Real Numbers
Category : 10th Class
Real 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 \[\frac{12}{5}=2.4\] and \[\frac{13}{9}=1.44444\]…, are decimal expansions of rational numbers.
For example,\[\sqrt{2}\]=1.41421356….,
\[\sqrt{3}=\]1.732058075....,\[\pi =\]3.14159265...... are irrational numbers since they can not be written in the form \[\frac{p}{q}\]
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) \[(a+\sqrt{b})\,\,and\,\,(a-\sqrt{b})\]
(ii) \[(a+b\sqrt{m})\,\,and\,\,(a-b\sqrt{m})\]
(iii) \[(\sqrt{m}+\sqrt{n})\,\,and\,\,(\sqrt{m}-\sqrt{n}\]
Are rationalising factors of each other, where a and b are integers and m and n are natural numbers.
between a and b are,
a + d, a + 2d, a + 3d,…., a + nd, where \[d=\frac{a-b}{n+1}\]
Some Results of Real Number
For all positive real numbers a and b,
(i) \[\sqrt{ab}=\sqrt{a}\times \sqrt{b}\]
(ii)\[\sqrt{\frac{a}{b}}=\sqrt{\frac{a}{b}}\]
(iii) \[(\sqrt{a}+\sqrt{b})\,\,(\sqrt{a}-\sqrt{b})=a-b\]
(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 \[\sqrt[n]{x}\]. The sign ‘ \[\sqrt[n]{{}}\]’ is called the radical sign the number under this sign ie. ‘x’ is called the radicand and n is called the order of the \[\sqrt{2}\],\[\sqrt{3},\]\[\sqrt{4},\]radical. For example etc. are radicals. Irrational radicals such as etc. \[\sqrt{2}\],\[\sqrt{3},\]\[\sqrt{4},\] are also known as surds.
Laws of Exponents for Real Numbers
If and n are rational numbers and a is a positive real number, 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) \[{{(\sqrt[n]{a})}^{n}}=a\] (vi) \[\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}\]
(vii) \[a{}^{p}/{}_{q}={{({}^{q}/{}_{a})}^{p}}={{\sqrt[q]{a}}^{p}}\] 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) \[\sqrt[6]{729}-\sqrt[11]{2048}+\sqrt[3]{1728}\]
Solution: (a) \[\sqrt[6]{729}-\sqrt[11]{2048}+\sqrt[3]{1728}\]
\[=\sqrt[6]{36}-\sqrt[11]{{{2}^{11}}}+\sqrt[3]{{{12}^{3}}}\]
\[=3-2+12=13\]
Express 3.07 8 in the form of \[\frac{p}{q}\], where p and q are integers and q # 0.
Solution: Here, we have 3.078 = 3.07888.....
let x=3.078
\[\therefore \] 100x=307.8888….
and 1000x=3078.8888….
\[\therefore \] 1000\[-\]100x= 3078.8888...... \[-\]307.888….
900x=2771
\[x=\frac{2771}{900}\]
Which is the required form.
Arrange the following numbers in their ascending order.
\[\sqrt[12]{32},\sqrt[6]{5},\sqrt[3]{3},\sqrt[4]{4}\]
(a) \[\sqrt[3]{3}<\sqrt[4]{4}<\sqrt[6]{5}<\sqrt[12]{32}\]
(b) \[\sqrt[3]{3}>\sqrt[4]{4}>\sqrt[12]{32}>\sqrt[6]{5}\]
(c) \[\sqrt[6]{5}<\sqrt[12]{32}<\sqrt[4]{4}>\sqrt[3]{3}\]
(d) \[\sqrt[12]{32}<\sqrt[6]{5}<\sqrt[4]{4}>\sqrt[3]{3}\]
(e) None of these
Answer (C)
Explanation: We have,
\[\sqrt[12]{32}={{(32)}^{1/12}},\sqrt[6]{5}={{5}^{{}^{1}/{}_{6}}},\sqrt[3]{3}={{3}^{1/3}},\sqrt[4]{4}={{4}^{1/4}}\]
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
\[\therefore \,\,{{32}^{1/12}}={{32}^{1/12}},{{5}^{1/6}}={{5}^{2/12}},{{3}^{1/3}}={{3}^{4/12}},{{4}^{1/4}}={{4}^{3/12}}\]Thus we now have numbers in their same exponents ie.
\[\sqrt[12]{32},\sqrt[12]{{{5}^{2}}},\sqrt[12]{{{3}^{4}}},\sqrt[12]{{{4}^{3}}}\]or \[\sqrt[12]{32},\sqrt[12]{25},\sqrt[12]{81},\sqrt[12]{64}\]
Clearly, the ascending order of these numbers are
\[\sqrt[12]{25}<\sqrt[12]{32}<\sqrt[12]{64}<\sqrt[12]{81}\]ie.\[\sqrt[6]{5}<\sqrt[12]{32}<\sqrt[4]{4}<\sqrt[3]{3}\]
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