Question

(a) If $a|b$ and $a|c$, prove that$a|\left(b+c\right)$.

(b) If $a|b$ and $a|c$, prove that$a|\left(br+ct\right)$for any $r,t\in \mathbb{z}$.

Short answer

(a) It is proved that If $a|b$and $a|c$, then $a|\left(b+c\right)$.

(b) It is proved that If $a|b$ and $a|c$then a $a|\left(br+ct\right)$for any$r,t\in \mathrm{\mathbb{Z}}$ .

Step-by-step solution

Apply divisibility definition 

As $a|b$ and $a|c$ , then by definition of divisibility, we have $b=ax$ and $c=ay$ for some integer $x$ and $y$.

Prove that if a|b  and  b|c then  a|b+c

Add $b=ax$ and $c=ay$ as:

$$\begin{gathered} b+c=ax+ay \\ b+c=a\left(x+y\right) \end{gathered}$$

This implies that $a|\left(b+c\right)$ for some integer $x+y$.

Thus, it is proved that if $a|b$ and $a|c$ then $a|\left(b+c\right)$.

Prove that if  a|b and b|c  then  a|b+c

From part (a), we have $b=ax$ and $c=ay$ for some integer $x$ and $y$.

Write $br+ct$ as:

$\begin{array}{rcl}br+ct& =& \left(ax\right)r+\left(ay\right)t\\ & =& axr+ayt\\ & =& a\left(xr+yt\right)\end{array}$

This implies that $a\left(br+ct\right)$ for some integer $xr+yt$.

Thus, it is proved that if $a|b$ and $a|c$ then $a|\left(br+ct\right)$ for any $r,t\in \mathbb{z}$.