Let √2= ab, where gcd(a,b)=1
so 2 = a2b2
or 2b2=a2
so 2|a2 →2|a
and 2|b2→2|b
which contradicts gcd(a,b) = 1
so its irrational;)
Let √2= ab, where gcd(a,b)=1
so 2 = a2b2
or 2b2=a2
so 2|a2 →2|a
and 2|b2→2|b
which contradicts gcd(a,b) = 1
so its irrational;)
If possible let √2 be rational
so, √2 = a/b where a and b are integers in the simplest form having no common factors other than 1
=> a2/b2 = 2
=> a2 = 2b2 ----- (i)
Now, 2 divides 2b2 => 2 divides a2
and, since 2 is prime and divides a2 => 2 divides a [result 1]
since, 2 divides a so, let a = 2c, for any integer c
putting, a = 2c in (i) we get,
(2c)2 = 2b2
=> 4c2 = 2b2
=> b2 = 2c2
Now, since 2 divides 2c2 => 2 divides b2
and since 2 is prime and divides b2 => 2 divides b [result 2]
But we get a contradiction from results 1 and 2 that, a and b are integers in the
simplest form having no common factors other than 1.
Hence, √2 is irrational.
Dhanyavad