# Euclidean Algorithm

## Algorithm

The **Euclidean algorithm** is a method for finding the greatest common divisor (GCD) of two integers $a$ and $b$.

Let $a, b \in \Z$ and $a \ne 0 \lor b \ne 0$.

The steps are:

- $(1): \quad$ Start with $\tuple {a, b}$ such that $\size a \ge \size b$. If $b = 0$ then the task is complete and the GCD is $a$.
- $(2): \quad$ If $b \ne 0$ then you take the remainder $r$ of $\dfrac a b$.
- $(3): \quad$ Set $a \gets b, b \gets r$ (and thus $\size a \ge \size b$ again).
- $(4): \quad$ Repeat these steps until $b = 0$.

Thus the GCD of $a$ and $b$ is the value of the variable $a$ after the termination of the algorithm.

In the words of Euclid:

*Given two (natural) numbers not prime to one another, to find their greatest common measure.*

(*The Elements*: Book $\text{VII}$: Proposition $2$)

### Variant: Least Absolute Remainder

Let $a, b \in \Z$ and $a \ne 0 \lor b \ne 0$.

The steps are:

- $(1): \quad$ Start with $\tuple {a, b}$ such that $\size a \ge \size b$. If $b = 0$ then the task is complete and the GCD is $a$.
- $(2): \quad$ If $b \ne 0$ then you take the least absolute residue $r$ such that:
- $a = q b + r: -\dfrac b 2 < r \le \dfrac b 2$

- $(3): \quad$ Set $a \gets b, b \gets r$ (and thus $\size a \ge \size b$ again).
- $(4): \quad$ Repeat these steps until $b = 0$.

## Proof 1

Suppose $a, b \in \Z$ and $a \lor b \ne 0$.

From the Division Theorem, $a = q b + r$ where $0 \le r < \size b$.

From GCD with Remainder, the GCD of $a$ and $b$ is also the GCD of $b$ and $r$.

Therefore, we may search instead for $\gcd \set {b, r}$.

Since $\size r < \size b$ and $b \in \Z$, we will reach $r = 0$ after finitely many steps.

At this point, $\gcd \set {r, 0} = r$ from GCD with Zero.

$\blacksquare$

## Proof 2

Suppose $a, b \in \Z$ and $a \lor b \ne 0$.

Let $d = \gcd \set {a, b}$.

By definition of common divisor:

- $d \divides a$

and:

- $d \divides b$

Hence from Common Divisor Divides Integer Combination:

- $d \divides \paren {a - q b}$

That is:

- $d \divides r$

Thus $d$ is a common divisor of $b$ and $r$.

Let $c$ be an arbitrary common divisor of $b$ and $r$.

Then:

- $c \divides \paren {q b + r}$

That is:

- $c \divides a$

Thus $c$ is a common divisor of $a$ and $b$.

Hence by definition of GCD:

- $c \le d$

Hence, again by definition of GCD: $d = \gcd \set {b, r}$

Then we work down the system of equations:

\(\ds \gcd \set {a, b}\) | \(=\) | \(\ds \gcd \set {b, r_1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \gcd \set {r_1, r_2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \cdots\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \gcd \set {r_{n - 1}, r_n}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \gcd \set {r_n, 0}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds r_n\) |

$\blacksquare$

## Euclid's Proof

Let $AB, CD$ be the two given (natural) numbers which are not coprime.

We need to find the greatest common divisor of $AB$ and $CD$.

Without loss of generality that $CD \le AB$.

We have that $CD$ is a divisor of itself.

If $CD$ is a divisor of $AB$ then $CD$ is a common divisor of $CD$ and $AB$.

It is clearly the greatest, because no number greater than $CD$ can be a divisor of $CD$.

Now, suppose $CD$ does not divide $AB$.

Then the less of the numbers $AB, CD$ being continually subtracted from the greater, some number will be left which will be a divisor of the one before it.

From Coprimality Criterion, this number will not be a unit, otherwise $AB$ and $CD$ will be coprime.

So some number will be left which is a divisor of the one before it.

Now let $CD$, dividing $BE$, leave $EA$ less than itself.

Let $EA$, dividing $DF$, leave $FC$ less than itself.

Let $CF$ divide $AE$.

Since then $CF$ divides $AE$, and $AE$ divides $DF$, then $CF$ will also divide $DF$.

But it also divides itself.

Therefore it will also divide the whole $CD$.

But $CD$ divides $BE$, therefore $CF$ divides $BE$.

But it also divides $EA$, therefore it will also divide the whole $BA$.

So $CF$ is a common divisor of $CD$ and $AB$.

Suppose $CF$ is not the greatest common divisor of $CD$ and $AB$.

Let $G > CF$ also be a common divisor of $CD$ and $AB$.

Since $G$ divides $CD$, while $CD$ divides $BE$, it follows that $G$ divides $BE$.

But $G$ also divides the whole $BA$, and so it also divides the remainder $AE$.

But $AE$ divides $DF$.

Therefore $G$ divides $DF$.

But $G$ also divides the whole $DC$.

Therefore it will also divide the remainder $CF$.

But $G$ is greater than $CF$, which is impossible.

Therefore no number greater than $CF$ divides the numbers $AB$ and $CD$.

Therefore $CF$ is the greatest common divisor of $AB$ and $CD$.

$\blacksquare$

## Demonstration

Using the **Euclidean Algorithm**, we can investigate in detail what happens when we apply the Division Theorem repeatedly to $a$ and $b$.

\(\ds a\) | \(=\) | \(\ds q_1 b + r_1\) | ||||||||||||

\(\ds b\) | \(=\) | \(\ds q_2 r_1 + r_2\) | ||||||||||||

\(\ds r_1\) | \(=\) | \(\ds q_3 r_2 + r_3\) | ||||||||||||

\(\ds \cdots\) | \(\) | \(\ds \) | ||||||||||||

\(\ds r_{n - 2}\) | \(=\) | \(\ds q_n r_{n - 1} + r_n\) | ||||||||||||

\(\ds r_{n - 1}\) | \(=\) | \(\ds q_{n + 1} r_n + 0\) |

From the Division Theorem, we know that the remainder is always strictly less than the divisor.

That is, in $a = q b + r$:

- $0 \le r < \size b$

So we know that:

- $b > r_1 > r_2 > \ldots > r_{n - 2} > r_{n - 1} > r_n > 0$

So the algorithm *has* to terminate.

$\blacksquare$

## Algorithmic Nature

It can be seen from the definition that the **Euclidean Algorithm** is indeed an algorithm:

- Finiteness: As has been seen, the algorithm always terminates after a finite number of steps.
- Definiteness: Each of the steps is precisely defined.
- The inputs are $a$ and $b$.
- The output is $\gcd \left\{{a, b}\right\}$.
- Effectiveness: Each operation is finite in extent and can be effectively performed.

## Formal Implementation

The **Euclidean algorithm** can be implemented as a computational method $\struct {Q, I, \Omega, f}$ as follows:

Let $Q$ be the set of:

- all singletons $\tuple n$
- all ordered pairs $\tuple {m, n}$
- all ordered quadruples:
- $\tuple {m, n, r, 1}$
- $\tuple {m, n, r, 2}$
- $\tuple {m, n, p, 3}$

where $m, n, p \in \Z_{> 0}$ and $r \in \Z_{\ge 0}$.

Let $I \subseteq Q$ be the set of all ordered pairs $\tuple {m, n}$.

Let $\Omega$ be the set of all singletons $\tuple n$.

Let $f: Q \to Q$ be defined as follows:

\(\ds \map f {\tuple {m, n} }\) | \(=\) | \(\ds \tuple {m, n, 0, 1}\) | ||||||||||||

\(\ds \map f {\tuple n}\) | \(=\) | \(\ds \tuple n\) | ||||||||||||

\(\ds \map f {\tuple {m, n, r, 1} }\) | \(=\) | \(\ds \tuple {m, n, \map \rem {m, n}, 2}\) | ||||||||||||

\(\ds \map f {\tuple {m, n, r, 2} }\) | \(=\) | \(\ds \begin{cases} \tuple n & : r = 0 \\ \tuple {m, n, r, 3} & : r \ne 0 \end{cases}\) | ||||||||||||

\(\ds \map f {\tuple {m, n, p, 3} }\) | \(=\) | \(\ds \tuple {n, p, p, 1}\) |

where $\map \rem {m, n}$ denotes the remainder of $m$ on division by $n$.

## Constructing an Integer Combination

Having determined the GCD of $a$ and $b$ using the Euclidean Algorithm, we are now in a position to find a solution to $\gcd \set {a, b} = x a + y b$ for $x$ and $y$.

By Bézout's Identity it is always possible to find such an $x, y \in \Z$.

Working back through the equations generated during the course of working the Euclidean Algorithm, we get:

\(\ds \gcd \set {a, b}\) | \(=\) | \(\ds r_n\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds r_{n - 2} - q_n r_{n - 1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds r_{n - 2} - q_n \paren {r_{n - 3} - q_{n - 1} r_{n - 2} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {1 + q_n q_{n - 1} } r_{n - 2} - q_n r_{n - 3}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {1 + q_n q_{n - 1} } \paren {r_{n - 4} - q_{n - 2} r_{n - 3} } - q_n r_{n - 3}\) |

... and so on. The algebra gets messier the further up the tree you go, and is not immediately enlightening.

Thus eventually we arrive at $\gcd \set {a, b} = x a + y b$ where $x$ and $y$ are numbers made up from some algebraic cocktail of the coefficients of the terms involving the remainders from the various applications of the Division Theorem.

Note that $a \divides b \implies \gcd \set {a, b} = a$, from GCD of Integer and Divisor.

## Also known as

The **Euclidean algorithm** is also known as **Euclid's algorithm** or the **Euclidean division algorithm**.

## Examples

### GCD of $341$ and $527$

The GCD of $341$ and $527$ is found to be:

- $\gcd \set {341, 527} = 31$

### GCD of $2190$ and $465$

The GCD of $2190$ and $465$ is found to be:

- $\gcd \set {2190, 465} = 15$

Hence $15$ can be expressed as an integer combination of $2190$ and $465$:

- $15 = 33 \times 465 - 7 \times 2190$

### GCD of $9 n + 8$ and $6 n + 5$

The GCD of $9 n + 8$ and $6 n + 5$ is found to be:

- $\gcd \set {9 n + 8, 6 n + 5} = 1$

Hence:

- $2 \paren {9 n + 8} - 3 \paren {6 n + 5} = 1$

### Solution of $31 x \equiv 1 \pmod {56}$

Let $x \in \Z$ be an integer such that:

- $31 x \equiv 1 \pmod {56}$

Then by using the Euclidean Algorithm:

- $x = -9$

is one such $x$.

## Also see

- Rational Numbers and Simple Finite Continued Fractions are Equivalent for an application of the
**Euclidean algorithm**in a slightly different context.

## Source of Name

This entry was named for Euclid.

## Historical Note

This proof is Proposition $2$ of Book $\text {VII}$ of Euclid's *The Elements*.

According to David M. Burton, in his *Elementary Number Theory, revised ed.* of $1980$, there exists historical evidence that the Euclidean Algorithm actually predates Euclid.

## Sources

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