Lecture 1 - Introducing Divisibility


Today we continued our discussion of divisibility and its basic properties. We saw some examples of how to put these properties into practice to prove exciting new results which might otherwise be quite difficult. Today's lecture culminated in the statement and proof of the division algorithm, one of the foundational results in number theory.

Divisibility Continued

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

Definition: An integer d is said to divide an integer a if there exists some integer q satisfying the equation $a = dq$.

Some Examples

We already saw proofs of $4 \mid 8$ and $2 \nmid 5$ in class on Wednesday. Most divisibility statements will seem pretty obvious to you just by inspection, but the one exception might be divisibility statements involving 0. Below we provide a few examples.

$0 \nmid 5$
since the equation $5 = q\cdot q$ doesn't have any solution; any value you plug in for q will still make the right hand side 0.
$0 \mid 0$
since the equation $0 = 0\cdot q$ is true for some integer-value of q (in fact, it's true for all q!).
$3 \mid 0$
since $0 = 3q$ does have a solution, namely $q=0$.

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.

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

\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

\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

\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 it might show up on your homework.

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.

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

\begin{align} 17 \mid 3(17a) - 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 $17 \mid 17b$, our result on integral linear combinations tells us that

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



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 $17 \mid 221$.

A Final Divisibility Result

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

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

Proof: We proved this result by induction, starting with the base case $n=1$. In this case it's easy to see that the statement is true: $5 \mid 1^5-1$.

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 just expand the term $(n+1)^5$, which gives us

\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

The following result, though it seems pretty basic, 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).

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 $a = qd + r$.

Proof of the division algorithm:

Part 1: Existence
We start by defining the set

\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 $a-ad = a(1-d)$. Now since d is positive by assumption we know that $1-d \geq 0$, and so the product $a(1-d)$ is either a positive number (if $1-d > 0$) or $0$ (if $1-d =0$). In either case we see that $a-ad$ is a non-negative element of 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

\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

\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

\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 see that 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 $q_1 = q_2$.$\square$

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