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
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Cancellation in Congruences Theorem 3 3 1 Suppose GCD(c, m) = 1 and ac ≡ bc(mod m) Then a ≡
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4 jan 2016 · If ac ≡ bc (mod m) and m ≥ 2 then a ≡ b (mod m) One way to look at this claim is to rewrite the first condition as (a − b)c ≡ 0 (mod m), and
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(7) (Reduction of Modulus Rule) If a ≡ b mod m and dm, d > 1, then a ≡ b mod d (8) (Scalar Multiplication Rule) If a ≡ b mod m and c > 0, then ac ≡ bc mod mc
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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
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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
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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 −
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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
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"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)
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