# Math Help - Division Problem

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.

2. Originally Posted by tttcomrader
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,
$a|(br+ct)$

Originally Posted by tttcomrader

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

6. Originally Posted by tttcomrader
Pity as I have completed Advanced Calculus, I thought I would have go over Modern Algerba easy.
Abstract Algebra, I think, is way more complicated then Advanced Calculus. Because the nice thing about analysis is that it can be visualized that is what makes it simpler. With Algebra there is absolutely no visualization at all. I would also say that Abstract Algebra is as difficult as courses get.