Lecture 10

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 $\begin{aligned} &{}\gcd(556,296)\\ &{}=\gcd(1\times296+260,296) {}=\gcd(260,296) {}=\gcd(296,260)\\ &{}=\gcd(1\times 260+36,260) {}=\gcd(36,260) {}=\gcd(260,36)\\ &{}=\gcd(7\times 36+8,36) {}=\gcd(8,36) {}=\gcd(36,8)\\ &{}=\gcd(4\times 8+4,8) {}=\gcd(4,8) {}=\gcd(8,4)\\ &{}=\gcd(2\times 4+0,4) {}=\gcd(0,4) {}= 4.\end{aligned}$

Here’s the general case:

Algorithm

[Euclid’s algorithm] Suppose we must calculate the greatest common divisor of two positive integers. Call them $a$ and $b$ with $a\geq b$. If they’re not in the right order, we can swap them over earlier.

By division with remainder, we can write $a = qb+r$ for some integers $q$ and $r$ with $0\leq r<b$.

But then we have $\gcd(a,b) {}= \gcd(qb+r,b) {}= \gcd(r,b) {}= \gcd(b,r),$ and since $b\leq a$ and $r<b$ we’ve made our numbers smaller.

If we keep doing this repeatedly, we’ll end up making one of the numbers zero and can stop (since $\gcd(d,0) = d$).

One might reasonably wonder just how fast Euclid’s algorithm really is. One good answer (not very hard to prove) is that if you’re trying to work out $\gcd(a,b)$ and $b\leq a$, then the number of steps you need is always less than five times the number of digits of $b$.

So working out $\gcd(123456789,987654321)$ will take less than $5\times 9 = 45$ divisions (actually, this one takes a lot less than $45$, if you try it). Compared with the other methods we discussed, this makes it seem really good.

Euclid’s algorithm is in fact even more useful than it looks: using Euclid’s algorithm, if we have $\gcd(a,b) = d$, that enables us to write $d$ in the form $ma+nb=d$ for some integers $m$ and $n$. (We say that we’re writing it as a linear combination of $a$ and $b$). This will be really useful later: I promise!

Let’s see how this works with an example. We saw earlier that $\gcd(126,70)=14$, so we expect to be able to find integers $m$ and $n$ such that $126m+70n=14$.

Along the way we found that: $\begin{aligned} {}126 &= 1\times 70 + 56, &\qquad(1)\\ {}70 &= 1\times 56 + 14. &\qquad(2)\end{aligned}$ Working through that backwards, we get that $\begin{aligned} {}14 &= 1\times 70 - 1\times 56 \qquad\text{(using (2))}\\ {}&= 1\times 70 - 1\times (1\times 126 - 1\times 70) \qquad\text{(using (1))}\\ {}&= 2\times 70 - 1\times 126.\end{aligned}$

Similarly, when we calculated that $\gcd(556,296)=4$, we found that: $\begin{aligned} {}556 &= 1\times 296+260, &\qquad(3)\\ {}296 &= 1\times 260+ 36, &\qquad(4)\\ {}260 &= 7\times 36+ 8, &\qquad(5)\\ {}36 &= 4\times 8+ 4. &\qquad(6)\end{aligned}$ This means that $\begin{aligned} {}4 &= 36-4\times 8\qquad\text{(using (6))}\\ {}&= 36-4\times(260-7\times 36)\qquad\text{(using (5))}\\ {}&= 29\times 36-4\times 260\\ {}&= 29\times(296-260)-4\times 260\qquad\text{(using (4))}\\ {}&= 29\times296-33\times260\\ {}&= 29\times296-33\times(556-296)\qquad\text{(using (3))}\\ {}&= 62\times296-33\times556.\end{aligned}$