Exercise 1.3

Question 1:

Prove that 3–√ is irrational .

Solution :

Let us assume ,that 3–√ is rational

i.e 3–√=xy (where,x and y are co-primes)

y3–√=x

Squaring both sides

We get,(y3–√)2=x2 3y2=x2……..(1)

x2 is divisible by 3

So, x is also divisible by 3

 therefore we can write x=3k (for some values of k)

Substituting ,x=3k in eqn 1

3y2=(3k)2 y2=3k2

y2 is divisible by 3 it means y is divisible by 3

 therefore x and y are co-primes.

Since ,our assumption about 3–√ is rational is incorrect .

Hence, 3–√ is irrational number.

Question 12:

Prove that  3+23–√ is irrational .

Solution :

Let us assume that  3+23–√ is rational .

So, x= 3+23–√ x2=(3+23–√)2 x2=21+123–√ 3–√=x2−2112…….(1)

 because x is a  rational number

So, the expression  x2−2112is also a rational number.This is a contradiction .Hence, 3+23–√ is irrational .

Question 15:

Prove that the following are irrational numbers.

i)13√

ii)73–√

iii)6+5–√

Solution :

i)13√ 13√=3√3√3√ ⇒ 13√=3√3 ⇒ 13√=13×3–√

Let, a=13√=13×3–√ be a rational number .

⇒ 3a=3–√

3a is a rational number .Since product of any two rational numbers is a rational number which will imply that 3√is a rational number .But,it contradicts since 3√ is a irrational number .

 therefore 3ais a irrational or ais irrational.

Hence, 13√ is irrational .

ii)73–√

Let, a=73–√ be a rational number .

⇒a7=3–√

a7 is a rational number .Since product of two rational number is a rational number. Which will imply that 7√

is a rational number .But,it contradicts since7√ is a irrational number .

 thereforea7is a irrational or ais irrational.

Hence,73–√ is irrational .

iii)6+5–√

Let, a=6+5–√ be a rational number .

Squaring , a2=(6+3–√)2 a2=36+23–√+3 a2=39+23–√ 3–√=a2−3912……………..(1)

Since, ais a rational number

So, the expression  a2−3912 is also rational number.

⇒3–√ is a rational number .

This is a contradiction.

Hence, 6+5–√ is irrational


Exercise 1.1