Class 9: Rational and Irrational Numbers – Exercise 1.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find two rational numbers between:

$\displaystyle \text{(i) } 1 \text{ and } 2$ $\displaystyle \text{(ii) } 3 \text{ and } 4$ $\displaystyle \text{(iii) } -2 \text{ and } 6$

$\displaystyle \text{(iv) } \frac{3}{5} \text{ and } \frac{4}{5}$ $\displaystyle \text{(v) } - \frac{2}{3} \text{ and } \frac{1}{4}$

Answer:

(i) $\displaystyle \text{The rational number between } 1 \text{ and } 2 \text{ is } (1 + 2) \div 2 = \frac{3}{2}$

$\displaystyle \text{Now a rational number between } 1 \text{ and } \frac{3}{2} \text{ is } (1 + \frac{3}{2} ) \div 2 = \frac{5}{4}$

$\displaystyle \text{Thus the required rational numbers are } 1 < \frac{5}{4} < \frac{3}{2} < 2$

(ii) $\displaystyle \text{The rational number between } 3 \text{ and } 4 \text{ is } (3 + 4) \div 2 = \frac{7}{2}$

$\displaystyle \text{Now a rational number between } 3 \text{ and } \frac{7}{2} \text{ is } (3 + \frac{7}{2} ) \div 2 = \frac{13}{4}$

$\displaystyle \text{Thus the required rational numbers are } 3 < \frac{13}{4} < \frac{7}{2} < 4$

(iii) $\displaystyle \text{The rational number between } -2 \text{ and } 6 \text{ is } (-2 + 6) \div 2 = 2$

$\displaystyle \text{Now a rational number between } -2 \text{ and } 2 \text{ is } (-2 + 2 ) \div 2 = 0$

$\displaystyle \text{Thus the required rational numbers are } -2 < 0 < 2 < 4$

(iv) $\displaystyle \text{The rational number between } \frac{3}{5} \text{ and } \frac{4}{5} \text{ is } ( \frac{3}{5} + \frac{4}{5} ) \div 2 = \frac{7}{10}$

$\displaystyle \text{Now a rational number between } \frac{3}{5} \text{ and } \frac{7}{10} ( \frac{3}{5} + \frac{7}{10} ) \div 2 = \frac{13}{20}$

$\displaystyle \text{Thus the required rational numbers are } \frac{3}{5} < \frac{7}{10} < \frac{13}{20} < \frac{4}{5}$

(v) $\displaystyle \text{The rational number between } - \frac{2}{3} \text{ and } \frac{1}{4} \text{ is } ( - \frac{2}{3} + \frac{1}{4} ) \div 2 = - \frac{5}{24}$

$\displaystyle \text{Now a rational number between } - \frac{2}{3} \text{ and } - \frac{5}{24} ( - \frac{2}{3} + - \frac{5}{24} ) \div 2 = - \frac{21}{48}$

$\displaystyle \text{Thus the required rational numbers are } - \frac{2}{3} < - \frac{5}{24} < - \frac{21}{48} < \frac{1}{4}$

$\displaystyle \\$

Question 2: Insert $\displaystyle 12$ rational numbers between $\displaystyle - \frac{4}{11} \text{ and } \frac{9}{11}$

Answer:

We know that $\displaystyle -4 < -3 < -2 < -1 < 0 < 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9$

Therefore, dividing each of these by $\displaystyle 11$ would give us:

$\displaystyle - \frac{4}{11} < - \frac{3}{11} < - \frac{2}{11} < - \frac{1}{11} < \frac{0}{11} < \frac{0}{11} < \frac{1}{11} < \frac{2}{11} < \frac{3}{11} < \frac{4}{11} < \frac{5}{11} < \frac{6}{11} < \frac{7}{11} < \frac{8}{11} < \frac{9}{11}$

Which gives us the $\displaystyle 12$ numbers between $\displaystyle - \frac{4}{11} \text{ and } \frac{9}{11}$ .

$\displaystyle \\$

Question 3: Insert $\displaystyle 100$ numbers between $\displaystyle - \frac{4}{11} \text{ and } \frac{9}{11}$

Answer:

Insert $\displaystyle 100$ numbers between $\displaystyle - \frac{4}{11} \text{ and } \frac{9}{11}$ is equivalent to inserting $\displaystyle 100$ numbers between $\displaystyle - \frac{40}{110} \text{ and } \frac{90}{110}$ . We just multiplied the numerator and denominator by $\displaystyle 10$ .

We know that:

$\displaystyle -40 < -39 < -38 < ... ... ... < -1 < 0 < 1 < ... ... < 57 < 58 < 59 <60$

Dividing by 110, we get the numbers

$\displaystyle - \frac{40}{110} < - \frac{39}{110} < - \frac{38}{110} <$ … $\displaystyle - \frac{1}{110} < \frac{0}{110} < \frac{1}{110} <$ … $\displaystyle \frac{57}{110} < \frac{58}{110} < \frac{59}{110} < \frac{60}{110}$ … $\displaystyle < \frac{90}{110}$

$\displaystyle \\$

Question 4: Express the following rational numbers as decimals:

$\displaystyle \text{(i) } \frac{42}{100}$ $\displaystyle \text{(ii) } \frac{327}{500}$ $\displaystyle \text{(iii) } \frac{15}{4}$ $\displaystyle \text{(iv) } \frac{2}{3}$ $\displaystyle \text{(v) } - \frac{4}{9}$ $\displaystyle \text{(vi) } - \frac{2}{15}$

$\displaystyle \text{(viii) } - \frac{22}{13}$ $\displaystyle \text{(ix) } \frac{437}{999}$ $\displaystyle \text{(x) } \frac{7}{8}$ $\displaystyle \text{(xi) } \frac{2157}{625}$

Answer:

$\displaystyle \text{(i) } \frac{42}{100}$

$\displaystyle 100 \ \overline{) \ 42 \ (}\ 0.42 \\ \underline {\ \ \ \ \ \ \ \ 400} \\ {\ \ \ \ \ \ \ \ \ 200} \\ \underline{\ \ \ \ \ \ \ \ \ 200} \\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

$\displaystyle \therefore \frac{42}{100} = 0.42$

$\displaystyle \text{(ii) } \frac{327}{500}$

$\displaystyle 500 \ \overline{) \ 327 \ (}\ 0.654 \\ \underline {\ \ \ \ \ \ \ \ 3000} \\ {\ \ \ \ \ \ \ \ \ 2700} \\ \underline{\ \ \ \ \ \ \ \ \ 2500} \\ {\ \ \ \ \ \ \ \ \ \ 2000} \\ \underline{\ \ \ \ \ \ \ \ \ \ 2000} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ 0}$

$\displaystyle \therefore \frac{327}{500} = 0.654$

$\displaystyle \text{(iii) } \frac{15}{4}$

$\displaystyle 4 \ \overline{) \ 15 \ (}\ 3.75 \\ \underline {\ \ \ \ \ 12} \\ {\ \ \ \ \ \ 30} \\ \underline{\ \ \ \ \ \ 28} \\ {\ \ \ \ \ \ \ 20} \\ \underline{\ \ \ \ \ \ \ 20} \\ {\ \ \ \ \ \ \ \ 0}$

$\displaystyle \therefore \frac{15}{4} = 3.75$

$\displaystyle \text{(iv) } \frac{2}{3}$

$\displaystyle 3 \ \overline{) \ 2 \ (}\ 0.666 \\ \underline {\ \ \ \ \ 18} \\ {\ \ \ \ \ \ 20} \\ \underline{\ \ \ \ \ \ 18} \\ {\ \ \ \ \ \ \ 20} \\ \underline{\ \ \ \ \ \ \ 18} \\ {\ \ \ \ \ \ \ \ 2}$

$\displaystyle \therefore \frac{2}{3} = 0.666 = 0.\overline{6}$

$\displaystyle \text{(v) } - \frac{4}{9}$

$\displaystyle 9 \ \overline{) \ 4 \ (}\ 0.444 \\ \underline {\ \ \ \ \ 36} \\ {\ \ \ \ \ \ 40} \\ \underline{\ \ \ \ \ \ 36} \\ {\ \ \ \ \ \ \ 40} \\ \underline{\ \ \ \ \ \ \ 36} \\ {\ \ \ \ \ \ \ \ 4}$

$\displaystyle \therefore - \frac{4}{9} = 0.444 = 0.\overline{4}$

$\displaystyle \text{(vi) } - \frac{2}{15}$

$\displaystyle 15 \ \overline{) \ 2 \ (}\ 0.133 \\ \underline {\ \ \ \ \ 15} \\ {\ \ \ \ \ \ 50} \\ \underline{\ \ \ \ \ \ 45} \\ {\ \ \ \ \ \ \ 50} \\ \underline{\ \ \ \ \ \ \ 45} \\ {\ \ \ \ \ \ \ \ 5}$

$\displaystyle \therefore - \frac{2}{15} = 0.133 = 0.1\overline{3}$

$\displaystyle \text{(viii) } - \frac{22}{13}$

$\displaystyle 13 \ \overline{) \ 22 \ (}\ 1.692307692307 \\ \underline {\ \ \ \ \ \ 13} \\ {\ \ \ \ \ \ 90} \\ \underline{\ \ \ \ \ \ 78} \\ {\ \ \ \ \ \ \ 120} \\ \underline{\ \ \ \ \ \ \ 117} \\ {\ \ \ \ \ \ \ \ \ 30} \\ \underline{\ \ \ \ \ \ \ \ \ 26} \\ {\ \ \ \ \ \ \ \ \ \ 40} \\ \underline{\ \ \ \ \ \ \ \ \ \ 39} \\ {\ \ \ \ \ \ \ \ \ \ \ 100} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ 91} \\ {\ \ \ \ \ \ \ \ \ \ \ \ 90} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ 78} \\ {\ \ \ \ \ \ \ \ \ \ \ \ 120} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ 117} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 30} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ 26} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 40}\\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 39} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 100} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 91} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 90} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 78} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 120} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 117} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 30}$

$\displaystyle \therefore - \frac{2}{15} = 1.692307692307 = 0.1\overline{692307}$

$\displaystyle \text{(ix) } \frac{437}{999}$

$\displaystyle 999 \ \overline{) \ 437 \ (}\ 0.437437 \\ \underline {\ \ \ \ \ \ \ \ 3996} \\ {\ \ \ \ \ \ \ \ \ 3740} \\ \underline{\ \ \ \ \ \ \ \ \ 2997} \\ {\ \ \ \ \ \ \ \ \ \ \ 7430} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ 6993} \\ {\ \ \ \ \ \ \ \ \ \ \ \ 437} \\ \underline {\ \ \ \ \ \ \ \ \ \ \ \ 3996} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ 3740} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ 2997} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ 7430} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ 6993} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 437}$

$\displaystyle \therefore - \frac{437}{999} = 0.437437 = 0.\overline{437}$

$\displaystyle \text{(x) } \frac{7}{8}$

$\displaystyle 8 \ \overline{) \ 7 \ (}\ 0.875 \\ \underline {\ \ \ \ \ 64} \\ {\ \ \ \ \ \ 60} \\ \underline{\ \ \ \ \ \ \ 56} \\ {\ \ \ \ \ \ \ \ 40} \\ \underline{\ \ \ \ \ \ \ \ 40} \\ {\ \ \ \ \ \ 0}$

$\displaystyle \therefore \frac{7}{8} = 0.875$

$\displaystyle \text{(xi) } \frac{2157}{625}$

$\displaystyle 625 \ \overline{) \ 2157 \ (}\ 3.4512 \\ \underline {\ \ \ \ \ \ \ \ 1875} \\ {\ \ \ \ \ \ \ \ \ 2820} \\ \underline{\ \ \ \ \ \ \ \ \ 2500} \\ {\ \ \ \ \ \ \ \ \ \ \ 3200} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ 3125} \\ {\ \ \ \ \ \ \ \ \ \ \ \ 750} \\ \underline {\ \ \ \ \ \ \ \ \ \ \ \ 625} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ 1250} \\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ 1250} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0}$

$\displaystyle \therefore \frac{2157}{625} = 3.4512$

$\displaystyle \\$

Question 5: Express each of the following decimals in the form $\displaystyle \frac{p}{q}$ :

$\displaystyle \text{(i) } 0.39$ $\displaystyle \text{(ii) } 0.750$ $\displaystyle \text{(iii) } 2.15$ $\displaystyle \text{(iv) } 7.010$ $\displaystyle \text{(v) } 9.90$

$\displaystyle \text{(vi) } 1.0001$ $\displaystyle \text{(vii) } 0.15$ $\displaystyle \text{(viii) } 0.675$ $\displaystyle \text{(ix) } 0.000026$ $\displaystyle \text{(x) } 8.0025$

Answer:

$\displaystyle \text{(i) } 0.39 = \frac{39}{100}$

$\displaystyle \text{(ii) } 0.750 = \frac{750}{1000} = \frac{3}{4}$

$\displaystyle \text{(iii) } 2.15 = \frac{215}{100} = \frac{43}{20}$

$\displaystyle \text{(iv) } 7.010 = \frac{701}{100}$

$\displaystyle \text{(v) } 9.90 = \frac{99}{10}$

$\displaystyle \text{(vi) } 1.0001 = \frac{10001}{10000}$

$\displaystyle \text{(vii) } 0.15 = \frac{15}{100} = \frac{3}{20}$

$\displaystyle \text{(viii) } 0.675 = \frac{675}{1000} = \frac{27}{40}$

$\displaystyle \text{(ix) } 0.000026 = \frac{26}{1000000} = \frac{13}{500000}$

$\displaystyle \text{(x) } 8.0025 = \frac{80025}{10000} = \frac{3201}{400}$

$\displaystyle \\$

Question 6: Express each of the following decimals in the form $\displaystyle \frac{p}{q}$ :

$\displaystyle \text{(i) } 0.\overline{4}$ $\displaystyle \text{(ii) } 0.\overline{37}$ $\displaystyle \text{(iii) } 0.\overline{54}$ $\displaystyle \text{(iv) } 0.\overline{621}$ $\displaystyle \text{(v) } 125.\overline{3}$

$\displaystyle \text{(vi) } 4.\overline{7}$ $\displaystyle \text{(vii) } 0.4\overline{7}$ $\displaystyle \text{(viii) } 0.3\overline{2}$ $\displaystyle \text{(ix) } 0.12\overline{3}$ $\displaystyle \text{(x) } 0.3\overline{2}$

Answer:

$\displaystyle \text{(i) } 0.\overline{4}$

Let $\displaystyle x = 0.\overline{4}$

$\displaystyle 10x = 4.\overline{4}$

$\displaystyle \text{Subtracting } 9x = 4 \Rightarrow x = \frac{4}{9}$

$\displaystyle \text{(ii) } 0.\overline{37}$

Let $\displaystyle x = 0.\overline{37}$

$\displaystyle 100x = 37.\overline{37}$

$\displaystyle \text{Subtracting } 99x = 37 \Rightarrow x = \frac{37}{99}$

$\displaystyle \text{(iii) } 0.\overline{54}$

Let $\displaystyle x = 0.\overline{54}$

$\displaystyle 100x = 54.\overline{54}$

$\displaystyle \text{Subtracting } 99x = 54 \Rightarrow x = \frac{54}{99} = \frac{6}{11}$

$\displaystyle \text{(iv) } 0.\overline{621}$

Let $\displaystyle x = 0.\overline{621}$

$\displaystyle 1000x = 621.\overline{621}$

$\displaystyle \text{Subtracting } 999x = 621 \Rightarrow x = \frac{621}{999} = \frac{23}{37}$

$\displaystyle \text{(v) } 125.\overline{3}$

Let $\displaystyle x = 125.\overline{3}$

$\displaystyle 10x = 1253.\overline{3}$

$\displaystyle \text{Subtracting } 9x = 1128 \Rightarrow x = \frac{1128}{9} = \frac{376}{3}$

$\displaystyle \text{(vi) } 4.\overline{7}$

Let $\displaystyle x = 4.\overline{7}$

$\displaystyle 10x = 47.\overline{7}$

$\displaystyle \text{Subtracting } 9x = 43 \Rightarrow x = \frac{43}{9}$

$\displaystyle \text{(vii) } 0.3\overline{2}$

Let $\displaystyle x = 0.3\overline{2}$

$\displaystyle 10x = 3.\overline{2}$

$\displaystyle \text{Subtracting } 9x = 3 \Rightarrow x = \frac{3}{9} = \frac{1}{3}$

$\displaystyle \text{(viii) } 0.12\overline{3}$

Let $\displaystyle x = 0.12\overline{3}$

$\displaystyle 100x = 12.\overline{3}$

$\displaystyle \text{Subtracting } 99x = 12 \Rightarrow x = \frac{12}{99} = \frac{4}{33}$

$\displaystyle \\$

Question 7: Without performing long division, state if the following rational numbers are terminating decimals or non-terminating decimals:

$\displaystyle \text{(i) } \frac{23}{8}$ $\displaystyle \text{(ii) } \frac{125}{441}$ $\displaystyle \text{(iii) } \frac{35}{50}$ $\displaystyle \text{(iv) } \frac{77}{210}$ $\displaystyle \text{(v) } \frac{129}{2^2 \times 5^7 \times 7^17}$ $\displaystyle \text{(vi) } \frac{987}{10500}$

Answer:

$\displaystyle \text{(i) } \frac{23}{8} = \frac{23}{2^3} = \frac{23 \times 5^3}{2^3 \times 5^3} = \frac{2875}{10^3}$ .

Since $\displaystyle 8$ is of the form $\displaystyle 2^m \times 5^n$ , hence the decimal expansion of $\displaystyle \frac{23}{8}$ is terminating.

$\displaystyle \text{(ii) } \frac{125}{441} = \frac{ 125}{3^2 \times 7^2}$

Since $\displaystyle 441$ is not of the form $\displaystyle 2^m \times 5^n$ , hence the decimal expansion of $\displaystyle \frac{125}{441}$ is a non-terminating repeating.

$\displaystyle \text{(iii) } \frac{35}{50} = \frac{7}{10} = \frac{7}{2^1 \times 5^1}$

Since $\displaystyle 10$ is of the form $\displaystyle 2^m \times 5^n$ , hence the decimal expansion of $\displaystyle \frac{35}{50}$ is terminating.

$\displaystyle \text{(iv) } \frac{77}{210} = \frac{11}{30} = \frac{7}{3^1 \times 2^1 \times 5^1}$

Since $\displaystyle 30$ is not of the form $\displaystyle 2^m \times 5^n$ , hence the decimal expansion of $\displaystyle \frac{77}{210}$ is a non-terminating repeating.

$\displaystyle \text{(v) } \frac{129}{2^2 \times 5^7 \times 7^17}$

Since $\displaystyle 2^2 \times 5^7 \times 7^{17}$ is not of the form $\displaystyle 2^m \times 5^n$ , hence the decimal expansion of $\displaystyle \frac{129}{2^2 \times 5^7 \times 7^{17}}$ is a non-terminating repeating.

$\displaystyle \text{(vi) } \frac{987}{10500} = \frac{21 \times 47}{2^2 \times 5^3 \times 21} = \frac{47}{2^2 \times 5^3}$

Since $\displaystyle 500$ is of the form $\displaystyle 2^m \times 5^n$ , hence the decimal expansion of $\displaystyle \frac{987}{10500} = \frac{47}{500}$ is terminating.

$\displaystyle \\$

Question 8: Write down the decimal expansion of the rational numbers by writing their denominators in the form of $\displaystyle 2^m \times 5^n$ given m and n are non-negative integers.

$\displaystyle \text{(i) } \frac{3}{8}$ $\displaystyle \text{(ii) } \frac{13}{25}$ $\displaystyle \text{(iii) } \frac{7}{80}$ $\displaystyle \text{(iv) } \frac{14588}{625}$ $\displaystyle \text{(v) } \frac{129}{2^2 \times 5^7}$ $\displaystyle \text{(vi) } \frac{15}{1600}$ $\displaystyle \text{(vii) } \frac{23}{2^3 \times 5^2}$ $\displaystyle \text{(viii) } \frac{17}{8}$

Answer:

$\displaystyle \text{(i) } \frac{3}{8} = \frac{3}{2^3 \times 5^0} = 0.375 \text{ (terminates are k = 3 places decimals) }$

$\displaystyle \text{(ii) } \frac{13}{25} = \frac{13}{2^0 \times 5^3} = 0.104 \text{ (terminates are k = 4 places decimals) }$

$\displaystyle \text{(iii) } \frac{7}{80} = \frac{7}{2^4 \times 5^1} = 0.0875 \text{ (terminates are k = 4 places decimals) }$

$\displaystyle \text{(iv) } \frac{14588}{625} = \frac{14588}{2^0 \times 5^4} = 23.3408 \text{ (terminates are k = 4 places decimals) }$

$\displaystyle \text{(v) } \frac{129}{2^2 \times 5^7} = \frac{129}{2^2 \times 5^7} = 0.0004128 \text{ (terminates are k = 7 places decimals) }$

$\displaystyle \text{(vi) } \frac{257}{5000} = \frac{257}{2^3 \times 5^4} = 0.0514 \text{ (terminates are k = 4 places decimals) }$

$\displaystyle \text{(vii) } \frac{15}{1600} = \frac{15}{2^6 \times 5^2} = 0.009375 \text{ (terminates are k = 6 places decimals) }$

$\displaystyle \text{(viii) } \frac{23}{2^3 \times 5^2} = \frac{23}{2^3 \times 5^2} = 0.115 \text{ (terminates are k = 3 places decimals) }$

$\displaystyle \text{(ix) } \frac{17}{8} = \frac{17}{2^3 \times 5^0} = 2.125 \text{ (terminates are k = 3 places decimals) }$

$\displaystyle \\$

Question 9: Show that the following numbers are irrational:

$\displaystyle \text{(i) } \frac{1}{\sqrt{2}}$ $\displaystyle \text{(ii) } 7\sqrt{5}$ $\displaystyle \text{(iii) } 6+\sqrt{2}$ $\displaystyle \text{(iv) } 3-\sqrt{5}$ $\displaystyle \text{(v) } \frac{2}{\sqrt{7}}$

$\displaystyle \text{(vi) } \frac{3}{2\sqrt{5}}$ $\displaystyle \text{(vii) } 4+\sqrt{2}$ $\displaystyle \text{(viii) } 5\sqrt{2}$ $\displaystyle \text{(ix) } 2-\sqrt{3}$ $\displaystyle \text{(x) } 3+\sqrt{2}$

$\displaystyle \text{(xi) } 4-5\sqrt{2}$ $\displaystyle \text{(xii) } 5-2\sqrt{3}$ $\displaystyle \text{(xiii) } 2\sqrt{3} -+ 1$ $\displaystyle \text{(xiv) } 2-3\sqrt{5}$

$\displaystyle \text{(xv) } \sqrt{5} + \sqrt{3}$ $\displaystyle \text{(xvi) } \sqrt{2}+ \sqrt{3}$

Answer:

$\displaystyle \text{(i) } \frac{1}{\sqrt{2}}$

Let us assume on the contrary that $\displaystyle \frac{1}{\sqrt{2}}$ is a rational number. Then there exist positive integers $\displaystyle a \text{ and } b$ such that:

$\displaystyle \frac{1}{\sqrt{2}} = \frac{a}{b}$

$\displaystyle \Rightarrow \frac{1}{2} = \frac{a^2}{b^2}$

$\displaystyle b^2 = 2 a^2$

$\displaystyle \Rightarrow 2 | b^2$

$\displaystyle \Rightarrow 2 | b$ … … … … … (i)

$\displaystyle \Rightarrow b = 2c$ for some integer $\displaystyle c$

$\displaystyle \Rightarrow b^2 = 4 c^2$

$\displaystyle \Rightarrow 2a^2 = 4 c^2$

$\displaystyle \Rightarrow a^2 = 2 c^2$

$\displaystyle \Rightarrow 2 | a^2$

$\displaystyle \Rightarrow 2 | a$ … … … … … (ii)

From (i) and (ii) , we observe that $\displaystyle a \text{ and } b$ have at least $\displaystyle 2$ as a common factor. But this contradicts the fact that $\displaystyle a \text{ and } b$ are co-primes. This means that our assumption is not correct.

Hence $\displaystyle \frac{1}{\sqrt{2}}$ is an irrational number.

$\displaystyle \text{(ii) } 7\sqrt{5}$

Let us assume on the contrary that $\displaystyle \sqrt{5}$ is a rational number. Then there exist positive integers $\displaystyle a \text{ and } b$ such that:

$\displaystyle \sqrt{5} = \frac{a}{b}$

$\displaystyle \Rightarrow 5 = \frac{a^2}{b^2}$

$\displaystyle 5 b^2 = a^2$

$\displaystyle \Rightarrow 5 | a^2$

$\displaystyle \Rightarrow 5 | a$ … … … … … (i)

$\displaystyle \Rightarrow a = 5c$ for some integer $\displaystyle c$

$\displaystyle \Rightarrow a^2 = 25 c^2$

$\displaystyle \Rightarrow 5b^2 = 25 c^2$

$\displaystyle \Rightarrow b^2 = 5 c^2$

$\displaystyle \Rightarrow 5 | b^2$

$\displaystyle \Rightarrow 5 | b$ … … … … … (ii)

From (i) and (ii) , we observe that $\displaystyle a \text{ and } b$ have at least $\displaystyle 5$ as a common factor. But this contradicts the fact that $\displaystyle a \text{ and } b$ are co-primes. This means that our assumption is not correct.

Hence $\displaystyle \sqrt{5}$ is an irrational number.

We know that the product of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 7\sqrt{5}$ is an irrational number.

$\displaystyle \text{(iii) } 6+\sqrt{2}$

Let us assume on the contrary that $\displaystyle \sqrt{2}$ is a rational number. Then there exist positive integers $\displaystyle a \text{ and } b$ such that:

$\displaystyle \sqrt{2} = \frac{a}{b}$

$\displaystyle \Rightarrow 2 = \frac{a^2}{b^2}$

$\displaystyle 2 b^2 = a^2$

$\displaystyle \Rightarrow 2 | a^2$

$\displaystyle \Rightarrow 2 | a$ … … … … … (i)

$\displaystyle \Rightarrow a = 2c$ for some integer $\displaystyle c$

$\displaystyle \Rightarrow a^2 = 4 c^2$

$\displaystyle \Rightarrow 2b^2 = 4 c^2$

$\displaystyle \Rightarrow b^2 = 2 c^2$

$\displaystyle \Rightarrow 2 | b^2$

$\displaystyle \Rightarrow 2 | b$ … … … … … (ii)

Hence $\displaystyle \sqrt{2}$ is an irrational number.

We know that the sum of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 6 + \sqrt{2}$ is an irrational number.

$\displaystyle \text{(iv) } 3-\sqrt{5}$

From Problem 9(ii) , we have proved that $\displaystyle \sqrt{5}$ is an irrational number.

Therefore $\displaystyle 3-\sqrt{5}$ is an irrational number.

$\displaystyle \text{(v) } \frac{2}{\sqrt{7}}$

Let us assume on the contrary that $\displaystyle \frac{1}{\sqrt{7}}$ is a rational number. Then there exist positive integers $\displaystyle a \text{ and } b$ such that:

$\displaystyle \frac{1}{\sqrt{7}} = \frac{a}{b}$

$\displaystyle \Rightarrow \frac{1}{7} = \frac{a^2}{b^2}$

$\displaystyle b^2 = 7 a^2$

$\displaystyle \Rightarrow 7 | b^2$

$\displaystyle \Rightarrow 7 | b$ … … … … … (i)

$\displaystyle \Rightarrow b = 7c$ for some integer $\displaystyle c$

$\displaystyle \Rightarrow b^2 = 49 c^2$

$\displaystyle \Rightarrow 7a^2 = 49 c^2$

$\displaystyle \Rightarrow a^2 = 7 c^2$

$\displaystyle \Rightarrow 7 | a^2$

$\displaystyle \Rightarrow 7 | a$ … … … … … (ii)

From (i) and (ii) , we observe that $\displaystyle a \text{ and } b$ have at least $\displaystyle 7$ as a common factor. But this contradicts the fact that $\displaystyle a \text{ and } b$ are co-primes. This means that our assumption is not correct.

Hence $\displaystyle \frac{1}{\sqrt{7}}$ is an irrational number.

We know that the product of a non-zero rational number and an irrational number is an irrational number. Hence $\displaystyle \frac{2}{\sqrt{7}}$ is an irrational number.

$\displaystyle \text{(vi) } \frac{3}{2\sqrt{5}}$

Let us assume on the contrary that $\displaystyle \frac{1}{\sqrt{5}}$ is a rational number. Then there exist positive integers $\displaystyle a \text{ and } b$ such that:

$\displaystyle \frac{1}{\sqrt{5}} = \frac{a}{b}$

$\displaystyle \Rightarrow \frac{1}{5} = \frac{a^2}{b^2}$

$\displaystyle b^2 = 5 a^2$

$\displaystyle \Rightarrow 5 | b^2$

$\displaystyle \Rightarrow 5 | b$ … … … … … (i)

$\displaystyle \Rightarrow b = 5c$ for some integer $\displaystyle c$

$\displaystyle \Rightarrow b^2 = 25 c^2$

$\displaystyle \Rightarrow 5a^2 = 25 c^2$

$\displaystyle \Rightarrow a^2 = 5 c^2$

$\displaystyle \Rightarrow 5 | a^2$

$\displaystyle \Rightarrow 5 | a$ … … … … … (ii)

Hence $\displaystyle \frac{1}{\sqrt{5}}$ is an irrational number.

We know that the product of a non-zero rational number and an irrational number is an irrational number. Hence $\displaystyle \frac{3}{2\sqrt{7}}$ is an irrational number.

$\displaystyle \text{(vii) } 4+\sqrt{2}$

In the above problem 9(iii) we proved that $\displaystyle \sqrt{2}$ is irrational.

We know that the sum of a non-zero rational number and an irrational number is an irrational number. Hence $\displaystyle 4+\sqrt{2}$ is an irrational number.

$\displaystyle \text{(viii) } 5\sqrt{2}$

In the above problem 9(iii) we proved that $\displaystyle \sqrt{2}$ is irrational.

We know that the product of a non-zero rational number and an irrational number is an irrational number. Hence $\displaystyle 5\sqrt{2}$ is an irrational number.

$\displaystyle \text{(ix) } 2-\sqrt{3}$

Let us assume on the contrary that $\displaystyle \sqrt{3}$ is a rational number. Then there exist positive integers $\displaystyle a \text{ and } b$ such that:

$\displaystyle \sqrt{3} = \frac{a}{b}$

$\displaystyle \Rightarrow 3 = \frac{a^2}{b^2}$

$\displaystyle 3 b^2 = a^2$

$\displaystyle \Rightarrow 3 | a^2$

$\displaystyle \Rightarrow 3 | a$ … … … … … (i)

$\displaystyle \Rightarrow a = 3c$ for some integer $\displaystyle c$

$\displaystyle \Rightarrow a^2 = 9 c^2$

$\displaystyle \Rightarrow 3b^2 = 9 c^2$

$\displaystyle \Rightarrow b^2 = 3 c^2$

$\displaystyle \Rightarrow 3 | b^2$

$\displaystyle \Rightarrow 3 | b$ … … … … … (ii)

From (i) and (ii) , we observe that $\displaystyle a \text{ and } b$ have at least $\displaystyle 3$ as a common factor. But this contradicts the fact that $\displaystyle a \text{ and } b$ are co-primes. This means that our assumption is not correct.

Hence $\displaystyle \sqrt{3}$ is an irrational number.

We know that the sum of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 2-\sqrt{3}$ is an irrational number.

$\displaystyle \text{(x) } 3+\sqrt{2}$

In the problem 9(iii) we proved that $\displaystyle \sqrt{2}$ is an irrational number.

We know that the sum of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 3 + \sqrt{2}$ is an irrational number.

$\displaystyle \text{(xi) } 4-5\sqrt{2}$

In the problem 9(iii) we proved that $\displaystyle \sqrt{2}$ is an irrational number.

We know that the product of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 5\sqrt{2}$ is an irrational number.

We know that the sum of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 4 - 5\sqrt{2}$ is an irrational number.

$\displaystyle \text{(xii) } 5-2\sqrt{3}$

In the problem 9(ix) we proved that $\displaystyle \sqrt{3}$ is an irrational number.

We know that the product of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 2\sqrt{3}$ is an irrational number.

We know that the sum of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 5-2\sqrt{3}$ is an irrational number.

$\displaystyle \text{(xiii) } 2\sqrt{3} - 1$

In the problem 9(ix) we proved that $\displaystyle \sqrt{3}$ is an irrational number.

We know that the product of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 2\sqrt{3}$ is an irrational number.

We know that the sum of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 2\sqrt{3} - 1$ is an irrational number.

(xiv) $\displaystyle 2-3\sqrt{5}$

In the problem 9(ii) we proved that $\displaystyle \sqrt{5}$ is an irrational number.

We know that the product of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 3\sqrt{5}$ is an irrational number.

We know that the sum of a non-zero rational number and an irrational number is an irrational number. Therefore $\displaystyle 2-3\sqrt{5}$ is an irrational number.

(xv) $\displaystyle \sqrt{5} + \sqrt{3}$

Let us assume on the contrary that $\displaystyle \sqrt{5} + \sqrt{3}$ is a rational number. Then there exist positive integers $\displaystyle a \text{ and } b$ such that:

$\displaystyle \sqrt{5} + \sqrt{3} = \frac{a}{b}$

$\displaystyle \Rightarrow \frac{a}{b} - \sqrt{3} = \sqrt{5}$

$\displaystyle \Rightarrow (\frac{a}{b} - \sqrt{3}) ^2 = (\sqrt{5}) ^2$

$\displaystyle \Rightarrow \frac{a^2}{b^2} - 2\sqrt{3} \frac{a}{b} + 3 = 5$

$\displaystyle \Rightarrow \frac{a^2}{b^2} - 2 = 2\sqrt{3} \frac{a}{b}$

$\displaystyle \Rightarrow \frac{a^2-2b^2}{2ab} = \sqrt{3}$

Since $\displaystyle a \text{ and } b$ are rational number which implies $\displaystyle \frac{a^2-2b^2}{2ab}$ means that $\displaystyle \sqrt{3}$ is a rational number.

This contradicts the fact that $\displaystyle \sqrt{3}$ is an irrational number. Hence our assumption is wrong. Therefore $\displaystyle \sqrt{5} + \sqrt{3}$ is irrational number.

$\displaystyle \\$

Question 10: What can you say about the prime factorization of the denominators of the following rational number:

$\displaystyle \text{(i) } 43.123456789$ $\displaystyle \text{(ii) } 43.\overline{123456789}$ $\displaystyle \text{(iii) } 327.781$

Answer:

$\displaystyle \text{(i) } 43.123456789$

This number is a rational terminating number. Therefore the denominator can be represented in the form of $\displaystyle 2^m \times 5^n$ where $\displaystyle m \text{ and } n$ are non-negative integers.

$\displaystyle \text{(ii) } 43.\overline{123456789}$

This number is a rational non-terminating repeating number. Therefore the denominator can not be represented in the form of $\displaystyle 2^m \times 5^n$ where $\displaystyle m \text{ and } n$ are non-negative integers.

$\displaystyle \text{(iii) } 327.781$

$\displaystyle \\$

Question 11: Identify the following as rational or irrational numbers. Give the decimal representation of rational numbers:

$\displaystyle \text{(i) } \sqrt{4}$ $\displaystyle \text{(ii) } 3\sqrt{18}$ $\displaystyle \text{(iii) } \sqrt{1.44} \sqrt{\frac{9}{27}}$ $\displaystyle \text{(iv) } -\sqrt{64}$ $\displaystyle \text{(vi) } \sqrt{100}$

Answer:

$\displaystyle \text{(i) } \sqrt{4} = 2$ . Therefore rational number.

$\displaystyle \text{(ii) } 3\sqrt{18} = 9\sqrt{2}$ . Therefore irrational number because we know, $\displaystyle \sqrt{2}$ is an irrational number.

$\displaystyle \text{(iii) } \sqrt{1.44} = 1.2$ . Therefore rational number.

$\displaystyle \text{(iv) } \sqrt{\frac{9}{27}} = \sqrt{\frac{1}{3}}$ . Therefore, irrational number.

$\displaystyle \text{(v) } -\sqrt{64} = -8$ . Therefore rational number.

$\displaystyle \text{(v) } \sqrt{100} = 10$ . Therefore rational number.

$\displaystyle \\$

Question 12: In the following equations, find which variables $\displaystyle x, y, z$ etc. represent rational or irrational numbers:

$\displaystyle \text{(i) } x^2=5$ $\displaystyle \text{(ii) } y^2=9$ $\displaystyle \text{(iii) } z^2 = 0.04$ $\displaystyle \text{(iv) } u^2 = \frac{17}{4}$ $\displaystyle \text{(v) } v^2 = 3$ $\displaystyle \text{(vi) } w^2 = 27$ $\displaystyle \text{(vii) } t^2 = 0.4$

Answer:

$\displaystyle \text{(i) } x^2=5$

$\displaystyle \Rightarrow x = \sqrt{5}$ . Therefore $\displaystyle x$ is an irrational number.

$\displaystyle \text{(ii) } y^2=9$

$\displaystyle \Rightarrow y = \pm 3$ . Therefore $\displaystyle y$ is a rational number.

$\displaystyle \text{(iii) } z^2 = 0.04$

$\displaystyle \Rightarrow z = 0.2$ . Therefore $\displaystyle z$ is a rational number.

$\displaystyle \text{(iv) } u^2 = \frac{17}{4}$

$\displaystyle \Rightarrow u = \frac{1}{2} \sqrt{17}$ . Therefore $\displaystyle u$ is an irrational number.

$\displaystyle \text{(v) } v^2 = 3$

$\displaystyle \Rightarrow v = \sqrt{3}$ . Therefore $\displaystyle v$ is an irrational number.

$\displaystyle \text{(vi) } w^2 = 27$

$\displaystyle \Rightarrow w = 3\sqrt{3}$ . Therefore $\displaystyle w$ is an irrational number.

$\displaystyle \text{(vii) } t^2 = 0.4$

$\displaystyle \Rightarrow t = \frac{2}{\sqrt{2} \times \sqrt{5}}$ . Therefore $\displaystyle t$ is an irrational number.

$\displaystyle \\$

Question 13: Find two irrational numbers between:

$\displaystyle \text{(i) } 0.5 \text{ and } 0.55$ (ii) ) $\displaystyle 0.1 \text{ and } 0.12$ $\displaystyle \text{(iii) } \frac{5}{7} \text{ and } \frac{9}{11}$ $\displaystyle \text{(iv) } 0.30300300003... \text{ and } 0.3010010001...$