Definition.
Let n be a fixed positive integer. Two integer are said to be congruent modulo n,symbolized by (pronounced a congruent to b modulo n)if n|a-b. If it's not,hence a is said to be not congruent to b modulo n.
Example:
Let say a=7,b=10 and n=3.
Since 7-10=-3 and -3|3,hence .
Theorem 14 - For any integers a and b, if and only if a and b leave the same nonnegative remainder when divided by n.
Theorem 15 - Len n be fixed and a,b,c and d can be any integers. The following properties hold
a)
b) If ,then
c) If ,and
,then
.
d) If ,and
,then
and
.
e) If ,then
and
.
f) If ,then
for any k>0.
Proof
a) Since a-a=0,and n|0,hence .
b) Suppose that . So from the definition,n|a-b. It follows that a-b=nk for some k. Since b-a=-nk,we multiply both side by -1,we'll ontain -b+a=a-b=nk. So,n|a-b,thus
.
c) Suppose that and
. So from the definition,n|a-b and n|b-c. It follows that a-b=nk for some k and b-c=nj for some j. By substituting b=a-nk into b-c=nj,we'll obtain a-nk-c=nj. So a-c=nj+nk. From that we get a-c|n and thus
.
d) Suppose that and
. From the definition,n|a-b and n|c-d. It follows that a-b=nk for some k and c-d=nj for some j.
a-b + (c-d) = nk + nj
a-b + c-d = n (k+j)
a+c - b-d = n (k+j)
a+c - (b+d) = n (k+j)
Because a+c - (b+d)|n,hence .
ac-bd=ac-bc+bc-bd
=c(a-b)+b(c-d)
=c(nk)+b(nj)
=n(ck-bj)
Because n|ac-bd,therefore .
e) Since ,we know n|a-b.
a-b=(a+c)-(b+c)
Hence n|(a+c)-(b+c). So,.
Since ,we know n|a-b.
ac-bc = c(a-b)
Hence n|c(a-b). So,.
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