NCERT Solutions for Class 9 Maths Exercise 1.1 @ Question 1. 1s zero a nal number? Can you write it in the form =, where p and q are integers and * 07 Solution : Consider the definition of a rational number. Arational number is the one that can be written in the form of 2 , where p and q are integers and q q=0 Zero can be written as. a 3 So, we arrive at the conclusion that 0 can be written in the form of , where q is any integer. @ Therefore, zero is a rational number. Question 2. Find six rational numbers between 3 and 4. Solution : We know that there are infinite rational numbers between any two numbers. Arational number is the one that can be written in the form of 2 , where p and q are q Integers and g #0. We know that the numbers 3.1,3.2,3.3,3.4.3.5 and 3.6 all lie between 3 and 4. We need to rewrite the numbers 3.1,3.2,3.3,3.4,3.5 and 3.6 in 2 form to get the rational numbers between 3 and 4. @ So, after converting, we get 31 32 33 34 36 poco and 10'10'10°10"10 “" 10 34 35 ana 36into lowest fractions. 10°10"10 "* 10 We can further convert the rational numbers On converting the fractions into lowest fractions, we get 16 177 438 Therefore, six rational numbers between 3 and 4 areQuestion 3. Find five rational numbers between : We know that there are infinite rational numbers between any two numbers. Arational number is the one that can be written in the form of, where p and q are q Integers andg +0 We know that the numbers 0.61,0.62, 0.63,0.64 and 0.65 can also be written as0.6 and 0.8. We can conclude that the numbers 0.61,0.62, 0.63, 0.64 and 0.65 all lie between0.6 and 0.8 We need to rewrite the numbers 0.61,0.62, 0.63, 0.64 and 0.65 in 2 form to get the rational numbers between 3 and 4. @ So, after converting, we get 1 8 8 64, 6 100'100'100'100 “100 We can further convert the rational numbers £2. ang ®° into lowest fractions. 100'100 “"* 700 (On converting the fractions, we get 16 ang 23, 30°25" 20 Therefore, six rational numbers between 3 and 4 are 61 31 63 16 | 13 100°50°100°25 “30 Question 4. State 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. (il) Every rational number is a whole number. Solution : (i) Consider the whole numbers and natural numbers separately. 4,5. We know that whole number series is 0.1.2, We know that natural number series is 1, 2.3.4.5. So, we can conclude that every number of the natural number series lie in the whole number series. Therefore, we conclude that, yes every natural number is a whole number.(ii) Consider the integers and whole numbers separately. We know that integers are those numbers that can be written in the form of = , where q Now, considering the series of integers, we have —4,-3,-2,-1,0,1,23.4 We know that whole number series is 0,1.2,3.4.5 We can conclude that all the numbers of whole number series lie in the series of integers. But every number of series of integers does not appear in the whole number series. Therefore, we conclude that every integer is not a whole number. (iii) Consider the rational numbers and whole numbers separately. We know that rational numbers are the numbers that can be written in the form , whereg = 0. We know that whole number series is 0,1,2,3.4,5 We know that every number of whole number series can be written in the form of 2as @ 4 1 We conclude that every number of the whole number series is a rational number. But, every rational number does not appear in the whole number series. Therefore, we conclude that every rational number is not a whole number. NCERT Solutions for Class 9 Maths Exercise 1.2 Question 1. State whether the following statements are true or false. Justify your answers. (i) Every irrational number is a real number. (ji) Every point on the number line is of the form fm , where m is a natural number. (ii) Every real number is an irrational number. Solution : (i) Consider the irrational numbers and the real numbers separately. We know that irrational numbers are the numbers that cannot be converted in the form, where p q and q are integers andg =0. We know that a real number is the collection of rational numbers and irrational numbers. Therefore, we conclude that, yes every irrational number is a real number,(ii) Consider a number line. We know that on a number line, we can represent negative as well as positive numbers. We know that we cannot get a negative number after taking square root of any number. Therefore, we conclude that not every number point on the number line is of the form Jn, where m is a natural number. (iii) Consider the irrational numbers and the real numbers separately. We know that irrational numbers are the numbers that cannot be converted in the form =, where p q and q are integers andg =0. We know that a real number is the collection of rational numbers and irrational numbers. ‘So, we can conclude that every irrational number is a real number. But every real number is not an irrational number. Therefore, we conclude that, every real number is not a rational number. Question 2, Are 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. Solution : We know that square root of every positive integer will not yield an integer. We know that ff is 2, which is an integer. But,./7 or 0 will give an irrational number. Therefore, we conclude that square root of every positive integer is not an irrational number. Question 3. Show how \5 can be represented on the number line Solution : Draw a number line and take point O and A on it such that OA = 1 unit, Draw BA | OAas BA=1 unit. Join OB = 2 units. Now draw BB1 1 OB such that BB1 =1 unit. Join OB1 = V3 units. Next, draw B1B21 OB1such that B1B2 = 1 unit. Join OB2 = units. Again draw B2B3 LOB2 such that B2B3 = 1 unit. Join OB3 = V5 units. oO A 2D ‘Take O as centre and OB3 as radius, draw an arc which cuts the number line at D. PointD represents 5 on the number line.NCERT Solutions for Class 9 Maths Exercise 1.3 Question 1. Write the following in decimal form and say what kind of decimal expansion each has: [5 Solution : oT On dividing 36 by 100, we get Therefore, we conclude tha 3S 36, which is a terminating decimal. a 1 wo) On dividing 1 by 11, we get=99 1 We can observe that while dividing 1 by 11, we got the remainder as 1, which will continue to be 1 Therefore, we conclude thats 33 _ 4.125, which is a non-terminating decimal and recurring decimal (ii) 44-38 8 8 On dividing 33 by 8, we get 4.125 sy) 33 32 We can observe that while dividing 33 by 8, we got the remainder as 0. 33 Therefore, we conclude thats 4.125, which is a terminating decimal. (iv) Ble On dividing 3 by 13, we get0.230769, 13 We can observe that while dividing 3 by 13 we got the remainder as 3, which will continue to be 3 after carrying out 6 continuous divisions. Therefore, we conclude that 230769... or 2 = (On dividing 2 by 11, we get 0.1818 ul 2 We can observe that while dividing 2 by 11, first we got the remainder as 2 and then 9, which will continue to be 2 and 9 alternately. Therefore, we conclude that 2 = 0.1818... or = =018 1, hich is a non-terminating decimal and recurring decimal, 329 Wi) 5 On dividing 329 by 400, we get 0.9225 400) 329 a 3290 We can observe that while dividing 329 by 400, we got the remainder as 0. Therefore, we conclude that * =0.8225, which is a terminating decimal. Question 2. You know that Z 0.142857...... Can you predict what the decimal expansions of are, without actually doing the long division? If so, how? [Hint: Study the remainders while finding the value of 4 carefully] Solution : We are given that 2 079857 or 1 =0.142887 axt 3x 7 On substituting value of Las 0.142857...., we get5x0.142857.....= 0.285714 3x0.142857.... = 0428571 4och = 4x0.142857.. = 0.571428 6x 0.142857. A = 0.857142 Therefore, we conclude that, we can predict the values of 2345 26 ani Question 3. Express the following in the form 2, where p and q are integers and q=0. q () 06 (li) 0.47 (it) 0.04 Solutio () Letx=08 > x=0.6666.....(a) We need to multiply both sides by 10 to get 10x = 6.6666. @) We need to subtract (a)from (b), to get 10x = 6.6666 ~x= 0.6666... 9x=6We can also write9x=6 as x 2 Therefore, on converting 9 in the= form, we get the answer a > q@ (li) Letxe= 0.47 => x=047777...(a) We need to multiply both sides by 10 to get 10x=47 @) We need to subtract (a)from (b), to get Therefore, on converting 47 in the= form, we get the answer as 2 q (iil) Let x=0.001 => x=0.001001......(@) We need to multiply both sides by 1000 to get 1000x=1.001001 0) We need to subtract (a)from (b), to get 1000x=1.001001 —x=0.001001... 999x=1 We can also write 999=1 as x Therefore, on convertingg 00] in the 2 form, we get the answer as 4 q Question 4. Express 0.99999... in the form . Are you surprised by your answer? Discuss q why the answer makes sense with your teacher and classmates. Solution : Let x=0.99999....(a) We need to multiply both sides by 10 to get10x =9.9999....() We need to subtract (a)from (b), to get 10x= 9.99999 = x= 0.99999. 9x=9 We can also write 9x=9 9 asx=— orx=1 9 Therefore, on converting 99999. in the= form, we get the answer as 1 q@ Yes, at a glance we are surprised at our answer. But the answer makes sense when we observe that 0.9999. goes on forever. SO there is not gap between 1 and 0.9999........ and hence they are equal. Question 5. What can the maximum number of digits be in the recurring block of digits in the decimal expansion ofl? Perform the division to check your answer. Solution : We need to find the number of digits in the recurring block oe. Let us perform the long division to get the recurring block of. We need to divide 1 by 17, to get We can observe that while dividing 1 by 17 we got the remainder as 1, which will continue to be 1 after carrying out 16 continuous divisions. Therefore, we conclude that 1 0.058823529411764 0.0588235294117647 , Which is a non-terminating decimal and recurring decimal. Question 6. Look at several examples of rational numbers in the form = (q #0), where p and q qare integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy? Solution : Let us consider the examples of the form= that are terminating decimals. qWe can observe that the denominators of the above rational numbers have powers of 2, 5 or both. Therefore, we can conclude that the property, which q must satisfy in , so that the rational q number is a terminating decimal is that q must have powers of 2, 5 or both. q Question 7. Write three numbers whose decimal expansions are non-terminating non- recurring. Solution : The three numbers that have their expansions as non-terminating on recurring decimal are given below. 0.04004000400004 0.07007000700007 .... 0.013001300013000013 5 Question 8. Find three different irrational numbers between the rational numbers = Solutio: 5 Let us convert = and xin decimal form, to get = 0.714285....and Fr osisisi Three irrational numbers that lie between0.714285.... and 0.818181....are: 0.73073007300073 .... 0.74074007400074 0,76076007600076 Question 9. Classify the following numbers as rational or irrational : (i) 23 (ii) 225 (ii) 0.3796(iv)7.478478 (v)1.101001000100001 We know that on finding the square root of 23, we will not get an integer. Therefore, we conclude that /33 is an irrational number. (ii) 235 We know that on finding the square root of 225, we get 15, which is an integer. Therefore, we conclude that 375 is a rational number. (ii) 0.3796 We know that 0.3796 can be converted into”. q While, converting 0.3796 into form, we get q 96 = 295 10000 3796 949 The rational number. can be converted into lowest fractions, to get —— . 10000 2300 We can observe that 0.3796 can be converted into a rational number. Therefore, we conclude that 0.3796 is a rational number. (iv) 7.478478... We know that 7.478478... is a non-terminating recurring decimal, which can be converted into= @ form. While, converting 7.478478... into” form, we get q x= 7AT8478 (a) 1000x=7478.478478......(5) While, subtracting (a) from (b), we get1000x=7478.478478. = 7.478478. 7471 999% We know that999x = 7471 can also be written asx a Therefore, we conclude that 7.478478... is a rational number. (v)1.101001000100001 We can observe that the number 1.101001000100001.... is a non-terminating on recurring decimal We know that non-terminating and non-recurring decimals cannot be converted into” form q Therefore, we conclude that 1.101001000100001.... is an irrational number. NCERT Solutions for Class 9 Maths Exercise 1.4 Question 1. Visualise 3.765 on the number line, using successive magnification. Solution : 3.765 lies between 3 and 4. 331 32 33 34 35 36 37 38 39 4 (i) 3.7 lies between 3 and 4 37.aM-TP 473 3.74 3.75 3763.77 3.78 7B. , PDR? AS OM BTS 378 3,77 3.78 37% G8 ii) 3.76 lies between 3.7 and 3.8 3.26 --"5.762 3.764 3,766 3.768 ~S3.77 37613763 3768 3767 3 769 (iii) 3.765 lies between 3.76 and 3.77 Question 2. Visualise 4.35 on the number line, upto 4 decimal places. Solution : 4.26 or 4.2626 lies between 4 and 5. 4 41 42 43 44 45 46 47 48 49 5 a a ee (i) 4.2 lies between 4 and 5 (ii) 4.26 lies between 4.2 and 4.3 426.--"4262 4.264 4266 4.268 ~~ 4.27 tet = (iii) 4.262 lies between 4.26 and 4.27 426° 42637-4265 4.267 4.269 4262, 4.2622, 42624 4.2626, 42626-4263 (iy) 4.2626 lies between 4.262 and 4.263 ee 4.2621 4.2623 4.2625 4.2627 4.2629NCERT Solutions for Class 9 Maths Exercise 1.5 Question 1. Classify the following numbers as rational or irrational: () 2-5 (ii) (3+ 23) V3 i) 07 ( a We know that 236...., which is an irrational number. =-0.236. which is also an irrational number. Therefore, we conclude that — J/§ is an irrational number. (i) (3+ 23) V3 Therefore, we conclude that(3 + 4/23 | - 23 is a rational number. ir «iy 287 Wi We can cancel, 7 in the numerator and denominator, as. well as denominator, to get ai WF is the common number in numerator aswi = is a rational number. Wi Therefore, we conclude that. 1 (iv) a We know that yi =1.414..... which is an irrational number We can conclude that, when 1 is divided by ,f2,, we will get an irrational number. 1 Therefore, we conclude that—= is an irrational number. (v) 2 We know that =3.1415..... which is an irrational number. We can conclude that 2 will also be an irrational number. Therefore, we conclude that 2 is an irrational number. Question 2. Simplify each of the following expressions () G+ 8) @ + V2) (i) 6 + 43) @ - V3) (iii) (V5 + V2) : (iv) (V5 - V2) (v5 + V2) Solutio (i) (3 + ¥3)(2 + V2) = 2(3 + V3) + V2(3 + V3) =6 + 2V3 +3\2+ V6 Thus, (3 + V3)(2 + V2) = 6 + 2\3 + 3y2 + V6 (ii) (3 + V3)(3 — V3) = (8)2 - (V3)2 =9-3=6 Thus, (3 + V3)(3 - ¥3) = 6 | ; (ii) (5 + V2)2 = (915)2 + (V2)2 + 2(V5)(v2) =5+242\10=7+2V10 Thus, (v5 + V2 )2=7 +210 (iv) (5 ~ V2)(V5 + V2) = (V5)2 - (V2)2 = 5-2=3 Thus, (v5 — V2) (V5 + V2) =3 Question 3. Recall, 1 is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is 1 = §. This seems to contradict the fact that n is irrational. How will you resolve this contradiction? Solution : When we measure the length of a line with a scale or with any other device, we only get an approximate ational value, i.e. c and d both are irrational :.$ is irrational and hence Tr is irrational. Thus, there is no contradiction in saying that itis irrationalQuestion 4. Represent /J.3on the number line. Solution : Draw a line segment AB = 9.3 units and extend it to C such that BC = 1 unit. Find mid point of AC and mark it as O. Draw a semicircle taking O as centre and AO as radius. Draw BD 1 AC. Draw an arc taking B as centre and BD as radius meeting AC produced at E such that BE = BD = /9Bunits. D BICE sl x Question 5. Ratfonatige the denominator of the following ‘a jy) 2 OF OF tele Wey Solution : tet 7 -_-2 5-2 _ 5-2 SBR 5-2 tv) 1,742 7a “W242 ~ V7 +2 ¥7+2_ 742 (7? -(2y NCERT Solutions for Class 9 Maths Exercise 1.6 Question 1. Find : (633We know that a? =a, wherea>0. We conclude that, can also be written as Therefore, the value of ¢ ,;will be 8. (ii) 35 We know that a® = 4a, where a >0. We conclude that»; can also be written as Therefore, the value of , will be 2. (i). 952 We know that Ya, where a>0. We conclude that, »,3can also be written as 5x5x5 Question 2. Find: (giWe know that a? =a, wherea>0. We know that Ya, where a>0. We conclude that, can also be written as =2x2 =4 Therefore, the value of ,:will be 4. (il) yg We know thatYa, where a>0. We conclude that, ,3 can also be written as im 7 an also be written as —r,, orf _1_ pst (as) We conclude that, 5. We know that Ya, where a>0. 1 *can also be written as We know that! = ns} Therefore, the value of , Question 3. Simplify : 93(i) (+) We know that” (7) We conclude that 3) can also be written as We know that We conclude that1 7 in 1 1 1 Therefore, the value of — will be; 3. iF We know that a™-b" =(axb)” We can conclude that 1 1 73.87 =(7x8)? =(56)3 Therefore, the value of 5 will be(56)3-