# Division Problem

• Jan 18th 2007, 12:32 PM
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.
• Jan 18th 2007, 12:41 PM
ThePerfectHacker
Quote:

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,
$a|(br+ct)$

Quote:

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,
$b|d$.
• Jan 18th 2007, 12:45 PM
CaptainBlack
[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
• Jan 18th 2007, 09:05 PM
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
• Jan 19th 2007, 04:38 AM
topsquark
I've found that, though these kinds of proof look easy, they do take some getting used to.

-Dan
• Jan 19th 2007, 06:20 AM
ThePerfectHacker
Quote:

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.