Prove that 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 $d = gcd(a,b)$.

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 $T$.

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