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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.

Please help thanks.

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  • $\begingroup$ Can you write down a suitable $c$? $\endgroup$ – Lord Shark the Unknown Nov 7 '18 at 16:58
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    $\begingroup$ Actually, $d|ca$ for all integers $c$... $\endgroup$ – Federico Nov 7 '18 at 16:58
  • $\begingroup$ 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 $d$? $\endgroup$ – JMoravitz Nov 7 '18 at 16:58
  • $\begingroup$ They way you asked it, it's trivial because you can just take $c=1$. $\endgroup$ – Federico Nov 7 '18 at 16:59
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    $\begingroup$ By taking $c=0$, you don't even need the hypothesis that $d\mid a$. $\endgroup$ – Barry Cipra Nov 7 '18 at 17:03
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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.

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  • $\begingroup$ 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. $\endgroup$ – JustinL Nov 7 '18 at 18:28
  • $\begingroup$ @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. $\endgroup$ – Don Thousand Nov 7 '18 at 18:54
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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 $d|ca$.

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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.

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