# Bézout's Lemma/Proof 2

## Contents

## Theorem

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 $b$.

Then:

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

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 $b$.

## Proof

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:

\(\displaystyle a > 0\) | \(\implies\) | \(\displaystyle \size a = 1 \times a + 0 \times b\) | |||||||||||

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

\(\displaystyle b > 0\) | \(\implies\) | \(\displaystyle \size b = 0 \times a + 1 \times b\) | |||||||||||

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

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 $0 < r$.

But:

\(\, \displaystyle \exists m, n \in \Z: \, \) | \(\displaystyle x\) | \(=\) | \(\displaystyle m a + n b\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle r\) | \(=\) | \(\displaystyle x - q d\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {m a + n b} - q \paren {u a + v b}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {m - q u} a + \paren {n - q v} b\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \) | \(\) | \(\displaystyle \paren {r \in S} \land \paren {r < d}\) |

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

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

In particular:

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

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

Thus:

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

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

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

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \gcd \set {a, b}\) | \(\divides\) | \(\displaystyle d\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \gcd \set {a, b}\) | \(\le\) | \(\displaystyle d\) |

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

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

$\blacksquare$

## Source of Name

This entry was named for Étienne Bézout.

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $3$: The Integers: $\S 11$. Highest Common Factor: Theorem $19$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 23$ - 1980: David M. Burton:
*Elementary Number Theory*(revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 3$