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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 $T$.

Define Greatest Common Divisor and Prove that if $a$ and $b$ are any integers, not both of them are zero. Then there exist integers $x$ and $y$ such that $gcd(a, b) =a \cdot x + b \cdot y$.

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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...

Definition of divisibility and Theorem on divisibility

  Definition: An integer $b$ is said to be divisible by an integer $a \not = 0$, if there exists some integer $c$ such that $ b = a \cdot c$. An integer $b$ is  divisible by an integer $a \not = 0$, is denoted by $a ~|~ b$.  We write $a \nmid b$ to indicate that $b$  is not divisible by $a$. Theorem: For integers $a, ~b, ~c$ the following hold: $1.$ $a ~|~ 0, ~ 1 ~|~ a, ~ a ~|~ a $. $2.$  $ a ~|~ 1 $ if and only if $a =  1$ or $a = -1$. $3.$  If $a ~|~ b$ and $c ~|~ d$, then $a \cdot c ~|~ b \cdot d$. $4.$   If $a ~|~ b$ and $b ~|~ c$, then $a ~|~ c$. $5.$  If $a ~|~ b$ and $b ~|~ a$ if and only if $a =  b$ or $a = -b$. $6.$  If $a ~|~ b$ and $b \not = 0$, then $|a| \leq |b|$. $7.$  If $a ~|~ b$ and $a ~|~ c$, then $a~ |~ (b \cdot x + c \cdot y)$ for arbitrary integers $x$ and $y$. Proof: $1.$ For any integer $a$, we can write $0 = a \cdot 0$, $a = 1 \cdot a$ and $a = a \cdot 1$. Theref...

If $a$ and $b$ are integers, with $b \not = 0$, then there exist unique integers $q$ and $r$ such that $a= b \cdot q + r$ where $0 \leq r <|b|$.

  Corollary: If $a$ and $b$ are integers, with $b \not = 0$, then there exist unique integers $q$ and  $r$ such that  $a= b \cdot q + r$ where  $0 \leq  r <|b|$. Proof: Consider the case in which $b$ is negative. So $|b| > 0$. By  Division Algorithm Theorem , there exist unique integers $q'$ and $r$ such that \linebreak $a=  |b| \cdot  q'+ r$ where  $0 \leq  r <|b|$. Note that $|b | = - b$, we can take $q = -q'$ to get $a= q \cdot b + r$,  where  $0 \leq  r <|b|$. Consider the case in which $b$ is positive. Proof follows from Division Algorithm Theorem .