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
∈
Z , m >1
"a is congruent to b modulo m" means m(ab).
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.
We Want: (ab) – (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)
≡
(b+d) (mod m)
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 + qm) ( d + rm) – bd
= bd + brm + qmd +qrm
2
– bd
= m (br + qd + qrm)
Divisible by m, so ac
≡
bd (mod m)
Corollary: If a
≡
b (mod m), then a
2
≡
b
2
(mod m)
Proof:
Case where a = c & b = d
3) If a
≡
b (mod m) and n
∈
N then an = bn (mod m)
Proof:
Use Mathematical Induction.
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 Spring '10
 Applebaugh
 Math, Congruence, Natural number, Prime number

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