Let $a, ~b$ and $c$ be given integers. Then 1. If $a \mid b$, then $ a \mid b \cdot c$. 2. If $a \mid b$ and $a \mid c$, then $ a^2 \mid b \cdot c$. 3. $a \mid b$ if and only if $a \cdot c \mid b \cdot c$, where $c \not = 0$.

 Let $a, ~b$ and $c$ be given integers. Then

1. If $a \mid b$, then $ a \mid b \cdot c$.

2. If  $a \mid b$ and $a \mid c$, then $ a^2 \mid b \cdot c$.

3.  $a \mid b$ if and only if $a \cdot c \mid b \cdot c$, where $c \not = 0$. 


Answer: $1.$ Suppose that $a \mid b$. Therefore there exists an integer $d$ such that $b = a \cdot d$. Multiply $b = a \cdot d$ by $c$, so we get $b \cdot c = a \cdot c \cdot d $, where $c \cdot d$ is an integer. This gives $a \mid b \cdot c$.

$2.$ Suppose that $a \mid b$ and $a \mid c$. So by $3$ of Theorem, we get $ a^2 \mid b \cdot c$.

$3.$ Suppose that $a \mid b$. Therefore there exists an integer $d$ such that $b = a \cdot d$. Put $b = a \cdot d$ in $b \cdot c$, where $c \not = 0$ so $b \cdot c = a \cdot d \cdot c = (a \cdot c ) \cdot d$. This gives that $a \cdot c \mid b \cdot c$. Conversely suppose that $a \cdot c \mid b \cdot c$, where $c \not = 0$. Therefore there exists an integer $d$ such that $b \cdot c = ( a \cdot c ) \cdot d$. Since $c \not = 0$, we can cancel $c$ from both side of $b \cdot c = ( a \cdot c ) \cdot d$ and we get $b = a \cdot d$ i.e. $a \mid b$.

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