Subject: Optional Maths

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.

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

**Addition**

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

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:

= \(\sqrt{4}\)

= \(\sqrt{2× 2}\)

= \(\sqrt{(2)^2}\)

= 2^{2× \(\frac{1}{2}\)}

= 2 (Rational) - \(\sqrt[3]{27}\)

= \(\sqrt[3]{3× 3× 3}\)

= \(\sqrt[3]{(3)^3}\)

= 3^{3× \(\frac{1}{3}\)}= 3 (Rational)

**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

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}

= a^{2} - b

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.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Solve

\(\sqrt{3}\)

Solution

\(\sqrt{3}\)

= \(\sqrt{3×3}\)

= 3 (rational)

Reduce the following surds into the simplest form.

\(\sqrt{9}\)

Solution

\(\sqrt{9}\)

= \(\sqrt{3×3}\)

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

=3

Divide

\(\sqrt{25}\)

Solution

= \(\sqrt{25}\)

= \(\sqrt{5×5}\)

= \(\sqrt{(5)}\)^{2}

= 5

Solve

6\(\sqrt{3}\) by 2\(\sqrt{3}\)

Solution

\(\frac{6\sqrt{3}}{2\sqrt{3}}\)

= \(\frac{6}{2}\)×\(\frac{\sqrt{3}}{\sqrt{3}}\)

= 3

Simplify the following

( 3 + \(\sqrt{3}\) ) ( 2 + \(\sqrt{2}\))

Solution

( 3 + \(\sqrt{3}\) ) ( 2 + \(\sqrt{2}\))

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

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

= 6 + 3 \(\sqrt{2}\) + 2 \(\sqrt{3}\) + \(\sqrt{ 3 × 2 }\)

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

Simplify

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

Solution

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

= ( 12 - 3 ) \(\sqrt{3}\)

= 9 \(\sqrt{3}\)

Simplify

8 \(\sqrt{3}\) ×2 \(\sqrt{3}\)

Solution

= 16 \(\sqrt{3×3}\)

=16 × \(\sqrt{(3)}\)^{2}

^{ }= 16 \(\times\) 3

= 48

Rationalize the denominator

\(\frac{1}{\sqrt{4}+1}\)

Solution

\(\frac{1}{\sqrt{4}+1}\)

Here, the denominator = \(\sqrt{4}\)+1

Rationalizing factor = \(\sqrt{4}\)- 1

Multiplying the numerator and denominator by \(\sqrt{4}\) -1 , we get

\(\frac{1}{\sqrt{4}+1}\) \(\times\)\(\frac{\sqrt{4}-1}{\sqrt{4}-1}\)

= \(\frac{\sqrt{4} - 1}{\sqrt{(4)}-(1)^2}\)

= \(\frac{\sqrt{4}-1}{3}\)

Hence, the denominator is rationalized.

Rationalize the denominator

\(\frac{1}{\sqrt{7}+1}\)

Solution

\(\frac{1}{\sqrt{4}+1}\)

Here, the denominator = \(\sqrt{7}\)+1

Rationalizing factor = \(\sqrt{7}\)- 1

Multiplying the numerator and denominator by \(\sqrt{7}\) -1 , we get

\(\frac{1}{\sqrt{7}+1}\) \(\times\) \(\frac{\sqrt{7}- 1}{\sqrt{7}-1}\)

= \(\frac{\sqrt{7} - 1}{\sqrt{(7)}-(1)^2}\)

= \(\frac{\sqrt{7}-1}{6}\)

Hence, the denominator is rationalized.

Rationalize the denominator

\(\frac{2}{\sqrt{5}}\)

Solution

\(\frac{2}{\sqrt{5}}\)

= Here, the denominator = \(\sqrt{5}\)

= Rationalizing factor (R.F) = \(\sqrt{5}\)

Multiplying the numerator and denominator by \(\sqrt{5}\), we get

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

= \(\frac{2\sqrt{5}}{5}\)

Hence , the denominator is rationalized.

Write the simplest rationalizing factor of:

\(\sqrt{50}\)

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 \(\sqrt{50}\).

Solve

\(\sqrt{2}\) + 2\(\sqrt{2}\)

Solution

\(\sqrt{2}\) + 2\(\sqrt{2}\)

= 3 \(\sqrt{2}\)

Multiply

\(\sqrt{5}\)× \(\sqrt{3}\)

Solution

\(\sqrt{5}\)× \(\sqrt{3}\)

= \(\sqrt{15}\)

Multiplication is possible with irrational numbers if the order of irrational number is same.

Solve

\(\sqrt{10}\) ÷ \(\sqrt{2}\)

Solution

\(\sqrt{10}\) ÷ \(\sqrt{2}\)

= \(\frac{\sqrt{ 10}}{\sqrt{2}}\)

= \(\sqrt{5}\)

What is a surd?

If the root of a rational number is an irrational number then the resulting number is a surd.

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