# A Lesson in Irrational Numbers & Surds

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

1. The product of a quadratic surd and its conjugate is always a rational number. This is because $\left( a + b\sqrt{k} \right)\left( a – b\sqrt{k} \right) = a^2 – kb^2$
2. A quadratic surd cannot be equal to the algebraic sum of the rational and irrational numbers.
3. 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.

### Exercise 1

August 12, 2020

This Exercise has no answers. The Submit button only takes you to Exercise 2.

1. Find five rational numbers between 1 and 2.
2. Find six rational numbers between 3/5 and 4/5.
3. Locate these on number-line: $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{13}$
4. Locate $$\sqrt{x}$$ on number-line.
5. Find irrational numbers between $$\frac{1}{7}$$ and $$\frac{2}{7}$$.
6. Find three different irrational numbers between $$\frac{5}{7}$$ and $$\frac{9}{11}$$.
7. Visualize $$4.\overline{{26}}$$ on number-line, upto 4 decimal points.
8. Insert a rational and an irrational number between (2, 3), (0, 0.1) and between $$\left( \sqrt{2}, \sqrt{3} \right)$$.
9. Show that $$0.1428571... = \frac{1}{7}$$

### Exercise 2

August 12, 2020

• This Exercise is ungraded. Write your answers on paper or in the boxes provided if you can. Submit to check.
• This Exercise is random. On revisit, you may get a partially renewed set of items.
1. Express into pure surds
1. $3\sqrt{2}$
2. $a\sqrt[n]{a}$
2. Of each pair, which is greater?
1. $$\sqrt{3}$$ and $$\sqrt{2}$$
2. $$\sqrt[3]{4}$$ and $$\sqrt{2}$$
3. Simplify $\sqrt[4]{\sqrt[3]{x^2}}$
4. Rationalize $\frac{2}{3\sqrt{3}}$
5. If $x = 2 + \sqrt{3},$ find the value of $x^3 + \frac{1}{x^3}$
6. Find the value of $$a$$ if $\frac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = 2 - a\sqrt{6}$
7. If $\frac{\sqrt{7} - \sqrt{3}}{x} = \frac{x}{\sqrt{7} + \sqrt{3}},$ find $$x$$.
8. Find the square: $\sqrt{x} - \sqrt{y}$
9. Transform into least equivalent order: $$\sqrt{2}$$, $$\sqrt[3]{5}$$
10. Simplify $\sqrt{20} + \sqrt{80} + \sqrt{45} - \sqrt{180}$

### Exercise 3

August 12, 2020

• This is a graded exercise. Tick or write the correct answers and press Submit to get scores.
• This exercise is random. Some or all of the items may change on a retake.
1. Simplify $\frac{8^{\frac{1}{3}} \times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$
2. If $$a = \frac{3 + \sqrt{5}}{2}$$, find $$a^2 + \frac{1}{a^2}$$
3. Evaluate $\left[ 5\left( 8^{\frac{1}{3}} + 27^{\frac{1}{3}} \right)^3 \right]^{\frac{1}{4}}$
4. What is the rationalising factor for $\frac{1}{\sqrt{2} + \sqrt{3}}?$
5. Evaluate$\frac{2^0 + 7^0}{5^0}$
6. Evaluate$(1^3 + 2^3 + 3^3)^{\frac{1}{2}}$
7. Solve $\left( \frac{x^b}{x^c} \right)^{b + c - a} \times \left( \frac{x^c}{x^a} \right)^{c + a - b} \times \left( \frac{x^a}{x^b} \right)^{a + b - c}$
8. Evaluate $\frac{4}{216^{-\frac{2}{3}}} + \frac{1}{256^{-\frac{3}{4}}} + \frac{2}{243^{-\frac{1}{5}}}$
9. Multiply $\sqrt{50} \times \sqrt{72}$
10. Simplify:$\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}$