Proof regarding division of integers

[srfriggen]

Homework Statement

Prove that for any a, b, c $$\in$$$$Z$$, if a l b (a divides b) and a does not divide c, then a does not divide b-c.

Homework Equations

The Attempt at a Solution

Using the contrapositive:

Prove that... if a l (b-c) then a does not divide b or a l c.

1. a l (b-c), equivalent to: b-c=ax (for some x in Z)
2. b=ax+c
3. c=b-ax

(Now I am trying to work out the equations to get c=an, for some n in Z).

b-c=(ax+c)-(b-ax)=ax+c-b+ax = 2ax+c-b


that's where I get stuck. I keep fooling around with the algebra but can't seem to get the desired result.

p.s. a fellow classmate claims the result can/should be obtained by using cases, where b-c is odd or even... For the former I suppose (b-c)=2n, for some n in Z... I can't seem to make use of this information, and I'm not even sure he's correct.

 

[mighty2000]

Hello, thank you for posting this question on the forum. Let me help you with solving this problem.

First, let's start with the given information: a l b (a divides b) and a does not divide c. This means that there exists some integer k such that b = ka and c is not equal to ka for any integer k.

Now, let's assume that a l (b-c). This means that there exists some integer m such that b-c = ma. We can rewrite this as b = ma + c.

Next, we can substitute the value of b into the given information: a l b. This gives us a l (ma + c). We can now factor out an 'a' from the right side of the equation: a l a(m + c/a). Since a does not divide c, we can rewrite the equation as a l a(m + c/a) implies a l a(m + c/a) implies a l m + c/a. This implies that a l m. However, this contradicts our initial assumption that a does not divide c. Therefore, our assumption that a l (b-c) is false and we can conclude that a does not divide b-c.

I hope this helps you in understanding how to prove this statement. Please let me know if you have any further questions.