# Number Theory: Problem 2

### Model solution

Let p, q, and r be integers (p non-zero). Suppose that $$p \mid 3q$$ and $$3q \mid r$$.

By the definition of divides, $$p | 3q$$ means pm = 3q for some integer m. Similarly, $$3q | r$$ means 3qn = r for some integer n.

Now substitute these two equations into the expression 3q+r: $$3q + r = 3q + 3qn = 3q(1 + n) = pm(1+n)$$

We know m(1+n) is an integer, since m and n are integers. So let s=m(1+n). Then 3q+r = ps. Therefore, by the definition of divides, $$p | (3q + r)$$, which is what we wanted to show.

### Also ok

It's ok to write the last paragraph without introducing the extra variable s. For example:
We know m(1+n) is an integer, since m and n are integers. So the above equation shows that 3q+r can be written as p times an integer. Therefore, by the definition of divides, $$p | (3q + r)$$, which is what we wanted to show.

### Self-check

At the start of the proof, did you use two different variables m and n when expanding the definition of "divides"?

When you invoke the definition of divides near the end, it's critical that m(1+n) is an integer. You need to make this clear to the reader, by specifying m and n as integers when you introduce them and briefly saying why m(1+n) is an integer later on.

### Poor style 1

Proofs like the following will lose points due to lack of connector words and brief explanations. How is the poor grader supposed to know what you are trying to do at each step?

Let p, q, and r $$\in \mathbb{Z}$$

$$p \mid 3q$$ and $$3q \mid r$$.

pm = 3q, $$m \in \mathbb{Z}$$

3qn = r, $$n \in \mathbb{Z}$$

$$3q + r = 3q + 3qn = 3q(1 + n) = pm(1+n)$$

$$m(1+n) \in \mathbb{Z}$$

So $$p | (3q + r)$$

### Poor style 2

Here's a more subtle example of poor style. It does have connector words, but the author has forgotten everything he learned in high school English class. Paragraphs, punctuation, and whitespace are critical to making technical arguments readable. Everyone makes occasional mistakes, and you'll make a few extras if your first language isn't English. But all of you can do a better job than this.

let p, q, and r be integers (p non-zero) suppose that $$p \mid 3q$$ and $$3q \mid r$$.

by definiton divides, $$p | 3q$$ means pm = 3q, m integer. $$3q | r$$ means 3qn = r for some integer n. now substitute these two equations into the expression 3q+r $$3q + r = 3q + 3qn = 3q(1 + n) = pm(1+n)$$ We know m(1+n) is an integer, since m and n are integers solet s=m(1+n) then 3q+r = ps therefore by the definiton of divides $$p | (3q + r)$$ QED