__EXERCISE 1.3__

__EXERCISE 1.3__

**Sol**. Let us assume, to the contrary, that _{} is rational.**Now, let ** _{} , where a and b are coprime and _{}

Squaring on both sides, we get

_{} => _{} … (1)

This shows that _{} is divisible by 5.

It follows that a is divisible by 5 … (2)

=> a = 5m for some integer m.

Substituting a = 5m in (1), we get

or

=> is divisible by 5

and hence b is divisible by 5 … (3)

From (2) and (3), we can conclude that 5 is a common factor of both a and b.

But this contradicts our supposition that a and b are coprime.

Hence, _{} is irrational.

**2. Prove that 3 + **_{}** is irrational.**

**Sol.** Let us assume, to the contrary, that 3 + _{} is a rational number.

Now, let 3 + _{} , where a and b are coprime and _{}

a and b are integers.

** _{}** is a rational number

=> _{} is a rational number.

But is an irrational number.

This shows that our assumption is incorrect.

So, 3 + 2_{} is an irrational number.

**3. Prove that the following are irrationals:**

**Sol.** (i) Let us assume, to the contrary, that _{}** **is rational.

That is, we can find co-prime integers *p* and *q* _{}

such that

Since p and q are integers, _{} is rational, and so_{} is rational.

But this contradicts the fact that _{} is irrational.

So, we conclude that _{} is irrational.

(ii) Let us assume, to the contrary, that _{} is rational.

That is, we can find co-prime integers *p *and *q*_{ } such that _{}

Since *p* and *q* are integers, _{}** ** is rational and so is _{}

But this contradicts the fact that is irrational.

So, we conclude that _{} is irrational.

(iii) Let us assume, to the contrary, that _{}is rational.

That is, we can find integers *p* and *q* _{} such that

Since p and q are integers, we get 6 – = _{}** **is rational, and so _{} is rational.

But this contradicts the fact that _{} is irrational.

So, we conclude that 6+ _{} is irrational.