Teaching fractions in elementary school: A manual for teachers


 Elisabeth Myrtle Johnston
 2 years ago
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1 Teaching fractions in eementary schoo: A manua for teachers H. Wu Apri 0, 998 [Added December, 200] I have decided to resurrect this fie of 998 because, as a reativey short summary of the basic eements of fractions, it may sti be of some interest. I have repaced some of the pictures in the origina fie which are definitey not adequate for the iustration of the discussion. Otherwise I have eft the origina manuscript undisturbed. A greaty expanded version of this materia wi appear as Part 2 of a forthcoming boo, Understanding Numbers in Eementary Schoo Mathematics, American Mathematica Society, to appear in May, 20. [Added, March 999] This is a preiminary draft of what is projected to be a monograph. Even as a draft, this eectronic version is incompete because some ine drawings and annotations to drawings are absent due to the author s incompetence with computer graphics. Nevertheess, I have decided to post this incompete version on the web because its centra idea of how to approach the teaching of fractions from grade 5 and up is sound, and at present nothing resembing it seems to be avaiabe. It is to be emphasized that () the primary audience of this artice is teachers of grades 5 8, not students, and (2) this approach to fractions is definitey not for grades 2 4 where fractions first mae their tentative appearances. The overriding concern here is that, after chidren s informa encounter with fractions in the eary grades, they reach the point in grade 5 where their haphazard nowedge needs consoidation and their initia foray into abstract mathematics (fractions) needs some structure for support.
2 This artice is designed to hep teachers face up to the chaenge of eading students to the next phase of mathematica achievement. This manua is essentiay the content of a worshop I conducted in March of 998 for teachers of grades 5 8. The main goa is to outine a different way to teach (positive) fractions to the 5th or 6th grade. This approach to fractions is a geometric one, where a fraction is identified unambiguousy as a point on the number ine. Thus a fraction becomes a ceary defined mathematica object. Moreover, the emphasis throughout is on mathematica reasoning as we progress from topic to topic; a reason is suppied for each and every step. Specia care has been given to avoid invoing the usua device of deus ex machina found in amost a schoo textboos, whereby new properties are periodicay conferred on fractions as if by heaveny decrees without any mathematica justification. To someone not famiiar with the mathematics of eementary schoo, the fact that we give a cearcut definition of a fraction must seem uttery trivia. After a, how can we as students to add and mutipy and divide fractions if we don t even te them what fractions are? Unfortunatey, it is the case that schoo texts usuay do not define fractions. Try ooing up the index of an eementary schoo mathematics text can be a very disconcerting experience. Under fractions one can find many items, such as add, mutipy, mixed numbers, or estimate, but never definition or meaning. This state of ambiguity not surprisingy spawns endess confusion about the concepts of ratio, proportion, and percent ; see the ast ast section ( 7) of this manua. It woud be reasonabe to conjecture that the Fear of Fractions (cf. [] and []) can aso be traced to this singuar pedagogica phenomenon. Certain mathematica and pedagogica aspects of the foowing exposition are worth mentioning. On the mathematica front, the difficuty with the teaching of fractions ies in the fact that the correct definition of fractions (rationa numbers) as equivaence casses of ordered pairs of integers is abysmay unsuitabe for teaching at the eve of the 5th or 6th grade. Furthermore, the forma definitions of the arithmetic operations as those dictated by the requirements of the fied axioms are aso grossy improper for instruction at this eve because of the ac of motivation. It has ong been recognized that a compromise between what is mathematicay correct and what is pedagogicay feasibe must be forged. For this reason, I have chosen 2
3 to define a fraction as a point on the number ine (whie dodging the expanation of what a number is), and have introduced addition, mutipication, and division on fractions by extending the intuitive idea geaned from experiences with natura numbers of what these operations ought to be. A this is part of an effort to eep the mathematics pausibe as much as possibe. There is a price, however. The expositon amost seems to prove that my definitions are the correct ones, whereas it merey motivates them. But I decided not to enter into a discussion of the subte distinction between the two on account of my beief that, at this eve, it is more important to buid upon students intuition and prior nowedge than to prematurey impose on them a rigid ogica scaffoding. Having said this, I am obiged to point out that whie this approach is by design mathematicay fauty, it is nevertheess a step forward compared with the methods used by most textboos to teach fractions because it woud be difficut to find any merit in the atter. On the pedagogica front, it shoud be pointed out that inasmuch as this manua is not a textboo on fractions but the content of a worshop for teachers, the eve of the presentation is higher than is appropriate for a 5th grade cass (say). exampe, at times I mae use of symboic notation in discussions (e.g., + m n ) in order to save time. Actua teaching in a 5th grade cassroom shoud probaby avoid that. I aso mae use of fairy arge numbers (in both denominators and numerators) for my first exampes on each topic, whereas in a cassroom it woud be advisabe to start off with, and the ies. In addition, since the worshop was designed aso to 2 increase the mathematica nowedge of the participants, I have suppied more proofs than woud be advisabe in a 5th grade cassroom. I hope teachers reading this woud mae judicious choices on what to give to their students. In the same breath, et me aso express my strong persona beief that such choices are best made ony after they have mastered everything in this manua. I mentioned in the preceding paragraph that I used fairy arge numbers for iustration. Part of the reason is that I was inspired to give my worshop on fractions by the foowing passage on p. 96 of the NCTM Standards [2]: The proficiency in the addition, subtraction, and mutipication of fractions and mixed numbers shoud be imited to those with simpe denominators that can be visuaized concretey and pictoriay and are apt to occur in reaword settings; such computation promotes conceptua un For
4 derstanding of the operations. This is not to suggest however that vauabe time shoud be devoted to exercises ie or 5 4 which are much harder to visuaize and uniey to occur in rea ife situations. Division of fractions shoud be approached conceptuay. I coud not envision what ind of schoo mathematics education woud, as a matter of principe, favor fractions with simpe denominators over those with arge ones, or how the socaed conceptua understanding of fractions coud be appicabe ony to the former ind of fractions. As to the trivia computations of and , they are the ind that every student who caims to now anything about fractions shoud be abe to do in his or her seep. There is ceary room for improvement in this part of the schoo mathematics curricuum. The present manua is my contribution in this direction. After I had finished writing this manua, I got to read the voume [9] on rationa numbers; it is apparenty somewhat difficut to find, but my coeague Leon Henin was ind enough to show me a copy. Its primary audience is aso schoo teachers and its Chapters 26 are devoted to what we ca fractions here. There are points of contact between [9] and this manua. In particuar, I was both surprised and gratified by the cose simiarity between the first two sections of this manua and Chapter 2 of [9] due to Henin. To begin the mathematics, the natura numbers 0,, 2,, 4,... wi be represented as points on a ine. So tae an infinite ine segment extending to the right from a fixed point. That point wi be designated as 0. Seect a unit of ength ( in., cm, etc.) and mar it off to the right as. Then mar off segments of unit ength successivey to the right to get 2,, 4, etc., as on a ruer. We create thereby an infinite ruer where the postion of each natura number n on the ine indicates the ength of the segment from 0 to n. We sha refer to it as the number ine etc. In this setting, each natura number is not ony a point on the number ine, but aso a ength. For exampe, the (ine) segment from 0 to the number 4 on this infinite 4
5 ruer has ength exacty 4. Incidentay, we sha denote such a segment by the symbo [0, 4]. (By the same toen, [m, n] wi denote the segment from m to n for any natura numbers m and n.) The doube meaning attached to a number woud not be a source of difficuty in students earning because they are aready used to a simiar situation with the ordinary ruer. Our immediate goa is to transate the usua four arithmetic operations on natura numbers into geometric terms. In a cassroom, this woud require giving students sufficient number of dris unti the geometric interpretations become instinctive with them. Let us start with addition: adding (say) 4 to a number is the same as shifting that number to the right of the infinite ruer by 4 units. In anaogy with (the ancient instrument of) a side rue, we can visuaize this as foows: Imagine siding the infinite ruer to the right unti the 0 is under 4, then the vaue of 4 + n can be obtained by ocating the number in the upper ruer that is directy above the number n in the ower ruer. In the figure above, we have indicated how 7 and are the vaues of 4 + and 4 + 7, respectivey. For us, a more usefu way to oo at addition is the foowing: = the ength of two abutting segments one of ength 4 and the other of ength 5 Note that here as ater, when we spea of abutting segments we mean that the segments are ined up in a straight ine from end to end. For exampe, the segments [0, 4] and [0, 5] may be used for this purpose: } {{ } tota ength = 9 5
6 In genera, for any two natura numbers and, we define: + = the ength of two abutting segments one of ength and the other of ength () Next, subtraction. One can obtain the answer to 5 2, for exampe, by cose anaogy with the above siding rue method we empoyed for addition: side the ower infinite ruer this time to the eft unti 2 is right under the 0 of the upper ruer, then the number (which is naturay ) on the upper ruer which is right above the 2 of the ower ruer provides the answer. But again, a more usefu way to thin of this is the foowing: for natura numbers and such that >, = the ength of the remaining segment when a segment of ength is removed from a segment of ength (2) Mutipication is more compicated. 2 is of course the ength of abutting segments each of ength 2, and in genera for natura number and, = the ength of abutting segments each of ength () It goes without saying that in the preceding statement, and can be interchanged. This amounts to the commutative aw of mutipcation, but we sha refer to this more forma aspect of fractions ony sparingy. This said, we turn to a more usefu version of the mutipication of natura numbers in terms of the concept of area of a rectange: 2 = the area of a rectange with sides and 2 This is seen by partitioning this rectange into unit squares (i.e., squares with sides of unit ength) and count that there are 6 of them. Since each unit square has area equa to, the area of the rectange is 6. 6
7 In genera, if n and are natura numbers, then n = the area of a rectange with sides n and (4) By the times students are in the 5th grade, they shoud be aready famiiar with this interpretation of mutipication. Just to be sure, one shoud mae them do many exampes, such as 6 = 8: Division among natura numbers of course appies ony to the situation where one number eveny divides another. When we say = n, we mean = n + n + + n ( times), or, can be divided into equa parts, and each part is of size n. In other words, = n exacty when = n (5) In geometric terms, we use (5) to interpret n as the ength of abutting segments each of ength n, so that for = n, 2 = the ength of a subdivision when a segment of ength is divided into equa parts (6) With the arithmetic of natura numbers out of the way, we are ready to tace fractions. Our first tas is to give meaning to 2, 4, or more generay, m where m and n 7 n are natura numbers. These wi be points on the number ine. However, to prepare students for this definition, it is important they get used to the idea of dividing a given We have purposey ignored the aternate interpretation of = n as the division of into n parts each of which has size. 7
8 ine segment into equa part (i.e., segments of equa ength), where is an arbitrary natura number. In practica terms, of course, this can be easiy accompished: if we are ased to divide (say) [0, ] into 7 equa parts, just use grid paper and choose a unit ength to consist of 7 or 4 grids (or for that matter, any mutipe of 7). The main point is that students shoud fee entirey comfortabe with the idea of equa subdivision of a segment because it wi serve as the foundation of this particuar approach to fractions. To this end, we introduce a geometric construction that routiney divides a given segment into any number of equa parts. This construction, stricty speaing, is not needed in our deveopment of fractions. However, it is a fun activity that students woud enjoy and it heps to put students psychoogicay at ease with the idea of arbitrary equisubdivisions of a given segment. Contrary to the prevaiing trend in education, it is aso a manua activity using ruer, compass, and trianges rather than an eectronic one. Thus suppose we have to divide the segment AB beow into equa parts. We draw a ray ρ issuing from A and, using a compass, mar off three points C, D, and E in succession on ρ so that AC = CD = DE the precise ength of AC being irreevant. Join BE, and through C and D draw ines parae to BE which intersect AB at C and D. A basic theorem of Eucidean geometry then guarantees that C and D are the desired subdivisions of AB, i.e., AC = C D = D B. ρ E D C A C D B It shoud be added that whie there is a standard Eucidean construction with ruer and compass of a ine through a given point parae to a give ine, the practica (and quic) way to draw such ines using a ruer and a (pastic) triange is of interest. So position such a triange so that one side is exacty over the points B and E. Keeping 8
9 the triange fixed in this position, now pace a ruer snugy aong another side of the triange. Since there are two other sides of the triange, choose one so that when the ruer is now hed fixed and the triange is aowed to gide aong the ruer, the side of the triange that was originay on top of B and E now passes over the points D and C as the triange sides aong the ruer in the direction of the point A. When this side is over D, hod the triange and draw the ine aong this side through D; this ine is then parae to BE. Same for the point C. There is nothing specia about the number in the preceding construction. We coud have ased for a division of AB into 7 equa parts, in which case we woud mar off 7 points A, A 2,..., A 7 in succession on the ray ρ so that AA = A A 2 = = A 6 A 7. Join A 7 B, and then draw ines through A,..., A 6 parae to A 7 B as before. The intersections of these parae ines with AB then furnish the desired subdivisions of AB. Be sure to devote sufficient cass time to this activity unti students can treat the division of a segment into any number of equa parts as a routine matter before proceeding any further. 2 Now we can give the definition of fractions. First a simpe case:. Divide [0, ] into equa parts, then by definition, 2 4 is the ength of the any one of these subdivisions, is the ength of 2 such abutting subdivisions, is the ength of such abutting subdivisions, is the ength of 4 such abutting subdivisions, etc. In anaogy with the way the natura numbers were mared down on our infinite ruer, we now add more marings to this ruer. The point to the right of 0 of distance from 0 wi be. Continuing on to the right, each point of distance wi from the preceding one wi give rise to 2,, 4,..., n,... Note that coincides with, 6 coincides with 2, and in genera n coincides with n for any natura number n. So writing / for, we have: 9
10 In genera, et, be natura numbers with > 0. Then by definition: = the ength of a subdivision when [0, ] is divided into equa parts (7) = the ength of abutting segments each of ength (8) For exampe, with = 5, and writing /5 for 5 we have: etc. 5 is caed a fraction of the natura numbers and. Sometime we aso refer informay to as a quotient of and, but the word quotient is used in many situations not necessariy connected with natura numbers. Our infinite ruer now has many more marers: in addition to the natura numbers, we aso have a the fractions, where and are natura numbers and 0. Moreover, in case is a mutipe of, say = n, then n = n, for a natura numbers n,, where > 0 (9) In particuar, n = n for any natura number n. This can be seen directy from our definition of a fraction: n is the ength of n abutting segments each being a subdivision of [0, ] into one part, and hence equas n. By its defintion, has the foowing interpretation: If we divide a segment of unit ength into equa parts, then the tota ength of of them is (0) 0
11 In common anguage, we woud say that is th s of, e.g., 2 is twosevenths of 7 and 4 is fourfifths of. Of course we do not aways dea with engths of segments 5 in rea ife, but more iey with a bag of candy, a box of crayons, a portion of a pie, etc. Nevertheess, it shoud not be difficut for students to draw the parae between twosevenths of a gaon of water and twosevenths of a unit segment, so that the consideration of segments is indeed reevant to the rea word. More wi be said on this topic ater. (See the end of this section and 4 and 6.) It is iey that teachers woud expose students to other pictoria representations of fractions at this point, such as a square divided into 4 equa parts, or a pie cut into 6 equa parts, etc., so a caveat is in order. There is certainy no harm in introducing these modes, but I woud suggest doing so ony after students have become proficient at woring with the number ine and the division of ine segments. One reason is that our reasoning throughout the deveopment of fractions is done with the hep of the number ine. But there is another reason: the pie representation of fractions, for instance, has the drawbac of being cumsy at representing fractions > because teachers and students aie ba at drawing many pies. Reasoning done with the pie mode therefore tends to accentuate the importance of sma fractions. Coud this be the expanation of the passage on p. 96 of the NCTM Standards ([2]) quoted in the introduction? On the other hand, the number ine automaticay puts a fractions, big or sma, on an equa footing so that they can a be treated in a uniform manner. An additiona advantage is the fexibiity of this mode in a inds of discussions, and this advantage woud become most apparent when we come to the mutipication of fractions. Exampe Describe roughy where 84 is on the number ine. Since 68 < 84 < 85 7 and 68 = 4 7 and 85 = 5 7, 84 is between 4 and 5, cose to 5. 7 We pause to note the convention that the symbo of a fraction automaticay assumes that > 0. Students probaby need some expanation as to why we do not consider : in definition (7), woud necessitate the consideration of dividing 0 0 [0, ] into 0 equa parts, which is meaningess. Or one coud wait ti the second
12 interpretation of a fraction in () is in pace before reminding them why we do not divide by 0. Our next assertion is of great importance in this approach to fraction. It is the canceation aw for fractions, and wi be seen to be the inchpin of amost everything ese that foows. It asserts that m =. The specia case where = is just (9) m and, athough it is not obvious, the reason why (9) is true is in fact the reason why the genera case must be true. As usua, et us begin with a simpe exampe: we sha show why = 2 Reca the definition of 2 : mar off to the right of 0 on the number ine 2, 2 2, 2, 4 2, ^ ^ ^ ^ Ca this the first subdivision. Then is the third division point of the first 2 subdivision. Each part of the first subdivision has ength. Now subdivide each of 2 these parts into 4 equa parts; ca this the second subdivision. Each part of the atter has ength. Put another way: 2 parts of the first subdivision fi out [0, ], but it 4 2 taes 4 2 parts of the second subdivision to fi out [0, ]. So the ength of each part of the second subdivision is /2 2 ^ ^ ^ ^ Now parts of the first subdivision fi out [0, ], so it taes 4 parts of the second 2 subdivision to fi out the same segment [0, ]. Thus, by (8), =
13 For a second exampe, et us oo at why 2 5 = is by definition the second marer when mutipes of are mared off on the number 5 5 ine. Ca these the first subdivision. 0 ^ ^ ^ ^ ^ Now subdivide each segment of the first subdivision into equa parts, and ca this the second subdivision. 0 ^ ^ ^ ^ ^ Each segment of the second subdivision has ength, because [0, ] has been 5 divided into 5 equa parts by the second subdivision. Since 2 is the 2nd marer 5 of the first subdivision, there wi be 2 parts of the second subdivision in [0, 2]. 5 Hence 2 = 2, by (8). 5 5 In genera, we have: if,, m are natura numbers with m 0,, m m = () The genera argument is briefy the foowing: Mar off mutipes of on the number ine. Ca this the first subdivision. The th marer is
14 then, by definition. [0, ] therefore contains exacty segments of the first subdvision. Divide each segment of the first subdivision into m equa parts; ca this the second subdivision. [0, ] now contains m segments of the second subdivision; the ength of each segment of the atter is therefore. Since the segment [0, ] contains abutting segments of the first m subdvision, it contains m abutting segments of the second subdivision and therefore has ength m m, by (8). Thus =. m m We now turn to the second interpretation of specia case: we caim mentioned earier. First oo at a 2 = the ength of any subdivision when a segment of ength 2 is subdivided into equa parts. Here is the reasoning: Divide [0, 2] into 2 equa parts; ca this the first subdivision. Note that each part of the first subdivision has unit ength. 0 2 Now further divide each of these 2 parts into equa parts; ca this the second subdivision. Each part of the second subdivision now has ength. 0 2 ^ ^ ^ ^ ^ ^ ^ [0, 2] has now been divided into 2 = 6 equa parts. If we tae 2 parts at a time, we get additiona marers which indicate where [0, 2] is divided into equa parts: 0 2 4
15 ^ ^ ^ ^ ^ ^ ^ We are being ased to show that the first marer after 0 is exacty at the 2 position. But this is so because this marer is the second division point of the second subdivision, and each part of the atter has ength. Let us run over the same argument for a different fraction:. We want to show 4 that 4 = the ength of any subdivision when [0, ] is divided into 4 equa parts. As before, we begin by dividing [0, ] into equa parts; ca this the first subdivision. Each part of the first subdvision is thus of unit ength. 0 2 Next divide each of these parts into 4 equa parts; ca this the second subdivision. Each part of the second subdivision therefore has ength ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ [0, ] has now been divided into 4 = 2 equa parts. If we tae parts at a time, we obtain additiona marers which indicate where [0, ] is divided into 4 equa parts: 0 2 ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 5
16 We must show that the first marer after 0 is exacty at the position, but this 4 is so because each part of the second subdivision has ength, and the marer under 4 inspection is at the third division point of the second subdivision and is therefore exacty. This competes the argument. 4 In a cassroom, more exampes of this ind shoud be discussed, preferaby some by students themseves. Use, for exampe, 5, 4, etc. Be sure to use for this purpose 7 some fractions with a arger numerator than denominator (improper fractions). The reasoning given above is perfecty genera, and we have therefore shown why = the ength of any subdvision when a segment of ength is divided into equa parts, > 0 (2) Here is the brief argument for the genera case. First divide [0, ] into equa parts so that each part has unit ength. Then subdivide each of these parts into equa parts, so that the ength of any part of the second subdivision is. The segment [0, ] has now been divided into equa parts; now tae these parts at a time, thereby obtaining a division of [0, ] into equa parts. The first division point of the atter must be shown to be at the postion, and this is so because it is the th of the division points of [0, ], and the ength of each of these subdvisions is. The second interpretation of then aows us to concude, in view of (6) on the division of natura numbers, that if = m, = (= m) () From now on, we write in pace of between natura number and, > 0. We emphasize that the content of (2) is reay quite nontrivia because our origina definition of is that it is the ength of the a segment consisting of abutting segments each of ength. In the rea cassroom, it is recommended that many more concrete exampes be done by the students, e.g., 5 =, 2 7 = 7, etc. It is very important that they are
17 exposed to ogica arguments of this sort eary. They woud then absorb the reasoning behind the caneation aw by by osmosis instead of just memorize this fact by rote. Whether or not the genera proof shoud actuay be given can ony be decided on a casebycase basis. We pauses to reate our definition of fractions to an everyday situation. Let us start with ony natura numbers but no fractions. If one box of crayons has 6 pieces and another has 2 pieces, how many crayons are there in both boxes together? To compute the answer, we can forget about crayons and thin of just natura numbers, in which case we can use the segment mode and put together two segments, one of ength 6 and another of ength 2. Putting them together produces a segment of ength 8: } {{ } 2 6 { }} { Then we go bac to crayons and concude that these two boxes have 8 crayons. The mora of this is that, whether we ie it or not, we have to resort to abstractions even when doing a simpe everyday probem. Now consider the foowing: how many pieces of candy are there in a fifth of a bag of 5 pieces? The first thing to note is that the everyday expression a fifth of a bag of 5 pieces means exacty the number the pieces in a portion when the 5 pieces are divided into 5 equa portions. This ind of everyday expression is a socia convention that is not unreasonabe, and in any case is one that chidren can easiy earn. Now, this is the same as ooing for the ength of a subdivision when a segment of ength 5 is divided into 5 equa parts. Thus the answer we are ooing for is the fraction 5 5, by virtue of (2). The canceation aw () then gives the expected answer, which is 7. Here is then an exampe of how a rea ife probem is soved by our precise concept of a fraction. We emphasize that the reasoning is the important thing. For exampe, if the bag (for some curious reason) happens to have 2709 pieces of candy and you are aowed to tae 2709 of it, then the number of pieces you tae woud be 2 2, by exacty the same reasoning regardess what the actua vaue of this fraction may be. (This vaue is actuay 29.) The arge number 2709 is purposey used here to stress the irreevance of the size of the numbers one encounters in woring with fractions, in contrast with the common advocacy among educators for the use of sma numbers 7
18 in teaching fractions. Exercise Assume that you can cut a pie into any number of equa parts, what is the most efficient way to cut pies in order to give equa portions to 7 ids? 7 pies to ids? Before we approach the addition of fractions, we first consider the more eementary concept of comparing two fractions. Which is the bigger of the two: 4 or? In terms 7 5 of segments, this shoud be rephrased as: which of [0, 4/7] and [0, /5] is onger? Now by definition: 4 7 is the ength of 4 abutting segments, each of ength 7 5 is the ength of abutting segments, each of ength 5 This comparison is difficut because the two fractions are expressed in terms of different units : and. Imagine, for exampe, if the preceding were the foowing: is the ength of 4 abutting segments, each of ength 6 5 is the ength of abutting segments, each of ength 6 Then we woud be abe to immediatey concude that 4 is the bigger of the two because 7 it incudes one more segment (of the same ength) than. This suggests that the 5 way to achieve the desired comparison is reduce both and to a common unit. 7 5 Students of this eve woud probaby understand this idea better by first considering a more common probem: which is onger, 500 yards or.2 m? In the atter case, everybody nows that we need to reduce both yard and m to a common unit, say, meter. One finds that yard = meter and m = 000 m., so that 500 yards = m.2 m = 200 m Concusion: 500 yards >.2 m. 8
19 In order to imitate this procedure, we have to decide on a common unit for 7 and. The canceation aw () suggests the use of because = = 5 5 = the ength of 5 abutting segments each of ength 5 5 = = 7 5 = the ength of 7 abutting segments each of ength 5 More generay then, 4 7 = = 20 5 = the ength of 20 abutting segments each of ength 5 5 = = 2 5 = the ength of 2 abutting segments each of ength 5 Concusion: 4 7 <. A coser oo of the preceding aso reveas that this concusion is 5 based on the inequaity 5 4 < 7 between the numerators and denominators of the two fractions. We are thus witnessing the socaed crossmutipication agorithm in a specia case. In picture: 0/7 /7 2/7 /7 4/7 5/7 6/7 7/7 0 ^ ^ ^ ^ ^ ^ 0/5 /5 2/5 /5 5/5 5/5 Students shoud be made to wor out for themseves the comparisons of 5 and 4, and, 4 and, 9 and 4, 9 and, 2 5 and, etc. Mae sure that in each case, it is not cear ahead of time which fraction is bigger. In genera, suppose we are to compare and m. The reasoning of the preceding n exampe tes us that we shoud use the canceation aw () to rewrite them as: = n n m n = m n 9
20 where for simpicity we sha henceforth use the agebraic notation of writing n for n, etc. It foows that and for the same reason, < m n exacty when n < m (4) = m n exacty when n = m (5) Either of the two is referred to as the crossmutipication agorithm. Exampe Which is greater: or? Do it both with and without computation. 25 because 2 25 = 575 < Using the crossmutipication agorithm, we see that 2 < = However, this resut coud aso be done by inspection: Both fractions are point on the unit segment [0, ]: 0 2/ /25 It suffices therefore to decide which is the onger segment: the one between 2 24 and the one between has ength 25 and. But the first segment has ength 24 ; ceary the atter is shorter. So again 2 24 < Exampe Do the same with the fractions 94 and The crossmutipication agorithm again shows that 94 < 95 4 and, whie the atter. But we can reason without computation as foows. As in the preceding exampe, we need ony compare which of the two is greater: or. However, we see that < 00 = 00 <
21 So 4 > We are now in a position to dea with the addition of fractions. Imitating the geometric definition of the addition of two natura numbers in (), we define: + m n = the ength of two abutting segments, one of ength and the other of ength m n (6) It foows directy from this definition that = ( times) (7) Stricty speaing, one cannot do repeated addition unti one proves the associative aw of addition, i.e., one must first prove ( + m ) + p n q = ( m + n + p ) q and this is true because both sides are equa to the ength of three abutting segments of respective engths, m, and p. This then aows us to write n q + m n + p q to denote either sum. The same argument extends to a sum of any number of terms so that repeated addition maes sense without specifying the order the additions must be carried out. Whie such coniderations woud not be suitabe for presentation in a 5th grade cass, the teacher must be aware of them in case a precocious youngster raises this issue. Aso immediate from the definition is the fact that m + = m + (8) 2
22 because both sides are equa to the ength of m + abutting segments each of ength. We woud ie to obtain a formua for the sum in (6). The previous consideration of comparing fractions prepares us we for this tas. In a concrete case such as 5 +, we see that we cannot add them as is, in the same way that we cannot add the sum ( m. + ft.) unti we reduce both to a common unit. For the atter, we have in fact ( m. + ft.) = 00 cm + (2 2.54) cm = 0.48 cm. For the fractions themseves, we foow (4) to arrive at 5 + = = = 8 5, by virtue of (). The genera case is just more of the same: given the canceation aw to rewrite + m n = n n + m n n + m =, n where the ast step uses (8). Thus we have the genera formua: and m, we use n + m n = n + m n This formua is different from the usua formua invoving the cm (east common mutipe) of the denominators and n, and we sha comment on the difference beow. Note however that this formua (9) has been obtained by a deductive process that is entirey natura, which shoud give teachers a soid foundation on which to answer students question about why addition doesn t tae the form of + m n = + m + n. The specia case of (9) when = shoud be singed out: since =, (9) + m n = n + m n (20) Thus, = 2 and = 47. A common abbreviation is to write 6 6 m n = + m n in case m < n. (2) 22
23 So 7 5 = 47 and 4 5 =. The entity on the eft side of (2) is usuay referred to as a mixed fraction or mixed number. Operations with mixed fractions are regarded by many teachers and students with fear. The purpose of being so meticuous with the definition in (2) is to aay such fears once and for a: a mixed fraction is merey an shorthand notation to denote the sum of a natura number and a fraction, so that once we expain the distributive aw for fractions ((26) beow), mixed fractions can be handed in an entirey routine fashion. Exampe Compute and = 2 + ( ) = = = Exampe Compute = (5 + 6) = = = = 8 24 = 9 2 which coud aso be written as 7. However, in this case one sees that it is not 2 necessary to go to as common unit of measurement of the fractions and, because we coud use instead: = and = 2. Hence we coud have computed this way: as before = = 9 2 = 7 2, The ast exampe brings us to the consideration of the usua formua for adding fractions. So suppose and m are given. Suppose we now that a natura number A n is a mutipe of both n and, then we have A = nn = L for some natura numbers L and N. One such exampe of A is of course A = n, in which case N = and L = n. Another exampe is A = the cm of n and (which was the case for 4 and 6 above, whose cm is 2). Then = L L = L A + m n = L + mn A and m n = mn nn 2 = mn A, so that where A = nn = L (22)
24 As remared earier, if A is taen to be the cm of n and, then this formua is the one found in amost a the textboos. On the rare occasion that cm does not mae an appearance, it is usuay because no formua for the addition of two fractions is given at a, on the ground that there shoud be a decreased emphasis on fractions (echoes of [2]). From a mathematica vantage point, however, nowing how to add two fractions is important, party because the underying mathematica reasoning (as we have seen) is instructive and party because it is an essentia mathematica technique. So the addition of fractions shoud aways be taught, and taught correcty, which then requires using formua (9) rather than (22). 2 The reason (22) shoud be avoided as the basic formua for addition is not ony that it is ess simpe than (9) and simpicity is a very important criterion in mathematics but aso that (22) says nowing how to get the cm of two natura numbers is a prerequisite to nowing how to add fractions. Formua (9) ays bare the fact that such is not the case. On the pedagogica eve, formua (22) is the ess desirabe of the two because the consideration of cm distracts students from the main idea (such as the expanation given preceding (9)) of how two fractions are added. I hope no reader of this manua wi ever again teach his or her students how to add fractions using (22). Finay, a few words about the subtraction of fractions. Suppose as usua that and m are given and that > m. Imitating the case of natura numbers in (2), we n n define the difference m as n m n = the ength of the remaining segment when a segment of ength m n is removed from a segment of ength 2 The ast thing I want is to mae mysef sound ie a crusader! The truth is that any mathematician who is at a competent woud advocate exacty the same thing. Such being the case, the fact that amost a the textboos use formua (22) for the addition of fractions bespeas ong years of negect of schoo mathematics education by professiona mathematicians. 24
25 Now > m means, by virtue of (4), that n > m. Hence the foowing maes sense: n m n = n n m n = the ength of the remaining segment when m abutting segments each of ength are removed n from n abutting segments each aso of ength n = the ength of n m abutting segments each of ength = n m, n where the ast equay is by definition of the fraction n m. This yieds the formua: n when > m n. Exercise m n = n m n n (2) Large numbers are used in (a) and (b) beow on purpose. You may use cacuator for them provided the answers are in fractions and not decimas. (a) or? =? (b) 0 67 =? (c) ? Do each of these with and without computations. or =? (d) Which is bigger: We now tace the mutipication of fractions. The first question is: what coud m n mean? If =, then m = m which, in view of (), shoud certainy mean n n m + m + + m ( times). When >, the meaning of m in an agebraic sense n n n n is ess cear. However, we reca that mutipication between natura numbers aso has a geometric interpretation in terms of the area of a rectange. See (4). This geometric interpretation maes sense even when the sides of the rectange are no onger natura numbers. Here then is a springboard to get at the mutipication of fractions. Thus define: m n = the area of a rectange with sides and m n If = n =, then this coincides with the ordinary product of the natura numbers and m, by (4). This at east gives us some confidence that such a definition is on the 25
26 right trac. In order to arrive at a formua for the product, we first prove: n = n First consider a simpe case: why is =? Tae a unit square and divide 2 6 one side into 2 equa parts and the other into equa parts. Joining corresponding points of the subdivision then partitions the square into 6 identica rectanges: (24) 2 By construction, each of the 6 rectanges has sides and an its area is therefore 2. However, the tota area of these 6 rectange is the area of the unit square, 2 which is, so each rectange has area, as to be shown. (In greater detai, computing 6 the area of such a rectange is the same as asing for the ength of a subdivision when a unit segment is divided into 6 equa parts. The answer is, by (7).) 6 Next, try showing =. Again we divide one side of a unit square into 6 8 equa parts and the other side 6 equa parts. The obvious connections of corresponding points ead to a partition of the unit square into 8 identica rectanges: 6 By construction, each rectange has sides and, so its area is. By since 6 6 these 8 rectanges are identica and they partition the unit square which has area equa to, the area of each rectanges is 8. The genera case in (24) can be handed in a simiar way. Divide the two sides of a unit square into equa parts and n equa parts, respectivey. Joining the corresponding division points creates a partition of the unit square into n identica rectanges. 26
27 copies n n copies Since each of these rectanges has sides and n by construction, its area is n by definition. But since these n identica rectanges partition a square of area equa to, each of them has area, in the same way that the ength of a subdivision when n a unit segment is divided into n equa parts is, by (7). Thus =, which n n n proves (24). Before attacing the genera case of m, et us consider 2. This is, by n 7 4 definition, the area of the rectange with the width 2 and ength. Again by definition, 7 4 the width consists of two abutting segments each of ength whie the ength consists 7 of three abutting segments each of ength. Joining the obvious corresponding points 4 on opposite sides yieds a partition of the origina rectange into 2 identica sma rectanges. 7 4 Now each of the sma rectanges has sides and and therefore, by (24), has 7 4 area. Since the big rectange is subdivided into 2 such identica rectanges, it 7 4 has area equa to This shows = We now prove in genera: m n = m n We construct a rectange with width and ength m, and our tas is to show that its n area is equa to m. By definition, its width consists of abutting segments each of n 27 (25)
28 ength and its ength m abutting segments each of ength. Joining corresponding n division points on opposite sides eads to a partition of the big rectange into m sma rectanges. copies n m copies Since each of these sma rectanges has sides equa to and, its area is by n n virtue of (24). But the origina rectange has been divided into m of such sma rectanges, so its area is m, as desired. n Observe that if we et = in (25), we woud have m n = m n = m n = (m n + m n + + m ) ( times) n where the ast is because of (8). We therfore see that our definition of the mutipication of fractions is consistent with the usua intuitive understanding of what mutipication by a natura number means. It woud be a good idea to caution students at this point that, whereas in the context of natura numbers mutipy by a number aways resuts in magnification, in the context of fractions this is no onger true. For exampe, if we start with 0, then mutipying 0 by gets, which is far smaer than Up to this point we have not stressed the more forma aspects of arithmetic, such as the commutative aw for addition and mutipication or the associative aw for both. A these facts are straightforward consequences of (9) and (25). However, we shoud singe out the distributive aw because it has nontrivia computationa consequences: ( m ± m ) ( 2 = n n 2 m ) ( ± n m ) 2 (26) n 2 28
29 This can be proved in two ways: agebraicay and geometricay. First the agebraic proof. By (9) and (25), the eft side of (26) equas m n 2 ± m 2 n = (m n 2 ± m 2 n ) n n 2 (n n 2 ) But by (22), we aso see that the right side of (26) equas m n ± m 2 n 2 = m n 2 ± m 2 n n n 2 = m n 2 ± m 2 n n n 2 So (26) hods. For the geometric proof, tae the case of + in (26); the case can be handed simiary. Consider a rectange with one side equa to, and such that the other side consists of two abutting segments of engths m n and m 2 n 2, respectivey. This gives rise to a partition of the rectange into two smaer rectanges as shown. m n + m 2 n 2 } {{ } } {{ } m m 2 n n 2 The distributive aw (26) is merey the statement that the area of the big rectange (eft side of (26)) is equa to the sum of the areas of the two smaer rectanges (right side of (26)). As an appication of the distributive aw, consider 8. It is equa to 6 7 Or more directy, (8 + 6 ) 7 = = = = = = We are now in a position to expain two everyday expressions. What does it mean when someone says twofifths of the peope in a room? The precise meaning is: the tota number of peope in 2 of the parts when the peope in the room are divided 29
30 into 5 equa parts. This is the universay accepted interpretation of this phrase, and there is no reason to ose seep over why this is so, any more than to do the same over why red was chosen as the coor of stop ights (cf. [5]). What we shoud do instead is to transate the atter phrase into precise mathematics. Let us say there are n peope in the room. Then the number of peope in each part when n is divided into 5 equa parts is the same as the ength of a subdivision when a segment of ength n is divided into 5 equa parts. By (2), the answer is n 5. So 2 parts woud be n 5 + n 5 = 2n 5 = 2 5 n. Thus we see that twofifths of n peope means ( ) 2 5 n peope (27) Simiary, twothirds of cars means ( 2 ) cars, etc. A reated common expression is 65% of the students are men or this product contains % fat. Let us oo at the first one. Again, we must accept that there is universa agreement on its meaning: If we divide the students into 00 equa parts, the the men constitute 65 of those parts. In terms of our understanding of fractions, suppose there are n students atogether and we divide them into n equa parts, then each part has n 00 subdivision when it is divided into 00 parts is students. That is, if a segment has ength n, then the ength of each of students in 65 of these parts is n + n + + n short, n, by (2). Thus the tota number 00 65n (65 times) = = 65 n. In ( ) 65 65% of n students means 00 n students (28) A simiar statement hods for every expression invoving percents. Exercise You may use cacuator for (d) and (e), but again be sure that the answers are in fractions rather than decimas. (a) =? (b) 2 7 ( ) =? (c) =? (d) =? (e) =? Before considering the division of fractions, et us oo at a seemingy unreated probem of constructing a rectange when one side and its area are prescribed. For exampe, is there a rectange with area equa to and with one side equa to 9 4? If 0
31 such a rectange exists and the the other side has ength s, then 9 s =. If s is a 4 fraction 9, then we want =, by (25). In view of the canceation aw (), we see 4 by visua inspection that = 4 and = 9 woud do. Thus the soution of our probem is a rectange of dimensions 4 and 9. The same reasoning yieds the fact that the 9 4 rectange with area equa to and one side equa to is one with dimensions and. Consider a more genera probem: suppose we want a rectange with one side prescribed (say 9) and its area aso presecribed (say ), how to find the other side? 4 7 Again, suppose the other side is, then we saw that = 4 and = 9 woud sove the probem if the area were equa to. Athough the area is now and not, 7 nevertheess, we see that we have the basis of a soution because: if a rectange has area equa to A and we mutipy one side by r, then the resuting rectange wi have area equa to ra. Since we aready have a rectange of area equa to (that with dimensions 4 9 and 9 4 ), to get a rectange with area 7 a we need to do is to mutipy one side by. Mutipying the side of ength 4 by 4, we get. So the desired rectange has dimensions 9 4 and In the same way, the rectange with prescribed area m and a prescribed side has n second side equa to m, as m = m, which is aso readiy confirmed by (25) and n n n (). We formaize the purey agebraic part of this statement as foows: given fractions A and C, with A 0, then there is a fraction B so that A B = C (29) This is so because if we et A = do. Put another way, and C = m m, then we have seen that B = woud n n given and m n, with 0, there is a fraction B so that B = m m, namey, B = (0) n n This may aso be the right moment to comment on the provision that A 0, i.e., 0. The atter impies 0, which is needed because otherwise the fraction m n woud have 0 as denominator. Another way to oo at it is that if A = 0, then A B woud aways be 0 no matter what B is and our probem woud have no soution. The property about fractions stated in (29) is of critica importance. It is the one property that distinguishes fractions from natura numbers, in the sense that whereas
32 in terms of arithmetic operations one can hardy te the difference between natura numbers and fractions, the property (29) is not shared by natura numbers. In other words, the statement that given natura numbers A and C with A 0, we can aways find a natura number B so that A B = C is ceary fase. For exampe, et A = 2 and C =. In fact, property (29) is the gateway to the division of fractions, as we now expain. As in the case of mutipication, we begin by asing, for fractions A and C, what coud it mean to divide C by A? Foowing the remar after (), we sha henceforth use the notation to stand for C divided by A. C A Let us retrace our steps and reexamine what it means, if A and C are natura numbers, to write C A = B. For exampe, 2 = 4 because we can divide 2 into equa parts (namey, 2 = ), and the size of each part is 4. Simiary, 4 = 2 because we can divide 4 into 7 equa parts (namey 7 4 = ) and the size of each part is 2. In other words, 2 = 4 because 2 = 4, and 4 = 2 because 4 = 7 2. In the same way then, for natura 7 numbers A. B, and C, C = B means C = B + B + + B (A times) = A B. This A viewpoint is entirey consistent with the meaning given to the quotient of natura numbers in (2). This then prompts us to give meaning to C = B for fractions A, B and C A with A 0 by defining: C A = B if B is the fraction satisfying C = A B () The importance of (29) is now evident: without it, this definition () may not mae sense because a priori there may not be such a B for the given A and C. In fact, using (0), we can mae () more expicit by deriving the expicit formua for the division of fractions: m/n / = m n for 0 (2) because, visiby, the fraction m n satisfies m n = m n. The specia case of (2) where m = n = = is of particuar interest: / = 2
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