A number that is not rational is irrational. So, what are rational numbers, in the first place? Well, a number is called a rational number, if it can be written in the form \( \frac{p}{q} \), where \(p\) and \(q\) are integers and \( q \neq 0 \). So, a number which cannot be expressed in the form \( \frac{p}{q} \), (where \(p\) and \(q\) are integers and \( q \neq 0 \)) is an irrational number. The sets of rational numbers and irrational numbers together make the set of real numbers.

Examples

Exercise 1

Exercise 2

Exercise 3

### Decimal Expansion

Rational Numbers:

- Terminating or Non-terminating.
- When non-terminating, it is recurring or repeating. (T/NTR)

Irrational Numbers:

- Always Non-terminating and Non-repeating (NTNR)

If \(r\) is a rational number and \(s\) is an irrational number, then the following are irrational:

- \( r + s \)
- \( r – s \)

Provided \( r \) is not zero, the following are also irrational:

- \( r \times s \)
- \( \frac{r}{s} \)

### Gist

\[

\sqrt{ab} = \sqrt{a}\sqrt{b}

\]

\[

\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

\]

\[

\left( \sqrt{a} + \sqrt{b} \right)\left( \sqrt{a} – \sqrt{b} \right) = a – b

\]

\[

\left( a + \sqrt{b} \right)\left( a – \sqrt{b} \right) = a^2 – b

\]

\[

\left( \sqrt{a} + \sqrt{b} \right)^2 = a + 2\sqrt{ab} + b

\]

\[

(a + b)^2 = a^2 + b^2 + 2ab

\]

\[

a^2 + b^2 = (a + b)^2 – 2ab

\]

\[

(a + b)(a – b) = a^2 – b^2

\]

### Laws of Indices

If \(p\) and \(q\) are rational numbers and \(a\) is positive real number, \[

a^p \times a^q = a^{p + q}

\]

\[

(a^p)^q = a^{pq}

\]

\[

(a^p)^{\frac{1}{q}} = a^{\frac{p}{q}}

\]

\[

\frac{a^p}{a^q} = a^{p – q}

\]

\[

a^{-p} = \frac{1}{a^p}

\]

\[

a^pb^p = (ab)^p

\]

\[

\left( \frac{a}{b} \right)^p = \frac{a^p}{b^p}

\]

## Surds

The \(n^{th}\) root of a rational number \(a\), i.e. \( \sqrt[n]{a} \) is known as surd, when \(a\) is not equal to the \(n^{th}\) power of any rational number. Here \(a\) is called the base and \(n\) the order of the surd. So, \( \sqrt[n]{a} \) is a surd of \(n^{th}\) order whose base is \(a\).

Thus, \(\sqrt{2}\), \(\sqrt{3}\) and \( \sqrt[3]{4} \) are surds, whereas \(\sqrt{4}\), \(\sqrt{9}\) and \( \sqrt[5]{32} \) are not surds. Further, we can convert surds of different order into surds of same order.

Note: All surds are irrational numbers whereas all irrational numbers are not surds.

##### Surds could be

- Simple Surd: a monomial e.g. \(\sqrt{5}\), \( \sqrt[3]{16} \)
- Compound: Two or more surds combined e.g. \( (\sqrt{5} – \sqrt{3} – \sqrt[7]{8}) \)
- Binomial Compound Surd: only two terms e.g. \( \sqrt{2} – \sqrt{3} \)
- Pure Simple Surd: when co-efficient of a monomial surd be unity e.g. \(\sqrt{12}\), \( \sqrt[3]{4} \)
- Mixed Simple Surd: when co-efficient of a monomial surd be a rational number e.g. \(2\sqrt{3}\), \(-\frac{4}{3}\sqrt[4]{5}\). Well, mixed surds can be converted into pure surds. Example: \(2\sqrt{3} = \sqrt{4} \times \sqrt{3} = \sqrt{12}\)
- Like and Unlike Surds: If two or more surds have the same irrational coefficients, they are called “like” surds, otherwise “unlike”. So, \(\sqrt{2}\), \(3\sqrt{2}\) and \(-\frac{4}{3}\sqrt{2}\) are like whereas \(\sqrt{3}\) and \(\sqrt[3]{2}\) are unlike surds.

### Conjugate Number

In the number of the form \(a + b\sqrt{k}\), \(a\) and \(b\) are rational numbers and \(\sqrt{k}\) is a quadratic surd. Note, a quadratic surd is a surd of second order. Now, for every number of the form \(a + b\sqrt{k}\), there is a number \(a – b\sqrt{k}\). The two are conjugate to each other.

#### Properties

- The product of a quadratic surd and its conjugate is always a rational number. This is because \[

\left( a + b\sqrt{k} \right)\left( a – b\sqrt{k} \right) = a^2 – kb^2

\] - A quadratic surd cannot be equal to the algebraic sum of the rational and irrational numbers.
- If two mixed surds are equal, their rational and irrational parts must be separately equal. That is, if \(a + \sqrt{b} = c + \sqrt{d},\) then \(a=c\) and \(b=d\).

### Examples

#1. Simplify: \[\sqrt[{2n}]{y^{3n}}\]

Solution: \[

\sqrt[{2n}]{y^{3n}} = y^{\frac{3n}{2n}} = y^{\frac{3}{2}} = \sqrt{y^3} = \sqrt{y^2 \cdot y} = y\sqrt{y}

\]

#2. Which of the following surds is greater? \[

\sqrt{5}\text{ or }\sqrt[3]{9}

\]

Steps: Write in the exponential form. Take LCM. Make the order equivalent. Compare base to get the answer.

Solution: \[

\sqrt{5} = 5^{\frac{1}{2}} = 5^{\frac{1}{2} \times \frac{3}{3}} = 5^{\frac{3}{6}} = (5^3)^{\frac{1}{6}} = 125^{\frac{1}{6}}

\]\[

\sqrt[3]{9} = 9^{\frac{1}{3}} = 9^{\frac{1}{3} \times \frac{2}{2}} = 9^{\frac{2}{6}} = (9^2)^{\frac{1}{6}} = 81^{\frac{1}{6}}

\]

Now, since \[

125 > 81

\] hence \[

\sqrt{5} > \sqrt[3]{9}

\]

#3. Find the square root of \[

7 + 4\sqrt{3}

\]

Solution: \[

7 + 4\sqrt{3} = 4 + 3 + 2 \times 2 \times \sqrt{3} = \left( 2 + \sqrt{3} \right)^2

\]\[

\therefore \sqrt{7 + 4\sqrt{3}} = \pm \left( 2 + \sqrt{3} \right)

\]

Now, do the exercises.