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)  

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 \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)  

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 \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.

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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

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.

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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.

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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.

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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...

Answer:

\displaystyle \text{(i)  } 0.5 \text{ and } 0.55

The two numbers could be \displaystyle 0.501001000100001.... \text{ and } 0.5101001000100001...

\displaystyle \text{(ii)  } 0.1 \text{ and } 0.12

The two numbers could be \displaystyle 0.101001000100001... \text{ and } 0.1101001000100001...

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

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

Therefore the number could be \displaystyle 0.7201001000100001... \text{ and } 0.7301001000100001...

\displaystyle \text{(iv)  } 0.30300300003... \text{ and } 0.3010010001...

The numbers could be \displaystyle 0.30201001000100001... \text{ and } 0.302501001000100001...