Real Numbers
Category : 10th 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
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 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.
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) \[{{(\sqrt{a}+\sqrt{b})}^{2}}=a+2\sqrt{ab}+b\]
The Radical Sign and Radicand
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
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) \[\frac{12\sqrt{125}+3\sqrt{245}+\sqrt{2205}}{\sqrt{405}}\]
\[=\frac{12\times 5\sqrt{5}+3\times 7\sqrt{5}+21\sqrt{5}}{9\sqrt{5}}=\frac{\sqrt{5}(60+21+21)}{9\sqrt{5}}=\frac{102}{9}=\frac{34}{3}\]
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.
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, \[\sqrt{\frac{2}{{{(512)}^{-2/3}}}+\frac{2}{{{(625)}^{-3/4}}}+\frac{21}{{{(729)}^{-1/6}}}}\]
= \[\sqrt{\frac{2}{{{[{{(18)}^{3}}]}^{-2/3}}}+\frac{2}{{{({{5}^{4}})}^{-3/4}}}+\frac{21}{{{({{3}^{6}})}^{-1/6}}}}\]
\[=\sqrt{\frac{2}{{{8}^{-2}}}+\frac{2}{{{5}^{-3}}}+\frac{21}{{{3}^{-1}}}}=\sqrt{2\times {{8}^{2}}+2\times {{5}^{3}}+21\times {{3}^{1}}}\]
\[=\sqrt{128+250+63}=\sqrt{441}=21\]
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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