Let
\[a, \; b\]
be positive integers, and let \[c, \; d\]
be any integers. Thena) If
\[a | c\]
then \[a | (c+na)\]
for any integer \[n\]
.b) If
\[c \neq 0\]
and \[a | c\]
then \[a \le |c|\]
.c) If
\[a | b\]
and \[b | a\]
then \[a=b\]
.d) If
\[a | b, \; b | c\]
then \[a | c\]
.e) If
\[a | c, \; a | d\]
then \[a | (mc+nd)\]
for any integers \[m, \; n\]
.Proof
a)
\[a |c\]
so there is an integer \[q\]
such that \[aq=c\]
. Then \[c+na=aq+na=a(q+n)=ax\]
where \[x=qn\]
is an integer, so \[a | (c+na)\]
.b) Let
\[c=aq\]
as in a) then \[| c |= | aq |= a | q |\]
and the result follows since \[| q | \ge 1\]
and \[q\]
is an integer. This result implies that \[c\]
has only a finite number of divisors, since any divisor must be less than or equal to \[| c |\]
.c) If
\[a | b, \; b | a \]
then from b) \[a \le b, \; b \le a\]
. The result follows.d) If
\[a | b, \; b | c \]
then there are integers \[s, \; t\]
such that \[b=as, \; c=bt\]
. Then \[c=ast=a(st)\]
so \[a | c\]
.e) If
\[a | c, \; a | d \]
then there are integers \[s, \; q\]
such that \[c=as, \; d=aq\]
. Then \[mc+nd=mas+naq=a(ms+nq)=ax\]
where \[x=ms+nq\]
is an integer so e) is proved.