# The Euclidean Algorithm

The Euclidean Algorithm is used to find the greatest common denominator (GCD) of two numbers.

Assume:

$\label{eqn:a} a = qb + r \tag{\star}$

\label{eqn:gcd} \begin{cases} \ gcd(a, 0) = a\\\
gcd(0, b) = b\\\
gcd(a, b) = gcd(b, r) \end{cases} \tag{$\dagger$}

The proof of the third equality in \ref{eqn:gcd} can be made by first proving that:

\begin{align*}\label{eqn:gcd-lemma} \tag{$\ddagger$} gcd(a, b) &= gcd(b, a - b) \\\
&= gcd(a - b, b) \end{align*}

Proof of \ref{eqn:gcd-lemma} 1:

\begin{align*} gcd(a, b) \lvert a & \implies \exists x \in \mathbb{Z} \textrm{ such that } x \cdot gcd(a, b) = a \\\
gcd(a, b) \lvert b & \implies \exists y \in \mathbb{Z} \textrm{ such that } y \cdot gcd(a, b) = b \\\
\end{align*}

Let

\begin{align*} c &= a - b \\\
&= x \cdot gcd(a, b) - y \cdot gcd(a, b) \\\
&= (x - y) \cdot gcd(a, b) \\\
\end{align*}

which implies that:

$$\label{eqn:abc} \tag{\circ} gcd(a, b) | c$$

Further:

\begin{align*} gcd(b, c) \lvert b & \implies \exists m \in \mathbb{Z} \textrm{ such that } m \cdot gcd(b, c) = b \\\
gcd(b, c) \lvert c & \implies \exists n \in \mathbb{Z} \textrm{ such that } n \cdot gcd(b, c) = c \\\
\end{align*}

So:

\begin{align*} c = a - b \iff a &= b + c \\\
&= m \cdot gcd(b, c) + n \cdot gcd(b, c) \\\
&= (m + n) \cdot gcd(b, c) \end{align*}

which implies that:

$$\label{eqn:bca} \tag{\diamond} gcd(b, c) | a$$

By definition, $$gcd(a, b) \lvert b$$, so together with \ref{eqn:abc} we get that $$gcd(a, b)$$ is a common divisor of both $$b$$ and $$c$$. But since $$gcd(b, c)$$ is the greatest such common divisor of $$b$$ and $$c$$ we must have:

$$\label{eqn:leq} \tag{1} gcd(a, b) \leqslant gcd(b, c)$$

Also by definition, $$gcd(b, c) \lvert b$$, so together with \ref{eqn:bca} we get that $$gcd(b, c)$$ is a common divisor of both $$b$$ and $$a$$. But since $$gcd(a, b) = gcd(b, a)$$ is the greatest such common divisor of $$b$$ and $$a$$ we must have:

$$\label{eqn:geq} \tag{2} gcd(a, b) \geqslant gcd(b, c)$$

The inequalities \ref{eqn:leq} and \ref{eqn:geq} give:

$$gcd(a, b) = gcd(b, c) = gcd(b, a - b)$$

thus finalizing the proof of \ref{eqn:gcd-lemma}.

Repeatedly applying \ref{eqn:gcd-lemma} gives:

\begin{align*} gcd(a, b) &= gcd(a - b, b) \\\
&= gcd(a - 2b, b) \\\
&= gcd(a - 3b, b) \\\
& \space \space \vdots \\\
&= gcd(a - qb, b) \\\
& \stackrel{(\ref{eqn:a})}{=} gcd(r, b) \\\
&= gcd(b, r) \qquad \square \end{align*}

1. The notation $$m \lvert n$$ means that $$m$$ evenly divides $$n$$. ↩︎