Cancellation in Congruences Theorem 3 3 1 Suppose GCD(c, m) = 1 and ac ≡ bc(mod m) Then a ≡
s
12 nov 2014 · same remainder upon division by m E g 3 ≡ 7 (mod 2) 9 ≡ 99 (mod 10) congruent modulo m, written a ≡ b (mod m), if and only if a – b is
modular
Integer a is congruent to integer b modulo m > 0, if a and b give the same remainder when divided by m Notation a ≡ b (mod m) Alternative definition: a ≡ b (mod
NumberTheory
If a ≡ b (mod m), then a+c ≡ b+c (mod m) and ac ≡ bc (mod m) for any c ∈ Z 6 If a ≡ b (mod m), then an ≡ bn (mod m) for any positive integer n 7 If ac ≡ bc
E textofChap Mod
Let m be a positive integer, and let a, b, c, d ∈ Z If a ≡ b mod m and c ≡ d mod m then ac ≡ bd mod m Proof By definition m (a −
modulus
22 sept 2013 · Reminder: a ≡ b (mod m) means a − b is divisible by m True or false? If a ≡ b ( mod m), then a + c ≡ b + c (mod m) If a ≡ b (mod m), then ac
number theory
"a is congruent to b modulo m" means m(a-b) Equivalently, a b leave the same remainder by division by m (for a,b≥0) 1) If a ≡ b (mod m) then (a+c) ≡ (b+d)
week
4 thg 1 2016 4.1.37 [2 points] Find counterexamples to the following claims: 1. If ac ? bc (mod m) and m ? 2 then a ...
Proof. Suppose GCD(c m) = 1 and ac ? bc(mod m). (We may assume m > 1 so that c = 0.) Then ac ? bc = c(a ? b) = km for some integer k. This implies c
then ac ? bd(mod m). (True). (8) If ac ? bc(mod m)
22 thg 9 2013 If ac ? bc (mod m)
Hence a+c ? b+c (mod m) and ac ? bc (mod m). Page 4. 4. Proof of Part 6: We prove an ? bn (mod m)
22 thg 2 2005 Prove all of the following statements except for the two that are ... (f) ac ? bc (mod n) implies a ? b (mod n) provided c ? 0 (mod n).
If a ? b (mod m) then ac ? bc (mod m) for any integer c. If a ? b (mod m) and c ? d (mod m)
(v) If a ? b (mod m) and c > 0 then ac ? bc (mod mc). Proof. Verification of these properties is straightforward. For instance
then ac ? bc(mod m.) You should be.
https://math.berkeley.edu/~mcivor/math115su12/lectures/lecture4.pdf
We have the following rules for modular arithmetic: Sum rule: IFa?b(modm) THENa+c?b+c(modm) (3) Multiplication Rule: IFa?b(modm) and if c?d(modm) THENac?bd(modm) (4) De?nitionAn inverse toamodulomis a integerbsuch that ab?1(modm) (5) By de?nition (1) this means that ab?1 =k· mfor some integer k
Proposition 1 For any integers abcm such that m > 1 and gcd(cm) = 1 if ac ? bc (mod m) then a ? b (mod m) Proof : If m[(a?b)c] and gcd(cm) = 1 by a property we proved in class we have m(a?b) Division of a real number x by another real y 6= 0 is the same as multiplying x by 1/y (also written as y?1) and that y?1 is also
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) ? 11999 ? 1999 ? 1 (mod 10) 9 ? –1 (mod 10) ? 9999 ? (–1)999 (mod 10) 7999 ? 49499 7 ? (–1)499 7 ? –7 ? 3 (mod 10)
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 rq ac – bd = (b + qm) ( d + rm) – bd = bd + brm + qmd +qrm2 – bd = m (br + qd + qrm) Divisible by m so ac ? bd (mod m) Corollary: If a ? b (mod m) then a2 ? b2 (mod m) Proof: Case where a = c & b = d 3) If a ? b
Suppose GCD(c;m) = 1 and ac bc(mod m) (We may assume m>1 so that c6= 0 ) Then ac bc= c(a b) = km for some integer k This implies cjkm: Since GCD(c;m) = 1 Lemma 2 from Module 5 2 asserts that if cjkm then cjk: Write k= cdfor some integer d and substitute for kin the equation above: c(a b) = km= (cd)m= c(dm): 3 APPLICATIONS OF NUMBER THEORY 165
Supposea=b (modm) andb=c (modm) Then thereare integersjandksuch that a?b=jm b?c=km Add the two equations: a?c= (j+k)m This implies thata=c (modm) Theorem Supposea=b (modm) andc=d (modm) Then: a+c=b+d (modm) ac=bd (modm) Note that you can use the second property and induction to show that ifa=b (modm) then an=bn (modm) for all n?1
Is a-c(b-c)(mod m)?
Is a-c? (b-c) (mod m)? Yes, this follows directly from the fact that it holds for addition, given that . However it does not hold for division - if and an integer with and both integers, it does not necessarily follow that . For example , but (. Furthermore, I am of the opinion that Quora should re-introduce the suggest edits feature.
Is a-c(b-c)(mod m) the same for subtraction?
We know that, if a?b (mod m) and c is an integer, then a+c? (b+c) (mod m). It is the same with multiplication too but is it the same for subtraction also? Is a-c? (b-c) (mod m)? Yes, this follows directly from the fact that it holds for addition, given that .
What does a b (mod m) mean?
1) If a ? b (mod m) then (a+c) ? (b+d) (mod m) & c ? (mod m) Proof: a ? b (mod m) means a – b = mq, some q ? Z Also c – d = mr some r ? Z.
What does a is congruent to b modulo m mean?
"a is congruent to b modulo m" means m|(a-b). Equivalently, a & b leave the same remainder by division by m (for a,b?0) 1) If a ? b (mod m) then (a+c) ? (b+d) (mod m) & c ? (mod m) Proof: a ? b (mod m) means a – b = mq, some q ? Z Also c – d = mr some r ? Z.
Homework 4 Solutions