Lecture 10

Coprimality

Now we’re going to introduce some very useful concepts. Rather than (as we were doing before) looking at one number at a time, it’s going to turn out to be really useful to consider two numbers and compare their factors.

Definition: Let $a$ and $b$ be integers. A common divisor of $a$ and $b$ is an integer $d$ such that $d\mid a$ and $d\mid b$. The greatest common divisor of $a$ and $b$, written $\gcd(a,b)$ (or sometimes as $\operatorname{hcf}(a,b)$ or sometimes even just $(a,b)$ for short) is the largest common divisor of $a$ and $b$.

That definition probably just says that a greatest common divisor is what you’d expect it to be, given the name!

That definition is dangerous, because it does something I’ve warned you against doing several times: it defines something that looks like a function, but it doesn’t prove that it is a function.

There are two reasons why the gcd might not exist; we need to satisfy ourselves that neither is a problem:

• There might be no common divisors at all (and hence no greatest common divisor): This is not a problem: we have observed before that $1$ is a divisor of every integer, and so will certainly be a common divisor.

• There may be lots of common divisors, but no largest one. For nonzero integers $a$, it’s easy to see that if $d\mid a$ then $|d|\leq|a|$, which means we can’t get arbitrarily large divisors…but $\gcd(0,0)$ doesn’t exist.

As happens quite often, the remark above, which looks like a slightly pedantic point at first, really says something practically important. Indeed, it gives us a way of finding the greatest common divisor of two numbers: to find $\gcd(a,b)$ we could just count down from $|a|$ and stop when we reach the first common divisor.

For example, $$\gcd(9,15) = {}3$$ and $$\gcd(-30,42) = {}6.$$

This approach to finding greatest common divisors is pretty terrible: imagine being asked to find $$\gcd(123456789,987654321)$$ by this approach!

Another way might be to work out all factors of one of the numbers ($a$, for example) and work out which of them are factors of $b$. That’s also a pretty terrible way, because factorising numbers is hard work: it seems like a lot of work to find all factors of $123456789$ still.

We will see a much better way soon, but, first, let’s spot some easy properties of greatest common divisors.

For all integers $a$ and $b$, we have $$\gcd(a,b) = \gcd(b,a),$$ because the definition is symmetric in $a$ and $b$.

Also, for all positive integers $a$, we have $$\gcd(a,a) = a,$$ and $$\gcd(a,1) = 1,$$ and $$\gcd(a,b) = \gcd(a,-b).$$

A slightly less obvious property is:

Proposition

Let $a,b$ and $k$ be integers. Then $$\gcd(a,b) = \gcd(a+kb,b).$$

Proof

We’ll show that the common divisors of $a$ and $b$ are the same as the common divisors of $a+kb$ and $b$.

Suppose first that $d$ is a common divisor of $a$ and $b$; in other words, $d\mid a$ and $d\mid b$. Then we can write $a=md$ and $b=nd$ for some integers $m$ and $n$. But then $$a+kb=md+knd=(m+kn)d,$$ so $d\mid a+kb$, so $d$ is a common divisor of $a+kb$ and $b$.

Similarly, if $d$ is a common divisor of $a+kb$ and $b$, then we can write $a+kb = ld$ and $b=nd$. But then $$a = a+kb-kb = ld-knd = (l-kn)d,$$ so $d\mid a$, so $d$ is a common divisor of $a$ and $b$.

Since we’ve now proved that $a$ and $b$ have the same common divisors as $a+kb$ and $b$, it follows that they have the same greatest common divisor.

We should also mention that the greatest common divisor has a close cousin:
Definition: Given two positive integers $a$ and $b$, the least common multiple $\operatorname{lcm}(a,b)$ is the smallest positive integer which is a multiple both of $a$ and $b$.

Given that $ab$ is a common multiple of $a$ and $b$, the least common multiple always exists (and is at most $ab$: we could find it by counting up from $1$ to $ab$, stopping on the first common multiple).

The last piece of terminology we might want is this:
Definition: Two integers $a$ and $b$ are said to be coprime, or relatively prime, if $\gcd(a,b)=1$.

Division with Remainder

The above Proposition looks slightly dry at first: so what if you can add multiples of one number to another number without changing their greatest common divisor?

It turns out this is the key step in a surprisingly efficient method for calculating greatest common divisors. We can use it to make the numbers smaller; the question is, how? It turns out that this is something familiar to you all:

Proposition

[Division with Remainder] Let $a$ and $b$ be integers, with $b>0$. One can write $$a = qb+r$$ for integers $q$ (the quotient) and $r$ (the remainder) such that $0\leq r<b$.

It is not too hard to prove this: one can do it with two inductions, for example, (one for the negative and one for the positive integers), but I won’t do so here.

It’s reasonable to ask why we had to take $b>0$. It’s true for $b<0$, too, we just have to say that the remainder $r$ satisfies $0\leq r<-b$ instead.

This observation gives us a really efficient way of computing greatest common divisors. Let’s illustrate it by an example.

Suppose we’re trying to compute $\gcd(126,70)$. If we divide $126$ by $70$ we get $1$ with remainder $56$; in other words $126 = 1\times 70+56$. That means that

$$\begin{aligned} \gcd(126,70) &= \gcd(56+1\times 70,70)\\ &= \gcd(56,70)\\ &= \gcd(70,56).\end{aligned}$$

That made the problem much smaller, and we can do the same trick repeatedly: $$\begin{aligned} \gcd(70,56) &= \gcd(14+1\times 56,56)\\ &= \gcd(14,56)\\ &= \gcd(56,14).\end{aligned}$$ That’s smaller still. Let’s see what happens next: $$\begin{aligned} \gcd(56,14) &= \gcd(0+14\times 4,14)\\ &= \gcd(0,14)\\ &= 14.\end{aligned}$$ As $56$ is a multiple of $14$, of course we get remainder $0$, so we stop here: the greatest common divisor is $14$.

For another example, let’s suppose we want the greatest common divisor of $556$ and $296$. We write $$\begin{aligned} &{}\gcd(556,296)\\ &{}=\gcd(1\times296+260,296) {}=\gcd(260,296) {}=\gcd(296,260)\\ &{}=\gcd(1\times 260+36,260) {}=\gcd(36,260) {}=\gcd(260,36)\\ &{}=\gcd(7\times 36+8,36) {}=\gcd(8,36) {}=\gcd(36,8)\\ &{}=\gcd(4\times 8+4,8) {}=\gcd(4,8) {}=\gcd(8,4)\\ &{}=\gcd(2\times 4+0,4) {}=\gcd(0,4) {}= 4.\end{aligned}$$