Integer Combination of Coprime Integers/Sufficient Condition/Proof 3

Proof
Let $a \perp b$.

Thus they are not both $0$.

Let $S$ be defined as:
 * $S = \set {\lambda a + \mu b: \lambda, \mu \in \Z}$

$S$ contains at least one strictly positive integer, because for example:
 * $a \in S$ (setting $\lambda = 1$ and $\mu = 0$)
 * $b \in S$ (setting $\lambda = 0$ and $\mu = 1$)

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

Let $d = \alpha a + \beta b$.

Let $c \in S$, such that $\lambda_0 a + \mu_0 b = c$ for some $\lambda_0, \mu_0 \in \Z$.

By the Division Algorithm:


 * $\exists \gamma, \delta \in \Z: c = \gamma d + \delta$

where $0 \le \delta < d$

Then:

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

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

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

Hence $\delta = 0$ and so $c = \gamma d$.

That is:
 * $d \divides c$

and so the smallest element of $S$ which is strictly positive is a divisor of both $a$ and $b$.

But $a$ and $b$ are coprime.

Thus it follows that, as $d \divides 1$:
 * $d = 1$

and so, by definition of $S$:


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