Mthemtil Proofs Tle of Contents Proof Stnr Pge(s) Are of Trpezoi 7MG. Geometry 8.0 Are of Cirle 6MG., 9 6MG. 7MG. Geometry 8.0 Volume of Right Cirulr Cyliner 6MG. 4 7MG. Geometry 8.0 Volume of Sphere Geometry 8.0 5 Algeri Ientities Alger I 0.0 6, 7, 8, 7 Irrtionlity of 7NS.4 9, 0, Alger I.0 Alger I 5.0 Division y 0 NS.6 Diviing y Frtion 6NS. 7NS. Slope of Line is Constnt 7AF. Alger I 6.0 Prout Property of Rils Alger I 0.0 4 Squre Roots Alger I.0 5 Pythgoren Theorem n Converse 7MG. 6, 7 Geometry 4.0 Definition of pi 6MG. 8 6MG. Cirumferene of Cirle 6MG. 8 6MG. Zero Prout Rule 7AF. Alger I.0 9 N is irrtionl if N is positive integer tht 7NS.4 0 is not perfet squre Alger I. Density A Property of Rtionl Numers Alger I. Squre n Tringulr Numers 7NS.0 4 Alger I. Multiplition of Negtive Numers 6NS. 5 7NS. Zero Property of Multiplition 7 NS.0 6 Alger I. Commuttive, Assoitive, n Distriutive 7NS.0 8 Properties Alger I 5.0 Terminting Deimls 7NS.5 0
Are of Trpezoi Proof of A h Trpezoi Region h Region h Region x x The proof of the re formul flls into ple s follows: The re of tringle Region : xh A l eft trin gle The re of retngle Region : h A retngle The re of tringle Region : A xh right tringle The omine re of the three Regions: A Trpezoi xh h xh Aing the three regions ATrpezoi xh h h h xh ATrpezoi h h h Aition Property ATrpezoi h h Simplify ATrpezoi h Q.E.D. Are of Cirle
Using the onept of the Are of Tringle to show tht Are of Cirle = r. r If we ut the irles long the rius of the isk n let them fn out to eome stright lines, we get tringle (euse the rtio of the irumferene of the irles to their imeters is onstnt). The se length of the resulting tringle is equl to the irumferene of the originl irle, n its height is equl to the rius of this irle. Thus, the re of irle is equl to hlf of the prout of the rius n the irumferene. A Tringle re of se height Sine, A re of se height A Tringle Tringle r r r r whih equls to the re of the DISK or CIRCLE. Volume of Right Cirulr Cyliner
Volume = (Are of Bse)(Height) V r h If you en retngulr sheet of pper, ringing two opposite sies together, you will get n open tue. r h h This is lle n open yliner. The surfe re of this open yliner is the re of its urve surfe given y the prout of the istne roun the rim n the height. rh r h If the top n ottom of the yliner re overe, we will hve lose yliner. Thus, the formul for volume of prism or yliner is given y V = Volume = (Are of Bse)(Height) V r h 4
Volumes Cvlieri s Priniple: Solis with the sme height n with ross-setions of equl re hve the sme volume; in prtiulr, prisms or yliners with equl ses n heights hve the sme volume. Volume of Sphere An Experimentl Approh: If we uil yliner with height r n the imeter r n fill the yliner with wter n then remove the sphere from the yliner, the wter will only tke up of the yliner. Thus, the volume of the sphere is the volume of the yliner hving the sme imeter s the sphere n height equl to the imeter. r r Volume of the Cyliner: V r h. Proven erlier Eqution h r. Given V r r. Sustitution So, V r. Simplify Eqution We know from the experiment tht: Volume of the Sphere the Volume of Cyliner Volume of the Sphere r Volume of the Sphere = 4 r Q.E.D. 5
Algeri Ientities Algeri mnipultion very often n e expline with the i of geometril figures. Ojetive : To show tht The figure on the right is squre me up of four Regions: I, II, III & IV. I A II Eh sie of the squre hs length units. III IV Are of the squre is: squre units. A A Are Region I: A squre units Are Region II: Are Region III: A squre units A squre units Are Region IV: A squre units Are of the Squre = Are Region I + Are Region II + Are Region III + Are Region IV, whih is Sum of the res in the lrge squre ( ) Equte the two expressions for the re 6
Ojetive : To show tht (see n lternte proof on pge 7) - I IV The figure on the left is me up of four Regions: I, II, III, n IV. II III Regions: I, II, III mke up SQUARE, whih hs n AREA of squre units. Region IV hs n re of units. squre Are Region I: A squre units Are Region II: A squre units Are Region III: A squre units Are Region IV: A squre units Are Regions I & IV: A squre units We know Region I + Region II + Region III = ( ) ( ) Sum of res in the squre Isolting ( ) ( ) ( ) ( ) Distriutive Property ( ) Assoitive n Commuttive Properties 7
Ojetive : To show tht II I II I Figure Figure A smll squre (in re), re squre units is remove from igger squre of re squre units s shown in the Figure ove. Thus, the re of regions I n II is squre units. In Figure, the two remining regions, I n II, re rerrnge to form the retngle shown. The re of the newly forme retngle is equl to Hene, Further justifition: Totl Are of Figure : Totl Are of Figure : squre units 8
Proof is irrtionl The Greeks isovere tht the igonl of squre whose sie is unit long hs igonl whose length nnot e rtionl. By the Pythgoren Theorem, the length of the igonl equls the squre root of. So the squre root of is irrtionl! The following proof is lssi exmple of proof y ontrition: We wnt to show tht A is true, so we ssume it's not, n ome to ontrition. Thus A must e true sine there re no ontritions in mthemtis! Proof: Suppose is rtionl. Tht mens n e written s the rtio of two integers p n q, where q 0. p Eqution # q Where we my ssume tht p n q hve no ommon ftors. (If there re ny ommon ftors we nel them in the numertor n enomintor.) Then, p Squring oth sies of Eq. # q This implies p q Isolting Thus, p p is even. This mens tht p itself must e even. woul lso e o. (An o numer times n o numer is lwys o.) So p is n even numer. Then p is times some other whole numer, or p k, where k is this other numer. We on't nee to know extly wht k is; it oesn't mtter. Soon we will get our ontrition: Why? Beuse if p ws o, then p p Then Hene p = 4k. So 4k = q, hene q = k. q is even n therefore q itself must e even. So p n q re BOTH even. They oth hve ommon ftor of. This ontrits our ssumption tht p n q hve no ommon ftors. The squre root of is NOT rtionl. 9
The Irrtionlity of Prove tht is n irrtionl numer. Solution: The numer,, is irrtionl, ie., it nnot e expresse s rtio of integers n. To prove tht this sttement is true, let us ssume tht is rtionl so tht we my write = /. for n = ny two integers. To show tht is irrtionl, we must show tht no two suh integers n e foun. We egin y squring oth sies of eq. : = /. or =. From eq., we must onlue tht (n, therefore, ) is even; (n, therefore, ) my e even or o. If is even, the rtio / my e immeitely reue y neling ommon ftor of. If is o, it is possile tht the rtio / is lrey reue to smllest possile terms. We ssume tht (n, therefore, ) is o. Now, we set = m, n = n +, n require tht m n n e integers (to ensure integer vlues of n ). Then = 4m. n = 4n + 4n + 4. Sustituting these expressions into eq., we otin (4n + 4n + ) = 4m 5. or 4n + 4n + = m 6. The L.H.S. of eq. 6 is n o integer. The R.H.S., on the other hn, is n even integer. There re no solutions for eq. 6. Therefore, integer vlues of n whih stisfy the reltionship = / nnot e foun. We re fore to onlue tht is irrtionl. 0
Division y 0 is Meningless Sine ivision is efine in terms of multiplition, the rules for the sign of prout le iretly to the orresponing rules for the sign of quotient, whih re given elow: Symols 0 ; 0 0 0 0 0 0 0 Divisor n ivien Both Positive Both Negtive One Positive One Negtive Divien 0 Divisor not 0 Quotient Positive Positive Negtive 0 Note: For 0, in the quotient 0 Q You must hve Q 0 Sine the prout Q Cn e 0 only if one of the ftors is 0. Therefore, 0 0 Also, for 0, if you h Q, 0 Then it woul follow tht But this is impossile, sine 0 0Q n 0Q 0 For exmple, suppose you wishe to fin the quotient Q in: Q 0 You woul hve to fin numer Q suh tht Q 0. But tht is impossile euse you hve lrey seen tht the prout of 0 n ny numer must e 0. Thus, we n onlue tht ivision y 0 is meningless.
Diviing y Frtion If n re rtionl numers with,, n 0, then. Let Q, where 0, 0, n 0 (you shoul orinrily sustitute prtiulr numers for,,, n ) Fining the quotient Q mens fining the missing ftor Q in Q Sine, Q n nme the sme numer The prout of this numer n n e written either s Q or s Tht is, Q Therefore, Q Assoitively Q Q Q.E.D.
Slope of Line is Constnt Prove tht the slope of nonvertil line oes not epen on whih two points re use to lulte it. Conept: Slope of line is CONSTANT n it oes NOT epen on whih two points re use to lulte it. Proof: Let A, B, C, D e ny four points on the line. Prove tht the slope otine using A n B equls the slope otine using C n D. D C x run y rise B A y rise x run Stuy Tip: If two figures re similr, then the rtios of the lengths of orresponing sies re equl. y y x Definition of similr tringles x y x x y Cross prout property of equlity y x x y xx x x Divie oth sies y x x y y x x Simplify Slope using points A n B = Slope using points C n D Q.E.D.
Prout Property of Rils Prove: Proof: n Definition of ue root Sustitution, Commuttivity, Assoitivity Definition of ue root Sustitution or trnsitive property of equlity Uniqueness of the rel ue root of numer Q.E.D. Prove:, 0 Proof: n Definition of ue root Definition. If 0, then Sustitution Definition of ue root n Sustitution or trnsitive property of equlity Uniqueness of the rel ue root of numer Q.E.D. n n 4
Squre Roots Prove: Any positive rel numer hs POSITIVE rel numer s squre root The re of SQUARE is otine y squring the length of sie. The sie of squre with n re of squre units must hve length whose squre is. Tht is, the sie hs length. Figure on the right shows squres with AREAS,,, n 4 n sies mesuring,,, n, respetively. Figure These squres show tht ny positive rel numer k hs POSITIVE rel numer s one of its squre roots. Speifilly, the squre whose re is equl to k squre units hs sie whose length is equl to the positive squre root of k. Sine every rel numer hs n itive inverse, n sine every positive rel numer lerly hs negtive squre root s well. Thus the sttement for eh rel numer, is onsequene of our efinition tht k, k 0, shll nme only nonnegtive numer. We lso hve the sttement This sttement simply resserts the efinition of k. This ie is of importne in lter mthemtis ourses when, for exmple, stuents re ske to solve equtions suh s: x 4. 5
Pythgoren Theorem Consier Figure, squre whose sie hs length +. As inite, there re four ientil right tringles n n interior qurilterl. The legs of eh tringle hve lengths n. The hypotenuse of eh tringle is. The mesures of the two ute ngles of eh tringle hve sum of 90. Now look t the point P. Sine the two ute ngles t tht point hve sum of 90, the interior ngle of the qurilterl must hve mesure of 90. The sme is true for eh of the other verties of the interior qurilterl. An the four sies of the interior qurilterl ll hve length. Thus the interior qurilterl is squre. Hene, the re of the lrge squre = the sum of the res of the 4 tringles plus the re of the interior squre 4 4. But the lrge squre n lso split into two squres n 4 tringles, s shown in Figure. The re of Figure is 4. These two expressions for the re of the lrge squre must e equl. Hene, 4 4 + = Figure Figure 6
Converse of Pythgoren Theorem Consier tringle ABC with sies of length,, n respetively. In the figure, ngle C is unknown. We ssume tht + =, n we wnt to prove tht ngle C is right ngle. Now onsier tringle A B C. As inite, this is right tringle with the right ngle t C n legs of length n. Sine we know tht + = n tringle A B C is right tringle, it follows tht the length of its hypotenuse must e. Thus in the two tringles, the three pirs of orresponing sies re equl in length. Hene these tringles must e ongruent y the Sie-Sie-Sie theorem of ongruent tringles. Hene their orresponing ngles re equl. Sine ngle C is right ngle, so is ngle C. B B? C A C A 7
Definition of π The numer π is efine to e the length of the perimeter (usully lle the irumferene) of irle whose imeter is unit. Its vlue is irrtionl ut is pproximtely equl to.459. Cirumferene of Cirle Notie tht ny two irles re similr to eh other. In similr geometri figures, the lengths of orresponing mesures re proportionl. Now onsier the two irles elow, with rii of ½ n r. Cll the imeter n irumferene of the seon irle n C respetively. Sine these irles re similr, the rtio of the irumferene of the first irle to its imeter is equl to the rtio of the irumferene of the seon irle to its imeter. This mens tht π/ = C/. When we simplify this eqution, we get C = π, or equivlently, sine = r, we hve C = πr. / C r 8
Are of Cirle Consier the following figure. The irle ontins n insrie regulr hexgon. As inite, the length of eh sie of the hexgon is s n the pothem (the istne of the perpeniulr segment from the enter of the irle to the sie of the hexgon) is. Notie tht the hexgon n e split into six ongruent tringles. Then the re of one of these tringles is (/)s. Hene the re of the entire hexgon is 6(/)s = (/)(6s) = (/) times the perimeter of the hexgon times the pothem. Now suppose inste of n insrie regulr hexgon, we hve n insrie regulr polygon with sies. Using the sme symols s ove, the re of this polygon is (/) times the perimeter of the polygon times the pothem. We n lso see tht the re of the polygon is loser to the re of the irle thn the re of the hexgon. Clerly, s the numer of sies of the polygon inreses, the re of the polygon gets loser n loser to the re of the irle. The formul for the re of this polygon is gin (/) times its perimeter times the pothem. Finlly, notie tht s the numer of sies inreses, the perimeter of the polygon pprohes the irumferene of the irle, πr, n the pothem pprohes the rius of the irle r. Thus the re of the irle is this limiting vlue, (/) times πr times r. When simplifie, this is just πr. S Zero Prout Rule: If = 0, then either = 0 or = 0. Suppose tht the prout is equl to zero. Suppose lso tht is not zero. Then the multiplitive inverse of, -, exists. So we hve: = 0 ( - ) = ( - )0 Multiply oth sies y - ( - ) = 0 Assoitivity of multiplition = 0 Definition of multiplitive inverse = 0 Definition of multiplitive ientity 9
A proof tht N is irrtionl if N is positive integer tht is not perfet squre The proof tht we re out to rener is epenent on the well orering priniple for positive integers. This priniple sttes tht ny nonempty set of positive integers ontins smllest positive integer. As in the usul proof tht the is irrtionl, we will proee y ontrition. Thus ssume tht the N is rtionl. Hene there re positive integers k n l with l 0 suh tht k N. Consier the set l k S { l : N for some positive integers k n l}. l The set S is nonempty set of positive integers; therefore, it ontins smllest positive integer y the well orering priniple. Denote the smllest positive integer in S y n. Thus there is positive m integer m suh tht N. Now ivie the integer m y the integer n to otin quotient q n n reminer r stisfying m nq r, where 0 r n; note tht the reminer r is greter thn zero sine the numer N is not perfet squre. Now from the eqution m nq r N n n we otin the eqution ( nq r) n q nqr r N. n n Hene Nn n q nqr r. Now rewrite this eqution to otin the equtions ( N q ) n nqr r ( N q ) n nqr nqr r n[( N q ) n qr] r( nq r) Now we otin the frtions ( N q ) n qr nq r N. r n Sine r n N re positive integers, we onlue tht the numertor ( N q ) n qr is positive integer. Therefore the positive integer r is in the set S. But r is stritly less thn n; thus we hve otine ontrition to the ft tht the integer n is the smllest positive integer in the set S. 0
A seon proof tht is irrtionl The proof tht we re out to rener is epenent on the well orering priniple for positive integers. This priniple sttes tht ny nonempty set of positive integers ontins smllest positive integer. As in the usul proof tht the is irrtionl, we will proee y ontrition. Thus ssume tht the is rtionl. Hene there re positive integers k n l with l 0 suh tht k. Consier the set l k S { l : for some positive integers k n l}. l The set S is nonempty set of positive integers; therefore, it ontins smllest positive integer y the well orering priniple. Denote the smllest positive integer in S y n. Thus there is positive m integer m suh tht. Sine, the integer m stisfies m n. Thus there is positive n integer k suh tht m n k. Furthermore, sine n k m n n ( n k) we onlue tht the positive integer k stisfies the inequlity k n. Now. Hene n n kn k n. Rewrite the previous eqution to otin the eqution n kn kn k. Therefore, n( n k) k( n k). Finlly we otin the eqution n k n k. k n The integer n k is positive integer sine k n. This inequlity together with the previous eqution shows tht the integer k is in the set S. But the positive integer k is less thn the integer n. We hve estlishe ontrition.
Density A Property of Rtionl Numers The set of rtionl numers hs the property tht etween ny two ifferent rtionl numers there re infinitely more rtionl numers. This property is referre to s ensity. One wy "etween ny two rel numers n there exists rtionl numer r suh tht r " then the proof is s follows: We first nee the Arhimeen Property: If x > 0 n y > 0 re rel numers, then there exists positive integer n suh tht n x y. This tells us tht even if is quite smll n is quite lrge, some integer multiple of will exee. Another wy to think of this is, "given enough time, one n empty swimming pool with te spoon." We nee to show tht given rel numers n we n onstrut m n n. We n ssume tht n > 0 without loss of generlity. If our frtion m n e the negtive numer. m for some integers n is to e negtive, just let m Sine n > 0 we n multiply everything y n n we on't hve to worry out flipping the inequlity. Thus we nee n m n. Sine we hve 0n so, y the Arhimeen property, there exists positive integer n suh tht n( ) (pik x n y in the sttement of the Arhimeen property ove; we n o this sine n re oth greter thn zero). Sine n n it is firly evient tht there is n integer m etween n n n. Does this mke sense? If the ifferene etween two numers is greter thn one unit (i.e., if two numers re more thn one unit prt), there must e n integer etween them, sine the integers re spe extly one unit prt. If this isn't ler, try to pik two numers tht re more thn one unit prt suh tht there isn't n integer etween them. I think you'll fin yourself unle to o so. Thus there exist integers m n n suh tht n m n, whih implies m whih ws to e shown. n Notie tht this proof oesn't tell us how to fin m n. All it oes is show tht m n exists. Suh proofs re very ommon in higher mthemtis.
Note: One wy of ientifying numer etween two given numers is to fin their verge. Tht the verge of two numers lies extly hlfwy etween them n e proven riefly s follows: if n re rtionl numers suh tht, ssume tht their verge,, oes not lie extly hlfwy etween them. Then Sine the lst sttement is flse, the ssumption is lso flse, n n, where. oes lie miwy etween
Squre n Tringulr Numers To ompute squre of given numer = 4, = 9, 4 = 6, n so on. Why is the opertion of multiplying numer y itself lle squring? The reson is est grspe from the following piture. A squre with sie of length n n e visulize s omprising gri of n n smller squres of size. Tringulr numers owe their nme to similr onstrution, now of tringles. Atully we lrey use tringulr numers in the proof of Lemm ove. Tringulr numers re in the form,, 6, nn ( ) 0, 5,... The generl formul is tht for every n the numer N is tringulr n esries the numer of points rrnge in tringulr shpe with n points on sie s shown elow. Of ourse it nees to e proven tht the totl numer of rosses rrnge s in the igrm is given nn ( ) y the formul N. However, we n oserve tht the numer of rosses in every tringle equls the sum + +... + n, where n is the numer of rosses on the sie of the tringle. Thus, long the wy, we re going to prove tht + +... + n = n(n+)/. Another igrm will mke this sttement ovious Inee, two tringles, eh with n rosses on the sie, together form n nn retngle. You my question whether pointing to the igrm proves nything. No, of ourse y itself the igrm oes not prove nything. But it oes help to visulize n perhps isern the proof. Inee, we seek formul for the sum of integers from through n. This sum represents the numer of rosses in tringle. Two tringles of whih one is rotte 80 o form retngle, i.e. shpe in whih ll rows hve the sme numer of rosses. Rows in the first tringle ount rosses,,... in the inresing orer wheres in the seon we hve rows with n, n-,... rosses in the eresing orer. Let us mimi this proeure in n lgeri mnner. First tke two sums (one in inresing, nother in eresing orer) ( +... + n) + (n +... + ). Now, regroup the terms y omining first terms in oth sums, then seon terms n so on: Quite rigorously. Q.E.D. [+n] + [+(n-)] + [+(n-)] +... + [(n-)+] + [n+] = n(n+) It is ovious tht ll Euli's perfet numers re tringulr. 4
Multiplition of Negtive Numers: (-)(-) = We know the ssoitive property of ition n the istriutive property of multiplition over ition. We lso know the ommuttive properties of ition n multiplition of rel numers. We know the ientity properties of ition n multiplition: 0 0 n for ll rel numers. Finlly, we know the inverse property of ition: For eh rel numer there is rel numer ( ) suh tht ( ) ( ) 0. Using these si rithmeti properties we re le to prove tht ( )( ). We hve lrey estlishe tht 00for ll rel numers ; thus ( ) 0 0. Also ( ) ( ) 0. So pplying the multiplitive ientity property n the istriutive property we otin the eqution ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) 0 0. So we hve shown tht ( ) ( )( ) 0. Now to oth sies of the ove eqution to otin [ ( ) ( )( ) ] 0 To this ltter eqution pply the ssoitive property of ition to otin ( )( ) 0 ( )( ) [ ( ) ] ( )( ) [ ( ) ( )( ) ]. 5
Zero Property of Multiplition: 0 = 0 = 0 for ll rel numers We know the ssoitive property of ition n the istriutive property of multiplition over ition. We lso know the ommuttive properties of ition n multiplition of rel numers. We know the ientity properties of ition n multiplition: 0 0 n for ll rel numers. Finlly, we know the inverse property of ition: For eh rel numer there is rel numer ( ) suh tht ( ) ( ) 0. Using these si rithmeti properties we re le to prove tht 0 0 0 for ll rel numers. Let e n ritrry rel numer. Then using the ientity n istriutive properties we otin the eqution () 0 (0 0) 0 0 By the inverse property for ition, there is rel numer ( 0) suh tht So from eqution () we otin 0 ( 0) ( 0) 0 0. [ ] 0 ( 0) 0 0 ( 0). Applying the ssoitive property of ition, we otin the eqution [ ] 0 ( 0) 0 0 ( 0) Applying the inverse property of ition, we otin the eqution 0 0 0 We onlue tht 0 0. Sine multiplition is ommuttive we lso hve tht 0 0. 6
To show tht (n lternte pproh see pge 7) No hrm is one if we ssume tht. - The figure on the right is squre me up of four Regions: I, II, III & IV. - I A II Eh sie of the lrge squre hs length units. III IV Are of the lrge squre, ompose of the four regions is squre units. A ( ) squre units Are Region I: Are Region II: A squre units Are Region III: A squre units Are Region IV: A squre units A A Are of the Squre = Are Region I + Are Region II + Are Region III + Are Region IV Therefore we otin the eqution ( ) ( ) ( ). Now simplify this eqution using ssoitive n ommuttive properties of ition to otin the eqution ( ). We n now solve for ( ) to otin ( ). 7
Use the ommuttive, ssoitive n istriutive properties n the four opertions to justify eh step of the following lultion: [ (4 7) + ( (4 ) + 8)] [( (( + 8) + 4) (7 )] = [ (4 7) + ( (4 ) + 8)] [( ((8 + ) + 4) (7 )] Commuttive Property = [ (4 7) + ( (4 ) + 8)] [( (8 + ( + 4)) (7 )] Assoitive Property = [ (4 7) + ( (4 ) + 8)] [( (8 + 7) (7 )] Aition = [[( 4) ( 7)] + ( (4 ) + 8)] [( (8 + 7) (7 )] Distriutive Property = [([( 4) ( 7)] + (4 )) + 8] [( (8 + 7) (7 )] Assoitive Property = [([( 4) ( 7)] + (4 )) + 8] [(( 8) + ( 7)) (7 )] Distriutive Property = [([ ] + (4 )) + 8] [(6 + 4) (7 )] Multiplition = [( 9 + (4 )) + 8] [(6 + 4) 4] Sutrtion = [( 9 + 4 (4 )) + 8] [(6 + 4) 4] Division = [( 9 + [(4 4) (4 )]) + 8] [(6 + 4) 4] Distriutive Property = [([(4 4) (4 )] + (-9)) + 8] [(6 + 4) 4] Commuttive Property = [[(4 4) (4 )] + ((-9) + 8)] [(6 + 4) 4] Assoitive Property = [[6 8] + ((-9) + 8)] [(6 + 4) 4] Multiplition = [[6 8] + ((-9) + 8)] [(4 + 6) 4] Commuttive Property = [8 + ((-9) + 8)] [(4 + 6) 4] Sutrtion = [8 + ((-9) + 8)] [0 4] Aition = [8 + (-)] [0 4] Aition = [8 + (-)] [5/] Division = 7 [5/] Aition = ½ Sutrtion 8
Prove tht multiplition is ommuttive y showing s typil exmple tht 9 58 = 58 9. Use only the multiplition lgorithm, expne form, ftoring, n the istriutive property. Proof. 9 58 = [0 + 9] 58 Expne form = [0 58] + [9 58] Distriutive property = [0 58] + [9 (50 + 8)] Expne form = [0 58] + [9 50] + [9 8] Distriutive property n ftoring = [0 (50 + 8)] + [9 50] + [9 8] Expne form = [0 50] + [0 8] + [9 5 0] + 7 Distriutive property, ftoring, n multiplition = [0 5 0] + [0 6] + [45 0] + 7 Multiplition n ftoring = [0 0 0] + 60 + 450 + 7 Multiplition = 000 + 60 + 450 + 7 Multiplition = 68 Aition 58 9 = [50 + 8] 9 Expne form = [50 9] + [8 9] Distriutive property = [50 9] + [8 (0 + 9)] Expne form = [0 5 9] + [8 0] + [8 9] Distriutive Property n ftoring = [0 5 (0 + 9)] + [8 0] + [8 9] Expne form = [0 5 0] + [0 5 9] + [8 0] + 7 Distriutive Property, ftoring, n multiplition = [0 5 0] + [0 45] + [6 0] + 7 Multiplition n ftoring = [0 0 0] + 450 + 60 + 7 Multiplition = 000 + 450 + 60 + 7 Multiplition = 68 Aition Sine oth prouts re equl to 68, they must e equl to eh other. Thus multiplition is ommuttive in this exmple. But sine there ws nothing speil out this exmple, we n onlue tht multiplition is lwys ommuttive. 9
Prove tht eiml is terminting eiml if n only if it is equivlent to frtion whose enomintor ontins no prime ftors other thn n 5. Proof. Suppose the enomintor of frtion ontins only the ftors, 5 or oth of them. Then we n multiply tht enomintor y itionl s n 5s so tht there is n equl numer of s n 5s. We lso multiply the numertor of the frtion y the sme numer of s n 5s. This proues frtion equivlent to the originl one whose enomintor onsists of n equl numer of s n 5s. But this mens tht the enomintor is power of 0. A frtion whose enomintor is power of 0 orrespons to terminting eiml. Thus, if the frtion with enomintor tht is power of 0 is the frtion. n /0 n, then the eiml tht is equivlent to this frtion is the terminting eiml 0.. n. Now suppose tht we egin with terminting eiml, sy 0.. n. Then this eiml is equivlent to the frtion. n /0 n. The enomintor of this frtion ontins only the prime ftors n 5. 0