Question

prove that root 5 is a inrational?

Answer 1

let us assume √5 is an rational number
√5=p/q (q≠0)
if p,q have a common factor,on cancelling the common factor let it be reduced to a/b, where a,b are co-primes
now,
√5=a/b, where HCF (a,b)=1
squaring on both sides we get,
(√5)^2=(a/b)^2
5=a^2/b^2
5b^2=a^2
5 divides a^2 and 5divides5
now take a=5c^2
then,a^2=25c^2
i.e.,5b^2=25c^2
b^2=5c^2
5 divides b^2 and by b
5divides both b and a
........this contradicts that a and b are co-primes .....hence our assumed is wrong and the given statement is correct.
i.e,√5 is an irrational number...
........ I hope this answer is useful to you..........

Answer 2

let
$\sqrt{5} = a \div b$
where a and b are coprime and b≠0
squaring on both sides,we get

$5 = {a }^{2} \div {b}^{2}$
eq1
$5b {}^{2} = {a}^{2}$
a is divisible by 5. eq2
$a {}^{2} is \: divisible by 5$
a=5m for some integer m.
sub. a=5 in (1)
$5b { }^{2} = (5m) {}^{2} = 25m {}^{2} or \: b {}^{2} = {5m}^{2}$

hence b is divisible by 5. eq2
5 is common factor of both a and b and a and b are coprimes