Real Numbers ⇒⇒ Exercise 1.2

Question 1

Prove that √5 is irrational.

Solution :

This question can be solved with the help of contradiction method.
Let's assume that √5 is rational.

If √5 is rational

By definition, a rational number can be written as a fraction $ {p \over q} $.

where p and q are integers and q ≠ 0

Let p and q have a common factor other than 1. Then we can divide them by the common factor, and assume that p and q are co-prime.

$${5\over 1} = {p\over q}$$

$$√5 q = p$$

Squaring both sides

$$5q^2 = p^2 ......... Eq. (1)$$

From the above equation, we can see that $ p^2$ is divisible by 5, Therefore p will also be divisible by 5 as 5 is a prime number.

Therefore p can be written as 5k, where k is an integer

${p\over 5}$ = k

p = 5 k

on squaring,

$${p^2 = 25k^2}$$

put the value of $ p^2$ in equation

$${5q^2 = 25k^2}$$

$${q^2 = 5k^2}$$

From the above equation, we can see that $ q^2$ is divisible by 5 , Therefore q will also be divisible by 5 .

Therefore,p and q have at least 5 as a common factor. But this contradicts the fact that p and q are coprime.

This contradiction has arisen because of our original assumption that √5 is a rational number must be false. Therefore, √5 is an irrational number.

Question 2

Prove that 3+2√5 is irrational.

Solution :

This question can be solved with the help of contradiction method.
Let's assume that 3+2√5 is rational.

If 3+2√5 is rational.

By definition, a rational number can be written as a fraction $ {p \over q} $.

where p and q are integers and q ≠ 0

Let p and q have a common factor other than 1. Then we can divide them by the common factor, and assume that p and q are co-prime.

$$ 3+2√5 = {p\over q}$$

$$2√5 ={p\over q}- 3$$

$$√5 = {{p-3q} \over 2}$$

As p and q are integers $ {{p-3q} \over 2}$ would be rational.

Since, $ {{p-3q} \over 2}$ is a rational number then √5 is also a rational number.

The ratio of two integers is, by definition, a rational number.So, our equation states that √5 ​ is equal to a rational number. However, we already know that √5 is an irrational number.

Therefore, our assumption was wrong that 3+2√5 is rational. So,3+2√5 is irrational.

Question 3 (i)

Prove that the following are irrationals: ${1\over √2}$

Solution :

This question can be solved with the help of contradiction method.
Let's assume that ${1\over √2}$ is rational.

If ${1\over √2}$ is rational.

By definition, a rational number can be written as a fraction $ {p \over q} $.

where p and q are integers and q ≠ 0

Let p and q have a common factor other than 1. Then we can divide them by the common factor, and assume that p and q are co-prime.

$${1\over √2} = {p\over q}$$

$${ √2p = q}$$

$$√2 = {q \over p}$$

As p and q are integers $ {q \over p}$ would be rational.

Since, $ {q \over p}$ is a rational number then √2 is also a rational number.

The ratio of two integers is, by definition, a rational number.So, our equation states that √2 ​ is equal to a rational number. However, we already know that √2 is an irrational number.

Therefore, our assumption was wrong that ${1\over √2}$ is rational. So, ${1\over √2}$ is irrational.

Question 3 (ii)

Prove that the following are irrationals: ${7√5}$

Solution :

This question can be solved with the help of contradiction method.
Let's assume that 7√5 is rational.

If ${7√5}$ is rational.

By definition, a rational number can be written as a fraction $ {p \over q} $.

where p and q are integers and q ≠ 0

Let p and q have a common factor other than 1. Then we can divide them by the common factor, and assume that p and q are co-prime.

$${7√5} = {p\over q}$$

$${√5} = {p\over 7q}$$

As p and q are integers $ {p\over 7q} $ would be rational.

Since, $ {p\over 7q}$ is a rational number then √5 is also a rational number.

The ratio of two integers is, by definition, a rational number.So, our equation states that √5 ​ is equal to a rational number. However, we already know that √5 is an irrational number.

Therefore, our assumption was wrong that ${7√5}$ is rational. So, ${7√5}$ is irrational.

Question 3 (iii)

Prove that the following are irrationals: ${6+√2}$

Solution :

This question can be solved with the help of contradiction method.
Let's assume that ${6+√2}$ is rational.

If ${6+√2}$ is rational.

By definition, a rational number can be written as a fraction $ {p \over q} $.

where p and q are integers and q ≠ 0

Let p and q have a common factor other than 1. Then we can divide them by the common factor, and assume that p and q are co-prime.

$${6+√2} = {p\over q}$$

$$ √2 = {p\over q} - 6$$

As p and q are integers ${p\over q} - 6$ would be rational.

Since, ${p\over q} - 6$ is a rational number then √2 is also a rational number.

The ratio of two integers is, by definition, a rational number.So, our equation states that √2 ​ is equal to a rational number. However, we already know that √2 is an irrational number.

Therefore, our assumption was wrong that ${6+√2}$ is rational. So, ${6+√2}$ is irrational.