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 $\frac{1}{4}$

(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 $\frac{9}{11}$

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 $\frac{9}{11}$

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) $\frac{2157}{625}$

(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) $8.0025$

(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) $0.3\overline{2}$

(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) $\frac{987}{10500}$

(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) $\frac{17}{8}$

(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) $\sqrt{2}+ \sqrt{3}$

(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) $327.781$

(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) $\sqrt{100}$

(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) $t^2 = 0.4$

(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) $0.30300300003... \ and \ 0.3010010001...$

(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 $0.302501001000100001...$