For any prime p , is an irrational number.
proof:-
We will prove this by using contradiction.
Let us assume that is a rational number. i.e. we can write as where a and b are integer and .and a and b are coprime.(gcd (a,b)=1). Now ,
squaring on both sides,
------------------ (1)
here we can say is multiple of p or p divides i.e. (as ,p is prime )
(we know that ,by Euclid's lemma, for any prime p if
for any integer a,b)
here Since p is a prime number ,so if p divdes then p divides a. we can write a=np where n is any constant
i.e. a is multiple of p
put this in equation (1) ,
This implies p divides , and as p is prime p also divides b (Euclid's lemma) We get prime p such that p divides a as well as b. hence p is common factor of a and b
Which contradicts that a and b are coprime.
This is due to our wrong assumption that is rational number. This proves is an irrational number.
But what if the number is composite. ???
Let us prove is an irrational number. proof :-
Contrary assume that is rational number. ( we will use Fundamental Theorem of Arithmetic, which states that Any positive integer >1 is itself a prime number or can be written as product of prime numbers.)
we write (from (1))
Which is not possible As we prove above, is a irrational number since 2 is prime number and As a and 2b are integer ( ) is a rational number.
Hence our assumption is wrong .
is an irrational number. Similarly we can prove the result for other composite irrational numbers.