml pproximtion of the point (tht is, getting within the nerest integer of the point, then the nerest tenth, then the nerest hundredth etc.). If ll we cre bout is nkedeye ccurcy, we might need two or three deciml points. For ccurcy with mgnifying glss, we;ll need more; with microscope, much more, nd so forth. No mtter how good our instrumenttion, point will pper s blot. Plto, who invented the concept of the idel, conceived of this process going on indefinitely: lthough mortls will lwys only see blot, the Gods cn ctully see the point. Aristotle ws more prgmtic: ll there ever is is blot, but it my get finer nd finer. Another wy to express n nswer is through deciml expnsions. The 7th grder is introduced to neverending decimls, so we strt the discussion there Given ny point, our construction of the deciml pproximtion to the point cn get us s close s we need to the point (s is close to 1/3, nd if we ppend more 3 s we continully get closer  but never there). To prphrse Plto s conception ( prphrse since deciml expnsions were not known t Plto s time)  on the contrry:.3 is there. Well tht s ok, since 1/3 is rtionl number. But wht bout ? Here... mens tht the pttern continues indefinitely; the ptter being tht ech 1 is followed by one more thn the preceding 1, going on indefinitely. Now rtionl number (s we hve seen in chpter 1) hs deciml expnsion tht is either terminting or repeting. So, if this represents number, it cnnot be rtionl. So mybe the nswer is tht ll points on the line re represented by deciml expnsions, nd every deciml expnsion does describe point on the line. It is extremely useful to discuss this ide through the understnding of deciml expnsion s recipe for getting s close s wee need to be to point on the line. The issue of mking mthemticl sense of Plto s idel ws not resolved until the 19th century. The focus ws on the questions: wht is number? wht does it men tht line consists of points? In the clssicl Greek dys, line nd point were undefined terms, nd number ws for counting. In the number line, number mens (signed) length: is the signed length of the intervl between nd (positive is is right of, negtive if is left of zero). Now, it ws known by the Pythgorens long before Plto tht there exists length (in the sense of constructble by strightedge nd compss) tht is not representble by nturl number. Let us begin with the problem s it ws understood t the time. The discovery of the Pythgorens is this: the digonl nd side of squre re incommensurble. Euclid (bout century fter Plto) defined this term in this wy. Tke two intervls A = [, ] nd B = [, b]. A nd B re commensurble if there is nother intervl R such tht both A nd B re integrl multiples of R; tht is, there re integers m nd n such tht A = mr nd B = nr. In prticulr, if B is the unit intervl, [, 1], then A corresponds to the rtionl number m/n. If no such intervl R exists, we sy tht A nd B re incommensurble. As mentioned erlier, the Pythgorens found out tht the side nd digonl of squre re incommensurble; Euclid, in his Elements, gve geometric proof of this fct, nd Aristotle gve n rithmetic proof (both these will be further discussed in 8th grde). Euclid bsed his rgument on test for commensurbility tht he invented, clled the Eucliden lgorithm. It goes like this: gin, let A nd B represent two (positive) lengths; let us suppose tht A > B. Now we mimic the deciml expnsion of A using B s the unit mesure: strt djoining copies of B inside the intervl A until we cnnot dd nother copy without going beyond A.If we end up right t the end of A, we re done, for we hve shown tht A is n integrl multiple of B. If not, let B 1 be the intervl remining fter this fill. Since B 1 < B, we cn strt filling B with copies of B 1. If we end precisely t the endpoint of B, we re done: both A nd B re integrl multiples of B 1. If not, let B 2 be the piece left over; since B 2 < B 1, we cn try to fill B 1 with n integrl multiple of B 2 s, nd so forth. If this process ends t ny stge, the lengths A nd B re commensurble; if not, they re incommensurble. 7MF216
17 As check on this procedure, we note tht integer lengths re commensurble, since they re both integrl multiples of the unit length. Thus, in this cse, Euclid s test must end. And it does: when we put copy of B inside A, since both A nd B re integers, wht remins is n integer lso; in fct the integer A B. So when we ve put s mny copies of B inside A s is possible, wht remins B 1 is n integer less thn B. Continuing this process, we get decresing sequence of integers, strting with B, so it must end. The remrkble thing is tht the integer R t which this process ends is the gretest common divisor of A nd B. How could Euclid possibly show tht the side S nd the digonl D re incommensurble; if he tried to keep repeting the process, he d still be there in his little ttic working wy, nd without proof. Well: he ws clever: he showed this: if we put the length S inside D, we hve piece P left over tht is less thn S. Now, he notices tht P cnnot fill S, nd tht the piece left over is smller thn P; but wht is more, the geometry of the originl squre hs been reproduced, only s smller version. So, since the geometry reproduces, the process cn never end. Let us give nother exmple: tht of the golden rectngle. A golden rectngle is rectngle tht is not squre, but hs this property: if we remove the squre of whose side is the length of the smller side of the rectngle, the remining rectngle is smller version of the originl. Clerly, if there is golden rectngle, its sides cnnot be commensurble. For, when we tke wy the squre bsed on the smller side, wht remins is reproduction of the originl, t smller scle. This tells us tht this process will never end. End Extension. Now, let s return to the representtion of rtionl number p/q on the number line. We hve been ble to geometriclly describe the rithmetic opertions, but still, the sum of two rtionl numbers requires some explntion. Since the expression /b is tken to men copies of 1/b, then it is esy to dd (or subtrct) rtionl numbers with the sme denomintor: copies of 1/b plus c copies of 1/b gives us + c copies of 1/b: b + c b = 1 b + 1 b c = 1 + c ( + c) =. b b Now, to dd /b + c/d with b, d, we proceed s follows. First, divide the unit intervl into bd equl pieces. Note tht 1/b consists of the first d copies of 1/bd, nd 1/d consists of the first b copies of 1/bd. In other words, the frctions 1/b nd d/bd represent the sme length, s do 1/d nd b/bd. Thus: Exmple 12. b + c d = 1 b + c 1 d = d bd + c b bd = d bd + cb d + bc =. bd bd Exmple 13. How much is 2/5 of $55.35? = = = Solution. There re severl wys to do this. Here is one: 2 35 ( ) = = = Let s tke second look t these computtions. In the first frction, divide 55 by 5 too get 11, nd then 7MF217
18 multiply by 2 to get 22 (dollrs). For the second frction, divide 35 by 5 to get 7 nd multiply 2 to get 14 (cents). Another wy is to observe tht 2/5 =.4, nd then multiply: = = = Solve relworld nd mthemticl problems involving the four opertions with rtionl numbers.(7.ns.3) Exmple 14. Our group shred 3 pizzs for lunch. The pepperoni pizz ws cut into 12 eqully sized pieces, the tomto pizz into 8 pieces nd the broccoli pizz into 6 pieces. I te one piece of ech. Wht frction of whole pizz did I et? Solution. My consumption ws 1/12 + 1/8 + 1/6. Now, I don t hve to multiply since ech denomintor is multiple of 24 : 24 = 2 12 = 3 8 = 4 6. Thus Exmple = = 9 24 = 3 8. Genoe left her house nd wlked to the Est down her street for mile nd qurter. She then turned round nd wlked to the West for two nd theeeights miles. How fr ws she from her home? Solution. We cn think of her street s the number line, with her house t the zero position. Genoe wlked in the positive direction to the point 1 1 4, nd then wlked in the negtive direction distnce of. Using the number line model, her new position is t the point ( 23 8 ). Mixed frctions re often convenient in everydy discourse, but re very ine So we rewrite the expression s cient in computtion Put both frctions over common denomintor, in this cse 8, nd subtrct for the nswer: = 9 8, so Genoe ended up mile nd n eighth west of her house. 7MF218
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