Mathematics

 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 \cdo...

 If $a \mid b$, show that $(- a ) \mid b, ~ a \mid (- b), $ and $ (- a ) \mid (- b)$.  Answer: Suppose that $a \mid b$. So there exists an integer $c$ such that $b = a \cdot c$.  We can write $b = a \cdot c$ as $b = (- a) \cdot (- c)$, where $(-a), ~ (-c)$ are integers. This gives $(- a ) \mid b$. Now multiply $b = a \cdot c$ by $(-1)$, we will have either $(- b) = a \cdot (- c)$, where $- c$ is an integer or $(- b) = (- a) \cdot c$, where $ c$ is an integer . This gives  $a  \mid (- b)$ or $ (- a ) \mid (- b)$. Therefore we have $(- a ) \mid b, ~ a \mid (- b), $ and $ (- a ) \mid (- b)$.

  Lemma:   If $a ~|~ b \cdot c$, with $ gcd(a, b) = 1$, then $a ~|~ c$. Proof:  Since $gcd(a, b) = 1$, by Theorem , there exist integers $x$ and $y$ such that $a \cdot x + b \cdot y = 1$. Multiplying this equation by $c$, we get  $c = c \cdot 1 = c~(a \cdot x + b \cdot y) = a \cdot c \cdot x + b \cdot c \cdot y$. As $ a ~|~ a \cdot c$ and  $ a ~|~ b \cdot c$, by part $7.$ of Theorem , $d ~|~ (a \cdot c \cdot x + b \cdot c \cdot y)$. Therefore $d ~|~ c$.

Corollary:  If $a ~|~ c$ and $b ~|~ c$, with $gcd(a, b) = 1$, then $a \cdot b ~|~ c$. Proof: Suppose that $a ~|~ c$ and $b ~|~ c$. Therefore there exist integers $r$ and $s$  such that $c = a \cdot r = b \cdot s$. Since $gcd(a, b) = 1$ By Theorem , there exist integers $x$ and $y$ such that $a \cdot x + b \cdot y = 1$. Multiplying this equation by $c$, we get $c = c \cdot 1 = c~(a \cdot x + b \cdot y) = a \cdot c \cdot x + b \cdot c \cdot y$. By putting appropriate value of $c$ on the right-hand side, we get $c = a~(b \cdot s)~x + b~(a \cdot r)~y = a \cdot b~(s \cdot x + r \cdot y)$. Since $s \cdot x+r \cdot y$ is an integer, we get $a \cdot b ~|~ c$. 

Corollary: If $gcd(a, b) = d$, then $gcd(\frac{a}{d}, \frac{b}{d}) = 1$. Proof:  Since $d = gcd(a,b)$, we have  $d ~|~ a$ and $d ~|~ b$. Hence $\frac{a}{d},~ \frac{b}{d}$ are integers. Since $d = gcd(a,b)$, Theorem , there exist integers $x$ and $y$ such that $d = gcd(a, b) = a \cdot x+ b \cdot y$. By dividing both side of this equation by $d$, we get $1 = \frac{a}{d} \cdot x + \frac{b}{d} \cdot y$. By Theorem , $gcd(\frac{a}{d}, \frac{b}{d}) = 1$.

  Definition:  Two integers $a$ and $b$, not both of which are zero, are said to be relatively prime integers   if and only if $gcd(a, b) = 1$.   Theorem:  If $a$ and $b$ are any integers, not both them are zero, then $a$ and $b$ are relatively prime integers if and only if there exist integers $x$ and $y$ such that $a \cdot x +  b \cdot y = 1$. Proof:  Suppose that $a$ and $b$ are relatively prime integers, so that $gcd(a, b) = 1$. By Theorem , there exist integers $x$ and $y$ such that $a \cdot x + b \cdot y = 1$. conversely, suppose that there exist integers $x$ and $y$ such that $a \cdot x + b \cdot y = 1$ and that $d = gcd(a, b )$. Since $d = gcd(a,b)$, we have  $d ~|~ a$ and $d ~|~ b$. By part $7.$ of Theorem ,  $d ~|~ (a  \cdot x + b \cdot y)$ for all integers $x, ~y$. So in particular  $d ~|~  (a \cdot x + b \cdot y) = 1$. Since $d$ is a positive  integer and $d ~|~ 1$,  $d$ must be equal to $1$. Therefor...