(i) 5√2
Let 5√2 is a rational number.
5√2 = a/b, b ≠ 0, (where a, b are co-prime numbers)
or √2 = a/5b ....(i)
Since, a, b are integer.
So, a/5b is a rational number. It is clear that from equation (i) √2 is a rational number which is contradict statement, since we know that √2 is irrational number.
So, our hypothesis is wrong.
Hence, 5√2 is a irrational number
(ii) 2/√7
Let 2/√7 is a rational number.
we find two integer such as
2/√7 = a/b(b ≠ 0) where a and be are co-primes
1/√7 = a/2b
∴ a and b is integers
∴ a/2b is a rational number
⇒ 1/7 also a rational number
But 1/7 is not rational number.
This is contradict.
So, our hypothesis is wrong,
Hence, 2/√7 is a irrational number.
(iii) 3/2√5
Let 3/2√5 is a rational number.
we find two integers such as
3/2√5 = a/b(b ≠ 0)
Where a and b are co-prime.
1/√5 = 2a/3b
∴ a and b integer r.
∴ 2a/3b is a rational number
⇒ 1/√5 will be also a rational number.
But 1/√5 is not rational number.
It is a irrational number.
This is contradict.
So, our hypothesis is wrong.
Hence, 3/2√5 is a irrational number.
(iv) 4 + √2
Let 4 + √2 is rational number.
Now we find two integers a and b such as
4 + √2 = a/b
⇒ √2 = a/b - 4
⇒ √2 = (a - 4b)/b
a, b, ab and 4 all are integers.
(a - 4b)/b is a rational number.
√2 will be a rational number
This contradicts.
our hypothesis is wrong.
So, 4 + √2 is an irrational number.
