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1 Factorisation of Integers

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1 Factorisation of Integers 1 1 Factorisation of Integers Definition N = f1; 2; 3;:::g are the natural numbers. Definition Z = f:::; −2; −1; 0; 1; 2;:::g are the integers. Closed under the binary operations +; ×; − Definition α 2 R then bαc is the greatest integer which is less than or equal to α. p Ex b3c = 3; 2 = 1, b−πc = −4 Then bαc 6 α < bαc + 1 Proposition 1 If a and b are two integers with b > 0 then there are integers q and r with 0 6 r < b and a = qb + r. a Proof. Let α = b . a a ) 0 6 b − b < 1 a ) 0 6 a − b b < b a a so if r = a − b b then a = qb + r with q = b . Definition If a = cb (a; b; c 2 Z) we say a is a multiple of b, or b divides a and write 2 bja. Proposition 2 If b 6= 0; c 6= 0 then (a) bja and cjb ) cja (b) bja ) bcjac (c) cjd and cje ) 8m; n 2 Z; cjdm + en: Proposition 3 Let a; b > 0. If bja and b 6= a then b < a. Definition If bja and b 6= 1 or a then we say b is a proper divisor of a. If b does not divide a write b - a. Definition P = fp 2 N : p > 1 and the only divisors of p are 1 and pg are the prime numbers. Then N n (P [ f1g) are the composite numbers. P = f2; 3; 5; 7; 11; 13; 17; 19; 23;:::g. Theorem 1 Every n > 1; n 2 N, is a product of prime numbers. Proof. If n 2 P we are done. If n is not prime, let q1 be the least proper divisor of n. Then q1 is prime (since otherwise, by Prop 3, it would have a smaller proper divisor). Let n = q1n1; 1 < n1 < n. If n1 is prime we are done. If not n1 = q2n2; 1 < n2 < n1 < n. This process must terminate in less than n steps. Hence n = q1q2 : : : qs with s < n. Ex 10725 = 3 · 5 · 5 · 11 · 13 3 In a prime factorization of n arrange the primes so that p1 < p2 < ··· < pk and exponents αi 2 N; 1 6 i 6 k so α1 α2 αk n = p1 p2 ··· pk k Y αj = pj j=1 is the standard factorisation of n. Prime Numbers We can use the sieve of Eratosthenes to list the primes 2 6 p 6 N. p p If n 6 N pand n is not prime, then n must be divisible by a prime p 6 N (if p1 > N and p2 > N ) p1p2 > N). List all of the integers between 2 and N 2; 3; 4; 5;:::;N successively remove (i) 4; 6; 8; 10;::: even integers from 22 on (ii) 9; 15; 21; 27;::: multiples of 3 from 32 on (iii) 25; 35; 55; 65;::: multiples of 5 from 52 on 4 etc. p i.e. remove all integers which are multiples of a prime p < N. We are left with all primes up to N. p Ex N = 16; N = 4 f2; 3; 64; 5; 66; 7; 68; 69; 106 ; 11; 126 ; 13; 146 ; 156 ;166 g Theorem 2 jPj = 1, i.e. there are an infinite number of primes. Qn Proof. Let P = fp1; p2; : : : ; png with p1 < p2 < ··· < pn and let q = j=1 pj + 1. Then Qn q > pj 8j ) q 62 P so q is composite. But pi j q ) pi j q − j=1 pj = 1 ) pi = 1 which is false. Hence jPj = 1. How many primes are there ? Note: 1 X 1 = 1 n n=1 1 X 1 π2 = < 1: n2 6 n=1 5 We can show 1 X 1 = 1 p j=1 j so the primes are denser than the squares. 2 p If x > 0, let S(x) = #fn 2 N : n 6 xg. Then S(x) = b xc. We can show π(x) = #fp 2 P : p 6 xg x ∼ log(x) Definition A modulus is a set of integers closed under ±. The zero modulus is just f0g. If a 2 Z then M = fna : n 2 Zg is a modulus. Proposition 4 If M is a modulus with a; b 2 M and m; n 2 Z then ma + nb 2 M. Proof. a 2 M ) a + a = 2a 2 M ) 2a + a = 3a 2 M etc. so ma 2 M and so is nb, thus ma + nb 2 M. Proposition 5 If M 6= f0g is a modulus, it is the set of multiples of a fixed positive integer. Proof. Let d be the least positive integer in M with 0 < d. Claim: every element of M is a multiple of d. If not (???) let n 2 M have d - n. Then n = dq + r with 1 6 r < d. But r = n − dq 2 M (!!!). 6 Definition Let a; b 2 Z and let M = fma + nb : m; n 2 Zg then M is generated by d in that M = fnd : n 2 Zg. We call d the greatest common divisor or GCD of a and b, and write (a; b) = d. Proposition 6 (i) 9x; y 2 Z so ax + by = (a; b) (ii) 8x; y 2 Z; (a; b)jax + by (iii) If eja and ejb then ej(a; b) Definition If (a; b) = 1 we say a and b are coprime. Ex The GCD (greatest common divisor) is normally computed using the Euclidean Algorithm. From Proposition 5: (a = 323; b = 221) 323 = 221 · 1 + 102 so 102 2 M 221 = 102 · 2 + 17 so 17 2 M 102 = 17 · 6 + 0 so 17 is the least positive integer in M ) (323; 221) = 17. Reading back: 17 = 221 − 2 · 102 = 221 − 2 · (323 − 221) = 3 · 221 − 2 · 323 so (a; b) = xa + yb ) x = −2; y = 3. 7 Proposition 7 If p 2 P and pjab then pja or pjb. Proof. If p - a then (a; p) = 1. By Prop 6(i) 9x; y 2 Z so xa + yp = 1 ) xab + ybp = b But pjab so ab = qp. Hence (xq + yb)p = b so pjb. Proposition 8 If c > 0 and (a; b) = d then (ac; bc) = dc. Proof. 9x; y 2 Z so xa + yb = d ) x(ac) + y(bc) = dc ) (ac; bc) j dc. Also d j a ) cd j ca (and similarly cd j cb) ) dc j (ac; bc). Hence dc = (ac; bc). Theorem 3 (Fundamental Theorem of Arithmetic) The standard factorisation of a number n 2 N is unique. Proof. If p j ab ··· m, by Proposition 7, p must divide one of the factors. If each of α1 αi β1 βj these is prime, then p must be one of them. If n = p1 ··· pi = q1 ··· qj are two standard factorizations of n, each p must be a q and each q a p. Hence i = j. Since p1 < p2 < ··· < pk and q1 < q2 < ··· < qk; p` = q` for 1 6 ` 6 k. If β1 < α1, divide n by β1 α1−β1 α2 β2 p1 to get p1 p2 ··· = p2 ···) α1 = β1 etc. 8 Proposition 9 Let a; b 2 N have non-standard factorisations m Y αj a = pj j=1 and m Y βj b = pj j=1 with αj > 0; βj > 0 then m Y min (αj ; βj ) (a; b) = pj : j=1 Ex a = 223451 b = 213051 ) (a; b) = 213051 Definition Let a; b 2 Z+ = f0; 1; 2;:::g = N [ f0g. The least common multiple or LCM of a and b is the smallest common multiple of a and b and is written fa; bg. Ex f3; 4g = 12 9 Proposition 10 With the same notation as for Proposition 9, m Y max (αj ; βj ) fa; bg = pj : j=1 Proposition 11 Any common multiple of a and b is a multiple of the least common multiple. Proposition 12 fa; bg (a; b) = ab Proof. m Y max (αj ; βj )+min (αj ; βj ) LHS = pj : j=1 But 8x; y max (x; y) + min (x; y) = x + y. Hence m m m Y αj +βj Y αj Y βj LHS = pj = pj · pj = ab: j=1 j=1 j=1 Alternative Characterisation of the GCD By Proposition 6 (ii), (a; b)jax + by. 10 Let x = 1; y = 0 ) (a; b)ja. Let x = 0; y = 1 ) (a; b)jb. So g = (a; b) is a common divisor of a and b. By Proposition 6 (iii), if eja and ejb then e j g i.e. g is divisible by every common divisor. Hence it is the greatest. This property: \being a common divisor divisible by every common divisor" characterises the GCD up to sign. Proof. If g1 and g2 satisfy this property, then g1 and g2 are both common divisors with g1 j g2 and g2 j g1. Hence g2 = αg1 = αβg2 ) αβ = 1 if g2 6= 0. Hence α = ±1. So g1 = ±g2. The GCD, so defined by the above property, is made unique by fixing the sign, g > 0. Ex divisors of 12 = {±1; ±2; ±3; ±4; ±6; ±12g = D12 divisors of 18 = {±1; ±2; ±3; ±6; ±9; ±18g = D18 common divisors = {±1; ±2; ±3; ±6g = D12 \ D18 So ±6 satisfies the property. Hence, fixing the sign, 6 = (12; 18). Linear Equations in Z Proposition 13 Given a; b; n 2 Z, the equation ax + by = n has an integer solution x; y , (a; b)jn. 11 Proof.(() By Proposition 6 (i) 9x; y such that ax + by = (a; b).
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