Mathematics

  Theorem:  There is an infinite number of primes.  Proof :  Suppose on the contrary that there are only finite number of primes. Let $p_{1} = 2, p_2 = 3, p_3 = 5, p_4 = 7, \cdots $ be  the primes in ascending order, and suppose that there is a last prime, called $p_n$· Now consider the positive integer $P = (p_{1} \cdot p_2 ~ \cdots ~ p_n)+ 1$. Since $P > 1$, by Theorem , $P$ is  divisible by some prime $p$. But $p_{1}, ~p_2, ~ p_3, ~\cdots~, p_n $ are the only prime numbers, so that $p$ must be equal to one of $p_{1}, ~p_2, ~ p_3, ~\cdots~, p_n $ and hence $p ~|~  p_{1} \cdot p_2 ~ \cdots ~ p_n$· Since $p ~|~  p_{1} \cdot p_2 ~ \cdots ~ p_n$ and $p ~|~ P$, we will get $p ~|~ (p_{1} \cdot p_2 ~ \cdots ~ p_n) - P$ i.e. $p ~|~ 1$. Since the only positive divisor of the integer $1$ is $1$ itself so we will get a contradiction to  $p > 1$. Hence there are  infinite number of primes.

  (Fundamental Theorem of Arithmetic): Every positive integer $n > 1$ is either a prime or can be written as a product of primes; this representation is unique, apart from the order in which the factors occur. Proof: If a positive integer $n > 1$ is a prime number, then we are done. So suppose that $n$ is a composite number. Therefore there exists an integer $d$ satisfying $d ~|~ n$ and $1 < d < n$. By Well-Ordering Principle , there exists smallest integer among all such integers $d$ (positive divisors), we will called this smallest integer (smallest positive divisor) as a $p_{1}$. We claim that $p_{1}$ is a prime number. Suppose that $p_{1}$ is not a prime number. So there exists  an integer $q$ satisfying $q ~|~ p_{1}$ and $1 < q < p_{1}$. Since $q ~|~ p_{1}$ and $p_{1} ~|~ n$, we will get $q ~|~ n$, a contradiction to choice of smallest positive divisor  $p_{1}$ of $n$. Therefore we can write $n$ as $n = p_{1} \cdot n_{1}$, where $p_{1}$ is a pri...