1.2 The Integers and Rational Numbers

Transcription

1 .2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl numers {, 2, 3, 4, 5,...}, II. The negtive integers {, 2, 3, 4, 5,...}, n III. The numer 0. The orering on the nturl numers extens to n orering of the integers:... < 5 < 4 < 3 < 2 < < 0 < < 2 < 3 < 4 < 5 <... ut there is no well-orere principle for the integers since mny susets of Z (incluing Z itself) hve no smllest element. Negtive numers my seem ovious toy, ut there ws long perio of time when only positive numers were use. The introuction of 0 is often cite s evience of the scientific superiority of the estern cultures uring the mile ges. It is remrkle tht we cn esily (if teiously) exten the opertions of ition n multipliction of nturl numers to inclue 0 n the negtive integers, mintining ll the funmentl lws of rithmetic. Negtive numers re necessry toy ecuse, in our society se upon the right to the pursuit of hppiness through creit cr et, we nee to e le to sutrct! Definition of Aition. + is efine on cse y cse sis: Cse I. n is positive. Then + n is efine y inuction. (i) + is the next numer fter, (ii) Ech + (n + ) is the next numer fter + n Cse II. n is negtive. Then + ( n) is lso efine y inuction. (i ) + ( ) is the numer immeitely efore, n (ii ) Ech + ( (n + )) is the numer immeitely efore + ( n) Cse III. 0. By efinition, + 0 The ssocitive n commuttive lws of ition cn now e prove for this new efinition of ition y the sme proof-y-inuction strtegy we use in. (ut it is teious, involving lots of ifferent cses, so we won t o it!) Since the first cse of the efinition of ition is ienticl to the efinition of ition of nturl numers from., the two itions give the sme result when pplie to two nturl numers.

2 2 CHAPTER. NUMBERS Before we tckle multipliction, we introuce: The Negtion Trnsformtion: Negtion is the function: : Z Z efine y: (n) n, (0) 0 n ( n) n. It is cler tht ( ) for ll integers (oule negtives cncel) n tht tking negtives reverses orer: if < then < Negtion hs the following three importnt properties, too: Proposition.2.. For ll integers, + ( ) 0 (n ecuse of this, we sy tht is n itive inverse of ). Proof: This is clerly true for 0 since 0 + ( 0) Otherwise, either or must e nturl numer, so it is enough y the commuttivity of ition to prove the itive inverse sentence n+( n) 0 for ll n, which we will now o y inuction: (i) + ( ) is the numer efore, y ition efinition (i ), which is 0. (ii) For ech n, once we know n + ( n) 0, then since (n + ) is the numer efore n, we lso know tht: (n + ) + ( (n + )) (n + ) + ( n + ( )) y ition efinition (i ), n then (n + ) + ( n + ( )) ( + ( )) + (n + ( n)) y the commuttive n ssocitive lws of ition. But now: + ( ) 0, n + ( n) 0 n llow us to conclue tht (n + ) + ( (n + )) 0, hence the inuction. Proposition.2.2. is the only itive inverse of. Proof: Suppose is nother itive inverse of. Then: + ( + ) + 0 ut using the ssocitive lw of ition, we lso hve: + ( + ) ( + ) so. Thus ny other itive inverse of is equl to, which is the sme thing s sying tht is the only itive inverse of.

3 .2. THE INTEGERS AND RATIONAL NUMBERS 3 Proposition.2.3. Negtion is liner trnsformtion, mening: ( + ) + ( ) Proof: By the lws of ition n Proposition.2.: ( + ) + ( + ( )) ( + ( )) + ( + ( )) 0 so + ( ) is n itive inverse of +. From Proposition.2.2, we know tht there is only one itive inverse of +, so + ( ) ( + ). Now we re finlly rey for the: Definition of Sutrction: For ll integers n : + ( ) (tht is, sutrction of is efine to e ition of the itive inverse of ) Finlly (for the integers), we use negtives to efine multipliction: Definition of Multipliction. is efine on cse-y-cse sis. Cse I. n is positive. Then n is efine y inuction: (i), (ii) Ech (n + ) n +. Cse II. n is negtive. Then ( n) is efine to e ( n). Cse III. 0. Then 0 0 Remrk: The efinitions in Cses II n III re force upon us, if we wnt multipliction to stisfy the istriutive lw! (see the exercises) Notice lso tht ( ) so the negtion trnsformtion is the sme s multipliction y. Agin, we will not go through the teious exercise of proving the rest of the sic lws of rithmetic, ut it cn e one with only inuction n these efinitions, if you re willing to work through ll the cses. Recp: The new numer 0 is the itive ientity, mening tht: + 0 for ll integers. Negtion tkes n integer to its itive inverse, llowing us to efine sutrction s ition of the itive inverse. Note tht is the multiplictive ientity, mening tht for ll integers, ut integer multiplictive inverses only exist for the integers n.

4 4 CHAPTER. NUMBERS The rtionl numers cn e thought of geometriclly s slopes of lines: Q {(slopes of) lines tht pss through (0, 0) n point (, )} where, Z n 0 (so the line isn t verticl.) The line L pssing through (0, 0) n (, ) hs eqution: y x n the slope is lso the y-coorinte of the intersection of L with the (verticl) line x. In prticulr, ifferent lines through the origin hve ifferent slopes. Mny ifferent points with integer coorintes will lie on the sme line L! If (, ) is nother point with integer coorintes, then y the eqution ove for the line L, we see tht (, ) is lso on L exctly when: Definition: An integer frction is symol of the form: where, Z n 0. Two integer frctions re equivlent, written: if (, ) n (, ) re on the sme line through the origin. Note: The symol is clle reltion, n it is esy to see tht: (i) is reflexive, mening tht (ii) is symmetric mening tht if then (iii) is trnsitive mening tht if n (A reltion stisfying (i)-(iii) is clle n equivlence reltion.) Definition: The equivlence clss then is the set of ll frctions tht re equivlent to. Notice tht: [ whenever, tht is, whenever (, ) n (, ) re on the sme line through the origin, or, s we notice ove, whenever.

5 .2. THE INTEGERS AND RATIONAL NUMBERS 5 Thus we cn reinterpret the rtionl numers s: Q {equivlence clsses of integer frctions} n this reinterprettion is very useful for seeing the rithmetic of Q. Before we o this, let s notice tht the rtionl numers re still orere: [ c < if the line through (0, 0) n (, ) intersects the verticl line x t point tht is elow the intersection of the line through (0, 0) n (, c). Unlike the integers, there is no such thing s the next rtionl numer fter rtionl numer, so there is no wy to use inuction to efine ition. Inste, we use the rule for ing frctions tht we lerne in greschool. Definition of Aition: Given rtionl numers n [ c, then: [ [ c + c + (using the rithmetic of integers to efine the numertor n enomintor). But we nee to check something! Is ition is well-efine? Here s the prolem. If [ [ [ c c n how o we know tht: [ + c + c? This isn t ovious, n it nees to e checke, ecuse if it weren t true, then this woul e efinition of ition, ecuse it woul only e n ition of integer frctions, n not of rtionl numers, which re equivlence clsses of integer frctions. This prolem will rise whenever we try to mke efinition involving equivlence clsses of frctions (or other things). So ewre! Proof tht ition is well-efine: Suppose: n c c This mens tht n c c. But then: ()( + c ) ( )( ) + (c )( ) ( )( ) + (c )( ) ( + c)( )

6 6 CHAPTER. NUMBERS (sustituting, n pplying the lws of rithmetic for integers). So: s esire. + c + c Now tht the efinition is OK, it is very useful! Aition is ssocitive. Proof: ( [ [ [ c e + c + + ) f + [ e f [ (()f + (c)f) + ()e ()f n + ( [ c + [ ) e + f [ [ cf + e (f) + ((cf) + (e)) f (f) n these re the sme ecuse of the ssocitive lws for integer rithmetic! Similrly, you cn prove tht ition is commuttive. Definition of Multipliction: [ c c Proof tht multipliction is well-efine: If n c c, then n c c, so ()( c ) ( )(c ) ( )(c ) (c)( ). But this gives us: c c which is just wht we neee to check. It is now esy to prove the ssocitive n commuttive lws for the multipliction of rtionl numers. To see tht multipliction istriutes with ition, though, we nee n extr: Proposition.2.4. Suppose f n f, in which cse we sy tht f is common fctor of oth n. Then: [ (i.e. we cn cncel common fctors of the numertor n enomintor.)

7 .2. THE INTEGERS AND RATIONAL NUMBERS 7 Proof: We nee to show tht. But: y sustituting, n likewise, ( f) ( f) so inee the two frctions re equivlent. Multipliction istriutes with ition. Proof: ( [ c ) [ [ [ [ e + c e ()e + (c)e + n f f ()f [ e [ c [ [ [ [ e e ce (e)(f) + (f)(ce) + + f f f f (f)(f) There is common fctor of f in the numertor n enomintor of the secon frction. Once we cncel it (Proposition.2.4), we see tht the two rtionl numers re the sme! Now for some extr gooies: The itive ientity is the rtionl numer: [ 0 (n it oesn t mtter wht is, s long s it isn t 0). Proof: 0 0 since 0 0 0, so ll choices of enomintor give the sme rtionl numer (nmely the slope of the x-xis!). Next: + [ 0 (using Proposition.2.4 gin) proves tht [ 0 is the itive ientity. Nottion: Mthemticins lwys enote the itive ientity y: so we will, too. Every rtionl numer hs n itive inverse. Proof: [ [ [ [ + ( ) + ( ) using Proposition.2.4. So [ is n itive inverse to. As in Proposition.2.2, this is the only itive inverse! 0

8 8 CHAPTER. NUMBERS Definition of Sutrction: [ c + [ c As lwys, sutrction is ition of the itive inverse The multiplictive ientity is the rtionl numer: [ (n it oesn t mtter wht is, s long s it isn t 0) Proof: since. So it oesn t mtter wht is, n: [ [ y Proposition.2.4. This is wht we neee to prove. Agin, it is esy to see tht this is the only multiplictive ientity. Nottion: Once gin, we follow mthemticl custom n write: for the multiplictive ientity. Every rtionl (except 0) hs multiplictive inverse. Proof: Every rtionl numer other thn 0 is of the form [ where 0. Then: [ so [ is the one n only multiplictive inverse (or reciprocl) of. Definition of Division (y nything other thn 0): [ c [ c (i.e. ivision is multipliction y the reciprocl) Finlly, I wnt to tlk out one lst efinition: Definition of Lowest Terms: An integer frction (not rtionl numer!) is in lowest terms if > 0 n n hve no common fctors other thn n (which re common fctors of ll integers, hence uninteresting!).

9 .2. THE INTEGERS AND RATIONAL NUMBERS 9 Proposition.2.5. Every rtionl numer contins exctly one frction in lowest terms (in the equivlence clss). Ie of Proof: Strt with the frction, which my not e in lowest terms. By cncelling out ll the (interesting) common fctors of n n lso if necessry (to mke > 0) we rrive t frction in lowest terms. This shows tht there is t lest one frction in lowest terms in the equivlence clss. To see tht there cnnot e more thn one, we will nee to know it more out prime numers, which we will work out lter in the course ( 2.2). This llows us to reefine one more time: Q {integer frctions tht re in lowest terms} In prticulr, we get n inclusion of sets: Z Q y ientifying ech integer with the frction, which is clerly in lowest terms. When we o this, we see something very importnt. Nmely: + [ [ + n [ so ition n multipliction re the sme regrless of whether we view n s integers, or s rtionl numers! Finlly, s in., we finish with nother gem from ncient Greece: Theorem (Pythgors): There is no squre root of 2 in Q. Proof: If there were rtionl numer squre root of 2, then: [ [ woul tell us tht: so tht 2 ivies 2. But then it woul follow tht 2 ivies since the squre of n o numer is o. Thus 2c for some c, n then: 2 2 (2c) 2 4c 2 so 2 2c 2 But then 2 ivies 2 so it woul follow s ove tht 2 ivies. In other wors, 2 woul e n interesting common fctor of n. All this woul e true no mtter wht frction / we chose to represent the rtionl squre root of 2. In other wors, there woul e no wy to put such rtionl numer in lowest terms! This contricts Proposition.2.5, so there cnnot e such rtionl numer.

10 20 CHAPTER. NUMBERS Recp: Rtionl numers re equivlence clsses of integer frctions, n they hve very stisfctory rithmetic, with itive inverses n multiplictive inverses (of everything except 0) llowing us to efine sutrction n exct ivision (y nything except 0). On the other hn, from the point of view of geometry, they re less stisfctory, since perfectly resonle length ( 2) cnnot e represente y rtionl numer..2. Integer n Rtionl Numer Exercises In the first three exercises, we consier rithmetic in n strct setting. The ie is tht mny of the results of this section re not specil properties of the integers or rtionl numers, ut rther follow from the lws of rithmetic themselves. 2- Suppose S is set n + is n ition rule for elements of S tht stisfies: the ssocitive lw: (s + t) + u s + (t + u) for ll s, t, u S, n the commuttive lw: s + t t + s for ll s, t S. () Prove tht there is t most one itive ientity element in S. Tht is, prove there is t most one element z S such tht: s + z s for ll s S (Mthemticins tell us to renme this element 0) Hint: If y is nother itive ientity, think out y + z in two wys. () Assuming tht there is n itive ientity element (renme 0), prove tht ech s S hs t most one itive inverse in S. Tht is, prove tht there is t most one element t S so tht: s + t 0 (Mthemticins tell us to renme this element s) (c) Prove tht if s hs n itive inverse s, then the itive inverse of s is s. Tht is, prove: ( s) s Definition: A set S with n ition rule + with 0 n itive inverses of everything is n Aelin group. 2-2 If S is n Aelin group with multipliction rule stisfying: the ssocitive lw: (s t) u s (t u) n the two-sie istriutive lw with ition: (s + t) u s u + t u n u (s + t) u s + u t (we will not, for now, ssume tht multipliction is commuttive!)

11 .2. THE INTEGERS AND RATIONAL NUMBERS 2 () Prove tht s 0 0 n 0 s 0 for ll s S. Hint: Consier s (0 + 0) n (0 + 0) s. () Prove tht for ll s, t S s ( t) (s t) n (s t) ( s) t Hint: Consier s (t + ( t)) n (s + ( s)) t. (c) Prove tht ( s) ( t) s t for ll s, t S. Definition: An Aelin group S with n ssocitive n two-sie istriutive multipliction rule is clle ring. Exmples: Z n Q re rings with commuttive multipliction rule. The n n mtrices (for n > ) with entries in Z or Q (or ny ring) re themselves ring with non-commuttive mtrix multipliction! 2-3 Suppose S is ring with commuttive multipliction rule. () Prove tht there is t most one multiplictive ientity in S. (Mthemticins tell us to renme this ) () Prove tht ech element s S hs t most one multiplictive inverse (reciprocl) element t S. (Mthemticins tell us to renme this /s.) (c) Prove tht 0 oes not hve multiplictive inverse (unless 0 ). Discuss wht the ring woul look like if 0. () If s hs multiplictive inverse, prove tht: /s s Hint: Exercise 2.3 is very similr to Exercise 2.. Definition: A ring S with commuttive multipliction n multiplictive ientity S, such tht every element of S (except 0) hs multiplictive inverse is clle fiel. Our Only Exmple of Fiel (so fr): Q is fiel. 2-4 Prove the following: () If, Z n 0, then either 0 or 0 (or oth). Hint: If n re oth nturl numers, then 0 ecuse it is nturl numer! Wht re the other possiilities for n? () If,, c Z n c n 0, then c. Hint: Fin wy to use ().

12 22 CHAPTER. NUMBERS 2-5 Recll tht: exctly when We ll check lgericlly tht this relly is n equivlence reltion. (i) Reflexive. This is the commuttive lw for multipliction! ecuse (ii) Symmetric. This is lso the commuttive lw for multipliction. If then ut then so (iii) Trnsitive. This is your exercise! You nee to explin why n together imply tht 2-6 Consier the two rtionl numers: [ [ n [ 3 Explin crefully why the fct tht [ [ shows tht um ition: [ [ c + c + is not well-efine on rtionl numers. 2-7 Prove Pscl s ientity. For nturl numers m < n, n! m!(n m)! + n! (m )!(n m + )! (n + )! m!(n m + )! Remrk (for your enjoyment): This proves tht the inomil coefficient: ( ) n n! m m!(n m)! is the m + st numer in the n + st row of Pscl s Tringle:

13 .2. THE INTEGERS AND RATIONAL NUMBERS The ncient Egyptins h some ies out frctions, though they pprently in t like to sutrct, in t like numertors, n in t like repetitions. The Egyptin frction expnsion of rtionl numer etween 0 n is sum of istinct frctions, ll of the form: n Here re some exmples (I m going to rop the cumersome rckets roun rtionl numers in this prolem n from now on!): , ( is not n Egyptin frction expnsion ecuse the 3 repets) (so sometimes there is more thn one possile expnsion). () Fin Egyptin frction expnsions for the numers: 5 7, 27, 9 49, 5 6 () Devise strtegy for fining n Egyptin frction for ny m/n (ssuming tht m < n). Hint: You might fin inuction useful. Apply your strtegy to the four numers ove (your clcultor will not give you enough ccurcy for the lst two...you will nee computer!).