\displaystyle \textbf{Exercise 1(A)}

\displaystyle \textbf{Question 1: }\text{Insert two rational numbers between:} \\ \quad \text{(i) }\frac38\text{ and }\frac7{12}\qquad \text{(ii) }\frac13\text{ and }\frac14
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have }\frac38=\frac{3\times3}{8\times3}=\frac9{24}\text{ and }\frac7{12}=\frac{7\times2}{12\times2}=\frac{14}{24}
\displaystyle \therefore\ \frac9{24}<\frac{10}{24}<\frac{11}{24}<\frac{14}{24}
\displaystyle \therefore\ \frac5{12}\text{ and }\frac{11}{24}\text{ are two rational numbers between }\frac38\text{ and }\frac7{12}.
\displaystyle \text{(ii) Since }\frac14<\frac13,\text{ we have }\frac14=\frac{1\times9}{4\times9}=\frac9{36}\text{ and }\frac13=\frac{1\times12}{3\times12}=\frac{12}{36}
\displaystyle \therefore\ \frac9{36}<\frac{10}{36}<\frac{11}{36}<\frac{12}{36}
\displaystyle \therefore\ \frac5{18}\text{ and }\frac{11}{36}\text{ are two rational numbers between }\frac14\text{ and }\frac13.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Insert three rational numbers between:} \\ \quad \text{(i) }\frac25\text{ and }\frac37\qquad \text{(ii) }\frac4{11}\text{ and }\frac9{16}
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have }\frac25=\frac{2\times28}{5\times28}=\frac{56}{140}\text{ and }\frac37=\frac{3\times20}{7\times20}=\frac{60}{140}
\displaystyle \therefore\ \frac{56}{140}<\frac{57}{140}<\frac{58}{140}<\frac{59}{140}<\frac{60}{140}
\displaystyle \therefore\ \frac{57}{140},\ \frac{29}{70}\text{ and }\frac{59}{140}\text{ are three rational numbers between }\frac25\text{ and }\frac37.
\displaystyle \text{(ii) We have }\frac4{11}=\frac{4\times16}{11\times16}=\frac{64}{176}\text{ and }\frac9{16}=\frac{9\times11}{16\times11}=\frac{99}{176}
\displaystyle \therefore\ \frac{64}{176}<\frac{65}{176}<\frac{66}{176}<\frac{67}{176}<\frac{99}{176}
\displaystyle \therefore\ \frac{65}{176},\ \frac38\text{ and }\frac{67}{176}\text{ are three rational numbers between }\frac4{11}\text{ and }\frac9{16}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{(i) Find three rational numbers between }5\text{ and }-2.\quad  \\ \text{(ii) Find three rational numbers between }-\frac34\text{ and }\frac12.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }-2<-1<0<1<5
\displaystyle \therefore\ -1,\ 0\text{ and }1\text{ are three rational numbers between }-2\text{ and }5.
\displaystyle \text{(ii) We have }\frac12=\frac24
\displaystyle \therefore\ -\frac34<-\frac24<-\frac14<0<\frac24
\displaystyle \therefore\ -\frac12,\ -\frac14\text{ and }0\text{ are three rational numbers between }-\frac34\text{ and }\frac12.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Insert 4 rational numbers between }5\text{ and }8.
\displaystyle \text{Answer:}
\displaystyle \text{We have }5=\frac{5\times5}{5}=\frac{25}{5}\text{ and }8=\frac{8\times5}{5}=\frac{40}{5}
\displaystyle \therefore\ \frac{25}{5}<\frac{26}{5}<\frac{27}{5}<\frac{28}{5}<\frac{29}{5}<\frac{40}{5}
\displaystyle \therefore\ \frac{26}{5},\ \frac{27}{5},\ \frac{28}{5}\text{ and }\frac{29}{5}\text{ are four rational numbers between }5\text{ and }8.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Insert 5 rational numbers between }\frac13\text{ and }\frac59.
\displaystyle \text{Answer:}
\displaystyle \text{The L.C.M. of the denominators }3\text{ and }9\text{ is }9.
\displaystyle \therefore\ \frac13=\frac39\text{ and }\frac59=\frac59
\displaystyle \text{There are not five rational numbers with denominator }9\text{ between }\frac39\text{ and }\frac59.
\displaystyle \text{Therefore, multiplying the numerator and denominator of each fraction by }4,\text{ we get}
\displaystyle \frac13=\frac{1\times12}{3\times12}=\frac{12}{36}\text{ and }\frac59=\frac{5\times4}{9\times4}=\frac{20}{36}
\displaystyle \therefore\ \frac{12}{36}<\frac{13}{36}<\frac{14}{36}<\frac{15}{36}<\frac{16}{36}<\frac{17}{36}<\frac{20}{36}
\displaystyle \therefore\ \frac{13}{36},\ \frac7{18},\ \frac5{12},\ \frac49\text{ and }\frac{17}{36}\text{ are five rational numbers between }\frac13\text{ and }\frac59.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Insert 6 rational numbers between }4.6\text{ and }8.4.
\displaystyle \text{Answer:}
\displaystyle \text{Since }4.6<4.7<4.8<4.9<5.0<5.1<5.2<8.4
\displaystyle \therefore\ 4.7,\ 4.8,\ 4.9,\ 5,\ 5.1\text{ and }5.2\text{ are six rational numbers between }4.6\text{ and }8.4.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Insert 7 rational numbers between }1\text{ and }2.
\displaystyle \text{Answer:}
\displaystyle \text{We have }1=\frac88\text{ and }2=\frac{16}{8}
\displaystyle \therefore\ \frac88<\frac98<\frac{10}{8}<\frac{11}{8}<\frac{12}{8}<\frac{13}{8}<\frac{14}{8}<\frac{15}{8}<\frac{16}{8}
\displaystyle \therefore\ \frac98,\ \frac54,\ \frac{11}{8},\ \frac32,\ \frac{13}{8},\ \frac74\text{ and }\frac{15}{8}\text{ are seven rational numbers between }1\text{ and }2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Insert 8 rational numbers between }1.8\text{ and }3.6.
\displaystyle \text{Answer:}
\displaystyle \text{We have }1.8<2.0<2.2<2.4<2.6<2.8<3.0<3.2<3.4<3.6
\displaystyle \therefore\ 2,\ 2.2,\ 2.4,\ 2.6,\ 2.8,\ 3,\ 3.2\text{ and }3.4\text{ are eight rational numbers between }1.8\text{ and }3.6.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Arrange }-\frac59,\ \frac7{12},\ -\frac23\text{ and }\frac{11}{18}\text{ in the ascending order of their magnitudes.}
\displaystyle \text{Also, find the difference between the largest and the smallest of these rational numbers.}
\displaystyle \text{Express this difference as a decimal fraction correct to one decimal place.}
\displaystyle \text{Answer:}
\displaystyle \left|-\frac59\right|=\frac59=\frac{20}{36},\qquad \left|\frac7{12}\right|=\frac7{12}=\frac{21}{36}
\displaystyle \left|-\frac23\right|=\frac23=\frac{24}{36},\qquad \left|\frac{11}{18}\right|=\frac{11}{18}=\frac{22}{36}
\displaystyle \text{Since }\frac{20}{36}<\frac{21}{36}<\frac{22}{36}<\frac{24}{36}
\displaystyle \therefore\ \text{The ascending order of their magnitudes is }-\frac59,\ \frac7{12},\ \frac{11}{18},\ -\frac23.
\displaystyle \text{Now, the largest rational number is }\frac{11}{18}\text{ and the smallest rational number is }-\frac23.
\displaystyle \therefore\ \text{Required difference}=\frac{11}{18}-\left(-\frac23\right)
\displaystyle =\frac{11}{18}+\frac{12}{18}=\frac{23}{18}=1.2777\ldots
\displaystyle \therefore\ \text{The required difference, correct to one decimal place, is }1.3.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Arrange }\frac58,\ -\frac3{16},\ -\frac14\text{ and }\frac{17}{32}\text{ in the descending order of their magnitudes.}
\displaystyle \text{Also, find the sum of the lowest and the largest of these rational numbers.}
\displaystyle \text{Express the result obtained as a decimal fraction correct to two decimal places.}
\displaystyle \text{Answer:}
\displaystyle \left|\frac58\right|=\frac58=\frac{20}{32},\qquad \left|-\frac3{16}\right|=\frac3{16}=\frac6{32}
\displaystyle \left|-\frac14\right|=\frac14=\frac8{32},\qquad \left|\frac{17}{32}\right|=\frac{17}{32}
\displaystyle \text{Since }\frac{20}{32}>\frac{17}{32}>\frac8{32}>\frac6{32}
\displaystyle \therefore\ \text{The descending order of their magnitudes is }\frac58,\ \frac{17}{32},\ -\frac14,\ -\frac3{16}.
\displaystyle \text{Now, the lowest rational number is }-\frac14\text{ and the largest rational number is }\frac58.
\displaystyle \therefore\ \text{Required sum}=-\frac14+\frac58
\displaystyle =-\frac28+\frac58=\frac38=0.375
\displaystyle \therefore\ \text{The required sum, correct to two decimal places, is }0.38.
\displaystyle \\

\displaystyle \textbf{Exercise 1(B)}

\displaystyle \textbf{Question 1: }\text{State which of the following decimal numbers are}
\displaystyle \text{pure recurring decimals and which are mixed recurring decimals:}
\displaystyle \text{(i) }0.\overline{083}\qquad \text{(ii) }0.0\overline{83}\qquad \text{(iii) }0.\overline{227}
\displaystyle \text{(iv) }3.\overline{54}\qquad \text{(v) }2.\overline{81}
\displaystyle \text{Answer:}
\displaystyle \text{A recurring decimal in which all the digits after the decimal point}
\displaystyle \text{are repeating is called a pure recurring decimal.}
\displaystyle \text{A recurring decimal in which some digits after the decimal point}
\displaystyle \text{do not repeat is called a mixed recurring decimal.}
\displaystyle \text{(i) }0.\overline{083}=0.083083083\ldots
\displaystyle \therefore\ 0.\overline{083}\text{ is a pure recurring decimal.}
\displaystyle \text{(ii) }0.0\overline{83}=0.083838383\ldots
\displaystyle \text{Here, the digit }0\text{ immediately after the decimal point is}
\displaystyle \text{non-recurring.}
\displaystyle \therefore\ 0.0\overline{83}\text{ is a mixed recurring decimal.}
\displaystyle \text{(iii) }0.\overline{227}=0.227227227\ldots
\displaystyle \therefore\ 0.\overline{227}\text{ is a pure recurring decimal.}
\displaystyle \text{(iv) }3.\overline{54}=3.54545454\ldots
\displaystyle \therefore\ 3.\overline{54}\text{ is a pure recurring decimal.}
\displaystyle \text{(v) }2.\overline{81}=2.81818181\ldots
\displaystyle \therefore\ 2.\overline{81}\text{ is a pure recurring decimal.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Represent the following as decimal numbers:}
\displaystyle \text{(i) }\frac4{15}\qquad \text{(ii) }\frac27\qquad \text{(iii) }\frac49\qquad \text{(iv) }\frac5{24}\qquad \text{(v) }\frac8{13}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac4{15}=0.26666\ldots=0.2\overline6

\displaystyle \text{(ii) }\frac27=0.285714285714\ldots=0.\overline{285714}

\displaystyle \text{(iii) }\frac49=0.44444\ldots=0.\overline4

\displaystyle \text{(iv) }\frac5{24}=0.2083333\ldots=0.208\overline3

\displaystyle \text{(v) }\frac8{13}=0.615384615384\ldots=0.\overline{615384}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Express each of the following as a rational number,}
\displaystyle \text{i.e. in the form }\frac ab,\text{ where }a,b\in\mathbb{Z}\text{ and }b\neq0:
\displaystyle \text{(i) }0.\overline{53}\qquad \text{(ii) }0.\overline{227}\qquad \text{(iii) }0.2\overline{104}
\displaystyle \text{(iv) }3.\overline{52}\qquad \text{(v) }2.2\overline{4689}\qquad \text{(vi) }0.\overline{572}
\displaystyle \text{(vii) }0.1\overline{58}\qquad \text{(viii) }0.0\overline{384}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }x=0.\overline{53}
\displaystyle \Rightarrow x=0.535353\ldots\qquad\ldots\text{(I)}
\displaystyle \Rightarrow 100x=53.535353\ldots\qquad\ldots\text{(II)}
\displaystyle \text{Subtracting (I) from (II), we get}
\displaystyle 100x-x=53
\displaystyle \Rightarrow 99x=53
\displaystyle \Rightarrow x=\frac{53}{99}
\displaystyle \therefore\ 0.\overline{53}=\frac{53}{99}

\displaystyle \text{(ii) Let }x=0.\overline{227}
\displaystyle \Rightarrow x=0.227227227\ldots\qquad\ldots\text{(I)}
\displaystyle \Rightarrow 1000x=227.227227227\ldots\qquad\ldots\text{(II)}
\displaystyle \text{Subtracting (I) from (II), we get}
\displaystyle 1000x-x=227
\displaystyle \Rightarrow 999x=227
\displaystyle \Rightarrow x=\frac{227}{999}
\displaystyle \therefore\ 0.\overline{227}=\frac{227}{999}

\displaystyle \text{(iii) Let }x=0.2\overline{104}
\displaystyle \Rightarrow x=0.2104104104\ldots\qquad\ldots\text{(I)}
\displaystyle \Rightarrow 10x=2.104104104\ldots\qquad\ldots\text{(II)}
\displaystyle \Rightarrow 10000x=2104.104104104\ldots\qquad\ldots\text{(III)}
\displaystyle \text{Subtracting (II) from (III), we get}
\displaystyle 10000x-10x=2104-2
\displaystyle \Rightarrow 9990x=2102
\displaystyle \Rightarrow x=\frac{2102}{9990}=\frac{1051}{4995}
\displaystyle \therefore\ 0.2\overline{104}=\frac{1051}{4995}

\displaystyle \text{(iv) Let }x=3.\overline{52}
\displaystyle \Rightarrow x=3.525252\ldots\qquad\ldots\text{(I)}
\displaystyle \Rightarrow 100x=352.525252\ldots\qquad\ldots\text{(II)}
\displaystyle \text{Subtracting (I) from (II), we get}
\displaystyle 100x-x=352-3
\displaystyle \Rightarrow 99x=349
\displaystyle \Rightarrow x=\frac{349}{99}
\displaystyle \therefore\ 3.\overline{52}=\frac{349}{99}

\displaystyle \text{(v) Let }x=2.2\overline{4689}
\displaystyle \Rightarrow x=2.246894689\ldots\qquad\ldots\text{(I)}
\displaystyle \Rightarrow 10x=22.46894689\ldots\qquad\ldots\text{(II)}
\displaystyle \Rightarrow 100000x=224689.46894689\ldots\qquad\ldots\text{(III)}
\displaystyle \text{Subtracting (II) from (III), we get}
\displaystyle 100000x-10x=224689-22
\displaystyle \Rightarrow 99990x=224667
\displaystyle \Rightarrow x=\frac{224667}{99990}=\frac{24963}{11110}
\displaystyle \therefore\ 2.2\overline{4689}=\frac{24963}{11110}

\displaystyle \text{(vi) Let }x=0.\overline{572}
\displaystyle \Rightarrow x=0.572572572\ldots\qquad\ldots\text{(I)}
\displaystyle \Rightarrow 1000x=572.572572572\ldots\qquad\ldots\text{(II)}
\displaystyle \text{Subtracting (I) from (II), we get}
\displaystyle 1000x-x=572
\displaystyle \Rightarrow 999x=572
\displaystyle \Rightarrow x=\frac{572}{999}
\displaystyle \therefore\ 0.\overline{572}=\frac{572}{999}

\displaystyle \text{(vii) Let }x=0.1\overline{58}
\displaystyle \Rightarrow x=0.1585858\ldots\qquad\ldots\text{(I)}
\displaystyle \Rightarrow 10x=1.585858\ldots\qquad\ldots\text{(II)}
\displaystyle \Rightarrow 1000x=158.585858\ldots\qquad\ldots\text{(III)}
\displaystyle \text{Subtracting (II) from (III), we get}
\displaystyle 1000x-10x=158-1
\displaystyle \Rightarrow 990x=157
\displaystyle \Rightarrow x=\frac{157}{990}
\displaystyle \therefore\ 0.1\overline{58}=\frac{157}{990}

\displaystyle \text{(viii) Let }x=0.0\overline{384}
\displaystyle \Rightarrow x=0.0384384384\ldots\qquad\ldots\text{(I)}
\displaystyle \Rightarrow 10x=0.384384384\ldots\qquad\ldots\text{(II)}
\displaystyle \Rightarrow 10000x=384.384384384\ldots\qquad\ldots\text{(III)}
\displaystyle \text{Subtracting (II) from (III), we get}
\displaystyle 10000x-10x=384
\displaystyle \Rightarrow 9990x=384
\displaystyle \Rightarrow x=\frac{384}{9990}=\frac{64}{1665}
\displaystyle \therefore\ 0.0\overline{384}=\frac{64}{1665}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the decimal representation of }\frac17\text{ and }\frac27.
\displaystyle \text{Deduce from the decimal representation of }\frac17,\text{ without actual}
\displaystyle \text{calculation, the decimal representation of }\frac37,\ \frac47,\ \frac57\text{ and }\frac67.
\displaystyle \text{Answer:}
\displaystyle \frac17=0.142857142857\ldots=0.\overline{142857}
\displaystyle \frac27=0.285714285714\ldots=0.\overline{285714}
\displaystyle \text{The repeating block in }\frac17\text{ is }142857.
\displaystyle \text{The decimal representations of the other fractions are obtained by cyclically rearranging these digits.}
\displaystyle \therefore\ \frac37=0.428571428571\ldots=0.\overline{428571}
\displaystyle \frac47=0.571428571428\ldots=0.\overline{571428}
\displaystyle \frac57=0.714285714285\ldots=0.\overline{714285}
\displaystyle \frac67=0.857142857142\ldots=0.\overline{857142}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Without doing any actual division, find which of the following}
\displaystyle \text{rational numbers have terminating decimal representations:}
\displaystyle \text{(i) }\frac7{16}\qquad \text{(ii) }\frac{23}{125}\qquad \text{(iii) }\frac9{14}\qquad \text{(iv) }\frac{32}{45}
\displaystyle \text{(v) }\frac{43}{50}\qquad \text{(vi) }\frac{17}{40}\qquad \text{(vii) }\frac{61}{75}\qquad \text{(viii) }\frac{123}{250}
\displaystyle \text{Answer:}
\displaystyle \text{A rational number }\frac pq\text{ in its lowest form has a terminating decimal}
\displaystyle \text{representation if }q=2^m\times5^n,\text{ where }m\text{ and }n\text{ are whole numbers.}

\displaystyle \text{(i) }\frac7{16},\qquad 16=2^4
\displaystyle \therefore\ \frac7{16}\text{ has a terminating decimal representation.}

\displaystyle \text{(ii) }\frac{23}{125},\qquad 125=5^3
\displaystyle \therefore\ \frac{23}{125}\text{ has a terminating decimal representation.}

\displaystyle \text{(iii) }\frac9{14},\qquad 14=2\times7
\displaystyle \text{Since the denominator contains the prime factor }7,
\displaystyle \therefore\ \frac9{14}\text{ does not have a terminating decimal representation.}

\displaystyle \text{(iv) }\frac{32}{45},\qquad 45=3^2\times5
\displaystyle \text{Since the denominator contains the prime factor }3,
\displaystyle \therefore\ \frac{32}{45}\text{ does not have a terminating decimal representation.}

\displaystyle \text{(v) }\frac{43}{50},\qquad 50=2\times5^2
\displaystyle \therefore\ \frac{43}{50}\text{ has a terminating decimal representation.}

\displaystyle \text{(vi) }\frac{17}{40},\qquad 40=2^3\times5
\displaystyle \therefore\ \frac{17}{40}\text{ has a terminating decimal representation.}

\displaystyle \text{(vii) }\frac{61}{75},\qquad 75=3\times5^2
\displaystyle \text{Since the denominator contains the prime factor }3,
\displaystyle \therefore\ \frac{61}{75}\text{ does not have a terminating decimal representation.}

\displaystyle \text{(viii) }\frac{123}{250},\qquad 250=2\times5^3
\displaystyle \therefore\ \frac{123}{250}\text{ has a terminating decimal representation.}
\displaystyle \therefore\ \frac7{16},\ \frac{23}{125},\ \frac{43}{50},\ \frac{17}{40}\text{ and}
\displaystyle \frac{123}{250}\text{ have terminating decimal representations.}
\displaystyle \\

\displaystyle \textbf{Exercise 1(C)}

\displaystyle \textbf{Question 1: }\text{State whether the following numbers are rational or not:}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }(2+\sqrt2)^2
\displaystyle =(2)^2+2(2)(\sqrt2)+(\sqrt2)^2
\displaystyle =4+4\sqrt2+2=6+4\sqrt2
\displaystyle \text{Since }\sqrt2\text{ is irrational, }6+4\sqrt2\text{ is irrational.}
\displaystyle \therefore\ (2+\sqrt2)^2\text{ is irrational.}

\displaystyle \text{(ii) }(3-\sqrt3)^2
\displaystyle =(3)^2-2(3)(\sqrt3)+(\sqrt3)^2
\displaystyle =9-6\sqrt3+3=12-6\sqrt3
\displaystyle \text{Since }\sqrt3\text{ is irrational, }12-6\sqrt3\text{ is irrational.}
\displaystyle \therefore\ (3-\sqrt3)^2\text{ is irrational.}

\displaystyle \text{(iii) }(5+\sqrt5)(5-\sqrt5)
\displaystyle =25-5=20
\displaystyle \therefore\ (5+\sqrt5)(5-\sqrt5)\text{ is rational.}

\displaystyle \text{(iv) }(\sqrt3-\sqrt2)^2
\displaystyle =3+2-2\sqrt6
\displaystyle =5-2\sqrt6
\displaystyle \text{Since }\sqrt6\text{ is irrational, }5-2\sqrt6\text{ is irrational.}
\displaystyle \therefore\ (\sqrt3-\sqrt2)^2\text{ is irrational.}

\displaystyle \text{(v) }\left(\frac3{2\sqrt2}\right)^2
\displaystyle =\frac9{8}
\displaystyle \therefore\ \left(\frac3{2\sqrt2}\right)^2\text{ is rational.}
\displaystyle \text{(vi) }\left(\frac{\sqrt7}{6\sqrt2}\right)^2
\displaystyle =\frac7{72}
\displaystyle \therefore\ \left(\frac{\sqrt7}{6\sqrt2}\right)^2\text{ is rational.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the square of:}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\left(\frac{3\sqrt5}{5}\right)^2
\displaystyle =\frac{9\times5}{25}
\displaystyle =\frac95

\displaystyle \text{(ii) }(\sqrt3+\sqrt2)^2
\displaystyle =3+2+2\sqrt6
\displaystyle =5+2\sqrt6

\displaystyle \text{(iii) }(\sqrt5-2)^2
\displaystyle =5-4\sqrt5+4
\displaystyle =9-4\sqrt5

\displaystyle \text{(iv) }(3+2\sqrt5)^2
\displaystyle =9+12\sqrt5+20
\displaystyle =29+12\sqrt5
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{State, in each case, whether True or False:}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\sqrt2+\sqrt3=\sqrt5
\displaystyle \textbf{False}

\displaystyle \text{(ii) }2\sqrt4+2=6
\displaystyle 2(2)+2=6
\displaystyle \textbf{True}

\displaystyle \text{(iii) }3\sqrt7-2\sqrt7=\sqrt7
\displaystyle (3-2)\sqrt7=\sqrt7
\displaystyle \textbf{True}

\displaystyle \text{(iv) }\frac27\text{ is an irrational number.}
\displaystyle \textbf{False, since }\frac27\text{ is a rational number.}

\displaystyle \text{(v) }\frac5{11}\text{ is a rational number.}
\displaystyle \textbf{True}

\displaystyle \text{(vi) All rational numbers are real numbers.}
\displaystyle \textbf{True}

\displaystyle \text{(vii) All real numbers are rational numbers.}
\displaystyle \textbf{False}
\displaystyle \text{Some real numbers are irrational.}

\displaystyle \text{(viii) Some real numbers are rational numbers.}
\displaystyle \textbf{True}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The given universal set is}
\displaystyle \left\{-6,-5\frac34,-\sqrt4,-\frac35,-\frac38,0,\frac45,2,1\frac23,\sqrt8,3.01,\pi,8.47\right\}.
\displaystyle \text{ From the given set, find:}
\displaystyle \text{(i) the set of rational numbers}
\displaystyle \text{(ii) the set of irrational numbers}
\displaystyle \text{(iii) the set of integers}
\displaystyle \text{(iv) the set of non-negative integers}
\displaystyle \text{Answer:}
\displaystyle \text{We know that }-\sqrt4=-2,\text{ which is an integer and hence rational.}
\displaystyle \text{Also, terminating decimals }3.01\text{ and }8.47\text{ are rational numbers.}

\displaystyle \text{(i) The set of rational numbers is}
\displaystyle \left\{-6,-5\frac34,-\sqrt4,-\frac35,-\frac38,0,\frac45,2,1\frac23,3.01,8.47\right\}

\displaystyle \text{(ii) The set of irrational numbers is }\left\{\sqrt8,\pi\right\}.

\displaystyle \text{(iii) The set of integers is }\left\{-6,-\sqrt4,0,2\right\}.

\displaystyle \text{(iv) The set of non-negative integers is }\left\{0,2\right\}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Use division method to show that }\sqrt3\text{ and }\sqrt5
\displaystyle \text{are irrational numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{By finding the square roots using the division method, we get}
\displaystyle \sqrt3=1.732050807568877\ldots
\displaystyle \text{The decimal expansion is non-terminating and non-recurring.}
\displaystyle \therefore\ \sqrt3\text{ is an irrational number.}
\displaystyle \text{Similarly, by the division method, we get}
\displaystyle \sqrt5=2.236067977499789\ldots
\displaystyle \text{The decimal expansion is non-terminating and non-recurring.}
\displaystyle \therefore\ \sqrt5\text{ is an irrational number.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Use the method of contradiction to show that }\sqrt3
\displaystyle \text{and }\sqrt5\text{ are irrational numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{First, let us prove that }\sqrt3\text{ is irrational.}
\displaystyle \text{Assume, on the contrary, that }\sqrt3\text{ is rational.}
\displaystyle \text{Then }\sqrt3=\frac ab,\text{ where }a,b\text{ are coprime positive integers.}
\displaystyle \Rightarrow 3=\frac{a^2}{b^2}
\displaystyle \Rightarrow a^2=3b^2
\displaystyle \therefore\ 3\mid a^2
\displaystyle \Rightarrow 3\mid a
\displaystyle \text{Let }a=3c,\text{ where }c\text{ is an integer.}
\displaystyle \therefore\ a^2=9c^2
\displaystyle \Rightarrow 9c^2=3b^2
\displaystyle \Rightarrow b^2=3c^2
\displaystyle \therefore\ 3\mid b^2
\displaystyle \Rightarrow 3\mid b
\displaystyle \text{Thus, }3\text{ is a common factor of }a\text{ and }b.
\displaystyle \text{This contradicts the fact that }a\text{ and }b\text{ are coprime.}
\displaystyle \therefore\ \sqrt3\text{ is an irrational number.}
\displaystyle \text{Now, let us prove that }\sqrt5\text{ is irrational.}
\displaystyle \text{Assume, on the contrary, that }\sqrt5\text{ is rational.}
\displaystyle \text{Then }\sqrt5=\frac ab,\text{ where }a,b\text{ are coprime positive integers.}
\displaystyle \Rightarrow 5=\frac{a^2}{b^2}
\displaystyle \Rightarrow a^2=5b^2
\displaystyle \therefore\ 5\mid a^2
\displaystyle \Rightarrow 5\mid a
\displaystyle \text{Let }a=5c,\text{ where }c\text{ is an integer.}
\displaystyle \therefore\ a^2=25c^2
\displaystyle \Rightarrow 25c^2=5b^2
\displaystyle \Rightarrow b^2=5c^2
\displaystyle \therefore\ 5\mid b^2
\displaystyle \Rightarrow 5\mid b
\displaystyle \text{Thus, }5\text{ is a common factor of }a\text{ and }b.
\displaystyle \text{This contradicts the fact that }a\text{ and }b\text{ are coprime.}
\displaystyle \therefore\ \sqrt5\text{ is an irrational number.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Write a pair of irrational numbers whose sum}
\displaystyle \text{is irrational.}
\displaystyle \text{Answer:}
\displaystyle \sqrt2\text{ and }\sqrt3\text{ are irrational numbers.}
\displaystyle \text{Their sum is }\sqrt2+\sqrt3,\text{ which is irrational.}
\displaystyle \therefore\ \text{The required pair is }\sqrt2\text{ and }\sqrt3.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write a pair of irrational numbers whose sum}
\displaystyle \text{is rational.}
\displaystyle \text{Answer:}
\displaystyle \sqrt2\text{ and }-\sqrt2\text{ are irrational numbers.}
\displaystyle \sqrt2+(-\sqrt2)=0
\displaystyle \text{Since }0\text{ is rational, the required pair is }\sqrt2\text{ and }-\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Write a pair of irrational numbers whose difference}
\displaystyle \text{is irrational.}
\displaystyle \text{Answer:}
\displaystyle \sqrt3\text{ and }\sqrt2\text{ are irrational numbers.}
\displaystyle \text{Their difference is }\sqrt3-\sqrt2,\text{ which is irrational.}
\displaystyle \therefore\ \text{The required pair is }\sqrt3\text{ and }\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Write a pair of irrational numbers whose difference}
\displaystyle \text{is rational.}
\displaystyle \text{Answer:}
\displaystyle 1+\sqrt2\text{ and }\sqrt2\text{ are irrational numbers.}
\displaystyle (1+\sqrt2)-\sqrt2=1
\displaystyle \text{Since }1\text{ is rational, the required pair is }1+\sqrt2\text{ and }\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Write a pair of irrational numbers whose product}
\displaystyle \text{is irrational.}
\displaystyle \text{Answer:}
\displaystyle \sqrt2\text{ and }\sqrt3\text{ are irrational numbers.}
\displaystyle \sqrt2\times\sqrt3=\sqrt6
\displaystyle \text{Since }\sqrt6\text{ is irrational, the required pair is }\sqrt2\text{ and }\sqrt3.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Write a pair of irrational numbers whose product}
\displaystyle \text{is rational.}
\displaystyle \text{Answer:}
\displaystyle \sqrt2\text{ and }\sqrt8\text{ are irrational numbers.}
\displaystyle \sqrt2\times\sqrt8=\sqrt{16}=4
\displaystyle \text{Since }4\text{ is rational, the required pair is }\sqrt2\text{ and }\sqrt8.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Write in ascending order:}
\displaystyle \text{(i) }3\sqrt5\text{ and }4\sqrt3
\displaystyle \text{(ii) }2\sqrt[3]{5}\text{ and }3\sqrt[3]{2}
\displaystyle \text{(iii) }6\sqrt5,\ 7\sqrt3\text{ and }8\sqrt2
\displaystyle \text{Answer:}

\displaystyle \text{(i) Since both numbers are positive, we compare their squares.}
\displaystyle (3\sqrt5)^2=9\times5=45
\displaystyle (4\sqrt3)^2=16\times3=48
\displaystyle \text{Since }45<48,\ 3\sqrt5<4\sqrt3.
\displaystyle \therefore\ \text{The ascending order is }3\sqrt5,\ 4\sqrt3.

\displaystyle \text{(ii) Since both numbers are positive, we compare their cubes.}
\displaystyle \left(2\sqrt[3]{5}\right)^3=2^3\times5=40
\displaystyle \left(3\sqrt[3]{2}\right)^3=3^3\times2=54
\displaystyle \text{Since }40<54,\ 2\sqrt[3]{5}<3\sqrt[3]{2}.
\displaystyle \therefore\ \text{The ascending order is }2\sqrt[3]{5},\ 3\sqrt[3]{2}.

\displaystyle \text{(iii) Since all the numbers are positive, we compare their squares.}
\displaystyle (6\sqrt5)^2=36\times5=180
\displaystyle (7\sqrt3)^2=49\times3=147
\displaystyle (8\sqrt2)^2=64\times2=128
\displaystyle \text{Since }128<147<180,
\displaystyle 8\sqrt2<7\sqrt3<6\sqrt5.
\displaystyle \therefore\ \text{The ascending order is }8\sqrt2,\ 7\sqrt3,\ 6\sqrt5.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Write in descending order:}
\displaystyle \text{(i) }2\sqrt[4]{6}\text{ and }3\sqrt[4]{2}
\displaystyle \text{(ii) }7\sqrt3\text{ and }3\sqrt7
\displaystyle \text{Answer:}

\displaystyle \text{(i) Since both numbers are positive, we compare their fourth powers.}
\displaystyle \left(2\sqrt[4]{6}\right)^4=2^4\times6=96
\displaystyle \left(3\sqrt[4]{2}\right)^4=3^4\times2=162
\displaystyle \text{Since }162>96,\ 3\sqrt[4]{2}>2\sqrt[4]{6}.
\displaystyle \therefore\ \text{The descending order is }3\sqrt[4]{2},\ 2\sqrt[4]{6}.

\displaystyle \text{(ii) Since both numbers are positive, we compare their squares.}
\displaystyle (7\sqrt3)^2=49\times3=147
\displaystyle (3\sqrt7)^2=9\times7=63
\displaystyle \text{Since }147>63,\ 7\sqrt3>3\sqrt7.
\displaystyle \therefore\ \text{The descending order is }7\sqrt3,\ 3\sqrt7.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Compare:}
\displaystyle \text{(i) }\sqrt[6]{15}\text{ and }\sqrt[4]{12}
\displaystyle \text{(ii) }\sqrt{24}\text{ and }\sqrt[3]{35}
\displaystyle \text{Answer:}

\displaystyle \text{(i) The L.C.M. of }6\text{ and }4\text{ is }12.
\displaystyle \left(\sqrt[6]{15}\right)^{12}=15^2=225
\displaystyle \left(\sqrt[4]{12}\right)^{12}=12^3=1728
\displaystyle \text{Since }225<1728,
\displaystyle \therefore\ \sqrt[6]{15}<\sqrt[4]{12}.

\displaystyle \text{(ii) The L.C.M. of }2\text{ and }3\text{ is }6.
\displaystyle \left(\sqrt{24}\right)^6=24^3=13824
\displaystyle \left(\sqrt[3]{35}\right)^6=35^2=1225
\displaystyle \text{Since }13824>1225,
\displaystyle \therefore\ \sqrt{24}>\sqrt[3]{35}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Insert two irrational numbers between }5\text{ and }6.
\displaystyle \text{Answer:}
\displaystyle 5=\sqrt{25}\text{ and }6=\sqrt{36}
\displaystyle \text{Choose the non-perfect squares }26\text{ and }27\text{ between }25\text{ and }36.
\displaystyle 25<26<27<36
\displaystyle \Rightarrow \sqrt{25}<\sqrt{26}<\sqrt{27}<\sqrt{36}
\displaystyle \Rightarrow 5<\sqrt{26}<\sqrt{27}<6
\displaystyle \therefore\ \sqrt{26}\text{ and }\sqrt{27}\text{ are two irrational numbers}
\displaystyle \text{between }5\text{ and }6.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Insert five irrational numbers between }2\sqrt5
\displaystyle \text{and }3\sqrt3.
\displaystyle \text{Answer:}
\displaystyle 2\sqrt5=\sqrt{20}\text{ and }3\sqrt3=\sqrt{27}
\displaystyle \text{Choose five non-perfect squares between }20\text{ and }27.
\displaystyle 20<21<22<23<24<26<27
\displaystyle \Rightarrow \sqrt{20}<\sqrt{21}<\sqrt{22}<\sqrt{23}<\sqrt{24}<\sqrt{26}<\sqrt{27}
\displaystyle \Rightarrow 2\sqrt5<\sqrt{21}<\sqrt{22}<\sqrt{23}<\sqrt{24}<\sqrt{26}<3\sqrt3
\displaystyle \therefore\ \sqrt{21},\ \sqrt{22},\ \sqrt{23},\ \sqrt{24}\text{ and }\sqrt{26}
\displaystyle \text{are five irrational numbers between }2\sqrt5\text{ and }3\sqrt3.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Write two rational numbers between }\sqrt2\text{ and }\sqrt3.
\displaystyle \text{Answer:}
\displaystyle \text{Choose two rational numbers between }2\text{ and }3\text{ which are perfect squares.}
\displaystyle \text{For example, }2.25\text{ and }2.56.
\displaystyle 2<2.25<2.56<3
\displaystyle \Rightarrow \sqrt2<\sqrt{2.25}<\sqrt{2.56}<\sqrt3
\displaystyle \Rightarrow \sqrt2<1.5<1.6<\sqrt3
\displaystyle \therefore\ 1.5\text{ and }1.6\text{ are two rational numbers between}
\displaystyle \sqrt2\text{ and }\sqrt3.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Write three rational numbers between }\sqrt3\text{ and }\sqrt5.
\displaystyle \text{Answer:}
\displaystyle \text{Choose three rational numbers between }3\text{ and }5\text{ which are perfect squares.}
\displaystyle \text{For example, }3.24,\ 4\text{ and }4.84.
\displaystyle 3<3.24<4<4.84<5
\displaystyle \Rightarrow \sqrt3<\sqrt{3.24}<\sqrt4<\sqrt{4.84}<\sqrt5
\displaystyle \Rightarrow \sqrt3<1.8<2<2.2<\sqrt5
\displaystyle \therefore\ 1.8,\ 2\text{ and }2.2\text{ are three rational numbers between}
\displaystyle \sqrt3\text{ and }\sqrt5.
\displaystyle \\

\displaystyle \textbf{Exercise 1(D)}

\displaystyle \textbf{Question 1: }\text{State, with reason, which of the following are surds}
\displaystyle \text{and which are not:}
\displaystyle \text{(i) }\sqrt{180}\qquad \text{(ii) }\sqrt[4]{27}\qquad \text{(iii) }\sqrt[5]{128}\qquad \text{(iv) }\sqrt[3]{64}
\displaystyle \text{(v) }\sqrt[3]{25}\cdot\sqrt[3]{40}\qquad \text{(vi) }\sqrt[3]{-125}\qquad \text{(vii) }\sqrt{\pi}\qquad \text{(viii) }\sqrt{3+\sqrt2}
\displaystyle \text{Answer:}
\displaystyle \text{A surd is an irrational root which cannot be simplified to a rational number.}

\displaystyle \text{(i) }\sqrt{180}=\sqrt{36\times5}=6\sqrt5
\displaystyle \text{Since }\sqrt5\text{ is irrational, }\sqrt{180}\text{ is a surd.}

\displaystyle \text{(ii) }\sqrt[4]{27}=\sqrt[4]{3^3}
\displaystyle \text{Since }27\text{ is not a perfect fourth power, }\sqrt[4]{27}\text{ is a surd.}

\displaystyle \text{(iii) }\sqrt[5]{128}=\sqrt[5]{2^7}=2\sqrt[5]{4}
\displaystyle \text{Since }\sqrt[5]{4}\text{ is irrational, }\sqrt[5]{128}\text{ is a surd.}

\displaystyle \text{(iv) }\sqrt[3]{64}=\sqrt[3]{4^3}=4
\displaystyle \text{Since it is rational, }\sqrt[3]{64}\text{ is not a surd.}

\displaystyle \text{(v) }\sqrt[3]{25}\cdot\sqrt[3]{40}=\sqrt[3]{1000}=\sqrt[3]{10^3}=10
\displaystyle \text{Since it is rational, }\sqrt[3]{25}\cdot\sqrt[3]{40}\text{ is not a surd.}

\displaystyle \text{(vi) }\sqrt[3]{-125}=-5
\displaystyle \text{Since it is rational, }\sqrt[3]{-125}\text{ is not a surd.}

\displaystyle \text{(vii) }\sqrt{\pi}
\displaystyle \text{Since }\pi\text{ is irrational, }\sqrt{\pi}\text{ is also irrational.}
\displaystyle \therefore\ \sqrt{\pi}\text{ is a surd.}

\displaystyle \text{(viii) }\sqrt{3+\sqrt2}
\displaystyle \text{Since }3+\sqrt2\text{ is irrational, }\sqrt{3+\sqrt2}\text{ is also irrational.}
\displaystyle \therefore\ \sqrt{3+\sqrt2}\text{ is a surd.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the lowest rationalising factor of:}
\displaystyle \text{(i) }5\sqrt2\qquad \text{(ii) }\sqrt{24}\qquad \text{(iii) }\sqrt5-3\qquad \text{(iv) }7-\sqrt7
\displaystyle \text{(v) }\sqrt{18}-\sqrt{50}\qquad \text{(vi) }\sqrt5-\sqrt2\qquad \text{(vii) }\sqrt{13}+3
\displaystyle \text{(viii) }15-3\sqrt2\qquad \text{(ix) }3\sqrt2+2\sqrt3
\displaystyle \text{Answer:}

\displaystyle \text{(i) }5\sqrt2
\displaystyle \text{The lowest rationalising factor is }\sqrt2.
\displaystyle \text{Since }5\sqrt2\times\sqrt2=10,\text{ which is rational.}

\displaystyle \text{(ii) }\sqrt{24}=2\sqrt6
\displaystyle \text{The lowest rationalising factor is }\sqrt6.
\displaystyle \text{Since }2\sqrt6\times\sqrt6=12,\text{ which is rational.}

\displaystyle \text{(iii) }\sqrt5-3
\displaystyle \text{The lowest rationalising factor is }\sqrt5+3.
\displaystyle (\sqrt5-3)(\sqrt5+3)=5-9=-4,\text{ which is rational.}

\displaystyle \text{(iv) }7-\sqrt7
\displaystyle \text{The lowest rationalising factor is }7+\sqrt7.
\displaystyle (7-\sqrt7)(7+\sqrt7)=49-7=42,\text{ which is rational.}

\displaystyle \text{(v) }\sqrt{18}-\sqrt{50}
\displaystyle =3\sqrt2-5\sqrt2=-2\sqrt2
\displaystyle \text{The lowest rationalising factor is }\sqrt2.
\displaystyle (-2\sqrt2)\times\sqrt2=-4,\text{ which is rational.}

\displaystyle \text{(vi) }\sqrt5-\sqrt2
\displaystyle \text{The lowest rationalising factor is }\sqrt5+\sqrt2.
\displaystyle (\sqrt5-\sqrt2)(\sqrt5+\sqrt2)=5-2=3,\text{ which is rational.}

\displaystyle \text{(vii) }\sqrt{13}+3
\displaystyle \text{The lowest rationalising factor is }\sqrt{13}-3.
\displaystyle (\sqrt{13}+3)(\sqrt{13}-3)=13-9=4,\text{ which is rational.}

\displaystyle \text{(viii) }15-3\sqrt2
\displaystyle =3(5-\sqrt2)
\displaystyle \text{The lowest rationalising factor is }5+\sqrt2.
\displaystyle 3(5-\sqrt2)(5+\sqrt2)=3(25-2)=69,\text{ which is rational.}

\displaystyle \text{(ix) }3\sqrt2+2\sqrt3
\displaystyle \text{The lowest rationalising factor is }3\sqrt2-2\sqrt3.
\displaystyle (3\sqrt2+2\sqrt3)(3\sqrt2-2\sqrt3)=18-12=6,
\displaystyle \text{which is rational.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Rationalise the denominators of:}
\displaystyle \text{(i) }\frac3{\sqrt5}\qquad \text{(ii) }\frac{2\sqrt3}{\sqrt5}\qquad \text{(iii) }\frac1{\sqrt3-\sqrt2}
\displaystyle \text{(iv) }\frac3{\sqrt5+\sqrt2}\qquad \text{(v) }\frac{2-\sqrt3}{2+\sqrt3}\qquad \text{(vi) }\frac{\sqrt3+1}{\sqrt3-1}
\displaystyle \text{(vii) }\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\qquad \text{(viii) }\frac{\sqrt6-\sqrt5}{\sqrt6+\sqrt5}\qquad \text{(ix) }\frac{2\sqrt5+3\sqrt2}{2\sqrt5-3\sqrt2}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac3{\sqrt5}
\displaystyle =\frac3{\sqrt5}\times\frac{\sqrt5}{\sqrt5}
\displaystyle =\frac{3\sqrt5}{5}

\displaystyle \text{(ii) }\frac{2\sqrt3}{\sqrt5}
\displaystyle =\frac{2\sqrt3}{\sqrt5}\times\frac{\sqrt5}{\sqrt5}
\displaystyle =\frac{2\sqrt{15}}5

\displaystyle \text{(iii) }\frac1{\sqrt3-\sqrt2}
\displaystyle =\frac1{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}
\displaystyle =\frac{\sqrt3+\sqrt2}{3-2}
\displaystyle =\sqrt3+\sqrt2

\displaystyle \text{(iv) }\frac3{\sqrt5+\sqrt2}
\displaystyle =\frac3{\sqrt5+\sqrt2}\times\frac{\sqrt5-\sqrt2}{\sqrt5-\sqrt2}
\displaystyle =\frac{3(\sqrt5-\sqrt2)}{5-2}
\displaystyle =\sqrt5-\sqrt2

\displaystyle \text{(v) }\frac{2-\sqrt3}{2+\sqrt3}
\displaystyle =\frac{2-\sqrt3}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}
\displaystyle =\frac{(2-\sqrt3)^2}{4-3}
\displaystyle =7-4\sqrt3

\displaystyle \text{(vi) }\frac{\sqrt3+1}{\sqrt3-1}
\displaystyle =\frac{\sqrt3+1}{\sqrt3-1}\times\frac{\sqrt3+1}{\sqrt3+1}
\displaystyle =\frac{(\sqrt3+1)^2}{3-1}
\displaystyle =\frac{4+2\sqrt3}{2}
\displaystyle =2+\sqrt3

\displaystyle \text{(vii) }\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}
\displaystyle =\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}
\displaystyle =\frac{(\sqrt3-\sqrt2)^2}{3-2}
\displaystyle =5-2\sqrt6

\displaystyle \text{(viii) }\frac{\sqrt6-\sqrt5}{\sqrt6+\sqrt5}
\displaystyle =\frac{\sqrt6-\sqrt5}{\sqrt6+\sqrt5}\times\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}
\displaystyle =\frac{(\sqrt6-\sqrt5)^2}{6-5}
\displaystyle =11-2\sqrt{30}

\displaystyle \text{(ix) }\frac{2\sqrt5+3\sqrt2}{2\sqrt5-3\sqrt2}
\displaystyle =\frac{2\sqrt5+3\sqrt2}{2\sqrt5-3\sqrt2}\times\frac{2\sqrt5+3\sqrt2}{2\sqrt5+3\sqrt2}
\displaystyle =\frac{(2\sqrt5+3\sqrt2)^2}{(2\sqrt5)^2-(3\sqrt2)^2}
\displaystyle =\frac{20+18+12\sqrt{10}}{20-18}
\displaystyle =\frac{38+12\sqrt{10}}2
\displaystyle =19+6\sqrt{10}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the values of }a\text{ and }b\text{ in each of the following:}
\displaystyle \text{(i) }\frac{2+\sqrt3}{2-\sqrt3}=a+b\sqrt3
\displaystyle \text{(ii) }\frac{\sqrt7-2}{\sqrt7+2}=a\sqrt7+b
\displaystyle \text{(iii) }\frac3{\sqrt3-\sqrt2}=a\sqrt3-b\sqrt2
\displaystyle \text{(iv) }\frac{5+3\sqrt2}{5-3\sqrt2}=a+b\sqrt2
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{2+\sqrt3}{2-\sqrt3}
\displaystyle =\frac{2+\sqrt3}{2-\sqrt3}\times\frac{2+\sqrt3}{2+\sqrt3}
\displaystyle =\frac{(2+\sqrt3)^2}{2^2-(\sqrt3)^2}
\displaystyle =\frac{4+3+4\sqrt3}{4-3}
\displaystyle =7+4\sqrt3
\displaystyle \text{Comparing with }a+b\sqrt3,\text{ we get}
\displaystyle \therefore\ a=7\text{ and }b=4.

\displaystyle \text{(ii) }\frac{\sqrt7-2}{\sqrt7+2}
\displaystyle =\frac{\sqrt7-2}{\sqrt7+2}\times\frac{\sqrt7-2}{\sqrt7-2}
\displaystyle =\frac{(\sqrt7-2)^2}{(\sqrt7)^2-2^2}
\displaystyle =\frac{7+4-4\sqrt7}{7-4}
\displaystyle =-\frac43\sqrt7+\frac{11}{3}
\displaystyle \text{Comparing with }a\sqrt7+b,\text{ we get}
\displaystyle \therefore\ a=-\frac43\text{ and }b=\frac{11}{3}.

\displaystyle \text{(iii) }\frac3{\sqrt3-\sqrt2}
\displaystyle =\frac3{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}
\displaystyle =\frac{3(\sqrt3+\sqrt2)}{3-2}
\displaystyle =3\sqrt3+3\sqrt2
\displaystyle =3\sqrt3-(-3)\sqrt2
\displaystyle \text{Comparing with }a\sqrt3-b\sqrt2,\text{ we get}
\displaystyle \therefore\ a=3\text{ and }b=-3.

\displaystyle \text{(iv) }\frac{5+3\sqrt2}{5-3\sqrt2}
\displaystyle =\frac{5+3\sqrt2}{5-3\sqrt2}\times\frac{5+3\sqrt2}{5+3\sqrt2}
\displaystyle =\frac{(5+3\sqrt2)^2}{5^2-(3\sqrt2)^2}
\displaystyle =\frac{25+18+30\sqrt2}{25-18}
\displaystyle =\frac{43}{7}+\frac{30}{7}\sqrt2
\displaystyle \text{Comparing with }a+b\sqrt2,\text{ we get}
\displaystyle \therefore\ a=\frac{43}{7}\text{ and }b=\frac{30}{7}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Simplify:}
\displaystyle \text{(i) }\frac{22}{2\sqrt3+1}+\frac{17}{2\sqrt3-1}
\displaystyle \text{(ii) }\frac{\sqrt2}{\sqrt6-\sqrt2}-\frac{\sqrt3}{\sqrt6+\sqrt2}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{22}{2\sqrt3+1}+\frac{17}{2\sqrt3-1}
\displaystyle =\frac{22(2\sqrt3-1)+17(2\sqrt3+1)}{(2\sqrt3)^2-1^2}
\displaystyle =\frac{44\sqrt3-22+34\sqrt3+17}{12-1}
\displaystyle =\frac{78\sqrt3-5}{11}

\displaystyle \text{(ii) }\frac{\sqrt2}{\sqrt6-\sqrt2}-\frac{\sqrt3}{\sqrt6+\sqrt2}
\displaystyle =\frac{\sqrt2(\sqrt6+\sqrt2)}{6-2}
\displaystyle \qquad-\frac{\sqrt3(\sqrt6-\sqrt2)}{6-2}
\displaystyle =\frac{2\sqrt3+2}{4}-\frac{3\sqrt2-\sqrt6}{4}
\displaystyle =\frac{2+2\sqrt3-3\sqrt2+\sqrt6}{4}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }x=\frac{\sqrt5-2}{\sqrt5+2}\text{ and }y=\frac{\sqrt5+2}{\sqrt5-2},\text{ find:}
\displaystyle \text{(i) }x^2\qquad \text{(ii) }y^2\qquad \text{(iii) }xy\qquad \text{(iv) }x^2+y^2+xy
\displaystyle \text{Answer:}
\displaystyle x=\frac{\sqrt5-2}{\sqrt5+2}\times\frac{\sqrt5-2}{\sqrt5-2}
\displaystyle =\frac{(\sqrt5-2)^2}{5-4}
\displaystyle =9-4\sqrt5
\displaystyle y=\frac{\sqrt5+2}{\sqrt5-2}\times\frac{\sqrt5+2}{\sqrt5+2}
\displaystyle =\frac{(\sqrt5+2)^2}{5-4}
\displaystyle =9+4\sqrt5

\displaystyle \text{(i) }x^2=(9-4\sqrt5)^2
\displaystyle =81+80-72\sqrt5
\displaystyle =161-72\sqrt5

\displaystyle \text{(ii) }y^2=(9+4\sqrt5)^2
\displaystyle =81+80+72\sqrt5
\displaystyle =161+72\sqrt5

\displaystyle \text{(iii) }xy=(9-4\sqrt5)(9+4\sqrt5)
\displaystyle =81-80=1
\displaystyle \text{(iv) }x^2+y^2+xy
\displaystyle =(161-72\sqrt5)+(161+72\sqrt5)+1
\displaystyle =323
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }m=\frac1{3-2\sqrt2}\text{ and }n=\frac1{3+2\sqrt2},\text{ find:}
\displaystyle \text{(i) }m^2\qquad \text{(ii) }n^2\qquad \text{(iii) }mn
\displaystyle \text{Answer:}
\displaystyle m=\frac1{3-2\sqrt2}\times\frac{3+2\sqrt2}{3+2\sqrt2}
\displaystyle =\frac{3+2\sqrt2}{9-8}
\displaystyle =3+2\sqrt2
\displaystyle n=\frac1{3+2\sqrt2}\times\frac{3-2\sqrt2}{3-2\sqrt2}
\displaystyle =\frac{3-2\sqrt2}{9-8}
\displaystyle =3-2\sqrt2
\displaystyle \text{(i) }m^2=(3+2\sqrt2)^2
\displaystyle =9+8+12\sqrt2
\displaystyle =17+12\sqrt2
\displaystyle \text{(ii) }n^2=(3-2\sqrt2)^2
\displaystyle =9+8-12\sqrt2
\displaystyle =17-12\sqrt2
\displaystyle \text{(iii) }mn=(3+2\sqrt2)(3-2\sqrt2)
\displaystyle =9-8=1
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }x=2\sqrt3+2\sqrt2,\text{ find:}
\displaystyle \text{(i) }\frac1x\qquad \text{(ii) }x+\frac1x\qquad \text{(iii) }\left(x+\frac1x\right)^2
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\frac1x=\frac1{2\sqrt3+2\sqrt2}
\displaystyle =\frac1{2(\sqrt3+\sqrt2)}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}
\displaystyle =\frac{\sqrt3-\sqrt2}{2(3-2)}
\displaystyle =\frac{\sqrt3-\sqrt2}{2}
\displaystyle \text{(ii) }x+\frac1x
\displaystyle =2\sqrt3+2\sqrt2+\frac{\sqrt3-\sqrt2}{2}
\displaystyle =\frac{4\sqrt3+4\sqrt2+\sqrt3-\sqrt2}{2}
\displaystyle =\frac{5\sqrt3+3\sqrt2}{2}
\displaystyle \text{(iii) }\left(x+\frac1x\right)^2
\displaystyle =\left(\frac{5\sqrt3+3\sqrt2}{2}\right)^2
\displaystyle =\frac{75+18+30\sqrt6}{4}
\displaystyle =\frac{93+30\sqrt6}{4}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }x=1-\sqrt2,\text{ find the value of}
\displaystyle \left(x-\frac1x\right)^3.
\displaystyle \text{Answer:}
\displaystyle \frac1x=\frac1{1-\sqrt2}\times\frac{1+\sqrt2}{1+\sqrt2}
\displaystyle =\frac{1+\sqrt2}{1-2}
\displaystyle =-1-\sqrt2
\displaystyle \therefore\ x-\frac1x=(1-\sqrt2)-(-1-\sqrt2)
\displaystyle =1-\sqrt2+1+\sqrt2
\displaystyle =2
\displaystyle \therefore\ \left(x-\frac1x\right)^3=2^3=8
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }x=5-2\sqrt6,\text{ find }x^2+\frac1{x^2}.
\displaystyle \text{Answer:}
\displaystyle \frac1x=\frac1{5-2\sqrt6}\times\frac{5+2\sqrt6}{5+2\sqrt6}
\displaystyle =\frac{5+2\sqrt6}{25-24}
\displaystyle =5+2\sqrt6
\displaystyle \therefore\ x+\frac1x=(5-2\sqrt6)+(5+2\sqrt6)=10
\displaystyle \text{Now, }\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2
\displaystyle \Rightarrow 10^2=x^2+\frac1{x^2}+2
\displaystyle \Rightarrow x^2+\frac1{x^2}=100-2
\displaystyle \therefore\ x^2+\frac1{x^2}=98
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Show that:}
\displaystyle \frac1{3-2\sqrt2}-\frac1{2\sqrt2-\sqrt7}+\frac1{\sqrt7-\sqrt6}
\displaystyle -\frac1{\sqrt6-\sqrt5}+\frac1{\sqrt5-2}=5.
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S.}=\frac1{3-2\sqrt2}-\frac1{2\sqrt2-\sqrt7}
\displaystyle \qquad+\frac1{\sqrt7-\sqrt6}-\frac1{\sqrt6-\sqrt5}+\frac1{\sqrt5-2}
\displaystyle =(3+2\sqrt2)-(2\sqrt2+\sqrt7)+(\sqrt7+\sqrt6)
\displaystyle \qquad-(\sqrt6+\sqrt5)+(\sqrt5+2)
\displaystyle =3+2\sqrt2-2\sqrt2-\sqrt7+\sqrt7+\sqrt6
\displaystyle \qquad-\sqrt6-\sqrt5+\sqrt5+2
\displaystyle =3+2
\displaystyle =5=\text{R.H.S.}
\displaystyle \therefore\ \frac1{3-2\sqrt2}-\frac1{2\sqrt2-\sqrt7}+\frac1{\sqrt7-\sqrt6}
\displaystyle -\frac1{\sqrt6-\sqrt5}+\frac1{\sqrt5-2}=5.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Rationalise the denominator of:}
\displaystyle \frac1{\sqrt3-\sqrt2+1}
\displaystyle \text{Answer:}
\displaystyle \frac1{\sqrt3-\sqrt2+1}
\displaystyle =\frac1{(\sqrt3+1)-\sqrt2}
\displaystyle =\frac1{(\sqrt3+1)-\sqrt2}
\displaystyle \qquad\times\frac{\sqrt3+\sqrt2+1}{\sqrt3+\sqrt2+1}
\displaystyle =\frac{\sqrt3+\sqrt2+1}{(\sqrt3+1)^2-(\sqrt2)^2}
\displaystyle =\frac{\sqrt3+\sqrt2+1}{2+2\sqrt3}
\displaystyle =\frac{\sqrt3+\sqrt2+1}{2(1+\sqrt3)}
\displaystyle \qquad\times\frac{\sqrt3-1}{\sqrt3-1}
\displaystyle =\frac{(\sqrt3+\sqrt2+1)(\sqrt3-1)}{2(3-1)}
\displaystyle =\frac{3-\sqrt3+\sqrt6-\sqrt2+\sqrt3-1}{4}
\displaystyle =\frac{2+\sqrt6-\sqrt2}{4}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\sqrt2=1.4\text{ and }\sqrt3=1.7,\text{ find the value of:}
\displaystyle \text{(i) }\frac1{\sqrt3-\sqrt2}\qquad \text{(ii) }\frac1{3+2\sqrt2}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\frac1{\sqrt3-\sqrt2}
\displaystyle =\frac1{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}
\displaystyle =\frac{\sqrt3+\sqrt2}{3-2}
\displaystyle =\sqrt3+\sqrt2
\displaystyle =1.7+1.4
\displaystyle =3.1
\displaystyle \text{(ii) }\frac1{3+2\sqrt2}
\displaystyle =\frac1{3+2\sqrt2}\times\frac{3-2\sqrt2}{3-2\sqrt2}
\displaystyle =\frac{3-2\sqrt2}{9-8}
\displaystyle =3-2\sqrt2
\displaystyle =3-2(1.4)
\displaystyle =3-2.8
\displaystyle =0.2
\displaystyle \\

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