Q: Question 3Show how 5\sqrt{5}5 can be represented on the number line.

Representing 5\sqrt{5}5 as the sum of squares of two natural numbers:5 = 4 + 1 = (2)2 + (1)2Let l be the number line. If point O represents number 0 and point A represents number 2 then draw a line segment OA = 2 units.At A, draw AC ⊥ OA. From AC, cut off AB = 1 unit.We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:(OB)2 = (OA)2 + (AB)2(OB)2 = (2)2 + (1)2(OB)2 = 4 + 1(OB)2 = 5OB = 5\sqrt{5}5 unitsWith O as centre and radius = OB, we draw an arc of a circle to meet the number line l at point P.As, OP = OB = 5\sqrt{5}5 units, the point P will represent the number 5\sqrt{5}5 on the number line as shown in the figure below:

Q: Question 2You know that 17\dfrac{1}{7}71 = 0.142857‾0.\overline{142857}0.142857. Can you predict what the decimal expansions of 27\dfrac{2}{7}72, 37\dfrac{3}{7}73, 47\dfrac{4}{7}74, 57\dfrac{5}{7}75, 67\dfrac{6}{7}76 are, without actually doing the long division? If so, how?[Hint : Study the remainders while finding the value of 17\dfrac{1}{7}71 carefully.]

Given,17\dfrac{1}{7}71 = 0.142857‾0.\overline{142857}0.142857(i) 27\dfrac{2}{7}72 = 2×172 \times \dfrac{1}{7}2×71= 2×0.142857‾2 \times 0.\overline{142857}2×0.142857= 0.285714‾0.\overline{285714}0.285714(ii) 37\dfrac{3}{7}73 = 3×173 \times \dfrac{1}{7}3×71= 3×0.142857‾3 \times 0.\overline{142857}3×0.142857= 0.428571‾0.\overline{428571}0.428571(iii) 47\dfrac{4}{7}74 = 4×174 \times \dfrac{1}{7}4×71= 4×0.142857‾4 \times 0.\overline{142857}4×0.142857= 0.571428‾0.\overline{571428}0.571428(iv) 57\dfrac{5}{7}75 = 5×175 \times \dfrac{1}{7}5×71= 5×0.142857‾5 \times 0.\overline{142857}5×0.142857= 0.714285‾0.\overline{714285}0.714285(v)67\dfrac{6}{7}76 = 6×176 \times \dfrac{1}{7}6×71= 6×0.142857‾6 \times 0.\overline{142857}6×0.142857= 0.857142‾0.\overline{857142}0.857142

Question 1

Question 1Is zero a rational number? Can you write it in the form pq\dfrac{p}{q}qp​, where p and q are integers and q ≠ 0?

Accepted Answer

Yes 0 (zero) is a rational number. We can write it in the form of pq\dfrac{p}{q}qp​ like 01\dfrac{0}{1}10​, 02\dfrac{0}{2}20​, 03\dfrac{0}{3}30​, 04\dfrac{0}{4}40​, ........where p = 0 , q ≠ 0

Question 2

Question 2Find six rational numbers between 3 and 4.

Accepted Answer

Since we want six rational numbers, we write 3 and 4 as rational numbers with denominator 3 + 4 = 7,i.e.,3=3×77=2173 = \dfrac{3 	imes 7}{7} = \dfrac{21}{7}3=73×7​=721​and4=4×77=2874 = \dfrac{4 	imes 7} {7} = \dfrac{28}{7}4=74×7​=728​∴ Rational numbers between 3 and 4 are, 227\dfrac{22}{7}722​, 237\dfrac{23}{7}723​, 247\dfrac{24}{7}724​, 257\dfrac{25}{7}725​, 267\dfrac{26}{7}726​, 277\dfrac{27}{7}727​

Question 3

Question 3Find five rational numbers between 35\dfrac{3}{5}53​ and 45\dfrac{4}{5}54​.

Accepted Answer

Since we want five rational numbers, we write 35\dfrac{3}{5}53​ and 45\dfrac{4}{5}54​ as rational numbers with denominator 5 + 1 = 6, i.e.,35=3×65×6=1830\dfrac{3}{5} = \dfrac{3 	imes 6} {5 	imes 6} = \dfrac{18}{30}53​=5×63×6​=3018​and45=4×65×6=2430\dfrac{4}{5} = \dfrac{4 	imes 6} {5 	imes 6} = \dfrac{24}{30}54​=5×64×6​=3024​∴ Rational numbers between 35\dfrac{3}{5}53​ and 45\dfrac{4}{5}54​ are, 1930\dfrac{19}{30}3019​, 2030\dfrac{20}{30}3020​, 2130\dfrac{21}{30}3021​, 2230\dfrac{22}{30}3022​, 2330\dfrac{23}{30}3023​

Question 4

Question 4State whether the following statements are true or false. Give reasons for your answers.(i) Every natural number is a whole number.(ii) Every integer is a whole number.(iii) Every rational number is a whole number.

Accepted Answer

(i) TrueReason — Whole numbers are = 0, 1, 2, 3, 4, 5,.......∞Natural numbers are = 1, 2, 3, 4, 5,......∞From above we see all natural numbers are whole numbers.(ii) FalseReason — Integers can be positive as well as negative i.e., integers are = 0, +(-1), +(-2), +(-3), +(-4)......∞ whereas whole number are 0 or positive only i.e., whole numbers are = 0, 1, 2, 3, 4,.......∞(iii) False,Reason — Rational numbers are pq\dfrac{p}{q}qp​ = 12\dfrac{1}{2}21​, 23\dfrac{2}{3}32​, 14\dfrac{1}{4}41​, 47\dfrac{4}{7}74​,.....Whole numbers are = 0, 1, 2, 3, 4,.......∞

Question 5

Question 1State whether the following statements are true or false. Justify your answers.(i) Every irrational number is a real number.(ii) Every point on the number line is of the form √m , where m is a natural number.(iii) Every real number is an irrational number.

Accepted Answer

(i) TrueReason — Every irrational number is a real number, because rational number, natural number, whole number all the numbers comes into real number.(ii) FalseReason — Every point on the number line represents a unique real number.(iii) FalseReason — For example 2\sqrt{2}2​ is real but not irrational.

Question 6

Question 2Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

Accepted Answer

No, square root of all positive integers are not always irrational.Example = 4=(2×2)\sqrt{4} = (\sqrt{2 	imes 2})4​=(2×2​) = 2, and 2 is a rational number

Question 7

Question 3Show how 5\sqrt{5}5​ can be represented on the number line.

Accepted Answer

Representing 5\sqrt{5}5​ as the sum of squares of two natural numbers:5 = 4 + 1 = (2)2 + (1)2Let l be the number line. If point O represents number 0 and point A represents number 2 then draw a line segment OA = 2 units.At A, draw AC ⊥ OA. From AC, cut off AB = 1 unit.We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:(OB)2 = (OA)2 + (AB)2(OB)2 = (2)2 + (1)2(OB)2 = 4 + 1(OB)2 = 5OB = 5\sqrt{5}5​ unitsWith O as centre and radius = OB, we draw an arc of a circle to meet the number line l at point P.As, OP = OB = 5\sqrt{5}5​ units, the point P will represent the number 5\sqrt{5}5​ on the number line as shown in the figure below:

Question 8

Question 1Write the following in decimal form and say what kind of decimal expansion each has :(i) 36100\dfrac{36}{100}10036​(ii) 111\dfrac{1}{11}111​(iii) 4184\dfrac{1}{8}481​(iv) 313\dfrac{3}{13}133​(v) 211\dfrac{2}{11}112​(vi) 329400\dfrac{329}{400}400329​

Accepted Answer

(i) 36100\dfrac{36}{100}10036​ = 0.36, terminating decimal expansion(ii) 111\dfrac{1}{11}111​11)0.090911)100‾11)22−99‾11x3−210011)x322−99‾11)x32222111)x3221111‾\begin{array}{l} \phantom{11)}{0.0909} \ 11\overline{\smash{\big)}\quad100\quad} \ \phantom{11)}\phantom{22}\underline{-99} \ \phantom{{11}x^3-2}100 \ \phantom{{11)}x^322}\underline{-99} \ \phantom{{11)}{x^32222}}1 \ \phantom{{11)}{x^322}}\overline{\phantom{1111}} \end{array}11)0.090911)10011)22−99​11x3−210011)x322−99​11)x32222111)x3221111​The remainder 1 keeps repeating.Hence, 111\dfrac{1}{11}111​ = 0.09‾0.\overline{09}0.09. This is a non-terminating repeating decimal expansion(iii) 418=3384\dfrac{1}{8} = \dfrac{33}{8}481​=833​8)4.1258)33‾8)−32 ‾8)−108−−8‾8)−8208−−16‾8)−82408−8−40‾8−82208−811111‾\begin{array}{l} \phantom{8)}{\enspace 4.125} \ 8\overline{\smash{\big)}\enspace 33\quad} \ \phantom{8)}\underline{-32\space} \ \phantom{8)-}10 \ \phantom{8-}\underline{-8} \ \phantom{8)-8}20 \ \phantom{8-}\underline{-16} \ \phantom{8)-82}40 \ \phantom{8-8}\underline{-40\enspace} \ \phantom{8-822}0 \ \phantom{8-8}\overline{\phantom{11111}} \end{array}8)4.1258)338)−32 ​8)−108−−8​8)−8208−−16​8)−82408−8−40​8−82208−811111​Hence, 4184\dfrac{1}{8}481​ = 4.125. This is a terminating decimal expansion.(iv) 313\dfrac{3}{13}133​8)0.23076923076913)30‾13)−26 ‾13)−408−−39‾8)−810013303−91‾13303333901330331−78‾13303333312013303333−117‾1330333333330133033333)−26‾1330333333340133033333311111‾\begin{array}{l} \phantom{8)}{\enspace 0.230769230769} \ 13\overline{\smash{\big)}\enspace 30\qquad\qquad\quad} \ \phantom{13)}\underline{-26\space} \ \phantom{13)-}40 \ \phantom{8-}\underline{-39} \ \phantom{8)-8}100 \ \phantom{13303}\underline{-91\enspace} \ \phantom{13303333}90 \ \phantom{1330331}\underline{-78\enspace} \ \phantom{133033333}120 \ \phantom{13303333}\underline{-117\enspace} \ \phantom{13303333333}30 \ \phantom{133033333)}\underline{-26\enspace} \ \phantom{13303333333}40 \ \phantom{1330333333}\overline{\phantom{11111}} \end{array}8)0.23076923076913)3013)−26 ​13)−408−−39​8)−810013303−91​13303333901330331−78​13303333312013303333−117​1330333333330133033333)−26​1330333333340133033333311111​Hence, 313\dfrac{3}{13}133​ = 0.230769‾0.\overline{230769}0.230769. This is a non-terminating repeating decimal expansion.(v) 211\dfrac{2}{11}112​8)0.181811)20‾11)−11 ‾11)−908−−88‾8)−8)201330)−11‾1330333)90133033−88‾13303333213303311111‾\begin{array}{l} \phantom{8)}{\enspace 0.1818} \ 11\overline{\smash{\big)}\enspace 20\qquad} \ \phantom{11)}\underline{-11\space} \ \phantom{11)-}90 \ \phantom{8-}\underline{-88} \ \phantom{8)-8)}20 \ \phantom{1330)}\underline{-11\enspace} \ \phantom{1330333)}90 \ \phantom{133033}\underline{-88\enspace} \ \phantom{13303333}2 \ \phantom{133033}\overline{\phantom{11111}} \end{array}8)0.181811)2011)−11 ​11)−908−−88​8)−8)201330)−11​1330333)90133033−88​13303333213303311111​Hence, 211=0.18‾\dfrac{2}{11} = 0.\overline{18}112​=0.18. This is a non-terminating repeating decimal expansion.(vi) 329400\dfrac{329}{400}400329​ = 0.8225, terminating decimal expansion400)0.8225400)3290‾400)−3200 ‾400)−900400)2−800‾400)2181000400)21−800‾400)212)2000400)21−2000‾400)211110400)211111111‾\begin{array}{l} \phantom{400)}{0.8225} \ 400\overline{\smash{\big)}\enspace 3290\quad} \ \phantom{400)}\underline{-3200\space} \ \phantom{400)-}900 \ \phantom{400)2}\underline{-800} \ \phantom{400)218}1000 \ \phantom{400)21}\underline{-800} \ \phantom{400)212)}2000 \ \phantom{400)21}\underline{-2000\enspace} \ \phantom{400)21111}0 \ \phantom{400)21}\overline{\phantom{1111111}} \end{array}400)0.8225400)3290400)−3200 ​400)−900400)2−800​400)2181000400)21−800​400)212)2000400)21−2000​400)211110400)211111111​Hence, 329400\dfrac{329}{400}400329​ = 0.8225. This is a terminating decimal expansion.

Question 9

Question 2You know that 17\dfrac{1}{7}71​ = 0.142857‾0.\overline{142857}0.142857. Can you predict what the decimal expansions of 27\dfrac{2}{7}72​, 37\dfrac{3}{7}73​, 47\dfrac{4}{7}74​, 57\dfrac{5}{7}75​, 67\dfrac{6}{7}76​ are, without actually doing the long division? If so, how?[Hint : Study the remainders while finding the value of 17\dfrac{1}{7}71​ carefully.]

Accepted Answer

Given,17\dfrac{1}{7}71​ = 0.142857‾0.\overline{142857}0.142857(i) 27\dfrac{2}{7}72​ = 2×172 	imes \dfrac{1}{7}2×71​= 2×0.142857‾2 	imes 0.\overline{142857}2×0.142857= 0.285714‾0.\overline{285714}0.285714(ii) 37\dfrac{3}{7}73​ = 3×173 	imes \dfrac{1}{7}3×71​= 3×0.142857‾3 	imes 0.\overline{142857}3×0.142857= 0.428571‾0.\overline{428571}0.428571(iii) 47\dfrac{4}{7}74​ = 4×174 	imes \dfrac{1}{7}4×71​= 4×0.142857‾4 	imes 0.\overline{142857}4×0.142857= 0.571428‾0.\overline{571428}0.571428(iv) 57\dfrac{5}{7}75​ = 5×175 	imes \dfrac{1}{7}5×71​= 5×0.142857‾5 	imes 0.\overline{142857}5×0.142857= 0.714285‾0.\overline{714285}0.714285(v)67\dfrac{6}{7}76​ = 6×176 	imes \dfrac{1}{7}6×71​= 6×0.142857‾6 	imes 0.\overline{142857}6×0.142857= 0.857142‾0.\overline{857142}0.857142

Question 10

Question 3Express the following in the form pq\dfrac{p}{q}qp​, where p and q are integers and q ≠ 0.(i) 0.6‾0.\overline{6}0.6(ii) 0.47‾0.4\overline{7}0.47(iii) 0.001‾0.\overline{001}0.001

Accepted Answer

(i) 0.6‾0.\overline{6}0.6Let x = 0.6‾0.\overline{6}0.6x = 0.666666........... (1)Here one digit 6 is repeated so we multiply both side by 10 in equation (1)10x = 6.6666.......we can write10x = 6.66666.......10x = 6 + 0.666666...........From equation (1)10x = 6 + x10x - x = 69x = 6x = 69\dfrac{6}{9}96​ = 23\dfrac{2}{3}32​Hence, 0.6‾0.\overline{6}0.6 = 23\dfrac{2}{3}32​(ii) 0.47‾0.4 \overline{7}0.47Let x = 0.47777777........ (1)Here one digit 7 is repeated so we multiply both side by 10 in equation (1)We can write10x = 4.77777..... (2)On subtracting equation (2) from equation (1)10x - x = 4.777777........ - 0.477777.........9x = 4.300000......x = 4.39\dfrac{4.3}{9}94.3​Multiplying numerator and denominator by 10 we get,x = 4.3×109×10\dfrac{4.3 	imes 10}{9 	imes 10}9×104.3×10​ = 4390\dfrac{43}{90}9043​Hence, 0.47‾0.4\overline{7}0.47 = 4390\dfrac{43}{90}9043​(iii) 0.001‾0.\overline{001}0.001Let x = 0.001001001...... (1)Here three digit 001 are repeating so we multiply both side by 1000 in equation (1)1000x = 0001.001001........ (2)On subtracting equation (2) from equation (1)1000x - x = 0001.001001...... - 0.001001......999x = 1.0000000.....x = 1999\dfrac{1}{999}9991​Hence, 0.001‾0.\overline{001}0.001 = 1999\dfrac{1}{999}9991​

Question 11

Question 4Express 0.99999 .... in the form pq\dfrac{p}{q}qp​. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Accepted Answer

Let x = 0.99999 ........ (1)Here one digit 9 is repeating so we multiply both side by 10 in equation (1)10 x = 9.99999.......... (2)By subtracting equation (2) - equation (1)10 x -x = 9.99999........ - 0.99999........9x = 9x = 99\dfrac{9}{9}99​x = 1Therefore, (0.99999...) is equivalent to (1), which might seem surprising at first glance, but it makes sense mathematically. This result can be demonstrated by the fact that any number infinitesimally close to 1 but less than 1, when infinitely added to itself, equals 1.For example, if we take (0.9) and add (0.09), we get (0.99), and if we add (0.009) to that, we get (0.999), and so on. As we keep adding these infinitesimal increments, we approach (1). So, in a way, (0.99999...) is just another way to represent the number (1).

Question 12

Question 5What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 117\dfrac{1}{17}171​ ? Perform the division to check your answer.

Accepted Answer

117\dfrac{1}{17}171​ = 0.0588235294117647‾0.\overline{0588235294117647}0.0588235294117647sixteen digit are of repeating blocks.17)0.0588235294117647‾17)100‾17))−85 ‾17)−15017))−136‾17))−14017)))−136‾13303333401330331−34‾1330333336013303333−51‾1330333333390133033333)−85‾133033333333)5013303333333−34‾1330333333333)160133033333333−153‾13303333333333337013303333333333)−68‾13303333333333333)201330333333333333)−17‾1330333333333333333)3013303333333333333))−17‾1330333333333333333)))1301330333333333333333)−119‾13303333333333333333)))11013303333333333333333)−102‾133033333333333333333333)8013303333333333333333333−68‾133033333333333333333333)12013303333333333333333333−119‾1330333333333333333333333)113303333333333333333333111111‾\begin{array}{l} \phantom{17)}{0.\overline{0588235294117647}} \ 17\overline{\smash{\big)}\enspace 100\qquad\qquad\qquad} \ \phantom{17))}\underline{-85\space} \ \phantom{17)-}150 \ \phantom{17))}\underline{-136} \ \phantom{17))-}140 \ \phantom{17)))}\underline{-136\enspace} \ \phantom{13303333}40 \ \phantom{1330331}\underline{-34\enspace} \ \phantom{133033333}60 \ \phantom{13303333}\underline{-51\enspace} \ \phantom{13303333333}90 \ \phantom{133033333)}\underline{-85\enspace} \ \phantom{133033333333)}50 \ \phantom{13303333333}\underline{-34\enspace} \ \phantom{1330333333333)}160 \ \phantom{133033333333}\underline{-153\enspace} \ \phantom{1330333333333333}70 \ \phantom{13303333333333)}\underline{-68\enspace} \ \phantom{13303333333333333)}20 \ \phantom{1330333333333333)}\underline{-17\enspace} \ \phantom{1330333333333333333)}30 \ \phantom{13303333333333333))}\underline{-17\enspace} \ \phantom{1330333333333333333)))}130 \ \phantom{1330333333333333333)}\underline{-119\enspace} \ \phantom{13303333333333333333)))}110 \ \phantom{13303333333333333333)}\underline{-102\enspace} \ \phantom{133033333333333333333333)}80 \ \phantom{13303333333333333333333}\underline{-68\enspace} \ \phantom{133033333333333333333333)}120 \ \phantom{13303333333333333333333}\underline{-119\enspace} \ \phantom{1330333333333333333333333)}1 \ \phantom{13303333333333333333333}\overline{\phantom{111111}} \end{array}17)0.058823529411764717)10017))−85 ​17)−15017))−136​17))−14017)))−136​13303333401330331−34​1330333336013303333−51​1330333333390133033333)−85​133033333333)5013303333333−34​1330333333333)160133033333333−153​13303333333333337013303333333333)−68​13303333333333333)201330333333333333)−17​1330333333333333333)3013303333333333333))−17​1330333333333333333)))1301330333333333333333)−119​13303333333333333333)))11013303333333333333333)−102​133033333333333333333333)8013303333333333333333333−68​133033333333333333333333)12013303333333333333333333−119​1330333333333333333333333)113303333333333333333333111111​

Question 13

Question 6Look at several examples of rational numbers in the form pq\dfrac{p}{q}qp​ (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Accepted Answer

Suppose, pq\dfrac{p}{q}qp​ = 25\dfrac{2}{5}52​ = 0.4or910\dfrac{9}{10}109​ = 0.9or32\dfrac{3}{2}23​ = 1.5q must contain factors of 2 or 5 or both.For example,In 78\dfrac{7}{8}87​, 8 = 2 x 2 x 2andin 710\dfrac{7}{10}107​, 10 = 2 x 5

Question 14

Question 7Write three numbers whose decimal expansions are non-terminating non-recurring.

Accepted Answer

π = 3.1415......2\sqrt{2}2​ = 1.414213......3\sqrt{3}3​ = 1.732050......

Question 15

Question 8Find three different irrational numbers between the rational numbers 57\dfrac{5}{7}75​ and 911\dfrac{9}{11}119​ .

Accepted Answer

57\dfrac{5}{7}75​ = 0.714285‾0.\overline{714285}0.7142857)0.7142857)50‾7)−49 ‾7)−108−−7‾8)−8301330−28‾13303332013303)−14‾13303333)601330333)−56‾13303333334013303333)−35‾13303333333513303333311111‾\begin{array}{l} \phantom{7)}{\enspace 0.714285} \ 7\overline{\smash{\big)}\enspace 50\qquad\quad} \ \phantom{7)}\underline{-49\space} \ \phantom{7)-}10 \ \phantom{8-}\underline{-7} \ \phantom{8)-8}30 \ \phantom{1330}\underline{-28\enspace} \ \phantom{1330333}20 \ \phantom{13303)}\underline{-14\enspace} \ \phantom{13303333)}60 \ \phantom{1330333)}\underline{-56\enspace} \ \phantom{1330333333}40 \ \phantom{13303333)}\underline{-35\enspace} \ \phantom{13303333333}5 \ \phantom{133033333}\overline{\phantom{11111}} \end{array}7)0.7142857)507)−49 ​7)−108−−7​8)−8301330−28​13303332013303)−14​13303333)601330333)−56​13303333334013303333)−35​13303333333513303333311111​911\dfrac{9}{11}119​ = 0.81‾0.\overline{81}0.8111)0.818111)90‾11)−88 ‾11)−2011)1−11‾8)−819013301−88‾133033320133033−11‾13303333913303311111‾\begin{array}{l} \phantom{11)}{\enspace 0.8181} \ 11\overline{\smash{\big)}\enspace 90\qquad} \ \phantom{11)}\underline{-88\space} \ \phantom{11)-}20 \ \phantom{11)1}\underline{-11} \ \phantom{8)-81}90 \ \phantom{13301}\underline{-88\enspace} \ \phantom{1330333}20 \ \phantom{133033}\underline{-11\enspace} \ \phantom{13303333}9 \ \phantom{133033}\overline{\phantom{11111}} \end{array}11)0.818111)9011)−88 ​11)−2011)1−11​8)−819013301−88​133033320133033−11​13303333913303311111​Irrational numbers between the rational numbers 57\dfrac{5}{7}75​ and 911\dfrac{9}{11}119​0.75075007500075000075. . .0.767076700767000767. . .0.808008000800008. . .

Question 16

Question 9Classify the following numbers as rational or irrational :(i) 23\sqrt{23}23​(ii) 225\sqrt{225}225​(iii) 0.3796(iv) 7.478478...(v) 1.101001000100001...

Accepted Answer

(i) 23\sqrt{23}23​ is a prime number. It is irrational because 23\sqrt{23}23​ does not have a complete root.(ii) 225\sqrt{225}225​ is not a prime number. 15 is root of 225\sqrt{225}225​ so it is rational.(iii) 0.3796 is terminating decimal expansion. Hence, it is rational.(iv) 7.478478... = 7.478‾7.\overline{478}7.478, it is non-terminating but repeating decimal expansion. Hence, it is rational.(v) 1.101001000100001... is non-terminating and non-repeating decimal expansion. Hence, it is irrational.

Question 17

Question 1Classify the following numbers as rational or irrational:(i) 2 − 5\sqrt{5}5​(ii) 3 + 23\sqrt{23}23​ − 23\sqrt{23}23​(iii) 2777\dfrac{2\sqrt{7}}{7\sqrt{7}}77​27​​(iv) 12\dfrac{1}{\sqrt{2}}2​1​(v) 2π

Accepted Answer

(i) 2 − 5\sqrt{5}5​If in any rational number we add, subtract, multiply or divide any irrational number it becomes irrational number.Hence, 2 − (5)\sqrt{5})5​) is an irrational number.(ii) 3 + 23\sqrt{23}23​ − 23\sqrt{23}23​= 3 + 23\sqrt{23}23​ − 23\sqrt{23}23​= 3Hence, (3 + 23\sqrt{23}23​) − 23\sqrt{23}23​ is a rational number(iii) 2777\dfrac{2\sqrt{7}}{7\sqrt{7}}77​27​​2777\dfrac{2\sqrt{7}}{7\sqrt{7}}77​27​​ = 27\dfrac{2}{7}72​It is in pq\dfrac{p}{q}qp​ where q ≠ 0.Hence, is a rational number(iv) 12\dfrac{1}{\sqrt{2}}2​1​12=rationalirrational\dfrac{1}{\sqrt{2}} = \dfrac{	ext{rational}}{\sqrt{	ext{irrational}}}2​1​=irrational​rational​ = irrational numberIf in any rational number we add, subtract, multiply or divide any irrational number it becomes irrationalHence, 12\dfrac{1}{\sqrt{2}}2​1​ is an irrational number(v) 2π2 x π = irrational number [∵ rational x irrational number = irrational number]Hence, 2π is an irrational number

Question 18

Question 2(i)Simplify the following expression:(3+3)(2+2)(3 + \sqrt{3}) (2 + \sqrt{2})(3+3​)(2+2​)

Accepted Answer

(3+3)(2+2)(3 + \sqrt{3}) (2 + \sqrt{2})(3+3​)(2+2​)= 3 x 2 + 3 x 2\sqrt{2}2​ + 2 x 3\sqrt{3}3​ + 3\sqrt{3}3​ x 2\sqrt{2}2​= 6 + 3 2\sqrt{2}2​ + 2 3\sqrt{3}3​ + 6\sqrt{6}6​Hence, (3+3)(2+2)=6+32+23+6(3 + \sqrt{3}) (2 + \sqrt{2}) = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}(3+3​)(2+2​)=6+32​+23​+6​

Question 19

Question 2(ii)Simplify the following expression:(3+3)(3−3)(3 + \sqrt{3}) (3 - \sqrt{3})(3+3​)(3−3​)

Accepted Answer

(3+3)(3−3)(3 + \sqrt{3}) (3 - \sqrt{3})(3+3​)(3−3​)= (3)2 - (3\sqrt{3}3​)2 [∵ (a + b)(a - b) = (a)2 - (b)2]= 9 - 3= 6Hence, (3+3)(3−3)(3 + \sqrt{3}) (3 - \sqrt{3})(3+3​)(3−3​) = 6

Question 20

Question 2(iii)Simplify the following expression:(5+2)(\sqrt{5} + \sqrt{2})(5​+2​) 2

Accepted Answer

(5+2)(\sqrt{5} + \sqrt{2})(5​+2​) 2= (a + b)2= a2 + 2ab + b2= (5)(\sqrt{5})(5​) 2 + 2 x (5)(\sqrt{5})(5​) x (2)(\sqrt{2})(2​) + (2)(\sqrt{2})(2​) 2 [∵ (a + b)2 = a2 + 2ab + b2]= 5 + 2 (10)(\sqrt{10})(10​) + 2= 7 + 2 (10)(\sqrt{10})(10​)Hence,(5+2)(\sqrt{5} + \sqrt{2})(5​+2​) 2 = 7 + 2 (10)(\sqrt{10})(10​)

Number Systems

Exercise 11

Exercise 12

Exercise 13

Exercise 14

Exercise 15