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 $gcd(a, b) =a \cdot x + b \cdot y$.

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 value of $u = 1$, when $a$ is positive or $u = -1$  when $a$ is negative. Since $S$ is nonempty subset of nonnegative integers, By Well-Ordering Principle, $S$ must contain a least element $d$. As $a \in  S$, there exist integers $x$ and $y$ such that  $d = a \cdot x + b \cdot y$. We claim that $d = gcd(a, b)$. We use Division Algorithm  for integers $a$ and $d$, so there exist integers $q$ and $r$ such that $a = d \cdot q + r$, where $0 \leq r <d$. We can write $ r$ in the form $ r = a - q \cdot d = a-q~(a \cdot x + b \cdot y) = a ~(1- q \cdot x) + b~(-(q \cdot y))$ .

If $r > 0$, then this representation imply that $r \in S$, a contradiction to the fact that $d$ is the least integer in $S$ (recall that $r < d$). Therefore, $r = 0$, and so $a = q \cdot d$, that is, $d ~|~ a$. By similar way, we can show that $d ~|~ b$. Hence $d$ is  a common divisor of $a$ and $b$. Let $c$ be any  arbitrary positive common divisor of the integers $a$ and $b$. By part $7.$ of Theorem  ,  $c ~|~ (a \cdot x + b \cdot y)$; that is, $c ~|~ d$. By part $6.$ of Theorem, we have $c = |c| \leq |d| = d$. Therefore $d$ is greater than every positive common divisor of $a$ and $b$. By the definition of greatest common divisor, we see that $d = gcd(a, b)$.





Popular posts from this blog

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...
Read more

The Division Algorithm / Prove that given integers $a$ and $b$, with $b > ~ 0$, there exist unique integers $q$ and $r$ such that $a = b \cdot q +r$, where $0 \leq r < b$. The integer $q$ is called the $\textbf{quotient}$ and the integer $r$ is called $\textbf{remainder}$.

Image
State and prove Division Algorithm theorem.  Theorem:  Given integers $a$ and $b$, with $b > ~ 0$, there exist unique integers $q$ and $r$ such that $a = b \cdot q +r$, where $0 \leq r < b$. The integer $q$ is called the $\textbf{quotient}$ and the integer $r$ is called $\textbf{remainder}$. Proof : Firstly, we will show that the set $S = \{a - x \cdot b \;|\; x \in \mathbb{Z}$ and $  a -x \cdot b \geq 0\}$ is nonempty.  For this purpose, we have to find value of $x$ such that $a -x \cdot b \geq 0$. Given that integer $b > 0$, i.e. $b \geq 1$. Multiply $b \geq 1$ by $|a|$, we get $|a|\cdot b \geq |a|$ and so $a - ( - |a|) \cdot b = a + |a| \cdot b \geq a + |a| \geq 0$. If we take $x = |a|$, then, $a -x \cdot b \in S$. This show that $S$ is nonempty subset of nonnegative integers. By the Well-Ordering Principle (for more details see  Well-Ordering Principle  ) , set $S$ contains a least integer; say it $r$. Since $r \in S$, there exists an i...
Read more