Integer Combination of Coprime Integers/Sufficient Condition/Proof 1
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Theorem
Let $a, b \in \Z$ be integers, not both zero.
Let $a$ and $b$ be coprime to each other.
Then there exists an integer combination of them equal to $1$:
- $\forall a, b \in \Z: a \perp b \implies \exists m, n \in \Z: m a + n b = 1$
Proof
\(\ds a\) | \(\perp\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \gcd \set {a, b}\) | \(=\) | \(\ds 1\) | Definition of Coprime Integers | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists m, n \in \Z: \, \) | \(\ds m a + n b\) | \(=\) | \(\ds 1\) | Bézout's Identity |
$\blacksquare$
Also known as
This result is sometimes known as Bézout's Identity, although that result is usually used for the more general result for not necessarily coprime integers.
Some sources refer to this result as the Euclidean Algorithm, but the latter as generally understood is the procedure that can be used to establish the values of $m$ and $n$, and for any pair of integers, not necessarily coprime.
Sources
- 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 $2 \text{-} 4$