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

(i) 1/√2 (ii) 7√5 (iii) 6 + √2

**Answer:**

**(i) **Let take that 1/√2 is a rational number.

So we can write this number as

1/√2 = a/b

Here a and b are two co prime number and b is not equal to 0

Multiply by √2 both sides we get

1 = (a√2)/b

Now multiply by b

b = a√2

divide by a we get

b/a = √2

Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it is contradict

Hence result is 1/√2 is a irrational number

**(ii)** Let take that 7√5 is a rational number.

So we can write this number as

7√5 = a/b

Here a and b are two co prime number and b is not equal to 0

Divide by 7 we get

√5) =a/(7b)

Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it is contradict

Hence result is 7√5 is a irrational number.

**(iii)** Let take that 6 + √2 is a rational number.

So we can write this number as

6 + √2 = a/b

Here a and b are two co prime number and b is not equal to 0

Subtract 6 both side we get

√2 = a/b – 6

√2 = (a-6b)/b

Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number But √2 is a irrational number so it is contradict

Hence result is 6 + √2 is a irrational number