Notes on Surds | Grade 7 > Optional Maths > Number System and Surds | KULLABS.COM

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### Surds

If the root of a rational number is an irrational number then the resulting number is called surds.
eg:$$\sqrt{4}$$
∴ Here, 2 is a rational number .When square root of 4 is taken out it becomes irrational.
Hence, $$\sqrt{4}$$ is a surd.

#### Operation of surds.

Here we can learn the basic operations addition, subtraction, multiplication, division, finding square root and finding cube roots of irrational numbers .

We can add irrational numbers if they have the same order and the same radicand .
eg :$$\sqrt{2}$$ + 3$$\sqrt{2}$$ = 4$$\sqrt{2}$$ [∴ a + 3a = 4a ]
Here,
$$\sqrt{2}$$ is added with 3$$\sqrt{2}$$ as the radicand is $$\sqrt{2}$$ as well as the order is 3.
Again,
$$\sqrt{3}$$ + $$\sqrt{2}$$ = $$\sqrt{3}$$ + $$\sqrt{2}$$ [∴ a + b = a + b ]
Then, we can not add with add $$\sqrt{3}$$ with $$\sqrt{2}$$ as the radicand is different even though the order is same .
Simillary,
$$\sqrt[3]{5}$$ + $$\sqrt{5}$$ = $$\sqrt[3]{5}$$ + $$\sqrt{5}$$
Here, addition is not positive as the order of $$\sqrt[3]{5}$$ is 3 and $$\sqrt{5}$$ is 2 even though the radicand are same .

Subtraction

Here, we subtract irrational numbers if they have the same order and the same radicand same as in addition .

eg :

• 4$$\sqrt {6}$$ - $$\sqrt{6}$$ = 4$$\sqrt {6}$$ - $$\sqrt{6}$$ [∴ a - b = a - b ]
• 4$$\sqrt{5}$$ - $$\sqrt{5}$$ = 3$$\sqrt{5}$$
• $$\sqrt[3]{2}$$ - $$\sqrt{2}$$ = $$\sqrt [3]{2}$$ - $$\sqrt{2}$$

Multiplication

Here, multiplication is is possible with irrational numbers if the order of irrational number is same $$\sqrt{a}$$ $$\times$$ $$\sqrt{b}$$ = $$\sqrt{ab}$$
eg :

• $$\sqrt{2}$$ $$\times$$ $$\sqrt{3}$$
= $$\sqrt{6}$$
• The order of the both surds is 2, multiplication is possible .As the order is same , the radicand aremultiplied togeather to give the result
$$\sqrt[3]{2}$$ x $$\sqrt[3]{7}$$
= $$\sqrt[3]{14}$$
$$\sqrt{2}$$ x $$\sqrt[3]{4}$$
This can not be multiplied as the order are different .
• $$\sqrt{2}$$ x $$\sqrt{2}$$
= $$\sqrt{4}$$

Division

We can divide the rationals if they have some order again

If p and q are real numbers .
$$\frac{\sqrt{p}}{\sqrt{q}}$$ = $$\sqrt{\frac{p}{q}}$$

eg;

$$\sqrt{8}$$÷ $$\sqrt{2}$$ = $$\sqrt{\frac{8}{2}}$$ = $$\sqrt{4}$$ = 2

#### Square and cube root of irrational number

Here the square and cube root of all irritational numbers are not possible .even we can find the square root and cube root of some numbers that may later turn to be rational or irrational.

eg;

• $$\sqrt{4}$$
Solution:
= $$\sqrt{4}$$
= $$\sqrt{2× 2}$$
= $$\sqrt{(2)^2}$$
= 22× $$\frac{1}{2}$$
= 2 (Rational)
• $$\sqrt[3]{27}$$
= $$\sqrt[3]{3× 3× 3}$$
= $$\sqrt[3]{(3)^3}$$
= 33× $$\frac{1}{3}$$
= 3 (Rational)

#### Rationalizing Factor . ( RF )

Let's see the example of following ;

Two irrationals are said to be rationalizing factors of each other if the product of these two irrationals is a rational number.

For example;

• 5$$\sqrt{3}$$ and 2$$\sqrt{3}$$
5$$\sqrt{3}$$× 2$$\sqrt{3}$$
= 10$$\sqrt{3×3}$$
= 10× $$\sqrt{3}^2$$
• 5$$\sqrt{3}$$ and $$\sqrt{3}$$
$$\sqrt{3}$$× $$\sqrt{3}$$
= $$\sqrt{3×3}$$
= 5 $$\sqrt{3^2}$$
= 5× 3
= 15

#### Rationalising factor for the surd of the form a - $$\sqrt{b}$$ .

As we know , for the surd of form a-$$\sqrt{b}$$ we need to multiply it with any another surd a + $$\sqrt{b}$$ so that it turns it intoa perfect square and the irrational turns into rational a - $$\sqrt{b}$$ and a + $$\sqrt{b}$$ or $$\sqrt{a}$$ + b and $$\sqrt{a}$$ - b are the rationalizing factors for each other. these terms are called conjugates.

eg:

(a - $$\sqrt{b}$$× (a + $$\sqrt{b}$$
= (a - $$\sqrt{b}$$ (a + $$\sqrt{b}$$
= (a)2 - ($$\sqrt{b}$$2
= a2 - b

#### Rationalization ;

The process of multiplying a given surd by its rationalizing factor to get a rational number as a product is known as a rationalization of a given surd.

eg :

$$\frac{1}{\sqrt{2}}$$
Solution:
= $$\frac{1}{\sqrt{2}}$$× $$\frac{\sqrt{2}}{\sqrt{2}}$$ (multiplying both numerator and denominator by $$\sqrt{2}$$
= $$\frac{\sqrt{2}}{\sqrt{2×2}}$$
= $$\frac{\sqrt{2}}{\sqrt{4}}$$
= $$\frac{\sqrt{2}}{\sqrt{2^2}}$$
= $$\frac{\sqrt{2}}{2}$$
Hence, the denominatior is rational number.

• If the root of a rational number is an irrational number then the resulting number is called surds.
• Two irrationals are said to be rationalizing factors of each other if the product of these two irrationals is a rational number.
• We can add irrational numbers if they have the same order and the same radicand.

.

#### Click on the questions below to reveal the answers

Solution

$$\sqrt{9}$$

= $$\sqrt{3×3}$$

= $$\sqrt{(3)}$$2

=3

Solution

= $$\sqrt{25}$$

= $$\sqrt{5×5}$$

=  $$\sqrt{(5)}$$2

= 5

Solution

$$\frac{6\sqrt{3}}{2\sqrt{3}}$$

= $$\frac{6}{2}$$×$$\frac{\sqrt{3}}{\sqrt{3}}$$

= 3

Solution

= ( 8 - 3 + 4 ) $$\sqrt{3}$$

= ( 12  - 3 ) $$\sqrt{3}$$

=  9 $$\sqrt{3}$$

Solution

$$\sqrt{50}$$

= $$\sqrt{5×5×2}$$

= $$\sqrt{(5)^2×2}$$

= 5$$\sqrt{2}$$

If we multiply 5$$\sqrt{2}$$ by $$\sqrt{2}$$ we will get the rational number. So, $$\sqrt{2}$$ is the simplest rationalizing factor of  $$\sqrt{50}$$.

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7(sqrt{3})
5(sqrt{3})
1(sqrt{3})
3(sqrt{3})

12
15
25
10

(sqrt{5})
(sqrt{7})
(sqrt{9})
(sqrt{11})

(sqrt{21})
(sqrt{24})
(sqrt{30})
(sqrt{16})
• ### Rationalize(frac{1}{sqrt{5}})

(frac{sqrt{5}}{7})
(frac{sqrt{5}}{5})
(frac{sqrt{5}}{3})
(frac{sqrt{5}}{9})
• ### The set of  all rational and irrational numbers form _____________________ .

Ratio
Natural numbers
Surds
Whole numbers

4
3
5
6

6
3
9
5

(sqrt{25})
(sqrt{21})
(sqrt{30})
(sqrt{31})

9 (sqrt{5})
7(sqrt{5})
12(sqrt{5})
11(sqrt{5})

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