If $c \cdot a \equiv c \cdot b ~(mod ~n)$, then Prove that $a \equiv b ~(mod ~\frac{n}{d} )$, where $d = gcd(c, n)$.

 In Theorem we saw that if $a \equiv b ~(mod ~n)$, then $c \cdot a \equiv c \cdot b ~(mod ~n)$ for any integer $c$. What about converse $?$ i.e. if $c \cdot a \equiv c \cdot b ~(mod ~n)$ for any integer $c$ does it implies  $a \equiv b ~(mod ~n) ~ ?$  we give answer to this question as No. Here is an example $2 \cdot 4  \equiv 2 \cdot 1 ~(mod~6)$, but $4 \not\equiv 1 ~(mod ~6)$. In brief: One cannot unrestrictedly cancel a common factor in the arithmetic of congruences. 

Theorem: If $c \cdot a \equiv c \cdot b ~(mod ~n)$, then $a \equiv b ~(mod ~\frac{n}{d} )$, where $d = gcd(c, n)$. 

Proof: Suppose that $c \cdot a \equiv c \cdot b ~(mod ~n)$. Therefore by definition of congruences $c \cdot a ~-~ c \cdot b = n \cdot k$ for some integer $k$. Since  $gcd(c, n) = d$, there exist relatively prime integers $r$ and $s$ such that  $c = d \cdot r,~ n = d \cdot s$. We put $c = d \cdot r,~ n = d \cdot s$ in $c \cdot a - c \cdot b = n \cdot k$ and cancel the common factor $d$, we will get  

$r \cdot (a - b) =k \cdot s$ This gives $s ~|~ r\cdot (a - b) $ but $gcd(r, s) = 1$. By Euclid's lemma, we will get  

$s ~|~ a - b$ i.e.  $a \equiv b ~(mod ~s)$. We put  $s = \frac{n}{d}$ in $a \equiv b ~(mod ~s)$, hence $a \equiv b ~(mod~ \frac{n}{d} )$.

Corollary: If $c \cdot a \equiv c \cdot b ~(mod ~n)$ and $gcd(c, n ) = 1$, \linebreak then $a \equiv b ~(mod ~ n )$. 

Corollary: If $c\cdot a \equiv c\cdot b ~(mod ~p) $  where $p$ is a prime number and $p \nmid c$, then $a \equiv b ~(mod ~p)$. 

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

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

Image
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...
Read more