disprove if ak ≡ bk (mod n) for some integer k then a ≡ b (mod n)

PDF	Lecture11 b = ms + (b mod m) for some integers q s – b = (mq + (a mod m)) – (ms + (b mod m)) = m(q – s) + (a mod m – b mod m) = m(q – s) since a mod m = b mod m Therefore m (a-b) and so a ≡ b (mod m) Let m be a positive integer If a ≡ b (mod m) and Let m be a positive integer If a ≡ b (mod m) and

PDF	Congruences and Modular Arithmetic Let n ∈ N and a b ∈ Z We say that a is congruent to b modulo n denoted a ≡ b (mod n) provided na − b

How do you disprove a positive integer?
Prove or disprove: For all positive integers n n and for all integers a a and b b, if a2 ≡b2 mod n a 2 ≡ b 2 mod n, then a ≡ b mod n a ≡ b mod n. If you disprove, you should give an explicit counterexample.
Does a b (mod 2) have the same parity?
According to Theorem 2, we find that a ≡ b (mod 2) iff a and b are both even or both odd. In this case we say a and b have the same parity. Every integer is congruent to either 0, 1 or 2 modulo 3 (and these options are mutually exclusive). Every integer is congruent to (exactly) one of the decimal digits modulo 10.
How do you find a positive integer with a mod?
c) – (b + d) = m(k + j). Now, re-applying the definition of mod gives us a + c ≡ b + d (mod m). Let m be a positive integer. If a ≡ b (mod m) and Let m be a positive integer. If a ≡ b (mod m) and Suppose a ≡ b (mod m) and c ≡ d (mod m). Unrolling definitions gives us some integer k such that
How do you proof a nonnegative integer?
If you are stuck, think about the how the proof will end & write that down. Sometimes it helps to work backwards. Prove or disprove: 2 n + 1 is prime for all nonnegative integer n. Prove if n is an integer, then n 2 has the same parity as n. Let n be an integer. Show that if n is odd, then n 2 is also odd.

Definition

Let n ∈ N and a, b ∈ Z. We say that a is congruent to b modulo n, denoted a ≡ b (mod n), provided na − b. ramanujan.math.trinity.edu

− b = nk

⇔ a = b + nk Our first result concerning congruences should be familiar from Intro to Abstract. ramanujan.math.trinity.edu

Theorem 1

Let n ∈ N. Then congruence modulo n is an equivalence relation on Z. Proof (Sketch). Let a, b, c ∈ Z. ✪ Reflexivity: Since n0, a ≡ a (mod n). Symmetry: If na − b, then n − (a − b) = b − a. So ✪ a ≡ b (mod n) implies b ≡ a (mod n). ✪ Transitivity: If na − b and nb − c, then n(a − b) + (b − c) = a − c. Thus a ≡ b (mod n) and b ≡ c (mod n) toget

Z/nZ = {a + nZ a ∈ Z}.

Before we give more examples, it will be convenient to give a complete description of Z/nZ. ramanujan.math.trinity.edu

Modular Arithmetic

One of the facts that makes congruences so useful in arithmetic is that they respect the operations of addition and multiplication. ramanujan.math.trinity.edu

Theorem 3

Let n ∈ N and a, b, c, d ∈ Z. If a ≡ b (mod n) and ramanujan.math.trinity.edu

c ≡ d (mod n), then:

+ c ≡ b + d (mod n); ac ≡ bd (mod n). Proof. Write a − b = nk and c − d = nl with k, l ∈ Z. Then ramanujan.math.trinity.edu

(a + c) − (b + d) = (a − b) + (c − d) = nk + nl = n(k + l),

so that a + c ≡ b + d (mod n). The proof that multiplication is respected is only slightly less straightforward: ac − bd = ac − ad + ad − bd = a(c − d) + (a − b)d = anl + nkd = n(al + kd), so that ac ≡ bd (mod n). ✪ ramanujan.math.trinity.edu

What does a ≡ b (mod n) mean? Basic Modular Arithmetic Congruence

[Discrete Mathematics] Direct Proofs Examples

PDF	Solutions to Homework Set 3 (Solutions to Homework Problems Prove that a ≡ b (mod n) if and only if a and b leave the same remainder when divided by n. Proof. ⇒. Suppose a ≡ b (mod n). Then by definition

PDF	Problem Set 4 Solutions Feb 22 2005 (i) a ≡ b (mod n) implies ak ≡ bk (mod n) for all k ≥ 0. Solution ... (a) An integer k is selfinverse if k·k ≡ 1 (mod p). Find all ...

PDF	SOLUTIONS TO PROBLEM SET 1 Section 1.3 Exercise 4. We see b(N−1)2 k ≡ 1 (mod N) and. N − 1 is divisible by 2k+1. Let N = 25. We have Then n = ab for some integers a b such that 1 < a

PDF	Congruences If a ≡ b ( mod m ) then ak ≡ bk ( mod m )

PDF	Problem Solving for Math Competitions Harm Derksen If k is an integer then m = 13 · (−m + 7k)+7 · (2m − 13k). For a suitable k we In particular

PDF	Congruences and Modular Arithmetic If a ≡ b (mod n) then a − b = nℓ for some ℓ ∈ Z. Multiplying through by k yields ak − bk = nkℓ

PDF	Carmens Core Concepts (Math 135) Corollary If a ≡ b (mod m) then ak ≡ bk (mod m) for k ∈ N. Carmen Bruni If the remainder is 0 then k = 2l for some integer l and hence x =2+6l and so ...

PDF	Number Theory If ak ≡ bk (mod m) and k is relatively prime to m then a ≡ b (mod m). Proof. Multiplying both sides by k−1

PDF	Lecture 2 Claim: If n is a positive integer then n 2 + 1 is not a perfect Then ∃k ∈ Z such that ak = b. Instructor's Comments: This is the 40 minute mark. 4. Page 111. Corollary If a ≡ b (mod m) then ak ≡ bk (mod m) for k ∈ N.

PDF	Solutions surjective because if b ∈ B then b = 7k for some integer k. Then 3k ∈ A If b is odd

PDF	Problem Set 4 Solutions 22/02/2005 Prove all of the following statements except for the two that are false; for ... (i) a ? b (mod n) implies ak ? bk (mod n) for all k ? 0.

PDF	Solutions to Homework Set 3 (Solutions to Homework Problems Suppose a ? b (mod n). Then by definition

PDF	Congruences and Modular Arithmetic If a ? b (mod n) then ak ? bk (mod n) for all k ? N. Proof. Use part 2 of Theorem 3 and induct on k. Daileda. Congruences

PDF	Congruences original document so there could be some minor mistakes. If a ? b ( mod m )

PDF	Number Theory If a ? b (mod m) then ak ? bk (mod m) for any integer k. If a ? b Theorem: An integer n is divisible by 11 i the di erence of the sums of the odd.

PDF	Number Theory assumption these are all the primes but N is seen not to be divisible by any of the pi If a ? b and c ? d (mod m)

PDF	Midterm Practice If n is even then n can be expressed in the form 2k for some integer k ? Z. Inductive Step: Suppose ak ? bk (mod m) for some k ? Z+. We must now.

PDF	SOLUTIONS TO PROBLEM SET 1 Section 1.3 Exercise 4. We see bk?ak k. ? m. ? for some m. ?. ? Z (note that m. ? is needed since b may also satisfies q1q2 ? 1 (mod 6) we conclude that N ? 1 (mod 6) which is a ...

PDF	Number Theory Homework. mod n. In particular for any nonnegative integer k a ? b mod n. =? ak ? bk mod n. Proof. This is just an easy induction on k. D. Proposition 5. If f(x)

PDF	Divisibility Memorize: If a and b are integers we say that a divides b integer k such that ak = b. Fact D1. Let a b and c be integers. Then. (i) if a

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Solutions to Homework Set 3 (Solutions to Homework Problems

2 1 3 If a, b are integers such that a ≡ b (mod p) for every positive prime p, prove that a = b But then by the definition of congruence modulo n a − b = nk for some k ∈ Z But this implies Prove or disprove: If ab = 0 in Zn, then a = 0 or b = 0

PDF	3 Congruence It replaces the con- gruence sign with an equality Theorem 3 3 If a ≡ b mod n then b = a + nq for some integer q, and conversely Proof: If a

PDF	Congruences - Mathtorontoedu third number m, then we say "a is congruent to b modulo m", and write a ≡ b ( mod m ) integer k Although some books give this as a lemma or theorem, it is always best to Prove or disprove that if a ≡ b ( mod m ), then a2 ≡ b2 ( mod m)

PDF	Solutions 31 Prove that if m is an integer, then m 2 ≡ 0(mod 4) Prove or disprove that if n is odd, then n2 ≡ 1(mod 8) Proof Let n be odd Then there exists an integer k ∈ Z such that n = 2k + 1 Thus we have

PDF	Number Theory If a ≡ b (mod m), then a+um ≡ b +vm (mod m) for every integers u and v The assertion of the Fermat' theorem is valid also for some composite numbers

PDF	Math 346 Final Review Problems Hints and Answers If you find a So p ≡ kr (mod 30) ⇔ p = kr + 30j for some integer j Then p = k(r + lj) so if k = 1, then k is a factor of p and so p is Prove or Disprove: If a c and b c then ab c

PDF	Homework 6 - Number Theory Homework Definition and some basic results and examples The following definition was For example if a ≡ b mod n and b ≡ c mod n, then n (b − a) and n (c − a) We can now do “arithmetic modulo n” by adding and multiplying integers and then

PDF	Workouts m for which both i ≡ j (mod m) and k ≡ l (mod m) hold while i k ≡ jl (mod Prove or disprove that: For all integers m and n, if m · n is even, then if n ≡ 1 ( mod p−1) then i n ≡ i (mod Btw This proof needs some care, so please revise your

PDF	Modular Arithmetic 12 nov 2014 · congruent modulo m, written a ≡ b (mod m), if and only if a – b i e if there is some integer k such that a – b = km If ab and ac, then a(b - c)