Mathematics

Corollary:  If $p$ is a prime and $p ~|~ a_{1} \cdot a_2 ~\cdots~ a_n$, then $p ~|~ a_k$ for some $k$, where $1 \leq k \leq n$. Proof:  Suppose that $p$ is a prime and $p ~|~ a_{1} \cdot a_2 ~\cdots~ a_n$, then by Theorem , $p ~|~ a_{1}$ or $p ~|~ a_2 \cdot a_3 ~\cdots~ a_n$. If $p ~|~ a_{1}$, then we are done. If $p \nmid a_{1}$, then $p ~|~ a_2 \cdot a_3 ~\cdots~ a_n$. We use Theorem , to $p ~|~ a_2 \cdot a_3 ~\cdots~ a_n$. So we will get either $p ~|~ a_2$ or $p ~|~ a_3 \cdot a_4 ~\cdots~ a_n$. If $p ~|~ a_2$, then we are done. If $p \nmid a_2$, then $p ~|~ a_3 \cdot a_4 ~\cdots~ a_n$. We will make use of Theorem   again and again. Since $n$ is finite, use of Theorem  ,  will stop after finitely many times. So we will get  $p ~|~ a_k$ for some $k$, where $1 \leq k \leq n$. Corollary:  If $p, q_{1}, q_2, ~\cdots~ q_n$ all are primes and $p ~|~ q_{1} \cdot q_2 ~\cdots~ q_n$, then $p = q_k$ for some $k$, where $1 \leq k \leq n$. Proof:   We w...

Definition: An integer $p > 1$ is called a prime number, or simply a prime, if its only positive divisors are $1$ and $p$. An integer greater than $1$ that is not a prime is termed composite. Example: We can observe that $2, 3, 5, 7$ etc. are prime numbers whereas $4, 6, 8, 9, 10$ etc. are composite numbers. Theorem: If $p$ is a prime and $p ~|~ ab$, then $p ~|~ a$ or $p ~|~ b$. Proof: If $p ~|~ a$, then we are done. So suppose that $p \nmid a$. Therefore $gcd(p,a) = 1$. Hence by Euclid Lemma , $p ~|~ b$..

Example:   If a cock is worth $5$ Rs., a hen $3$ Rs., and three chicks together $1$ Rs. How many cocks, hens and chicks totaling $100$ can be bought for $100$ Rs. Proof:  Suppose that $x$ is the number of cocks, $y$ is the number of hen and $z$ is the number of chicks totaling $100$ bought for $100$ Rs. Therefore we will get following equations: $x + y + z = 100$ and  $5x +3y + \dfrac{1}{3} z = 100$ We put $z = 100 - x - y$ in $5x +3y + \dfrac{1}{3} z = 100$. So $5x + 3y + \dfrac{1}{3} (100 - x - y) = 100$ Multiply both side by $3$ $15x +9y +100 - x -y = 300$ $14x +8y = 200$ $7x + 4y = 100$. So we got the linear equation $7x + 4y = 100$. Since $gcd(7,4) = 1$ divides $100$, linear equation $7x + 4y = 100$ is a Diophantine equation and it has a solution. General solution of the equation $7x + 4y = 100$ is $x =4t, ~ y = 25 - 7t$ and hence $z= 75 + 3t$, where $t$ is an arbitrary integer.  Therefore $x = 4,~y = 18 $ and...

 Definition:  An equation in one or more variables is said to be a Diophantine equation if it has integer  solution. Example:  Consider the linear equation  $2 x + 8 y = 16$. Linear equation $2 x + 8 y = 16$ has $x = 4, y = 1$ as a integers solution. Therefore $2 x + 8 y = 16$ is a linear Diophantine equation in two variables $x, y$. The Diophantine equation can have a number of solutions. As $x = 8, y = 0$ and $x = 0, y = 2 $ are also integers solution of equation $2 x + 8 y = 16$. Example:  Consider the linear equation  $5 x + 8 y = 16$. Linear equation $5 x + 8 y = 16$ has no integers solution (Why $?$ (We will see in the following Theorem)). Therefore $5 x + 8 y = 16$ is  not a  Diophantine equation. Theorem: The linear  equation $ax +by = c$ has a integers solution if and only if $d$ divides $c$, where $d = gcd(a,b)$.  Moreover, if $x_{0}, ~y_{0}$ is any particular solution of this equation, then all other solutions are...

  Lemma:   If $a ~|~ b \cdot c$, with $ gcd(a, b) = 1$, then $a ~|~ c$. Proof:  Since $gcd(a, b) = 1$, by Theorem , there exist integers $x$ and $y$ such that $a \cdot x + b \cdot y = 1$. Multiplying this equation by $c$, we get  $c = c \cdot 1 = c~(a \cdot x + b \cdot y) = a \cdot c \cdot x + b \cdot c \cdot y$. As $ a ~|~ a \cdot c$ and  $ a ~|~ b \cdot c$, by part $7.$ of Theorem , $d ~|~ (a \cdot c \cdot x + b \cdot c \cdot y)$. Therefore $d ~|~ c$.