1. Division Problem

1. Let a, b, c, r, and t be integers. Prove that if a divides both b and c, then a divides br + ct.

2. Let b, c and d be integers. If b divides c and c divides d, then the equation bx = d has an integer solution. [Note: x is a variable.]

Thank you.

1. Let a, b, c, r, and t be integers. Prove that if a divides both b and c, then a divides br + ct.
.
$a|b,a|c$
Thus,
$b=ai,c=aj$ for some $i,j$.

Then,
$a(ri+tj)=ari+atj=br+ct$
Thus,
$a|(br+ct)$

2. Let b, c and d be integers. If b divides c and c divides d, then the equation bx = d has an integer solution. [Note: x is a variable.]
$b|c$ and $c|d$ thus, $b|d$.*
Thus,
$d=bk$ for some $k$.
Thus,
$b(k)=d$ solves the equation.

*)Proof.
$b|c$ thus, $c=bi$ and $c|d$ thus, $d=cj$
Thus,
$d=cj=(bi)j=b(ij)$
Thus,
$b|d$.

3. [quote=tttcomrader;35535]1. Let a, b, c, r, and t be integers. Prove that if a divides both b and c, then a divides br + ct.[\quote]

If $a|b$ then there exists an integer $\alpha$ such that $b=\alpha \, a$.

If $a|c$ then there exists an integer $\beta$ such that $c=\beta \, a$.

So $b\, r+c\,t = \alpha \, a \, r + \beta \, a \, t = a(\alpha \, r + \beta \, t)$.

Hence $a|(b\, r+c\, t)$

RonL

4. Thanks, guys, it looks so easy now that I see the solution, wonder why I couldn't see that at first... Pity as I have completed Advanced Calculus, I thought I would have go over Modern Algerba easy.

Thank you!

KK

5. I've found that, though these kinds of proof look easy, they do take some getting used to.

-Dan