Real Numbers : Exercise 1.4 (Mathematics NCERT Class 10th)



CLICK HERE to watch second part

Q.1     Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non - terminating repeating decimal expansion :
          (i) {{13} \over {3125}}                                   (ii) {{17} \over 8}
          (iii) {{64} \over {455}}                                   (iv) {{15} \over {1600}}
          (v) {{29} \over {343}}                                    (iv) {{23} \over {{2^3}{5^2}}}
         (vii) {{129} \over {{2^2}{5^7}{7^5}}}             (viii) {6 \over {15}}
         (ix) {{35} \over {50}}                                      (x) {{77} \over {210}}

Sol.     We know that if the denominator of a rational number has no prime factors other than 2 or 5, then it is expressible as a terminating, otherwise it has non - terminating  repeating decimal representation. Thus, we will have to check the prime factors of the denominators of each of the given rational numbers.  
         
(i) In {{13} \over {3125}}, the denominator is 3125.
19
          We have, 3125 = 5 × 5 × 5 × 5 × 5.
          Thus, 3125 has 5 as the only prime factor.
           Hence, {{13} \over {3125}} must have a terminating decimal representation.

          (ii) In {{17} \over 8}, the denominator is 8.
20
          We have, 8 = 2 × 2 × 2
          Thus, 8 has 2 as the only prime factor.
          Hence, {{17} \over 8} must have a terminating decimal representation.

         (iii) In {{64} \over {455}}, denominator is 455. We have, 455 = 5 × 7 × 13
         Clearly, 455 had prime factors other than 2 and 5. So, it will not have a terminating decimal representation.

         (iv) In {{15} \over {1600}}, the denominator is 1600.
22
        We have, 1600
        = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
       Thus, 1600 has only 2 and 5 as prime factors.
       Hence, {{15} \over {1600}} must have a terminating decimal representation.

      (v) In {{29} \over {343}}, the denominator is 343.
23

      We have, 343 = 7 × 7 × 7
      Clearly, 343 has prime factors other than 2 and 5.
      So, it will not have terminating decimal representation.

      (vi) In {{23} \over {{2^3}{{.5}^2}}} Clearly, the denominator {2^3}{.5^2} has only 2 and 5 as prime factors.
       Hence, {{23} \over {{2^3}{{.5}^2}}} must have a terminating decimal representation.

       (vii) In {{129} \over {{2^2}{{.5}^7}{{.7}^5}}} Clearly, the denominator {{2^2}{{.5}^7}{{.7}^5}} has prime factors other than 2 and 5.So, it will not have terminating decimal representation.

       (viii) In {6 \over {15}}, we have 15 = 3 × 5
       Clearly, 15 has prime factors other than 2 and 5. So, it will not have terminating decimal representation.

      (ix) In {{35} \over {50}} , we have 50 = 2 × 5 × 5 The denominator has only 2 and 5 as prime factors. Hence, {{35} \over {50}} must have a terminating decimal representation.

      (x) In {{77} \over {210}}, the denominator is 210.
24
      We have, 210 = 2 × 3 × 5 × 7
      Clearly, 210 has prime factors other than 2 and 5.
      So, it will not have terminating decimal representation.


Q.2     Write down the decimal expansion of those rational numbers in Question 1 above which have terminating decimal expansions.
Sol.      (i) {{13} \over {3125}} = {{13} \over {5 \times 5 \times 5 \times 5 \times 5}}
             = {{13 \times 2 \times 2 \times 2 \times 2 \times 2} \over {5 \times 2 \times 5 \times 2 \times 5 \times 2 \times 5 \times 2 \times 5 \times 2}}
            = {{13 \times 32} \over {10 \times 10 \times 10 \times 10 \times 10}} = {{416} \over {100000}} = 0.00416

           (ii) {{17} \over 8} = {{17 \times {5^3}} \over {{2^3} \times {5^3}}} = {{17 \times {5^3}} \over {{{10}^3}}} = {{17 \times 125} \over {{{10}^3}}}
            = {{2125} \over {1000}} = 2.125

          (iii) Non - terminating.

          (iv) {{15} \over {1600}} = {{15} \over {{2^6} \times {5^2}}} = {{15} \over {{2^4} \times {2^2} \times {5^2}}}
         = {{15} \over {{2^4} \times {{10}^2}}} = {{15 \times {5^4}} \over {{2^4} \times {5^4} \times {{10}^2}}}
         = {{15 \times 625} \over {{{10}^4} \times {{10}^2}}} = {{9375} \over {1000000}} = 0.009375

          (v) Non - terminating.

          (vi) {{23} \over {{2^3}{{.5}^2}}} = {{23} \over {{{2.2}^2}{{.5}^2}}} = {{23} \over {{{2.10}^2}}} = {{23 \times 5} \over {2 \times 5 \times {{10}^2}}}
        = {{115} \over {10 \times {{10}^2}}} = {{115} \over {1000}} = 0.115

         (vii) Non - terminating.

         (viii) {6 \over {15}} = {2 \over 5} = {4 \over {10}} = 0.4

         (ix) {{35} \over {50}} = {{35 \times 2} \over {50 \times 2}} = {{70} \over {100}} = 0.70

         (x) Non - terminating.


Q.3     The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form {p \over q}, what can you say about the prime factors of q ?
         (i) 43.123456789      (ii) 0.120120012000120000.......      (iii)
43.\overline {123456789}

Sol.     (i) 43.123456789 is terminating.
           So, it represents a rational number.
          Thus, 43.123456789 = {p \over q}, where q = {10^9}.

         (ii) 0.12012001200012000... is non - terminating and non-repeating. So, it is irrational.

         (iii) 43.\overline {123456789} is non - terminating but repeating. So, it is rational.
         Thus, 43.\overline {123456789}  = {p \over q}, where q = 999999999.



Contact Us

Call us: 8287971571,0261-4890016

Or, Fill out the form & get a call back.