Foundations of Arithmetic

Transcription

1 Foundatons of Arthmetc Notaton We shall denote the sum and product of numbers n the usual notaton as a 2 + a 2 + a a = a, a 1 a 2 a 3 a = a The notaton a b means a dvdes b,.e. ac = b where c s an nteger, and a b means a does not dvde b. If ac = b ( b > 1) mples a = ±1 or a = ±b then b s a prme number. A number b ( b > 1) that s not prme s sad to be composte. Let a (a > b) max(a, b) = { b (b > a) It s obvous that b (a > b), mn(a, b) = {, max(a, a) = mn(a, a) = a. a (b > a) max(a, b) + mn(a, b) = a + b. (1) The Fundamental Theorem of Arthmetc The Fundamental Theorem of Arthmetc states that any number m 1 s expressed unquely as a product of powers of prme numbers. Denote the th prme number by p,.e. p 1 = 2, p 2 = 3, p 3 = 5,, p 8 = 19, etc. Then m = p j (j 0). (2) If p, say, s not a prme factor of m then j = 0 so that the factor p j = 1. If p M s the largest prme factor of m then j = 0 for > M so that all the factors nvolvng prmes greater than p M reduce to 1. If m = 1 then j = 0 for 1 The representaton (2) can be proved by nducton. Assume t s true for 1 m n 1. When m = n, n can be ether prme, n whch case the representaton (2) s trvally true, or t s composte and can therefore be expressed as n = ab, where 1 < a n 1 and 1 < b n 1. But by the nducton hypothess both a and b can be expressed as a product of prmes as n (2) so that the product ab can also be expressed as a product of prmes. It follows that (2) s true for m = n = ab and by nducton the result s proved for all m. It remans to be shown that (2) s unque. Suppose there are two dfferent representatons of m expressed n the form of (2),.e.

2 m = p j = p (j 0, 0). (3) Consder a prme factor p s for whch max(j s, s ) 0 and j s s and assume for the sae of argument that j s > s. Then (3) can be wrtten as s 1 j p s s j s p j p =s+1 s 1 = p p =s+1 (j 0, 0). (Note f s > j s, a factor p s s j s would be taen out of the rght-hand sde nstead.) If js s the left-hand sde of ths equaton s dvsble by p s but the rght-hand sde s not, whch s mpossble. Thus j s = s and snce ths apples to all prme factors p whose exponents are not already equal, we deduce that the two products n (3) are dentcal. Therefore, the representaton of a number m as a product of prmes s unque. Let us now consder a couple of consequences of the Fundamental Theorem. () The hghest common factor and lowest common multple of two numbers By analogy wth (2), another number n 1 can be represented as n = p ( 0) where = 0 for > N when p N s the largest prme factor of n. The Hghest Common Factor (HCF) of two numbers m and n s the largest dvsor of both numbers (whch s why t s sometmes called the Greatest Common Dvsor or GCD). Clearly the HCF must nclude all the common prme factors of both numbers. If p s a common factor rased to the powers j and respectvely, then the largest factor common to both m and n s p mn(j, ). Note that f p s not a common factor but occurs only n the representaton of m (say) then t s necessary that = 0 n order to exclude t from the representaton of n, as requred. In ths case mn(j, ) = 0 so that p s omtted from the HCF as well, as ndeed t should be. Thus all cases are ncluded n the formal defnton mn(j HCF(m, n) = p, ) The Lowest Common Multple (LCM) s the smallest number that s dvsble by both m and n. Ths tme we must choose p rased to the greater of the two powers j and n order for both m and n to be dvsors of the LCM. Thus we defne max(j LCM(m, n) = p, )

3 Usng (1) we obtan Hence mn(j p, ) max(j p, ) max(j = p, )+mn(j, ) j = p + j = p p HCF(m, n) LCM(m, n) = mn. Thus once the HCF s nown the LCM s easly found. A famlar way of calculatng the HCF dates bac to Eucld. Let h = HCF(m, n) and suppose m n. Then m = c 1 n + r 1 (0 r 1 < n). (4) Here r 1 s the remander left after dvdng m by n. If n m then r 1 = 0 and the HCF s smply n tself. Otherwse r 1 must be smaller than n because c 1 s the maxmum number of tmes n goes nto m. Snce h m and h n t s obvous from (4) that h r 1 as well, and because n > r 1 we may therefore wrte by analogy wth (4) n = c 2 r 1 + r 2 (0 r 2 < r 1 ). (5) Agan we have h r 2 because t also dvdes both n and r 1 n the above equaton. The procedure contnues n ths way, the next step yeldng r 1 = c 3 r 2 + r 3 (0 r 3 < r 2 ) where h r 3 and so on untl we reach the ( 1) th and th steps r 2 = c r 1 + r (0 r < r 1 ) (6) r 1 = c +1 r + r +1 (0 r +1 < r ). The sequence of postve ntegers r 1 > r 2 > > r 1 > r s decreasng so must eventually termnate n 0. Let r +1 = 0 so that the last equaton above reduces to where h r whch mples h r. r 1 = c +1 r (0 < r ) (7) Reversng the argument, we see from (7) that r r 1 and r r 1 r r 2 by (6). Lewse r r 2 r r 3 r r 4 r r 2 r r 1 r n r m the last two steps followng from (5) and (4). Thus r s a common factor of both m and n, but because h s the hghest common factor t must satsfy h r. We have now proved that h satsfes both h r and h r from whch we conclude h = r, the fnal remander n Eucld s algorthm. As a numercal example of ts applcaton, let us calculate the HCF of 2472 and 9216:

4 9216 = = = = = = 9 24 Thus 24, the last remander, s the HCF of 2472 and Fnally, we derve from ths algorthm a property of the HCF that s not partcularly obvous from ts defnton, namely that there exst ntegers a and b (one of them beng negatve) such that h = am + bn. For, from (6) we have h = r = r 2 c r 1 = r 2 c (r 3 c 1 r 2 ) = ur 2 c r 3 = ur 4 vr 3 = wr 4 vr 5 = where u = 1 + c c 1, v = c + uc 2, w = u(1 + c 2 c 3 ) + c c 3. Note that c, u, v, w, etc. are all postve numbers so that one term n each step s postve and the other negatve. The procedure contnues as we wor bacwards through the algorthm to the fnal two equatons, whch accordng to (4) and (5), wll tae the form h = = cn + ar 1 = am + bn wth a, b and c representng ntegers to be determned. Usng the same numercal example (m = 9216, n = 2472) to llustrate the theory, we fnd that successve stages of the calculaton gve m = 3n m 3n = 1800 n = 1 (m 3n) n m = 672 m 3n = 2 (4n m) m 11n = 456 4n m = 1 (3m 11n) n 4m = 216 3m 11n = 2 (15n 4m) + h h = 11m 41n. Thus a = 11 and b = 41 n ths example. Checng we see that = 101, ,352 = 24 whch verfes the stated property of the HCF.

5 () Analytcal form of the fundamental theorem Now consder the expresson f (s) = 1 1 p (s > 1). Snce (1 x) 1 = Σ r=0 x r for x < 1, and snce p 2, the expresson under the product sgn can be expanded n a convergent seres,.e. Suppose = 2 for smplcty, then f (s) = (p ) r (s > 1). r=0 f 2 (s) = (p 1 ) r (p 2 ) t = (p r t 1 p 2 r=0 t=0 r=0 t=0 where p 1 = 2 and p 2 = 3 of course. Clearly all numbers that have prme factors 2 and 3 rased to all possble combnatons of powers wll be ncluded wthn the bracets of ths double summaton. For example, 1 = ( ), 2 = ( ), 3 = ( ), 72 = ( ), (124,416) = ( ) are fve such terms n the sum defnng f 2 (s). Thus we may wrte f 2 (s) = n where the notaton mples that summaton s over all numbers n whose prme factors comprse every possble combnaton of powers of p 1 and p 2. By the Fundamental Theorem each n s unquely expressed and can therefore only appear once n the summaton. Lewse, f 3 (s) wll be the sum of all numbers wth prme factors 2, 3 and 5 rased to all possble combnatons of powers. In general, we have p 1,p 2 f (s) = n > n p 1,p 2 p the nequalty resultng from the fact that the numbers from 1 to p are already ncluded n the frst summaton, as ndcated by the frst three terms n the example above for f 2 (s). Snce s > 1, the nfnte seres Σ n=1 n s convergent. It s n fact the well-nown Remann zeta functon ζ(s). It follows from the nequalty above that p n=1 )

6 whch, on rearrangement, becomes p f (s) > n = n n > 0 n=1 n=1 n=p +1 0 < ζ(s) f (s) < n. n=p +1 Now let whch means p as well. Then the rght-hand sde of ths equaton tends to 0 so that f (s) = lm f (s) = ζ(s), or wth reference to the orgnal defnton of f (s), ζ(s) = 1. 1 p Hardy and Wrght (An Introducton to the Theory of Numbers, 4 th Edton, Oxford Unversty Press, 1960) call ths an analytcal expresson of the Fundamental Theorem of Arthmetc. It s an mportant result n the theory of prmes as t relates them to the zeta functon whch has been extensvely analysed. The product of n consecutve postve ntegers s dvsble by n! Ths s a smple result but t provdes a good example of proof by nducton. We want to prove n! m(m + 1)(m + 2) (m + n 1), for m 1 and n 1. In one sense ths s obvous because m(m + 1)(m + 2) (m + n 1) = n! (m + n 1)! (m 1)! n! m+n 1 = C n m+n 1 where C n s the famlar notaton for the number of ways of selectng n objects from a collecton of m + n 1 unle objects. Snce ths s a countable number t must be an nteger and the result s verfed. A more formal proof that doesn t rely on a practcal nterpretaton s by nducton. Defne = m + n and P() = ( n)( n + 1)( n + 2) ( 1), and let S() represent the statement n! P() for 2 and 1 n 1 whch s equvalent to the statement to be proved. Note that each value of > 2 covers several cases, e.g. S(9) ncludes 5! and 3! among sx other possbltes. The case = 2 corresponds to m = n = 1. Thus P(2) = 1 and S(2) s trvally true. Assume that S() s true for 2 K so that n partcular S(K) s true,.e. n! P(K) for 1 n K 1, or wrtten out n full n! (K n)(k n + 1)(K n + 2) (K 1) (1 n K 1). (8) Snce 1 n 1 < n K 1 for n 2, we also have

7 Now (n 1)! (K n + 1)(K n + 2) (K 1) (n 2). (9) P(K + 1) = (K n + 1)(K n + 2) (K 1)K = (K n)(k n + 1) (K 1) + n(k n + 1)(K n + 2) (K 1). By (8) the frst term on the rght-hand sde s dvsble by n! for 1 n K 1 and because of ts addtonal factor n, the second term s also dvsble by n! for n 2 accordng to (9). Thus n! P(K + 1) for 2 n K 1. Furthermore, for n = K the statement n! P(K + 1) becomes K! K whch s obvously true and t reduces to the trval result 1! K for n = 1. Thus n! P(K + 1) for 1 n K and therefore S(K + 1) s true f S() s true for 2 K. Snce we have shown S(2) to be true t follows by nducton that S() s true for all 2 and the result s proved. A number s dvsble by nne f and only f the sum of ts dgts s nne Ths famlar arthmetcal trc s nown to many school students who otherwse have lttle nterest n mathematcs. Its proof, however, serves as an elementary ntroducton to modular arthmetc and congruences. If a, r and m are ntegers, then the statement a s congruent to r modulo m means m (a r) and s wrtten formally as a r (mod m). In other words, a r s some multple of m, that s a = r + m where 0 r < m and s a postve nteger. Ths last expresson shows that r s the remander when a s dvded by m, but there are other numbers, namely those that dffer from a by a multple of m, that have the same remander r when dvded by m. Modular arthmetc doesn t dstngush between such numbers; they are all regarded as equvalent. For example 10 1 (mod 9) but so s 19 1 (mod 9), and also 28, 37 and so on. Moreover, f 10 n 1 (mod 9) then 10 n+1 1 = 10(10 n 1) + 9 s also dvsble by 9, showng that 10 n+1 1 (mod 9). Hence by nducton 10 n 1 (mod 9) for all postve n snce t s true for n = 0. Clearly f a r (mod m) and b s (mod m), then a + b r + s (mod m) because both terms a r and b s are dvsble by m. Smlarly, a r (mod m) because m (a r). We proved n the last paragraph that 10 n 1 (mod 9) so t follows that a10 n a (mod 9) and hence a10 n + b10 q a + b (mod 9). Now let a number c be wrtten n the usual way as a n a n 1 a 1 a 0, e.g. f c = 9704 we have a 0 = 4, a 1 = 0, a 2 = 7, a 3 = 9. Ths standard notaton s, of course, shorthand for the expresson c = a 0 + a a a n 10 n. By vrtue of the result n the precedng paragraph we deduce that c a 0 + a 1 + a a n (mod 9) or 9 [c (a 0 + a 1 + a n )]. Thus f c s dvsble by 9 then the sum of ts dgts must also be dvsble by 9 and conversely, f 9 dvdes the sum of the dgts of c then c tself s dvsble by 9.