**Exercise 1.3**

**Q1. Prove that √5 is irrational.**

**Answer: **Let take √5 as rational number

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

We can write √5 = a/b

Multiply by b both side we get

b√5 = a

To remove root, Squaring on both sides, we get

5b2 = a2 … (i)

Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p which is divides a2 then it will divide a also.

That means 5 will divide a. So we can write

a = 5c

Putting value of a in equation (i) we get

5b2 = (5c)2

5b2 = 25c2

Divide by 25 we get

b2/5 = c2

Similarly, we get that b will divide by 5

and we have already get that a is divide by 5

but a and b are co prime number. so it contradicts.

Hence √5 is not a rational number, it is irrational.

**Q2. Prove that 3 + 2√5 is irrational.**

Answer:

Let take that 3 + 2√5 is a rational number.

So we can write this number as

3 + 2√5 = a/b

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

Subtract 3 both sides we get

2√5 = a/b – 3

2√5 = (a-3b)/b

Now divide by 2, we get

√5 = (a-3b)/2b

Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.

Hence, 3 + 2√5 is a irrational number.

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 contradicts.

Hence, 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 contradicts.

Hence, 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 contradicts.

Hence, 6 + √2 is a irrational number.