Let
\[n\]
be a fixed positive integer and let \[a, \; b, \; c, \; d\]
be any integers. Thena)
\[a \equiv a \; (mod \; n)\]
b)
\[a \equiv b \; (mod \; n) \rightarrow b \equiv a \; (mod \; n)\]
c)
\[a \equiv b \; (mod \; n), \; b \equiv c \; (mod \; n) \rightarrow a \equiv c \; (mod \; n)\]
d)
\[a \equiv b \; (mod \; n), \; c \equiv d \; (mod \; n) \rightarrow a+c \equiv b+d \; (mod \; n)\]
e)
\[a \equiv b \; (mod \; n), \; c \equiv d \; (mod \; n) \rightarrow a+c \equiv ac=bd \; (mod \; n)\]
f)
\[a \equiv b \; (mod \; n), \; r \in \mathbb{Z}^{{}+{}} \rightarrow a^r \equiv b^r \; (mod \; n)\]
Proof
a)
\[a-a= 0 \times n \rightarrow a \equiv a \; (mod \; n)\]
b)
\[a \equiv b \; (mod \; n) \rightarrow a=b+kn \rightarrow b=a-kn \rightarrow b \equiv a \; (mod \; n)\]
c)
\[a \equiv b \; (mod \; n) \rightarrow a=b+rn\]
\[b \equiv c \; (mod \; n) \rightarrow b=c+sn\]
Then
\[a=c+rn+sn=c+(r+s)n \rightarrow a \equiv c \; (mod \; n)\]
d)
\[a \equiv b \; (mod \; n) \rightarrow a=b+rn\]
\[c \equiv d \; (mod \; n) \rightarrow c=d+sn\]
\[a+c=b+d+rn+sn=b+d+(r+s)n \rightarrow a+b=c+d \; (mod \; n)\]
e)
\[a \equiv b \; (mod \; n) \rightarrow a=b+rn\]
\[c \equiv d \; (mod \; n) \rightarrow c=d+sn\]
\[ac=(b+rn(d+sn)=bd+(dr+bs+rsn)n \rightarrow ac=bd \; (mod \; n)\]
f) The case
\[r=1\]
is trivial. Suppose the result is true for \[r=k\]
so that \[a^k=b^k \; (mod \; n)\]
then \[a^{k+1}=aa^k \equiv bb^k \; (mod \; n)* \equiv b^{k+1} \; (mod \; n)\]
*by e)
Hence proved by induction.