Example 1.
\(3\mid 6\) since \(6=2\cdot 3\)
\(5\mid 15\) since \(15=3\cdot 5\)
\(2\nmid 7\) since \(7\neq 2\cdot k, k\in\Z\)
\(y\) is said to be a multiple of \(x\)
\(x\) is said to be a divisor or factor of \(y\)
\(3\mid 6\) since \(6=2\cdot 3\)
\(5\mid 15\) since \(15=3\cdot 5\)
\(2\nmid 7\) since \(7\neq 2\cdot k, k\in\Z\)
Are the following true or false?
\(2\mid 0\)
True since \(0=0\cdot 2\)
\(3\mid 11\)
False since \(11\neq 3\cdot k\) for any integer \(k\)
\(5\mid (20 + 15)\)
True since \(20+15=35\) and \(35=5\cdot 7\)
\(20\mid 4\)
False since \(4\neq 20\cdot k\) for any integer \(k\)
\(-3\mid 9\)
True since \(9=(-3)(-3)\)
Let \(x\text{,}\) \(y\text{,}\) and \(z\) be integers. If \(x\mid y\) and \(x\mid z\text{,}\) then \(x\mid (sy+tz)\) for any integers \(s\) and \(t\text{.}\)
\(3\mid 6\) and \(3\mid 9\text{.}\) What about \(6+9\) and \(2\cdot 6+3\cdot 9\text{?}\)
We have \(3\mid (6+9)\) since \(3\mid 15\text{.}\) Also, \(2\cdot 6+3\cdot 9 = 12+27=39\text{,}\) and we also have \(3\mid 39\text{.}\)