Bézout's Lemma/Proof 2

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Let $a, b \in \Z$ such that $a$ and $b$ are not both zero.

Let $\gcd \set {a, b}$ be the greatest common divisor of $a$ and $b$.


$\exists x, y \in \Z: a x + b y = \gcd \set {a, b}$

That is, $\gcd \set {a, b}$ is an integer combination (or linear combination) of $a$ and $b$.

Furthermore, $\gcd \set {a, b}$ is the smallest positive integer combination of $a$ and $b$.


Let $a, b \in \Z$ such that $a$ and $b$ are not both zero.

Let $S$ be the set of all positive integer combinations of $a$ and $b$:

$S = \set {x \in \Z, x > 0: x = m a + n b: m, n \in \Z}$

First we establish that $S \ne \O$.

We have:

\(\displaystyle a > 0\) \(\implies\) \(\displaystyle \size a = 1 \times a + 0 \times b\) $\quad$ $\quad$
\(\displaystyle a < 0\) \(\implies\) \(\displaystyle \size a = \paren {-1} \times a + 0 \times b\) $\quad$ $\quad$
\(\displaystyle b > 0\) \(\implies\) \(\displaystyle \size b = 0 \times a + 1 \times b\) $\quad$ $\quad$
\(\displaystyle b < 0\) \(\implies\) \(\displaystyle \size b = 0 \times a + \paren {-1} \times b\) $\quad$ $\quad$

As it is not the case that both $a = 0$ and $b = 0$, it must be that at least one of $\size a \in S$ or $\size b \in S$.

Therefore $S \ne \O$.

As $S$ contains only positive integers, $S$ is bounded below by $0$ and therefore $S$ has a smallest element.

Call this smallest element $d$: we have $d = u a + v b$ for some $u, v \in \Z$.

Let $x \in S$.

By the Division Theorem:

$x = q d + r, 0 \le r < d$

Suppose $d \nmid x$.

Then $x \ne q d$ and so $0 < r$.


\(\, \displaystyle \exists m, n, \in \Z: \, \) \(\displaystyle x\) \(=\) \(\displaystyle m a + n b\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle r\) \(=\) \(\displaystyle x - q d\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {m a + n b} - q \paren {u a + v b}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {m - q u} a + \paren {n - q v} b\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(\) \(\displaystyle \paren {r \in S} \land \paren {r < d}\) $\quad$ $\quad$

which contradicts the choice of $d$ as the smallest element of $S$.

Therefore $\forall x \in S: d \divides x$.

In particular:

$d \divides \size a = 1 \times a + 0 \times b$
$d \divides \size b = 0 \times a + 1 \times b$


$d \divides a \land d \divides b \implies 1 \le d \le \gcd \set {a, b}$

However, note that as $\gcd \set {a, b}$ also divides $a$ and $b$ (by definition), we have:

\(\displaystyle \gcd \set {a, b}\) \(\divides\) \(\displaystyle \paren {u a + v b} = d\) $\quad$ Common Divisor Divides Integer Combination $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \gcd \set {a, b}\) \(\divides\) \(\displaystyle d\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \gcd \set {a, b}\) \(\le\) \(\displaystyle d\) $\quad$ $\quad$

Since $d$ is the smallest number in $S$:

$\gcd \set {a, b} = d = u a + v b$


Source of Name

This entry was named for Étienne Bézout.