(i) let's assume that √6 is rational.

By definition, that means there are two integers a and b with no common divisors where:

⇒ a/b = √6

So let's take square of both the sides -

⇒ (a/b)(a/b) = (√6)( √6)
⇒ a²/b² = 6
⇒ a² = 6b²

Last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least 6) is even. So a² must be even. But any odd number times itself is odd, so if a² is even, then a is even.

Since a is even, there is some integer c such that:

⇒ 2c = a.

Now let's replace a with 2c:

⇒ a² = 6b²
⇒ (2c)² = (2)(3)b²
⇒ 2c² = 3b²

But now we can argue the same thing for b, because the LHS is even, so the RHS must be even and that means b is even.

Now this is the contradiction: if a is even and b is even, then they have a common divisor (2). Then our initial assumption must be false, so the square root of 6 cannot be rational.

Thus, proved that √6 is irrational.

(ii) assume that 2 - √3 is rational
2 - √3 = a/b , where a and b are integers .
⇒ - √3 = a/b - 2
⇒ √3 = 2 - a/b
⇒ √3 = 2b/b - a/b
⇒ √3 = (2b – a) / b
we know that a, b and 2 are integers and they are also rational {i.e RHS is rational}
therefore √3 will be rational.
but we know that √3 is irrational.
there is a contradiction
so, 2 - √3 is an irrational number

(iii) assume that 3 + √2 is rational
3 + √2 = a/b , where a and b are integers .
⇒ √2 = a/b - 3
⇒ √2 = (a - 3b)/b
we know that a, b and 3 are integers and they are also rational {i.e RHS is rational}
therefore √2 will be rational.
but we know that √2 is irrational.
there is a contradiction
so, 3 + √2 is an irrational number

(iv) assume that 2 + √5 is rational
2 + √5 = a/b , where a and b are integers .
⇒ √5 = a/b - 2
⇒ √5 = a/b - 2b/b
⇒ √5 = (a - 2b)/ b
we know that a, b and 2 are integers and they are also rational {i.e RHS is rational}
therefore √5 will be rational.
but we know that √5 is irrational.
there is a contradiction
so, 2 + √5 is an irrational number

(v) assume that 5 + 3√2 is rational
5 + 3√2 = a/b , where a and b are integers .
⇒ 3√2 = a/b - 5
⇒ 3√2 = a/b - 5b/b
⇒ √2 = (a - 5b)/3b
we know that a, b, 3 and 5 are integers and they are also rational {i.e RHS is rational}
therefore √2 will be rational.
but we know that √2 is irrational.
there is a contradiction
so, 5 + 3√2 is an irrational number.

(vi) assume that 3√7 is rational
⇒ 3√7 = a/b , where a and b are integers .
⇒ √7 = a/3b
we know that a, b and 3 are integers and they are also rational {i.e RHS is rational}
therefore √7 will be rational.
but we know that √7 is irrational.
there is a contradiction
so, 3√7 is an irrational number

(vii) {Rationalising}

Now, let us assume that 3√5 / 5 is a rational number

⇒ 3√5/5 = p/q (where, p & q are integers, q not equal to 0)

⇒ 3√5 = 5p/q

⇒ √5 = 5p/3q

Here, LHS is an irrational number whereas RHS is a rational number.

So by contradiction 3√5 / 5 is an irrational number

(viii) assume that 2 - 3√5 is rational
2 - 3√5 = a/b , where a and b are integers .
⇒ - 3√5 = a/b - 2
⇒ 3√5 = 2 - a/b
⇒ 3√5 = 2b/b - a/b
⇒ √5 = (2b – a) / 3b
we know that a, b, 2 and 3 are integers and they are also rational {i.e RHS is rational}
therefore √5 will be rational.
but we know that √5 is irrational.
there is a contradiction
so, 2 - 3√5 is an irrational number

(ix) Multiplying both sides by (√5 - √3).
(√5 - √3) (√5 + √3) = 5 - 3 = 2

⇒ 2 = p/q × (√5 - √3)
⇒ (√5 - √3) = 2q/p,

∴ √5 - √3 is rational = 2q/p
⇒ √5 + √3 = p/q
⇒ √5 - √3 = 2q/p

Adding the equations -

, which is a rational number.

But we know that √5 is IRRATIONAL.

Therefore, the assumption is wrong and √3 + √5 is irrational.