Mathematics

Corollary: If $a$ and $b$ are any integers, not both them are zero, then the set  $T = \{a \cdot x+ b \cdot y \;|\; x, y \in \mathbb{Z}\}$ is the set of all multiples of $d = gcd(a,b)$. Proof:  Since $d = gcd(a,b)$, we have  $d ~|~ a$ and $d ~|~ b$. By part $7.$ of Theorem ,  $d ~|~ (a \cdot x + b  \cdot y)$ for all integers $x, ~y$. Thus, every member of $T$ is a multiple of $d$. Conversely, Since $d = gcd(a,b)$,With the help of   Theorem , there exist integers $x'$ and $y'$ such that $gcd(a, b) =a \cdot x'+ b \cdot y'$. So that any multiple $n \cdot d$ of $d$ is of the form $n \cdot d = n~(a \cdot x' + b \cdot y') = a~(n \cdot x') + b~(n \cdot y')$. Hence, $n\cdot d$ is a linear combination of $a$ and $b$, and hence lies in $T$.

Definition: Let $a$ and $b$ be any integers, with at least one of them is not zero. The greatest common divisor of $a$ and $b$, denoted by $gcd(a, b)$, is the positive integer $d$ satisfying the following: $1.$  $d ~|~ a$ and $d ~|~ b$. $2.$ If $c ~|~ a$ and $c ~|~ b$, then $c \leq d$. Theorem:  If $a$ and $b$ are any integers, not both of them are zero. Then there exist integers $x$ and $y$ such that $gcd(a, b) =a \cdot x + b \cdot y$. Proof:   Let  $a$ and $b$ be any integers, not both of them are zero. Consider the set $S = \{a \cdot u + b \cdot  v \;|\; a \cdot  u + b \cdot  v > 0$ and $ u, v \in \mathbb{Z}\}$, that is  $S$ is the set of all positive linear combinations of $a$ and $b$. Firstly, we will show  that $S$ is nonempty set. We have taken integers $a$ and $b$ such that not both of them are zero. Without loss of generality, suppose that $a \not = 0$. Then the integer $|a| = a \cdot u + b \cdot 0 \in S$ for the v...