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.
$\blacksquare$