[PDF] [PDF] Congruences (Part 1) - Mathtorontoedu - University of Toronto

[PDF] Congruence and Congruence Classes

Theorem 11 10 If a ≡ b (mod n) and c ≡ d (mod n), then (i) a + c ≡ b + d (mod n) (ii) ac ≡ bd (mod n) The congruence class of a modulo n, denoted [a], is the set of all integers that are congruent to a modulo n; i e , [a] = {z ∈ Z a − z = kn for some k ∈ Z}

[PDF] Math 371 Lecture  §21 - BYU Math Department

metic of Z Theorem 2 2 If a ≡ b (mod n) and c ≡ d (mod n), then (1) a + c ≡ b + d (mod n), and (2) ac ≡ bd (mod n) Proof We suppose that a − b = nk for 

[PDF] 3 Congruence

Theorem 3 3 If a ≡ b mod n then b = a + nq for some integer q, and conversely Similarly, multiplying, we get bd = (a + nq1)(c + nq2) = ac + naq2 + ncq1 + n 2 Now suppose that we wish to solve the congruence ax ≡ b mod n where d 

[PDF] Congruences (Part 1) - Mathtorontoedu - University of Toronto

Eg 7 ≡ 3 (mod 4) -2 ≡ 6 (mod 4) 5 ≡ 9 (mod 4) 2) If a ≡ b (mod m) c ≡ d ( mod n) then ac ≡ bd (mod m) Proof: a = b + qm c = d + rm, some r,q ac – bd = (b  

[PDF] MTHSC 412 Section 21 --Congruence and Congruence Classes

a is congruent to b modulo n and write a ≡ b (mod n) when Suppose that a ≡ b (mod n) and c ≡ d (mod n) Then a + c ≡ b + d (mod n) and ac ≡ bd (mod n)

[PDF] Number Theory

If a ≡ b (mod m) and c ≡ d (mod m), then ac ≡ bd (mod m) If a ≡ b (mod m), Theorem: An integer n is divisible by 11 i the di erence of the sums of the odd

[PDF] Modular Arithmetic - Cornell CS

12 nov 2014 · Let a, b ∈ ℤ, m ∈ ℕ a and b are said to be congruent modulo m, written a ≡ b ( mod m), if and only if a – b is If a ≡ b (mod m) and c ≡ d (mod m), then – a + c ≡ b + d (mod m) – ac ≡ bd (mod m) E g 11 ≡ 1 (mod 10) ⇒ 

[PDF] Congruences

integers a, b are congruent mod n, which is written as a ≡ b (mod n), if nb − a Example 2 For all a, b, c ∈ Z, if a ≡ b (mod n) and b ≡ c (mod n), then a ≡ c ( mod n) multiplication by a, since if ab ≡ ac (mod n), then multiply by x to get x( ab) ≡ x(ac) ax ≡ b (mod n) has a solution if and only if d = gcd(a, n) divides b

[PDF] READING TO ACCOMPANY CSU-P MATH 319 - poritznet

17 fév 2014 · (9) If a ≡ b (mod n) and c ≡ d (mod n) then ac ≡ bd (mod n) Proof (1) Given a , b, c ∈ Z and n ∈ N, define d = gcd(a, n) ∈ N If ab ≡ ac

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Congruences (Part 1)

Original Notes adopted from September 25, 2001 (Week 3) © P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics typed by A. Ku Ong

Modular Arithmetic

a, b Î Z , m >1 "a is congruent to b modulo m" means m|(a-b). Equivalently, a & b leave the same remainder by division by m (for a,b³0)

1) If a ºº b (mod m) then (a+c) ºº (b+d) (mod m) & c ºº (mod m)

Proof: a º b (mod m) means a - b = mq, some q Î Z

Also c - d = mr some r Î Z.

We Want: (a-b) - (b+d) is a multiple of m

a = b + mq c = d + mr (a+c) - (b+d) = ( b + nq + (d + mr)) - b = b + d + mq + mr - b - d = mq + mr = m( q + r) , a multiple \ (a + c) º (b+d) (mod m)

Eg. 7 º 3 (mod 4)

-2 º 6 (mod 4)

5 º 9 (mod 4)

2) If a ºº b (mod m) & c ºº d (mod n) then ac ºº bd (mod m)

Proof:

a = b + qm c = d + rm, some r,q ac - bd = (b + qm) ( d + rm) - bd = bd + brm + qmd +qrm2 - bd = m (br + qd + qrm)

Divisible by m, so ac º bd (mod m)

Corollary: If a ºº b (mod m), then a2 ºº b2 (mod m)

Proof: Case where a = c & b = d

3) If a ºº b (mod m) and n ÎÎ N then an = bn (mod m)

Proof: Use Mathematical Induction.

Case n = 1 True

Assume true for n = k.

Induction Hypothesis: ak º bk (mod m)

Given a º b (mod m)

By 2) aak º bbk (mod m) or ak + 1 = b k +1 (mod m).

Does 7/ (229 + 3)?

23 º 1 (mod 7)

(23) 8 º 18 (mod 7)

224 º 1 (mod 7)

23 º 1 (mod 7)

23224 º 1 * 1 (mod 7)

227 º 1 (mod 7)

22227 º 4 * 1 (mod 7)

229º 4 (mod 7)

229 + 3 º (4+3)(mod 7)

229 + 3 º 0 (mod 7)

229 + 3 is divisible by 7.

Eg. What is the remainder when 3202 + 59 is divided by 8?

32 º 1 (mod 8) 52 º 1 (mod 8)

(32) 101 º 1101(mod 8) (52) 4 º 14(mod 8)

3202 º 1 (mod 8) 54 º5 (mod 8)

3302 + 59 º 6 (mod 8). The remainder when 3202 + 59 is divisible by 8 is 6.

Eg. 52 ºº 32 (mod 8)

52 º 1 (mod 8)

32 º 1 (mod 8 ) But 5 ¹ 3 (mod 8)

Eg. To tell if 3, 974, 279 is divisible by 3, see if 3+ 9 + 7 + 4+ 2 + 7+ 9 is divisible by 3.

3, 974, 279 = 9 + 7 x 10 + 2 x 102 + 4 x 103 + 7 x 104 + 9 x 105 + 3 x 106

10 º 1 (mod 3) \a x 10k º a (mod 3) for all k

102 º 1 (mod 3)

10k º 1 (mod 3) for all k Î N

7 x 10 º 7 (mod 3)

2 x 102 º 2 (mod 3) etc.

9 + 7*10 + 2*102 + 4*103 + 7*104 + 9*105 + 3*106

= 9 + 7 + 2+ 4 + 7 + 9 + 3

Thm: The natural number an10n + an-110n-1 + an-210 n-2 + ... + a110 + a0 whose each ai is a number from

{ 0,1,2,3... 9} is congruent to an + an-1 +... + a1 + a0 (mod 3)

Proof: 10m º 1(mod 3) for all m

\an10m º an (mod 3) for all an for all m. \an10n + an-110n-1 + ... an10 + a0 º an + an-1 + ... + a1 + a0. Eg. 7,230, 591, 006 leaves remainder 0 upon division by 3 (its divisible by 3)

Thm: The natural number an10n + an-110n-1 + an-210 n-2 + ... + a110 + a0 whose each ai is a number from

{ 0,1,2,3... 9} is congruent to an + an-1 +... + a1 + a0 (mod 9)

Proof : As in case mod 3;

10m º 1 (mod 9) for all m.

What about mod 11?

10 º (-1)(mod 11)

10 2 º (-1) 2(mod 11) º 1 (mod 11)

10 3 º -1(mod 11)

10 m º 1(mod 11) if m even

10 m º -1 (mod 11) if m odd.

Eg. What is the remainder when 7, 224, 689 is divisible by 11?

7224689 = 9 + 8 *10 + 6*10 2 + 4*10 3 + 2*10 4 + 2*10 5 + 7-10 6

= 9 ± 8(odd) + 6(even) ± 4 + 2 ±2 + 7 (mod 11) = 10 (mod 11)

Question : Is 3 729 = 7 104 ?

Recall: We proved that every natural number other than 1 is a product of prime numbers.

Thm: The Fundamental Thm of Arithmetic

Every natural number ¹1 is a product of primes & the primes in the product are unique (including multiplicity) except for the order in which they occur.

Eg. 48 = 16 * 3 = 4*4*3 = 2*2*2*3

48 = 24*2 = 3*8*2 = 3*2*2*2

Proof: We know that every natural number ¹1 is a product of primes. We must show the uniqueness of

the factorization into primes. (Using contradiction)

If there are natural numbers with two distinct factorizations into primes, then there is a smallest one, say k.

We"ll show that this is impossible. We have k = p1, p2, p3... pm = q1, q2,... qn where all pi of all q, are primes

and a difference in the occurrence of primes in p1,... pn to q other than order.

Since k is the smallest, the factorization have no primes in common for if since p1 were a qj, dividing both

sides by it, getting a smaller number than k that has distinct prime factorizations.

Either p > q or p < q.

Suppose p1 < q1 ( Etc. if p > q)

Consider L = k ± p1q2 q3... qn Î N

= p1p2... pm ± p1q2... q (I)= p1 (p2...pm-q1...qm)

But also L = q1q2... qn ± p1q2.... qn

(II) = (q1-p1)(q2... qn)

L < k, so it has unique factorization

(I) Þ p1 occurs in factorization of L ( II) Þ p1 divides into (q1 ± p1) q1 ± p1 = tp1 \ q1 = p1 (1+t) q1 ¹ p, p, q primes, impossible.quotesdbs_dbs17.pdfusesText_23