## What Is Irrational Number

- A number is irrational if and only if its decimal representation is non-terminating and non-repeating. e.g.√2, √3, π ……………. etc.

- Rational number and irrational number taken together form the set of real numbers.
- If a and b are two real numbers, then either

(i) a > b or (ii) a = b or (iii) a < b - Negative of an irrational number is an irrational number.
- The sum of a rational number with an irrational number is always irrational.
- The product of a non-zero rational number with an irrational number is always an irrational number.
- The sum of two irrational numbers is not always an irrational number.
- The product of two irrational numbers is not always an irrational number.
- In division for all rationals of the form \(\frac { p }{ q } \)(q ≠ 0), p & q are integers, two things can happen either the remainder becomes zero or never becomes zero.

**Type (1) ****Example: **\(\frac { 7 }{ 8 } \) = 0.875

This decimal expansion 0.875 is called **terminating**.

∴ If remainder is zero then decimal expansion ends (terminates) after finite number of steps. These decimal expansion of such numbers terminating.

**Type (2) Example: **\(\frac { 1 }{ 3 } \) = 0.333……… = \(0 . \overline{3}\)

or \(\frac { 1 }{ 7 } \) = 0.142857142857….. = \(0 . \overline{142857}\)

In both examples remainder is never becomes zero so the decimal expansion is never ends after some or infinite steps of division. These type of decimal expansions are called **non terminating.**

In above examples, after I^{st} step & 6 steps of division (respectively) we get remainder equal to dividend so decimal expansion is repeating (recurring).

So these are called **non terminating recurring decimal expansions**.

Both the above types (1 & 2) are rational numbers.

**Types (3) Example: **The decimal expansion 0.327172398……is not ends any where, also there is no arrangement of digits (not repeating) so these are called **non terminating not recurring**. These numbers are called **irrational numbers**.

**Example:**

0.1279312793 rational terminating

0.1279312793…. rational non terminating

or \(0 . \overline{12793}\) recurring

0.32777 rational terminating

or \(0 . 32\overline{7}\) rational non terminating

0.32777……. & recurring

0.5361279 rational terminating

0.3712854043…. irrational non terminating non recurring

0.10100100010000 rational terminating

0.10100100010000…. irrational non terminating non recurring.

**Irrational Number Example Problems With Solutions**

**Example 1: **Insert a rational and an irrational number between 2 and 3.

**Sol. ** If a and b are two positive rational numbers such that ab is not a perfect square of a rational number, then \(\sqrt { ab } \) is an irrational number lying between a and b. Also, if a,b are rational numbers, then \(\frac { a+b }{ 2 } \) is a rational number between them.

∴ A rational number between 2 and 3 is

\(\frac { 2+3 }{ 2 } \) = 2.5

An irrational number between 2 and 3 is

= \(\sqrt { 2\times 3 } \) = \(\sqrt { 6 } \)

**Example 2: **Find two irrational numbers between 2 and 2.5.

**Sol.** If a and b are two distinct positive rational numbers such that ab is not a perfect square of a rational number, then is an irrational number lying between a and b.

∴ Irrational number between 2 and 2.5 is

= \(\sqrt { 2\times 2.5 } \) = \(\sqrt { 5 } \)

Similarly, irrational number between 2 and \(\sqrt { 5 } \) is \(\sqrt { 2\times \sqrt { 5 } }\)

So, required numbers are \(\sqrt { 5 } \) and \(\sqrt { 2\times \sqrt { 5 } }\)

**Example 3: **Find two irrational numbers lying between \(\sqrt { 2 } \) and \(\sqrt { 3 } \) .

**Sol.** We know that, if a and b are two distinct positive irrational numbers, then \(\sqrt { ab } \) is an irrational number lying between a and b.

∴ Irrational number between \(\sqrt { 2 } \) and \(\sqrt { 3 } \) is = \(\sqrt { \sqrt { 2 } \times \sqrt { 3 } } \) = 6^{1/4}

Irrational number between \(\sqrt { 2 } \) and 6^{1/4 }is \(\sqrt { \sqrt { 2 } \times { 6 }^{ \frac { 1 }{ 4 } } } \) = 2^{1/4} × 6^{1/8}.

Hence required irrational number are 6^{1/4 }and

2^{1/4} × 6^{1/8}.^{ }

**Example 4: **Find two irrational numbers between 0.12 and 0.13.

**Sol. ** Let a = 0.12 and b = 0.13. Clearly, a and b are rational numbers such that a < b.

We observe that the number a and b have a 1 in the first place of decimal. But in the second place of decimal a has a 2 and b has 3. So, we consider the numbers

c = 0.1201001000100001 ……

and, d = 0.12101001000100001…….

Clearly, c and d are irrational numbers such that a < c < d < b.

**Example 5: **Prove that is \(\sqrt { 2 } \) irrational number

**Sol. **Let us assume, to the contrary, that \(\sqrt { 2 } \) is rational. So, we can find integers r and s (≠0) such that \(\sqrt { 2 } =\frac { r }{ s } \). Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get \(\sqrt { 2 } =\frac { a }{ b } \) where a and b are coprime.

So, b\(\sqrt { 2 } \) = a.

Squaring on both sides and rearranging, we get 2b^{2} = a^{2}. Therefore, 2 divides a^{2}. Now, by Theorem it following that 2 divides a.

So, we can write a = 2c for some integer c.

Substituting for a, we get 2b^{2} = 4c^{2}, that is,

b^{2} = 2c^{2}.

This means that 2 divides b^{2}, and so 2 divides b (again using Theorem with p = 2).

Therefore, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that \(\sqrt { 2 } \) is rational.

So, we conclude that \(\sqrt { 2 } \) is irrational.

**Example 6: **Prove that is \(\sqrt { 3 } \) irrational number.

**Sol. **Let us assume, to contrary, that is rational. That is, we can find integers a and b (≠0) such that \(\sqrt { 2 } =\frac { a }{ b } \). Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b\(\sqrt { 3 } \) = a.

Squaring on both sides, and rearranging, we get 3b^{2} = a^{2}.

Therefore, a^{2} is divisible by 3, and by Theorem, it follows that a is also divisible by 3.

So, we can write a = 3c for some integer c.

Substituting for a, we get 3b^{2} = 9c^{2}, that is,

b^{2} = 3c^{2}.

This means that b^{2} is divisible by 3, and so b is also divisible by 3 (using Theorem with p = 3).

Therefore, a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that \(\sqrt { 3 } \) is rational.

So, we conclude that \(\sqrt { 3 } \) is irrational.

**Example 7: **Prove that \(7-\sqrt { 3 } \) is irrational

**Sol.** **Method I :**

Let \(7-\sqrt { 3 } \) is rational number

∴ \(7-\sqrt { 3 } \) = \(\frac { p }{ q } \) (p, q are integers, q ≠ 0)

∴ 7 – \(\frac { p }{ q } \) = \(\sqrt { 3 } \)

⇒ \(\sqrt { 3 } \) = \(\frac { 7q-p }{ q } \)

Here p, q are integers

∴ \(\frac { 7q-p }{ q } \) is also integer

∴ LHS = \(\sqrt { 3 } \) is also integer but this \(\sqrt { 3 } \) is contradiction that is irrational so our assumption is wrong that \(7-\sqrt { 3 } \) is rational

∴ \(7-\sqrt { 3 } \) is irrational proved.

**Method II :**

Let \(7-\sqrt { 3 } \) is rational

we know sum or difference of two rationals is also rational

∴ \(7-7-\sqrt { 3 } \)

= \(\sqrt { 3 } \) = rational

but this is contradiction that \(\sqrt { 3 } \) is irrational

∴ \(7-\sqrt { 3 } \) is irrational proved.

**Example 8: **Prove that \(\frac { \sqrt { 5 } }{ 3 } \) is irrational.

**Sol. **Let \(\frac { \sqrt { 5 } }{ 3 } \) is rational

∴ \(3\left( \frac { \sqrt { 5 } }{ 3 } \right) \) = \(\sqrt { 5 } \) is rational

(∵ Q product of two rationals is also rational)

but this is contradiction that \(\sqrt { 5 } \) is irrational

∴ \(\frac { \sqrt { 5 } }{ 3 } \) is irrational proved.

**Example 9: **Prove that \(2\sqrt { 7 } \) is irrational.

**Sol. **Let is rational

∴ \(2\sqrt { 7 } \times \left( \frac { 1 }{ 2 } \right) \) = \(\sqrt { 7 } \)

(∵ Q division of two rational no. is also rational)

∴ \(\sqrt { 7 } \)is rational

but this is contradiction that is irrational

∴ \(2\sqrt { 7 } \) is irrational

**Example 10: **Find 3 irrational numbers between 3 & 5.

**Solution: **∵ 3 and 5 both are rational

The irrational are 3.127190385……………

3.212325272930………

3.969129852937…………