Exercise 1.1

Q.1     Is zero a rational number ? Can you write it in the form pq, where p and q are integers and q≠0?
Sol.

Yes, zero is a rational number.
Zero can be written in any of the following forms :
01,0−1,02,0−2 and so on.
Thus, 0 can be written as pq, where p = 0 and q is any non- zero integer.
Hence , 0 is a rational number.

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Q.2      Find six rational numbers between 3 and 4.
Sol.

We know that between two rational numbers x and y, such that x < y, there is a rational number x+y2. That is , x<x+y2<y
There are two rational numbers 3 and 4 , such that 3 < 4 , a rational number between 3 and 4 is 12(3+4)i.e.,72
Therefore 3<72<4
Now ,two rational numbers 3 and72 such that, a rational number between 3 and 72is
12(3+72)=12×6+72=134
There are two rational numbers72 and 4 , a rational number between 72 and 4 is
12(72+4)=12×7+82=154
Therefore 3<134<72<154<4
There are two rational numbers 3 and 134  ,a rational number between 3 and 134is
12(3+134)=12×12+134=258
There are two rational numbers154 and 4 , a rational number between 154 and 4 is
12(154+4)=12×15+164=318
There are two rational numbers 318 and 4 , a rational number between 318 and 4 is
12(318+4)=12×31+328=6316
Therefore 3<258<134<72<154<318<6316<4
Hence, six rational numbers between 3 and 4 are :
258,134,72,154,318and6316
ALTERNATIVE METHOD
Since we want 6 rational number between 3 and 4, so we write
3=3×71×7=217 and4=4×71×7=287
We know that 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28
⇒ 217<227<237<247 <257<267<277<287
Hence, six rational numbers between 3=217and4=287are
227,237,247,257,267 and277

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Q.3      Find five rational numbers between 35and45.
Sol.

Since we want 5 rational numbers between 35and45, so we write.
35=3×65×6=1830 and 45=4×65×6=2430
We know that 18 < 19 < 20 < 21 < 22 < 23 < 24
⇒    1830<1930<2030<2130<2230 <2330<2430
Hence , 5 rational numbers between 35=1830and45=2430are:
1930,2030,2130,2230,2330and2430

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Q.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.
                (iii) Every rational number is a whole number.
Sol.

(i) True : Every natural number lies in the collection of whole numbers.
(ii) False : -3 is not a whole number.
(iii) False 35 is not a whole number.

Exercise 1.2

Q.1      State 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.
Sol.

(i)     Every irrational number is a real number – True
         Justification : – A real number is either rational or irrational.
(ii)   Every point on the number line is of the form m−−√  – False
         Justification : – Numbers  of other types also lie on the number line.
(iii)  Every real number is an irrational number – False
          Justification : – Rational numbers are also real numbers.

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Q.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 numbers.
Sol.

No, the square roots of all positive integers are not irrational.
Example : –  4 is a positive integer but 4–√=2 is a  natural number.

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Q.3       Show how 5–√ can be represented on the number line.
Sol.

We shall now show how to represent 5–√ on the number line.
We first represent 5–√ on the number line l. We construct a right – angled Δ OAB, right – angled at A such that OA = 2 and AB = 1  unit (see figure)

Then,  OB=OA2+AB2−−−−−−−−−−√=4+1−−−−√=5–√
Now, we cut off a length OC=OB=5–√ on the number line.
Then the point C represents the irrational number 5–√.

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Q.4      Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP1 of unit  length (see figure). Now draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3. Continuing in the manner, you can get the line segment Pn−1Pn by drawing a line segment of unit length perpendicular to OPn−1. In this manner, you will have created the points P2,P3....Pn,.... and joined them to create a beautiful spiral depicting 2–√,3–√,4–√,.....

Sol.        Classroom activity – Do as directed.

Exercise 1.3

Q.1     Write the following in decimal form and say what kind of decimal expansion each has :
           (i) 36100                           
           (ii) 111            
           (iii) 418
           (iv) 313                          
           (v) 211             
           (vi) 329400
Sol. 

(i)    36100 in decimal form –
         36100=0.36,, terminating decimal.
(ii)   By long division,we have

                            Therefore 111=0.090909.....=0.09¯¯¯¯¯, non- terminating and repeating decimal.
                  (iii)   418=4×8+18=338 By long division , we have

                               Therefore 338=4.125, terminating decimal.
                       (iv)  By long division, we have

                                   Therefore 313=0.23076923..=0.230769¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯,
                                   non- terminating and repeating decimal.  
                           (v)   By long division we have

                                    Therefore 211=0.181818..=0.18¯¯¯¯¯,
                                    non- terminating and repeating decimal.                 
                            (vi)  By long division, we have

                                        Therefore 329400=0.8225, terminating decimal.

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Q.2         You know that 17=0.142857¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ . Can you predict what the decimal expansions of 27,37,47,57,67 are , without actually doing the long division? If so, how?
Sol.

Yes. All of the above will have repeating decimals which are permutations of 1, 4, 2, 8, 5, 7.
For example, here

To find 27, locate when the remainder becomes 2 and respective quotient (here it is 2), then write the new quotient beginning from there (the arrows in the above division) using the repeating digits 1, 4, 2, 8, 5, 7. Therefore 27=0.285714¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Similarly,      37=0.428571¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯,47=0.571428¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
57=0.714285¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯,67=0.857142¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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Q.3      Express the following in the form pq, where p and q are integers and q≠0..
            (i) 0.6¯¯¯                
            (ii) 0.47¯¯¯              
            (iii) 0.001¯¯¯¯¯¯¯¯
Sol.

(i)    Let x=0.6¯¯¯
        Then, x = 0.666 … … (1) Here , we have only one repeating digit.
        So, we multiply both sides of (1) by 10 to get
        10x = 6.66…. … (2)
        Subtracting (1) from (2) we get
        10x – x = (6.66…) – (0.66…) ⇒ 9x = 6 ⇒ x=23
         Hence, 0.6¯¯¯=23
(ii)   Let x=0.47¯¯¯
         Clearly, there is just one digit
         on the right side of the decimal point which is without bar.
         So, we multiply both sides by 10.
         So that only the repeating decimal is left on the right side of the decimal point.
         Therefore 10x=4.7¯¯¯
         ⇒ 10x=4+0.7¯¯¯
         ⇒ 10x=4+79
         For example : 0.7¯¯¯=79,0.35¯¯¯¯¯=3599 etc.
         ⇒ 10x=4×9+79 ⇒ 10x=439
         ⇒ x=4390
         Hence 0.47¯¯¯=4390
         ALITER
         Let x=0.47¯¯¯=0.4777…
         Therefore 10x = 4.777… … (1)
          and 100x = 47.777… … (2)
          Subtracting (1) from (2) we get
          100 x – 10 x = (47.777 … ) – (4.777… )
          ⇒ 90x = 43 ⇒ x=4390
          Hence, 0.47¯¯¯=4390
(iii)   Let x=0.001¯¯¯¯¯¯¯¯
          ⇒ x = 0.001001001 …. … (1)
          Here , we have three repeating digits after the decimal point. So we multiply (1) by 103=1000,
          we get
          1000x = 1.001001… … (2)
          Subtracting (1) from (2) we get –
          1000 x – x = (1.001001…) – (0.001001…)
          ⇒ 999x = 1 ⇒ x=1999
          Hence, 0.001¯¯¯¯¯¯¯¯=1999

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Q.4      Express 0.99999… in the form pq. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Sol.

Let x = 0.9999 … … (1)
Here , we have only one repeating digit. So, we multiply both sides of (1) by 10 to get.
10x = 9.999… … (2)
Subtracting (1) from (2) we get  
10 x – x = (9.999…) – (0.999…)
⇒ 9x = 9 ⇒ x = 1
Hence, 0.9999 … = 1
Since , 0.9999… goes on forever. So, there is no gap between 1 and 0.9999… and hence they are equal.

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Q.5     What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 117? Perform the division to check your answer.
Sol.         We have

Thus, 117=0.588235294117647¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Therefore the maximum number of digits in the quotient while computing 117 are 15.

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Q.6       Look at several examples of rational numbers in the form pq(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?
Sol. 

Consider several rational numbers in the form pq(q≠0), where p and q are integers with common factors other than 1 and having terminating decimal representation.
Let the various such rational numbers be 12,14,78,3725,
8125,1720,3116etc.
In all cases, we think of the natural number which when multiplied  by their respective denominators gives 10 or a power of 10.
12=1×52×5=510=0.5 [Since 2 × 5 = 10]
14=1×254×25=25100=0.25 [Since 4 ×25 = 100]
78=7×1258×125=8751000=0.875 [Since 8 × 125 = 1000]
3725=37×425×4=128100=1.28 [Since 25 × 4 = 100]
8125=8×8125×8=641000=0.064 [Since 125 × 8 = 1000]
1720=17×520×5=85100=0.85 [Since 20 × 5 = 100]
3116=31×62516×625=1937510000=1.9375 [Since 16 × 625 = 10000]
We have seen that those rational numbers whose denominators when multiplied by a suitable integer produce a power of 10 are expressible in the finite decimal form. But this can always be done only when the denominator of the given rational number has either 2 or 5 or both of them as the only prime factors. Thus, we obtain the following property :
If the denominator of a rational number in standard form has no prime factors other than 2 or 5, then and only then it can be represented as a terminating decimal.

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

Three numbers whose decimal representations are non- terminating non- recurring are
2–√,3–√and5–√ or we can say 0.100100010001…, 0.20200200020002… and .003000300003.

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Q.8       Find three different irrational numbers between the rational numbers 57and911.
Sol.          We have

Therefore 57=0.714285¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯and911=0.81¯¯¯¯¯
Thus , three different irrational numbers between 57and911 are.
0.75075007500075000075…. , 0.767076700767000… and 0.80800800080000…

 

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Q.9          Classify the following numbers as rational or irrational :
                (i) 23−−√                      
                (ii) 225−−−√                            
                (iii) 0.3796
                (iv) 7.478478 …                          
                (v) 1.101001000100001…
Sol.

(i) 23−−√ is an irrational number as 23 is not a perfect square.
(ii) 225−−−√=3×3×5×5−−−−−−−−−−−√ 
=3×5
=15
Thus, 15 is a rational number.
(iii) 0.3796 is a rational number as it is terminating decimal.
(iv) 7.478478 …. is non- terminating but repeating , so, it is a rational number.
(v) 1.101001000100001 … is non – terminating and non- repeating so, it is an irrational number.

Exercise 1.4

Q.1       Visualise 3.765 on the number line, using successive magnification.
Sol.

We know that 3.765 lies between 3 and 4, that is, in the interval [3, 4] to have a rough idea where it is. Now, we divide the interval [3,4] into 10 equal parts and look at [3.7, 3.8] through a magnifying glass and realize that 3.765 lies between 3.7 and 3.8 (see figure (i)). Now, we imagine that each of the new  intervals [3.1, 3.2], [3.2, 3.3], … [3.9, 4] has been sub divided into 10 equal parts. As before , we can now visualize through the magnify glass that 3.765  lies in the interval [3.76, 3.77] (see figure (ii)).

So, we have seen that it is possible by sufficient successive magnifications to visualize the position (or representation of a real number with a terminating decimal expansion on the number line. Let us now try and visualize the position (or representation) of a real number with a non- terminating recurring  decimal expansion on the number line. We can look at appropriate intervals through a magnifying glass and by successive magnifications visualize the  position of the number on the number line.

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Q.2       Visualise 4.26¯¯¯¯¯ on the number line, upto 4 decimal places.
Sol.

We proceed by successive magnifications, and successively decrease the lengths of the intervals in which 4.26¯¯¯¯¯ is located. 4.26¯¯¯¯¯  is located in the interval [4, 5] of length 1. We further locate 4.26¯¯¯¯¯ in the interval [4.2, 4.3] of length 0.1. To get more accurate visualization of the representation, we divide even this interval into 10 equal parts and use a magnifying glass to visualize that 4.26¯¯¯¯¯ lies in the interval [4.26, 4.27] of length 0.01. To visualize 4.26¯¯¯¯¯ in an interval of length 0.001, we again divide each of the new intervals  into 10 equal parts and visualize the representation of 4.26¯¯¯¯¯  in the interval [4.262, 4.263] of length 0.001. Notice that 4.26¯¯¯¯¯ is located closer to 4.263 than to 4.262.
Note : – We can proceed endlessly in this manner and simultaneously imagining the decrease in the length of the interval in which  4.26¯¯¯¯¯ is located.

Exercise 1.5

Q.1      Classify the following numbers as rational or irrational :
               (i) 2−5–√                                      
            (ii) (3+23−−√)−23−−√
               (iii) 27√77√
            (iv) 12√
               (v) 2π
Sol.

(i) 2−5–√ is an irrational number being a difference between a rational and an irrational.
(ii) (3+23−−√)−23−−√=3+23−−√−23−−√=3, which is a rational number.
(iii) 27√77√=27, which is a rational number.
(iv) −12√ is an irrational number being the quotient of a rational and an irrational number .
(v) 2π is irrational being the product of rational and irrational number .

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Q.2      Simplify each of the following expressions :
             (i) (3+3–√)(2+2–√)
             (ii) (3+3–√)(3−3–√)
             (iii) (5–√+2–√)2
             (iv) (5–√−2–√)(5–√+2–√)
Sol.

(i)   (3+3–√)(2+2–√)
        =3×2+32–√+23–√+2–√×3–√
        =6+32–√+23–√+6–√
(ii)  (3+3–√)(3−3–√)
        = (3)2−(3–√)2
        =9−3=6
(iii) (5–√+2–√)2
         (5–√)2+25–√2–√+(2–√)2
         5+210−−√+2=7+210−−√
(iv)   (5–√−2–√)(5–√+2–√)
         = (5–√)2−(2–√)2
         =5−2=3

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Q.3      Recall , π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is π=cd. This seems to  contradict the fact that π is irrational. How will you resolve this contradiction?
Sol.         There is no contradiction as either c or d irrational and hence π is an irrational number.

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Q.4      Represent 9.3−−−√ on the number line.
Sol.

Mark the distance 9.3 units from a fixed point A on a given line to obtain a point B such that AB = 9.3 units. From B, mark a distance of 1 unit and mark the new point as C. Find the mid- point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D.

Then BD=9.3−−−√. To represent 9.3−−−√ on the number line. Let us treat the line BC as the number line, with B as zero, C as 1 and so on. Draw an arc with centre B and radius BD, which intersects the number line in E. Then E represent 9.3−−−√

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Q.5      Rationalise the denominators of the following :
            (i) 17√
            (ii) 17√−6√
               (iii) 15√+2√
            (iv) 17√−2
Sol.

(i)    17√=1×7√7√×7√=7√7
(ii)   17√−6√ =17√−6√×7√+6√7√+6√ =7√+6√(7√)2−(6√)2
         =7√+6√7−6=7–√+6–√
(iii)  15√+2√ =15√+2√×5√−2√5√−2√ =5√−2√(5√)2−(2√)2
          =5√−2√5−2=5√−2√3
(iv) 17√−2 =17√−2×7√+27√+2
        =7√+2(7√)2−(2)2
        =7√+27−4 =7√+23

Exercise 1.6

Q.1     Find
              (i) 6412   (ii) 3215   (iii)12513
Sol.

(i)    6412 =(8×8)12 =(82)12 =82×12=81=8
(ii)   3215 =(2×2×2×2×2)15=(25)15 =25×15=21=2
(iii)  12513 =(5×5×5)13=(53)13 =53×13=51=5

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Q.2       Find :
             (i) 932   (ii) 3225   (iii) 1634   (iv) 125−13
Sol.

(i)    932 =(3×3)32=(32)32=32×32 =33=3×3×3=27
(ii)   3225=(2×2×2×2×2)25=(25)25 25×25=22=2×2=4
(iii)  1634=(2×2×2×2)34=(24)34 =24×34=23=2×2×2=8
(iv)    125−13=112513 =1(5×5×5)13=15

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Q.3      Simplify :
            (i) 223.215    
            (ii) (133)7
            (iii) 11121114
            (iv) 712.812
Sol.

(i)    223.215=223+15=210+315=21315
(ii)   (132)7=(1)732×7=1321
(iii)  11121114=1112−14=1114
(iv)   712.812=(7×8)12=5612 = (4×14)12=(56)12