Question 1: Find two rational numbers between:

(i) 1 \ and \  2        (ii) 3 \ and \  4        (iii)  -2 \ and \ 6        (iv) \frac{3}{5} and \frac{4}{5}                (v) - \frac{2}{3} and \frac{1}{4} 

Answer:

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

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

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

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

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

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

(iii)   The rational number between -2 \ and \  6 is (-2 + 6) \div 2 = 2

Now a rational number between -2 \ and \  2 is (-2 + 2 ) \div 2 = 0

Thus the required rational numbers are -2 < 0 < 2 < 4

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

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

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

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

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

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

\\

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

Answer:

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

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

- \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 12 numbers between - \frac{4}{11} and \frac{9}{11} .

\\

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

Answer:

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

We know that:

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

Dividing by 110, we get the numbers

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

\\

Question 4: Express the following rational numbers as decimals:

(i) \frac{42}{100}        (ii) \frac{327}{500}        (iii) \frac{15}{4}        (iv) \frac{2}{3}        (v) - \frac{4}{9}        (vi) - \frac{2}{15}         (viii) - \frac{22}{13}         (ix) \frac{437}{999}        (x) \frac{7}{8}        (xi) \frac{2157}{625}

Answer:

(i) \frac{42}{100}

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

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

(ii) \frac{327}{500}

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

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

(iii) \frac{15}{4}

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

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

(iv) \frac{2}{3}

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

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

(v) - \frac{4}{9}

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

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

(vi) - \frac{2}{15}

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

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

(viii) - \frac{22}{13}

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}

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

(ix) \frac{437}{999}

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

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

(x) \frac{7}{8}

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

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

(xi) \frac{2157}{625}

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

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

\\

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

(i) 0.39        (ii) 0.750         (iii) 2.15        (iv) 7.010        (v) 9.90        (vi) 1.0001          (vii) 0.15        (viii) 0.675        (ix) 0.000026         (x) 8.0025

Answer:

(i) 0.39 = \frac{39}{100}

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

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

(iv) 7.010 = \frac{701}{100}

(v) 9.90 = \frac{99}{10}

(vi) 1.0001 = \frac{10001}{10000}

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

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

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

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

\\

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

(i) 0.\overline{4}        (ii) 0.\overline{37}        (iii) 0.\overline{54}        (iv) 0.\overline{621}        (v) 125.\overline{3}        (vi) 4.\overline{7}        (vii) 0.4\overline{7}        (viii) 0.3\overline{2}        (ix) 0.12\overline{3}           (x) 0.3\overline{2}

Answer:

(i) 0.\overline{4}

Let x = 0.\overline{4}

10x = 4.\overline{4}

Subtracting 9x = 4 \Rightarrow x = \frac{4}{9}

(ii) 0.\overline{37}

Let x = 0.\overline{37}

100x = 37.\overline{37}

Subtracting 99x = 37 \Rightarrow x = \frac{37}{99}

(iii) 0.\overline{54}

Let x = 0.\overline{54}

100x = 54.\overline{54}

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

(iv) 0.\overline{621}

Let x = 0.\overline{621}

1000x = 621.\overline{621}

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

(v) 125.\overline{3}

Let x = 125.\overline{3}

10x = 1253.\overline{3}

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

(vi) 4.\overline{7}

Let x = 4.\overline{7}

10x = 47.\overline{7}

Subtracting 9x = 43 \Rightarrow x = \frac{43}{9}

(vii) 0.3\overline{2}

Let x = 0.3\overline{2}

10x = 3.\overline{2}

Subtracting 9x = 3 \Rightarrow x = \frac{3}{9} = \frac{1}{3}

(viii) 0.12\overline{3}

Let x = 0.12\overline{3}

100x = 12.\overline{3}

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

\\

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

\\

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

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

Answer:

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

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

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

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

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

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

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

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

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

\\

Question 9: Show that the following numbers are irrational:

(i) \frac{1}{\sqrt{2}}        (ii) 7\sqrt{5}        (iii) 6+\sqrt{2}        (iv) 3-\sqrt{5}      (v) \frac{2}{\sqrt{7}}        (vi) \frac{3}{2\sqrt{5}}             (vii) 4+\sqrt{2}        (viii) 5\sqrt{2}        (ix) 2-\sqrt{3}        (x) 3+\sqrt{2}        (xi) 4-5\sqrt{2}           (xii) 5-2\sqrt{3}        (xiii) 2\sqrt{3} -+ 1        (xiv) 2-3\sqrt{5}        (xv) \sqrt{5} + \sqrt{3}                  (xvi) \sqrt{2}+ \sqrt{3}

Answer:

(i)  \frac{1}{\sqrt{2}}

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

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

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

b^2 = 2 a^2

\Rightarrow 2 | b^2

\Rightarrow 2 | b … … … … … (i)

\Rightarrow b = 2c for some integer c

\Rightarrow b^2 = 4 c^2

\Rightarrow 2a^2 = 4 c^2

\Rightarrow a^2 = 2 c^2

\Rightarrow 2 | a^2

\Rightarrow 2 | a … … … … … (ii)

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

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

(ii) 7\sqrt{5}

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

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

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

5 b^2 = a^2

\Rightarrow 5 | a^2

\Rightarrow 5 | a … … … … … (i)

\Rightarrow a = 5c for some integer c

\Rightarrow a^2 = 25 c^2

\Rightarrow 5b^2 = 25 c^2

\Rightarrow b^2 = 5 c^2

\Rightarrow 5 | b^2

\Rightarrow 5 | b … … … … … (ii)

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

Hence \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 7\sqrt{5} is an irrational number.

(iii) 6+\sqrt{2}

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

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

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

2 b^2 = a^2

\Rightarrow 2 | a^2

\Rightarrow 2 | a … … … … … (i)

\Rightarrow a = 2c for some integer c

\Rightarrow a^2 = 4 c^2

\Rightarrow 2b^2 = 4 c^2

\Rightarrow b^2 = 2 c^2

\Rightarrow 2 | b^2

\Rightarrow 2 | b … … … … … (ii)

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

Hence \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 6 + \sqrt{2} is an irrational number.

(iv) 3-\sqrt{5}

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

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

(v) \frac{2}{\sqrt{7}}

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

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

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

b^2 = 7 a^2

\Rightarrow 7 | b^2

\Rightarrow 7 | b … … … … … (i)

\Rightarrow b = 7c for some integer c

\Rightarrow b^2 = 49 c^2

\Rightarrow 7a^2 = 49 c^2

\Rightarrow a^2 = 7 c^2

\Rightarrow 7 | a^2

\Rightarrow 7 | a … … … … … (ii)

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

Hence \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 \frac{2}{\sqrt{7}} is an irrational number.

(vi) \frac{3}{2\sqrt{5}}

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

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

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

b^2 = 5 a^2

\Rightarrow 5 | b^2

\Rightarrow 5 | b … … … … … (i)

\Rightarrow b = 5c for some integer c

\Rightarrow b^2 = 25 c^2

\Rightarrow 5a^2 = 25 c^2

\Rightarrow a^2 = 5 c^2

\Rightarrow 5 | a^2

\Rightarrow 5 | a … … … … … (ii)

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

Hence \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 \frac{3}{2\sqrt{7}} is an irrational number.

(vii) 4+\sqrt{2}

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

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

(viii) 5\sqrt{2}

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

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

(ix) 2-\sqrt{3}

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

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

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

3 b^2 = a^2

\Rightarrow 3 | a^2

\Rightarrow 3 | a … … … … … (i)

\Rightarrow a = 3c for some integer c

\Rightarrow a^2 = 9 c^2

\Rightarrow 3b^2 = 9 c^2

\Rightarrow b^2 = 3 c^2

\Rightarrow 3 | b^2

\Rightarrow 3 | b … … … … … (ii)

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

Hence \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 2-\sqrt{3} is an irrational number.

(x) 3+\sqrt{2}

In the problem 9(iii) we proved that \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 3 + \sqrt{2} is an irrational number.

(xi) 4-5\sqrt{2}

In the problem 9(iii) we proved that \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 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 4 - 5\sqrt{2} is an irrational number.

(xii) 5-2\sqrt{3}

In the problem 9(ix) we proved that \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 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 5-2\sqrt{3} is an irrational number.

(xiii) 2\sqrt{3} - 1

In the problem 9(ix) we proved that \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 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 2\sqrt{3} - 1 is an irrational number.

(xiv) 2-3\sqrt{5}

In the problem 9(ii) we proved that \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 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 2-3\sqrt{5} is an irrational number.

(xv) \sqrt{5} + \sqrt{3}

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

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

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

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

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

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

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

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

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

\\

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

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

Answer:

(i) 43.123456789

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

(ii) 43.\overline{123456789}

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

(iii) 327.781

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

\\

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

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

Answer:

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

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

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

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

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

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

\\

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

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

Answer:

(i) x^2=5

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

(ii) y^2=9 

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

(iii) z^2 = 0.04

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

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

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

(v) v^2 = 3

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

(vi) w^2 = 27

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

(vii) t^2 = 0.4

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

\\

Question 13: Find two irrational numbers between:

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

Answer:

(i) 0.5 \ and \ 0.55

The two numbers could be 0.501001000100001.... and 0.5101001000100001...

(ii) ) 0.1 \ and \ 0.12

The two numbers could be 0.101001000100001... and 0.1101001000100001...

(iii) \frac{5}{7} \ and \  \frac{9}{11}

We have a = \frac{5}{7} = 0.\overline{714285} and b = \frac{9}{11} = 0.\overline{81}

Therefore the number could be 0.7201001000100001... and 0.7301001000100001...

(iv) 0.30300300003... \ and \  0.3010010001...

The numbers could be 0.30201001000100001... and 0.302501001000100001...