# Existence of Greatest Common Divisor

## Theorem

Let $a, b \in \Z$ be integers such that $a \ne 0$ or $b \ne 0$.

Then the greatest common divisor of $a$ and $b$ exists.

## Proof 1

### Proof of Existence

This is proved in Greatest Common Divisor is at least $1$.

$\Box$

### Proof of there being a Largest

Without loss of generality, suppose $a \ne 0$.

First we note that from Absolute Value of Integer is not less than Divisors:

- $\forall c \in \Z: \forall a \in \Z_{>0}: c \divides a \implies c \le \size c \le \size a$

The same applies for $c \divides b$.

Now we have three different results depending on $a$ and $b$:

\(\ds a \ne 0 \land b \ne 0\) | \(\implies\) | \(\ds \gcd \set {a, b} \le \min \set {\size a, \size b}\) | ||||||||||||

\(\ds a = 0 \lor b = 0\) | \(\implies\) | \(\ds \gcd \set {a, b} = \max \set {\size a, \size b}\) | ||||||||||||

\(\ds a = b = 0\) | \(\implies\) | \(\ds \forall x \in \Z: x \divides a \land x \divides b\) |

So if $a$ and $b$ are *both* zero, then *any* $n \in \Z$ divides both, and there is no greatest common divisor.

This is why the proviso that $a \ne 0 \lor b \ne 0$.

So we have proved that common divisors exist and are bounded above.

Therefore, from Set of Integers Bounded Above by Integer has Greatest Element there is always a **greatest** common divisor.

$\blacksquare$

## Proof 2

By definition of greatest common divisor, we aim to show that there exists $c \in \Z_{>0}$ such that:

\(\ds c\) | \(\divides\) | \(\ds a\) | ||||||||||||

\(\ds c\) | \(\divides\) | \(\ds b\) |

and:

- $d \divides a, d \divides b \implies d \divides c$

Consider the set $S$:

- $S = \set {s \in \Z_0: \exists x, y \in \Z: s = a x + b y}$

$S$ is not empty, because by setting $x = 1$ and $y = 0$ we have at least that $a \in S$.

From the Well-Ordering Principle, there exists a smallest $c \in S$.

So, by definition, we have $c > 0$ is the smallest such that $c = a x + b y$ for some $x, y \in \Z$.

Let $d$ be such that $d \divides a$ and $d \divides b$.

Then from Common Divisor Divides Integer Combination:

- $d \divides a x + b y$

That is:

- $d \divides c$

We have that:

\(\ds \exists t, u \in \Z: \, \) | \(\ds a\) | \(=\) | \(\ds c t + u:\) | \(\ds 0 \le u < c\) | Division Theorem | |||||||||

\(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds a x t + b y t + u\) | Definition of $c$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds u\) | \(=\) | \(\ds r \paren {1 - x t} + b \paren {-y t}\) | rearranging | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds u\) | \(=\) | \(\ds 0\) | as $u < c$ and the definition of $c$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds c\) | \(\divides\) | \(\ds a\) | Definition of Divisor of Integer |

- $c \divides b$

Now suppose $c'$ is such that:

\(\ds c'\) | \(\divides\) | \(\ds a\) | ||||||||||||

\(\ds c'\) | \(\divides\) | \(\ds b\) |

and:

- $d \divides a, d \divides b \implies d \divides c'$

Then we have immediately that:

- $c' \divides c$

and by the same coin: $c \divides c'$

and so:

- $c = c'$

demonstrating that $c$ is unique.

$\blacksquare$

## Proof 3

From Integers form Integral Domain, we have that $\Z$ is an integral domain.

From Euclidean Domain is GCD Domain, $a$ and $b$ have a greatest common divisor $c$.

This proves existence.

From Ring of Integers is Principal Ideal Domain, we have that $\Z$ is a principal ideal domain.

Suppose $c$ and $c'$ are both greatest common divisors of $a$ and $b$.

From Greatest Common Divisors in Principal Ideal Domain are Associates:

- $c \divides c'$

and:

- $c' \divides c$

and the proof is complete.

$\blacksquare$

## Proof 4

From the Euclidean Algorithm, we have calculated a sequence $\tuple {r_1, r_2, \ldots r_{n - 2}, r_{n - 1}, r_n}$ such that:

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

We have that:

- $r_{n - 1} \divides a$

and:

- $r_{n - 1} \divides b$

Working backwards from the final equation, we see that:

- $r_k \divides r{k - 1}$

for all $k$ such that $1 < k \le n$.

Hence, if $d \divides a$ and $d \divides b$, we can use induction to proceed through the Euclidean Algorithm and see that $d$ divides $r_1, r_2, \ldots, r_{n - 2}, r_{n - 1}$.

Thus we see that $r_{n - 1}$ fulfils the criteria to be the greatest common divisor of $a$ and $b$.

$\Box$

Suppose $c_1$ and $c_2$ are both greatest common divisors of $a$ and $b$.

Then by definition there exist $g, h \in \Z_{>0}$ such that:

- $g c_1 = c_2$
- $h c_2 = c_1$

Hence:

- $c_2 = g h c_2$

and so:

- $g h = 1$

That is:

- $g = h = 1$

and so:

- $c_1 = c_2$

That is, the greatest common divisor of $a$ and $b$ is unique.

$\blacksquare$