# Prove if d | a, then d | ca for any integer c. Correction [duplicate]

I am frustrated because this is literally exercise 3 from my textbook and I still can not get it. I already failed my first midterm. I am wondering why I am bad with discrete mathematics, but love calculus?

Anyway, how do I solve this? I have tried looking for the same problem but to no avail.

• Can you write down a suitable $c$? – Lord Shark the Unknown Nov 7 '18 at 16:58
• Actually, $d|ca$ for all integers $c$... – Federico Nov 7 '18 at 16:58
• Look at the definition. $d\mid a$ means that there exists some integer $k$ such that $a = d\cdot k$ (equivalently phrased, $d$ divides $a$, $a$ is divisible by $d$, $a$ is an integer multiple of $d$). Now. Knowing that $a= d\cdot k$ what can you say about $ac$ and its relation to $d$? – JMoravitz Nov 7 '18 at 16:58
• They way you asked it, it's trivial because you can just take $c=1$. – Federico Nov 7 '18 at 16:59
• By taking $c=0$, you don't even need the hypothesis that $d\mid a$. – Barry Cipra Nov 7 '18 at 17:03

Looking at the relevant definitions is always an important first step to take in introductory problems in proof writing.

For nonzero integers $$d,a$$ the following are equivalent statements:

• $$d\mid a$$
• $$d$$ divides $$a$$
• $$a$$ is divisible by $$d$$
• $$a$$ is an integer multiple of $$d$$
• There exists some integer $$k$$ such that $$a = d\times k$$
• $$\frac{a}{d}$$ is an integer
• $$\vdots$$

(There are still several more equivalent statements, but those can come later in your studies and don't need to be mentioned now)

Suppose that $$d\mid a$$. We wish to show that for any integer $$c$$ it follows that $$d\mid ac$$.

Since $$d\mid a$$, it follows that there is some integer $$k$$ such that $$a = d\times k$$.

By multiplying both sides of that equation by $$c$$ and slight rearranging we have $$ac = d\times (kc)$$

Now... we ask, is there an integer that we can fill in the blue square with in the following $$ac = d\times \color{blue}{\square}$$ to make the equality true? Yes, we can, we can fill it in with $$kc$$ which is also an integer since our earlier work already showed that $$ac = d\times (kc)$$.

Since $$ac$$ is an integer multiple of $$d$$, by definition then $$d\mid ac$$

Side note: you would notice that when we talked about how $$d\mid a$$ that means there is some $$k$$ such that $$a=d\times k$$. If we were to also talk about how $$d\mid b$$ and rewrite this using another equality, it does not have to be "the same $$k$$" in the equality... so we would probably want to use a different letter., say for example $$\ell$$ such that $$b = d\times \ell$$. We don't care what the integer looks like or how it is written, all we care about is that it is an integer.

• Thank you. The trick that i was looking for was the multiply both sides by c. My question now is, how do i not pick up on these things easily? This seem like a very basic proof. – JustinL Nov 7 '18 at 18:28
• @JustinL I've always found that whenever in doubt, refer to the definitions. If you had done that here, I'm pretty sure you would have gotten it. – Don Thousand Nov 7 '18 at 18:54

It is always worth thinking to yourself "can I write the problem in a different way ?".

In this case $$d|a$$ means that $$a=kd$$ for some integer $$k$$. But if $$a=kd$$ then $$ca = ckd$$. And since $$c$$ and $$k$$ are both integers, $$d$$ is clearly a divisor of $$ckd$$, which is just $$ca$$ in disguise. So $$d|ca$$.

Um... not just for some $$c$$ but for all $$c$$.

$$d|a$$ means, by definition that $$\frac ad = m$$, an integer.

So me must prove there is some integer $$c$$ where $$\frac {ca}d = k$$, and integer.

Well, if there were then $$\frac {ca}d = c\frac ad = cm$$.

So this is asking is there some integer $$c$$ where $$cm$$ is an integer? And the answer is EVERY integer $$c$$ will have $$cm$$ being an integer.