# 9th Class Mathematics Real Numbers Real Numbers

## Real Numbers

Category : 9th Class

Real Numbers

In the previous classes, we have learnt about rational and irrational numbers. In this chapter we will learn about real numbers. A real number can be any positive or negative numbers. All the rational and irrational numbers are real numbers. In other words we can say that real numbers are the set of rational and irrational numbers.

Important Points Related to Real Numbers

•                 A rational number is a real number which can be written as a simple fraction (i.e., in a ratio of two integers). In other words, a number r is called  a rational number when it can be written in the form $\frac{p}{q}$where p and q are integers and q is not equal to zero. For example$\frac{3}{5}$, 0, 3, $\frac{\mathrm{1}}{\mathrm{100}}$ are rational numbers
•            The decimal expansion of a rational number is either terminating or non-terminating recurring. For example $\frac{\mathrm{12}}{\mathrm{5}}=2.4$ and $\frac{13}{9}=1.44444$???are decimal expansion of rational numbers.
•                  An irrational number is a real number which cannot be written as a simple fraction. In other words, a number s is called an irrational number when it cannot be written in the form$\frac{\mathrm{p}}{\mathrm{q}}$, where p and q are integers and q is not equal to zero.

For example, $\sqrt{\mathrm{2}}\mathrm{=1}\mathrm{.41421356}.........\mathrm{,}\sqrt{\mathrm{3}}\mathrm{=1} \mathrm{.7320508075}.........\mathrm{, }\!\!\pi\!\!\text{ =3}\mathrm{.14159265}........$ are irrational numbers since they cannot be written in the form$\frac{\mathrm{p}}{\mathrm{q}}$.

Note:

(i)         We use  $\pi =\frac{22}{7}$, which is its approximate value but not accurate.

(ii)         The decimal expansion of irrational number is non-terminating non-recurring. For example 1.002000200002 ??. is an irrational number.

•                For any rational number r and irrational number $s,$ $r+s,$ $r-s$ are irrational numbers and for any non-zero rational number r and irrational number $s,$$r\,.\,s$ and $\frac{r}{s}$are irrational numbers.
•                 When product of two irrational numbers is rational then each one of these factors is called the rationalizing factor of other.

For example

(i)  $\mathrm{(a+}\sqrt{\mathrm{b)}}\mathrm{ and (a-}\sqrt{\mathrm{b)}}$

(ii) $\mathrm{(a+b}\sqrt{\mathrm{m)}}\mathrm{ and (a-b}\sqrt{\mathrm{m)}}$

(iii) $\mathrm{(}\sqrt{\mathrm{m}}\mathrm{+}\sqrt{\mathrm{n)}}\mathrm{ and (}\sqrt{\mathrm{m}}\mathrm{-}\sqrt{\mathrm{n)}}$

are rationalizing factors of each other, where a and b are integers and m and n are natural numbers.

•                 For any two rational numbers a and b (when a < b), $\frac{\mathrm{a+b}}{\mathrm{2}}$is a rational number lying between a and b.
•                 For any two rational numbers a and b (when a < b), n number of rational numbers between a and b are,

a + d, a + 2d, a+3d, ??.. , a + nd, where$\mathrm{d =}\frac{\mathrm{b-a}}{\mathrm{n+1}}$.

Some Results of Real Numbers

For all positive real numbers a and b

(i) $\sqrt{\mathrm{ab}}\mathrm{=}\sqrt{\mathrm{a}}\times \sqrt{\mathrm{b}}$

(ii) $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

(iii) $\mathrm{(}\sqrt{\mathrm{a}}\mathrm{+}\sqrt{\mathrm{b}}\mathrm{)(}\sqrt{\mathrm{a}}\mathrm{-}\sqrt{\mathrm{b}}\mathrm{)=a-b}$                  (iv) $(a+\sqrt{b})(a-\sqrt{b})={{a}^{2}}-b$

(v)  ${{(\sqrt{a}+\sqrt{b})}^{2}}=a+2\sqrt{ab}+b$

A radical expression is an expression of the type$^{n}\sqrt{x}$. The sign ?$^{n}\sqrt{x}$? is called the radical sign, the number under this sign i.e. ?x? is called the radicand and n is called the order of the radical. For example $\sqrt{2}, \sqrt{3}, \sqrt{4}$etc. are radicals. Irrational radicals such as $\sqrt{2}, \sqrt{3}, \sqrt{5}$etc. are also known as surds.

Laws or Exponents for Real Numbers

If m and n are rational numbers and a is a positive real numbers, then

(i)  ${{a}^{m}} . {{a}^{n}} = {{a}^{m+n}}$

(ii)  $\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$

(iii)  ${{({{a}^{m}})}^{n}}={{a}^{mn}}$

(iv)   ${{a}^{m}}{{b}^{m}} = {{(ab)}^{m}}$

(v)   ${{{{(}^{n}}\sqrt{a})}^{n}}=a$

(vi)   $^{m}\sqrt{^{n}\sqrt{a}}{{=}^{mn}}\sqrt{a}$

(vii)   ${{a}^{p/q}}={{{{(}^{q}}\sqrt{a})}^{p}}{{=}^{q}}\sqrt{{{a}^{p}}}$; where p and q are integers, $q>o$ and there is no common factors between p and q other than 1

•          Example:

Simplify the following:

(a) $^{6}\sqrt{729}{{-}^{11}}\sqrt{2048}{{+}^{3}}\sqrt{1728}$                   (b)   $\frac{12\sqrt{125}+3\sqrt{245}+\sqrt{2205}}{\sqrt{405}}$

Solution:

(a) $^{6}\sqrt{729}{{-}^{11}}\sqrt{2048}{{+}^{3}}\sqrt{1728}{{=}^{6}}\sqrt{{{3}^{6}}}{{-}^{11}}\sqrt{{{2}^{11}}}{{+}^{3}}\sqrt{{{12}^{3}}}=3-2+12=13$

(b)  $\frac{12\sqrt{125}+3\sqrt{245}+\sqrt{2205}}{\sqrt{405}}$

$=\frac{12\times 5\sqrt{5}+3\times 7\sqrt{5}+21\sqrt{5}}{9\sqrt{5}}=\frac{\sqrt{5}(60+21+21)}{9\sqrt{5}}=\frac{102}{9}=\frac{34}{3}$

•          Example:

Express $\mathbf{3}\mathbf{.07}\overline{\mathbf{8}}$ in the form of $\frac{\mathbf{p}}{\mathbf{q}}$, where p and q are integers and q # 0.

Solution: Here, we have $3.07\overline{8}$= 3.07888

Let x = $3.07\overline{8}$

$\therefore 100x = 307.8888.......$   and $1000x = 3078.8888....$

$\therefore 1000x -100x= 3078.8888....-307.888....$

$900x=2771$

$x=\frac{2771}{900}$

Which is the required form.

•           Example:

Solve:  $\sqrt{\frac{\mathbf{2}}{{{\mathbf{(512)}}^{\mathbf{-2/3}}}}\mathbf{+}\frac{\mathbf{2}}{{{\mathbf{(625)}}^{\mathbf{-3/4}}}}\mathbf{+}\frac{\mathbf{21}}{{{\mathbf{(729)}}^{\mathbf{-1/6}}}}}$

(a) 17                                                                (b) 21

(c) 25                                                                (d) 28

(e) None of these

Ans.     (b)

Explanation: We have,  $\sqrt{\frac{2}{{{(512)}^{-2/3}}}+\frac{2}{{{(625)}^{-3/4}}}+\frac{21}{{{(729)}^{-1/6}}}}$

= $\sqrt{\frac{2}{{{[{{(18)}^{3}}]}^{-2/3}}}+\frac{2}{{{({{5}^{4}})}^{-3/4}}}+\frac{21}{{{({{3}^{6}})}^{-1/6}}}}$

$=\sqrt{\frac{2}{{{8}^{-2}}}+\frac{2}{{{5}^{-3}}}+\frac{21}{{{3}^{-1}}}}=\sqrt{2\times {{8}^{2}}+2\times {{5}^{3}}+21\times {{3}^{1}}}$

$=\sqrt{128+250+63}=\sqrt{441}=21$

•           Example:

Arrange the following numbers in their ascending order.

$^{\mathbf{12}}\sqrt{\mathbf{32}}{{\mathbf{,}}^{\mathbf{6}}}\sqrt{\mathbf{5}}{{\mathbf{,}}^{\mathbf{3}}}{{\sqrt{\mathbf{3,}}}^{\mathbf{4}}}\sqrt{\mathbf{4}}$

(a) $^{3}\sqrt{3}{{<}^{4}}\sqrt{4}{{<}^{6}}\sqrt{5}{{<}^{12}}\sqrt{32}$                        (b)  $^{3}\sqrt{3}{{>}^{4}}\sqrt{4}{{>}^{12}}\sqrt{32}{{>}^{6}}\sqrt{5}$

(c) $^{6}\sqrt{5}{{<}^{12}}\sqrt{32}{{<}^{4}}\sqrt{4}{{<}^{3}}\sqrt{3}$                        (d)  $^{12}\sqrt{32}{{<}^{6}}\sqrt{5}{{<}^{4}}\sqrt{4}{{<}^{3}}\sqrt{3}$

(e) None of these

Ans.     (c)

Explanation: We have, $^{12}\sqrt{32}={{(32)}^{1/12}}{{,}^{\text{6}}}\sqrt{5}={{5}^{1/6}}{{,}^{3}}\sqrt{3}={{3}^{1/3}}{{,}^{4}}\sqrt{4}={{4}^{1/4}}$

Now we will express these surds in their same exponents by taking LCM of the denominator of these exponents.

So, LCM of 12, 6, 3 and 4 is 12

$\therefore 3{{2}^{1/12}}=3{{2}^{1/12}},{{5}^{1/6}}={{5}^{2/12}}$$,{{3}^{1/3}}={{3}^{4/12}},{{4}^{1/4}}={{4}^{3/12}}$

Thus we now have numbers in their same exponents i.e.

$^{12}\sqrt{32}{{,}^{12}}\sqrt{{{5}^{2}}}{{,}^{12}}\sqrt{{{3}^{4}}}{{,}^{12}}{{\sqrt{{{4}^{3}}}}^{{}}}or{{,}^{12}}{{\sqrt{32,}}^{12}}\sqrt{25}{{,}^{12}}{{\sqrt{81,}}^{12}}\sqrt{64}$

Clearly, the ascending order of these numbers are

$^{12}\sqrt{25}{{<}^{12}}\sqrt{32}{{<}^{12}}\sqrt{64}{{<}^{12}}{{\sqrt{81}}^{{}}} ie{{.}^{6}}\sqrt{5}{{<}^{12}}\sqrt{32}{{<}^{4}}\sqrt{4}{{<}^{3}}\sqrt{3}$

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