Lecture 1: Introducing Divisibility

Summary

Today we continued our discussion of divisibility and its basic properties. This culminated in the division algorithm. We also got a preview of the topic for next class: prime numbers.

Divisibility Continued

Last class previous we defined the notion of divisibility in the integers as follows:

We have plenty of other language that is associated with divisibility. For instance, if $d \mid a$ then we say that d is a factor of a, or that a is a multiple of d. When $d\mid a$ and we find an integer c so that $a = dc$, then we call this a factorization of a.

Some Examples

We already saw proofs of $2 \mid 18$ and $2 \nmid 5$ in class on Monday. As a few more examples, we have

$2 \mid -8$ since -8 = 2(-4)

The take away message is that you shouldn't be nervous about dividing into negative numbers

$2 \mid 0$ since 0 = 2(0)

The take away message here was that you shouldn't be nervous about dividing into 0 by any integer

$0 \nmid 2$ since any multiple of 0 is again 0

The take away message is that 0 isn't a factor of any number — except the number 0 itself.

The case of evens and odds

We also single out a special case of divisibility by using the terms even and odd. Specifically, we have

Definition: An integer a is even if $2 \mid a$, and an integer a is odd if $2 \nmid a$.

You might also be used to thinking of even integers as those numbers a which satisfy a = 2k for some integer k. Indeed, we have the following

Lemma: An integer a is even if and only if there exists $k \in \mathbb{Z}$ so that $a = 2k$.

Our proof of this result will require us to simply recall the definitions of divisiblity and evenness.

Proof:
We know that an integer a is even if and only if $2 \mid a$; this is just the definition of evenness. We also know that $2 \mid a$ if and only if there exists an integer k so that $a = 2k$; this is just the definition of divisibility. Hence we have

(1)

\begin{align} a\mbox{ is even} \Longleftrightarrow 2\mid a \Longleftrightarrow a = 2k \mbox{ for some }k \in \mathbb{Z} \end{align}

as desired. $\square$

You'll extend this problem in your homework when you show that all odd numbers can be written in the form $2k+1$.

Properties of Divisibility

There are a handful of properties of divisibility which are handy to remember; basically, these are good tools to use when you want to try to divide one integer into another. You can also think of these lemmas as good exercise for the definitions we've encountered in the class: none of the proofs require much more than writing down definitions, so they are a good chance for you to get used to the new terminology we've covered.

Lemma: For $a,b,c \in \mathbb{Z}$, if $a \mid b$ and $b \mid c$ then $a \mid c$.

Proof: We're told that $a \mid b$ and $b \mid c$. By the definition of divisibility, this means we have

• an integer d so that $b = ad$ (using the first divisibility condition), and
• an integer e so that $c = be$ (using the second divisibility condition).

Substituting appropriately, this means that

(2)

\begin{equation} c = be = (ad)e = a(de) \end{equation}

Since de is an integer, this equation tells us that $a \mid c$ as desired. $\square$

Lemma: For $a,b,c,m,n \in \mathbb{Z}$, if $a \mid b$ and $a \mid c$, then $a \mid mb+nc$.

Proof: Again, we start by just writing down the definitions. In this case, we're told that $a \mid b$ and $a \mid c$, which means we have

• an integer d so that $b = ad$ (using the first divisibility condition), and
• an integer e so that $c = ae$ (using the second divisibility condition).

Hence we have

(3)

\begin{equation} mb+nc = m(ad) + n(ae) = a(md+ne) \end{equation}

Since md+ne is an integer, this equation tells us that $a \mid mb+nc$.$\square$

There was another basic property of division we mentioned that allowed us to compare the size of a divisor to the size of the integer it is dividing. The statement of this result is

Lemma: If $d \mid a$ for a nonzero integer a, then $|d| \leq |a|$.

We didn't prove this result, but instead left it as an exercise you could do to get a little practice with proofs.

A Neat Trick

One of the examples of divisibility we gave in class was a rule for determining when an integer is divisible by 17. You can think of this as a cousin of the old "casting our nines" rule that you use to determine whether a given integer is divisible by 9. This new rule says

Theorem: An integer $10a+b$ is divisible by 17 if and only if $a-5b$ is divisible by 17.

I sort of screwed up the proof in class, so let me redeem myself by giving a proof here in the coursenotes.

Proof: First, assume that $17 \mid (10a+b)$. Since $17 \mid 17(3a)$ is obvious, our result on integral linear combinations tells us that

(4)

\begin{align} 17 \mid 17(3a) - 5(10a+b) = 51a-50a -5b = a-5b. \end{align}

In the other direction, assume that we are told that $17 \mid a-5b$, and we want to prove $17 \mid 10a+b$. Now since we know that $17 \mid 17(3b)$, our result on integral linear combinations tells us that

(5)

\begin{align} 17 \mid 10(a-5b) + 17(3b) = 10a-50b+51b = 10a+b. \end{align}

$\square$

Example

To see this result in practice, notice that we have 221 = 22(10)+1. Since $17 \mid 22-5(1) = 17$, we can conclude that $17 \mid 221$.

For a more complicated result, notice that 2142 = 214(10)+2. Hence $17 \mid 2142$ if and only if $17 \mid 214-5(2) = 204$. To check this divisibility condition, notice that it is equivalent to showing $17 \mid 20 - 5(4) = 0$, which we know to be true.

A Final Divisibility Result

We finished off with one last example of a divisiblity proof, when we showed that

For every integer n, we have $5 \mid n^5 - n$.

Proof: We proved this result by induction. We said it was sufficient to prove the result for positive integers n

• if n happens to be negative, then -n is positive, in which case we would know $5 \mid (-n)^5 -(-n) = -(n^5-n)$. This is equivalent to the desired statement (do you know why?).
• if n happens to be 0, then we're certainly in the clear: $5 \mid 0^5-0 = 0$ was (basically) proven in class already.

So to get this proof off the ground, we need to start with a base case. We'll use the case $n=1$, in which case we certainly have $5 \mid 1^5-1$.

For the inductive step, we'll assume we know that $5 \mid n^5-n$, and we'll try to use this to prove that $5 \mid (n+1)^5-(n+1)$. In order to do this, we'll try to simplify the expression $(n+1)^5 - (n+1)$ into something more user friendly; we decided the bast way to do this was to use the binomial theorem, which gives us

(6)

\begin{align} \begin{equation*}\begin{split}(n+1)^5-(n+1) &= (n^5 + 5n^4+10n^3+10n^2+5n+1) - n - 1\\&= (n^5-n) + 5(n^4+2n^3+2n^3+n).\end{split}\end{equation} \end{align}

Since $5 \mid n^5-n$ by induction and $5 \mid 5(n^4+2n^3+2n^3+n)$ due to our clever factorization, our result on integral linear combinations tells us that $5 \mid (n+1)^5-(n+1)$ as well.$\square$

The Division Algorithm

Having spent plenty of time talking about what it means for one integer to divide another, we now give a little time to those unfortunate scenarios when $d \nmid a$. In this case, it will be convenient to have some measure of the failure of d to divide a, for which we'll use the notion of "remainder after division" which you might have learned way back in grade school. We'll see that this notion is actually extremely powerful, giving rise not just to a method for finding greatest common divisors (Section 1.3) but also laying the foundation for the notion of modular arithmetic (Chapter 2).

With that as motivation, here's the statement of

The Division Algorithm: For a positive integer d and an arbitrary integer a, there exist unique integers q and r with $0 \leq r < d$ and $a = qd + r$.

As a quick example before proceeding to the proof, notice that when d = 13 and a = 268, we have

(7)

\begin{align} 268 = 20\cdot 13 + 8. \end{align}

Proof of the division algorithm:

Part 1: Existence
We start by defining the set

(8)

\begin{align} S = \{a - nd : n \in \mathbb{Z}\}, \end{align}

and we claim that S has at least one non-negative element. To back up this claim, notice that

• if $a>0$ then we can take $n=0$ and find that $a \in S$;
• otherwise $a < 0$, in which case taking $n = a$ shows that $a-ad = -a(d-1) \in S$.

In either case we see that S contains a non-negative element, and hence the well ordering principle tells us that S contains a least non-negative element. We'll call this element r, and notice that r takes the form

(9)

\begin{align} r = a-qd \mbox{ for some }q \in \mathbb{Z}. \end{align}

Hence we get $a = qd + r$. To show this satisfies the conditions of the division algorithm, we simply need to show that $0 \leq r < d$. The condition $0 \leq r$ is satisfied since r is chosen to be non-negative, so we only need to verify $r < d$.

To see that $r < d$, assume to the contrary that $r \geq d$, and we'll derive a contradiction. In this case we have that

(10)

\begin{align} \tilde r = a - (q+1)d = a-qd -d = r-d \geq 0 \end{align}

Since $r \geq d$ by assumption we have $\tilde r$ is a non-negative element of S which is smaller than r. This is a contradiction to the selection of r as the smallest non-negative element of S, so we must conclude that $r < d$ as desired.

Part 2: Uniqueness
To finish the proof we need to show that the q and r we found in the previous part of the theorem are, indeed, unique. Hence suppose we have

(11)

\begin{equation} a = q_1d+r_1 = q_2d+r_2. \end{equation}

This tells us that
$r_1-r_2 = d(q_2-q_1)$, and therefore that $d \mid r_1 - r_2$. But since we also have $-d < -r_2 \leq r_1-r_2 < r_1 < d$ by our conditions $0 \leq r_1,r_2 < d$, we are in the scenario where the divisor d has larger absolute value than the number it is dividing into — namely, $r_2-r_1$. This tells us that we must have $r_2 - r_1 = 0$, and hence $r_2 = r_1$.

With this in hand, we the equation $q_1d + r_1 = q_2d + r_2$ then becomes $q_1d = q_2 d$. Using the cancellation law of multiplication, we therefore have $q_1 = q_2$.$\square$

A sneak peak of primes

Now that we're masters of divisibility, we'll spend the next class period talking about the possible factorizations of numbers. In particular, we'll give a special name to those integers which admit no non-trivial factorization.

Definition: An integer $p>1$ is said to be prime if any factorization $p = ab$ has either a or b equal to 1.

There are other ways to characterize primes as well, so we'll mention them next class period. We'll also start asking questions about properties of prime numbers: how many are there? how spread out can they get? how close together can they get? how fast do they grow? Indeed, asking questions about prime numbers is what number theory is all about, so this means that Friday marks our formal introduction to the most important objects in number theory.