If a and b are integers, a * divides* b if there
is an integer c such that

The notation means that a divides b.

For example, , since . And , since . Also, , since .

* Remarks.* (a) Be careful not to confuse
" " with " " or " ". The notation
" " is read "a divides b",
which is a * statement* --- a complete sentence
which could be either true or false. On the other hand, " " is read "a divided by b". This is an
expression, not a complete sentence. Compare "6 divides 18"
with "18 divided by 6" and be sure you understand the
difference.

(b) By this definition, " " ("0 divides 0") is true, since (for example) . Does this violate the rule that "you can't divide by 0"?

This is not a problem, and the reason has to do with a subtle
difference in terminology. The rule that "you can't divide by
0" means that *0 does not have a multiplicative inverse*.
In general, "dividing by x" means "multiplying by the
multiplicative inverse" --- for instance, dividing by 3 is
multiplying by .

To see that 0 can't have a multiplicative inverse , suppose toward a contradiction that it did. A number and its multiplicative inverse (by definition) multiply to 1:

But any number multiplied by 0 gives 0, so

The contradiction " " shows that is undefined.

The definition we gave above implies, as we noted, that "0
divides 0", but this is not the same as saying "you can
divide 0 by 0". The wording is close, *but different*.
The definition in this section defines *divisibility* in terms
of multiplication; it is not the definition of *dividing* in
term of multiplying by the multiplicative inverse.

This is probably more than you wanted to know about this. But if you are still bothered by it, you can adjust the definition, so that " " is only defined if . The reason I haven't done this is because I would need to check the condition or make an assumption whenever I used the notation.

The properties in the next proposition are easy consequences of the definition of divisibility; see if you can prove them yourself.

* Proposition.*

(a) Every number divides 0.

(b) 1 divides everything. So does -1.

(c) Every number is divisible by itself.

* Proof.* (a) If , then , so .

(b) To take the case of 1, note that if , then , so .

(c) If , then , so .

* Definition.* An integer is * prime* if its only positive
divisors are 1 and itself. An integer is *
composite* if it isn't prime.

The first few primes are

The first few composite numbers are

Prime numbers play an important role in number theory.

* Proposition.* Let .

(a) If and , then .

(b) If , , and , then

(c) If and , then .

(In case you were wondering, mathematicians have different names for
results which are intended to indicate their relative importance. A
* Theorem* is a very important result. A * Proposition* is a result of less importance. A * Lemma* is a result which is primarily a step in the
proof of a theorem or a proposition. Of course, there is some
subjectivity involved in judging how important a result is.)

* Proof.* (a) Suppose and . This means that there are
numbers d and e such that and . Substituting the first equation into the second, I
get , or . This implies
that .

(b) Suppose and . This means that there are numbers d and e such that and . Then

(c) means for some e, and means for some f. Therefore,

Part (b) says, in words, that if an integer a divides integers b and
c, then a divides any * linear combination* of b
and c.

* Corollary.* Suppose and .

(a) .

(b) .

(c) for all .

In words, (a) says that if a number divides two other numbers, it divides their sum.

(b) says that if a number divides two other numbers, it divides their difference.

(c) says that if a number divides another number, it divides any multiple of the other number.

* Proof.* All three parts follow from part (b) of
the Proposition. For (a), take and . For (b), take and . And for (c), take .

* Example.* Prove that if x is even, then is divisible by 4.

x is even means that .

and implies that by part (c) of the proposition.

and implies that by part (c) of the proposition.

Obviously, .

Then by part (b) of the proposition, so , again by part (b) of the proposition.

Here is an important result about division of integers. It will have
a lot of uses --- for example, it's the key step in the * Euclidean algorithm*, which is used to compute * greatest common divisors*.

* Theorem.* (* The Division
Algorithm*) Let a and b be integers, with .

(a) There are unique integers q and r such that

(b) .

Of course, this is just the "long division" of grade school, with q being the quotient and r the remainder.

* Proof.* (a) The idea is to find the remainder r
using Well-Ordering. What is division? Division is successive
subtraction. You ought to be able to find r by subtracting b's from a
till you can't subtract without going negative. That idea motivates
the construction which follows.

Look at the set of integers

In other words, I take a and subtract *all possible multiples*
of b.

If I choose (as I can --- there's always an
integer less than any number), then , so . This choice of n produces a positive integer in S. So the subset T consisting of nonnegative
integers in S is *nonempty*.

Since T is a nonempty set of nonnegative integers, I can apply Well-Ordering. It tells me that there is a smallest element . Thus, , and for some q (because , , and everything in S has this form).

Moreover, if , then , so

So , but . This contradicts my assumption that r was the smallest element of T.

All together, I now have r and q such that

To show that r and q are unique, suppose and also satisfy these conditions:

Then

But r and are two nonnegative numbers less than b, so they are less than b units apart. This contradicts the last equation, which says they are units apart --- unless . Since , this means , or . In addition, , so . This proves that r and q are unique.

(b) Assuming with , I want to show that .

This shows that q is an integer less than or equal to . Hence, . I have to show that this is actually equality.

Suppose on the contrary that . The next integer larger than q is , and must be at least as big. So

Since , the last inequality gives

This contradicts . Since is ruled out, I must have .

* Example.* (a) Apply the Division Algorithm to
divide 59 by 7.

(b) Apply the Division Algorithm to divide -59 by 7.

(a) The quotient is , the remainder is 3, and . I have

(b) The quotient is , the remainder is 4, and . I have

* Example.* Prove that if , then does not leave a remainder of 2 or
3 when it's divided by 5.

It is easier to do this using * modular
arithmetic*, but I'll do this using the Division Algorithm as an
illustration.

If n is divided by 5, the remainder r satisifies . Thus, . Hence, n can have one of the following 5 forms:

Check each case:

In all cases, dividing by 5 gave a remainder of 0, 1, or 4. I never got a remainder of 2 or 3.

As an illustration, can't be a perfect square, because it leaves a remainder of 3 when it's divided by 5.

Copyright 2019 by Bruce Ikenaga