Integer Combination of Coprime Integers/Proof 2

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Two integers are coprime if and only if there exists an integer combination of them equal to $1$:

$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$


Sufficient Condition

Let $a, b \in \Z$ be such that $\exists m, n \in \Z: m a + n b = 1$.

Let $d$ be a divisor of both $a$ and $b$.


$d \divides m a + n b$

and so:

$d \divides 1$


$d = \pm 1$

and so:

$\gcd \set {a, b} = 1$

Thus, by definition, $a$ and $b$ are coprime.


Necessary Condition

Let $a \perp b$.

Thus they are not both $0$.

Let $S$ be defined as:

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

$S$ contains at least one strictly positive integer, because for example $a^2 + b^2 \in S$.

By Set of Integers Bounded Below has Smallest Element, let $d$ be the smallest element of $S$ which is strictly positive.

Let $d = a x + b y$.

It remains to be shown that $d = 1$.

By the Division Theorem:

$a = d q + r$ where $0 \le r < d$


\(\displaystyle r\) \(=\) \(\displaystyle a - d q\)
\(\displaystyle \) \(=\) \(\displaystyle a - \paren {a x + b y} q\)
\(\displaystyle \) \(=\) \(\displaystyle a \paren {1 - x q} + b \paren {- y q}\)
\(\displaystyle \) \(\in\) \(\displaystyle S\)

But we have that $0 \le r < d$.

We have defined $d$ as the smallest element of $S$ which is strictly positive

Hence it follows that $r$ cannot therefore be strictly positive itself.

Hence $r = 0$ and so $a = d q$.

That is:

$d \divides a$

By a similar argument:

$d \divides b$

and so $d$ is a common divisor of both $a$ and $b$.

But the GCD of $a$ and $b$ is $1$.

Thus it follows that, as $d \in S$:

$\exists m, n \in \Z: m a + n b = 1$