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.

Theorem:  The number $\sqrt{2}$ is an irrational.  Proof:   Suppose on the contrary that $\sqrt{2}$ is rational number.   Therefore there exist integers $a$ and $b$ such that $\sqrt{2} = \frac{a}{b}$ with $gcd(a, b) = 1$. By Theorem ,  there must exist integers $r$ and $s$ such that $ ar + bs = 1$. Hence $ \sqrt{2} = \sqrt{2}~ (ar + bs) = \sqrt{2} ar + \sqrt{2} bs =  2br +as $. This show that $\sqrt{2}$ is an integer, which is absurd. Therefore  The number $\sqrt{2}$ is an irrational.