Class 9 Maths Ganita Manjari
Chapter 3: The World of Numbers
Hello students! I have shared all the solutions for Chapter 3: The World of Numbers below. Click on any section to jump directly to it:
Long before humanity built cities, formulated laws, or studied the stars, there existed a fundamental, practical necessity: the need to keep count. Mathematics did not begin in a classroom with equations on a board; it began in the dirt, on the bark of trees, and on bones.
Core Concepts
- Natural Numbers (N): The basic counting numbers {1, 2, 3, ...}.
- Integers (Z): These include zero and negative numbers {..., -2, -1, 0, 1, 2, ...}.
- Rational Numbers (Q): These include all fractions \( \dfrac{p}{q} \), where p, q are integers and q ≠ 0.
- Irrational Numbers (I): Numbers that cannot be written as fractions (e.g., \( \sqrt{2} \), π).
- Real Numbers (R): Together, the rational and irrational numbers make up the entire real number line.
Exercise Set 3.1
1. A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?
Solution
\[ \begin{aligned} \text{Ingots for 2 bags} & =15 \text{ ingots} \\[5pt] \text{Ingots for 1 bags} & = \frac{15}{2} \\[5pt] \text{Ingots for 12 bags} & = \frac{15}{\cancel2_1} \times \cancel{12}^6 \\ & = 15 \times 6 \\ \color{green} \text{Total ingots} & = \color{green} 90 \end{aligned} \]
Answer The merchant will leave with 90 copper ingots.
2. Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.
Answer
These numbers are consecutive prime numbers starting from 11.
The next three numbers that fit this pattern are: 23, 29, 31.
3. We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.
Answer No, Natural numbers are not closed under subtraction.
Examples:
\( \color{green} 1 - 5 = -4 \) (-4 is not a natural number)
\( \color{green} 10 - 18 = -8 \) (-8 is not a natural number)
4. Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?
Answer
On one hand, excluding the thumb (which is used for counting), you have 4 fingers and each has 3 joints.
\( \color{magenta} 4 \text{ fingers} \times 3 \text{ joints} = \color{magenta} 12 \text{ joints} \). (So you can count upto 12 on one hand.)
This relates perfectly to ancient base-12 counting systems, as early humans could easily count up to 12 on a single hand using their thumb.
Dozen = 12
1 feet = 12 inches
Exercise Set 3.2
1. The temperature in the high-altitude desert of Ladakh is recorded as 4 °C at noon. By midnight, it drops by 15 °C. What is the midnight temperature?
Solution
\[ \begin{align*} \text{Temperature at noon} &= 4^\circ \text{C} \\ \text{Drop in temperature} &= 15^\circ \text{C} \\ \text{Midnight temperature} &= 4^\circ \text{C} - 15^\circ \text{C} \\ &= \color{red} -11^\circ \text{C} \end{align*} \]
Answer Midnight temperature -11 °C
2. A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing.
Solution
\[ \begin{align*} \text{Loan (Debt)} & = -850 \\ \text{Profit (Fortune)} & = +1200 \\ \text{Loss (Debt)} & = -450 \\[5pt] \textbf{Equation} & = \color{red} -850 + 1200 - 450 \\[5pt] \textbf{Financial standing} & = - 850 + 1200 - 450 \\ & = 350 - 450 \\ &= \color{red} -100 \end{align*} \]
Answer Equation: -850 + 1200 - 450 , Financial standing = ₹100.
3.
Calculate the following using Brahmagupta's laws:
(i) (-12) × 5
(ii) (-8) × (-7)
(iii) 0 - (-14)
(iv) (-20) ÷ 4
Answer
(i) The product of a debt and a fortune is a debt.
\( { \color{magenta} \implies} (-12) \times 5 = \color{red} -60\)
(ii) The product of two debts is a fortune. \( { \color{magenta} \implies} (-8) \times (-7) = \color{red} 56\)
(iii) Zero minus debt is a fortune. \( { \color{magenta} \implies} 0 - (-14) = 0 + 14 = \color{red} 14\)
(iv) Debt divided by a fortune is a debt. \( { \color{magenta} \implies} (-20) \div 4 = \color{red} -5\)
4. Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 - (-5) = 15).
Answer
Imagine you have ₹10 in your pocket, but you owe a friend ₹5 (a debt of -5). Your total net worth is effectively ₹5.
If your friend forgives your debt (meaning the -5 debt is subtracted or taken away from your liabilities), you no longer owe that money. Your net worth goes back up to the full ₹10 you hold, plus the ₹5 value you gained by not having to pay. Removing a debt increases your net wealth exactly the same way receiving cash does!
Exercise Set 3.3
1.
Prove that the following rational numbers are equal:
(i) \(\dfrac{2}{3}\) and \(\dfrac{4}{6}\)
(ii) \(\dfrac{5}{4}\) and \(\dfrac{10}{8}\)
(iii) \(-\dfrac{3}{5}\) and \(-\dfrac{6}{10}\)
(iv) \(\dfrac{9}{3}\) and 3
Solution
Two rational numbers \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\) are equal if \(ad = bc\).
\(
\begin{aligned}
\textbf{(i) }\frac{2}{3} \ &\quad and \quad \ \frac{4}{6} \\ \\
2 \times 6 \ &\quad \quad \ 3 \times 4 \\ \\
12 \ &\quad = \quad \ 12 \\
\end{aligned}
\)
\( \therefore \) The given rational numbers are equal.
\(
\begin{aligned}
\textbf{(ii) }\frac{5}{4} \ &\quad and \quad \ \frac{10}{8} \\ \\
5 \times 8 \ &\quad \quad \ 10 \times 4 \\ \\
40 \ &\quad = \quad \ 40 \\
\end{aligned}
\)
\( \therefore \) The given rational numbers are equal.
\(
\begin{aligned}
\textbf{(iii) } -\frac{3}{5} \ &\quad and \quad \ -\frac{6}{10} \\ \\
-3 \times 10 \ &\quad \quad \ -6 \times 5 \\ \\
-30 \ &\quad = \quad \ -30 \\
\end{aligned}
\)
\( \therefore \) The given rational numbers are equal.
\(
\begin{aligned}
\textbf{(iv) } \frac{9}{3} \ &\quad and \quad \ \frac{3}{1} \\ \\
9 \times 1 \ &\quad \quad \ 3 \times 3 \\ \\
9 \ &\quad = \quad \ 9 \\
\end{aligned}
\)
\( \therefore \) The given rational numbers are equal.
2.
Find the sum:
(i) \(\dfrac{2}{5} + \dfrac{3}{10}\)
(ii) \(\dfrac{9}{3} + 3\)
(iii) \(-\dfrac{4}{7} + \dfrac{3}{14}\)
(iv) \(\dfrac{7}{12} + \dfrac{5}{8}\)
Solution
\[ \begin{aligned} & \textbf{(i) } \ \frac{2}{5} + \frac{3}{10} \\[8pt] & = \frac{2 \times 2}{5 \times 2} + \frac{3}{10} \\[8pt] &= \frac{4}{10} + \frac{3}{10} \\[8pt] &= \color{red} \frac{7}{10}\\\\ \hline \\ & \textbf{(ii) } \ \frac{9}{3} + 3 \\[8pt] &= 3 + 3 \\[8pt] &= \color{red} 6 \\\\ \hline \\ & \textbf{(iii) } \ -\frac{4}{7} + \frac{3}{14} \\[8pt] &= \frac{-4 \times 2}{7 \times 2} + \frac{3}{14} \\[8pt] & = \frac{-8}{14} + \frac{3}{14} \\[8pt] & = \color{red} -\frac{5}{14} \\ \\ \hline \\ & \textbf{(iv) } \ \frac{7}{12} + \frac{5}{8} \\[8pt] &= \frac{7 \times 2}{24} + \frac{5 \times 3}{24} \\[8pt] &= \frac{14}{24} + \frac{15}{24} \\[8pt] &= \color{red} \frac{29}{24} \end{aligned} \]
3.
Find the difference:
(i) \(\dfrac{5}{6} - \dfrac{1}{4}\)
(ii) \(\dfrac{11}{8} - \dfrac{3}{4}\)
(iii) \(-\dfrac{7}{9} - \left(-\dfrac{2}{3}\right)\)
Solution
\[ \begin{aligned} & \textbf{(i) } \ \frac{5}{6} - \frac{1}{4} \\[8pt] & = \frac{5 \times 2}{6 \times 2} - \frac{1 \times 3}{4 \times 3} \\[8pt] &= \frac{10}{12} - \frac{3}{12} \\[8pt] &= \color{red} \frac{7}{12}\\\\ \hline \\ & \textbf{(ii) } \ \frac{11}{8} - \frac{3}{4} \\[8pt] &= \frac{11}{8} - \frac{3 \times 2}{4 \times 2} \\[8pt] &= \frac{11}{8} - \frac{6}{8} \\[8pt] &= \color{red} \frac{5}{8} \\\\ \hline \\ & \textbf{(iii) } \ -\frac{7}{9} - \left(-\frac{2}{3}\right) \\[8pt] &= -\frac{7}{9} + \frac{2}{3} \\[8pt] &= -\frac{7}{9} + \frac{2 \times 3}{3 \times 3} \\[8pt] & = -\frac{7}{9} + \frac{6}{9} \\[8pt] & = \color{red} -\frac{1}{9} \end{aligned} \]
4.
Find the product:
(i) \(\dfrac{2}{3} \times \dfrac{3}{10}\)
(ii) \(\dfrac{7}{11} \times \dfrac{5}{8}\)
(iii) \(-\dfrac{4}{7} \times \dfrac{5}{14}\)
Solution
\[ \begin{aligned} & \textbf{(i) } \ \frac{\cancel2^1}{\cancel3_1} \times \frac{\cancel3^1}{\cancel{10}_5} \\[8pt] &= \color{red} \frac{1}{5}\\\\ \hline \\ & \textbf{(ii) } \ \frac{7}{11} \times \frac{5}{8} \\[8pt] &= \color{red} \frac{35}{88} \\\\ \hline \\ & \textbf{(iii) } \ -\frac{\cancel4^2}{7} \times \frac{5}{\cancel{14}_7} \\[8pt] & = \color{red} -\frac{10}{49} \end{aligned} \]
5.
Find the quotient:
(i) \(\dfrac{2}{3} \div \dfrac{3}{10}\)
(ii) \(\dfrac{7}{11} \div \dfrac{5}{8}\)
Solution
\[ \begin{aligned} & \textbf{(i) } \ \frac{2}{3} \div \frac{3}{10} \\[8pt] & = \frac{2}{3} \times \frac{10}{3} \\[8pt] &= \color{red} \frac{20}{9}\\\\ \hline \\ & \textbf{(ii) } \ \frac{7}{11} \div \frac{5}{8} \\[8pt] &= \frac{7}{11} \times \frac{8}{5} \\[8pt] &= \color{red} \frac{56}{55} \end{aligned} \]
6. Show that: \(\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}\)
Solution
\[ \begin{aligned} \textbf{LHS:} \ & \left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3} \\[8pt] & = \left(\frac{1 \times 2}{2 \times 2} + \frac{3}{4}\right) \times \frac{8}{3} \\[8pt] &= \left(\frac{2}{4} + \frac{3}{4}\right) \times \frac{8}{3} \\[8pt] &= \frac{5}{\cancel4_1} \times \frac{\cancel8^2}{3} \\[8pt] &= \color{magenta} \frac{10}{3} \\\\ \textbf{RHS:} \ & \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3} \\[8pt] &= \frac{8}{6} + \frac{24}{12} \\[8pt] &= \frac{4}{3} + \frac{6}{3} \\[8pt] &= \color{magenta} \frac{10}{3} \\\\ \textbf{LHS} &= \textbf{RHS} \end{aligned} \]
7. Simplify the following using the distributive property: \(\dfrac{7}{9}\left(\dfrac{6}{7} - \dfrac{3}{4}\right)\).
Solution
\[ \begin{aligned} & \frac{7}{9}\left(\frac{6}{7} - \frac{3}{4}\right) \\[8pt] &= \left(\frac{\cancel7}{\cancel9_3} \times \frac{\cancel6^2}{\cancel7}\right) - \left(\frac{7}{\cancel9_3} \times \frac{\cancel3^1}{4}\right) \\[8pt] &= \frac{2}{3} - \frac{7}{12} \\[8pt] &= \frac{8-7}{12} \\[8pt] &= \color{red} \frac{1}{12} \end{aligned} \]
8. Find the rational number \(x\) such that: \(\dfrac{5}{6}\left(x + \dfrac{3}{5}\right) = \dfrac{5}{6}x + \dfrac{1}{2}\).
Solution
\[ \begin{aligned} \frac{5}{6}\left(x + \frac{3}{5}\right) &= \frac{5}{6}x + \frac{1}{2} \\[8pt] \frac{5}{6}x + \left(\frac{\cancel5}{\cancel6_2} \times \frac{\cancel3^1}{\cancel5}\right) &= \frac{5}{6}x + \frac{1}{2} \\[8pt] \frac{5}{6}x + \frac{1}{2} &= \frac{5}{6}x + \frac{1}{2} \\ \end{aligned} \]
Answer Both the sides are identical . Therefore, 𝑥 can be any rational number.
Exercise Set 3.4
1. Represent the rational numbers \(\dfrac{2}{3}\) and \(-\dfrac{5}{4}\) and \(1\dfrac{1}{2}\) on a single number line.
Solution
\(\dfrac{2}{3} \) (Lies between 0 and 1)\(-\dfrac{5}{4} = -1\dfrac{1}{4}\) (Lies between -1 and -2)
\(1\dfrac{1}{2} \) (Lies between 1 and 2)
2. Find three distinct rational numbers that lie strictly between \(-\dfrac{1}{2}\) and \(\dfrac{1}{4}\).
Solution
\[ \begin{aligned} -\frac{1}{2} \text{ and } & \frac{1}{4} \\[8pt] \textbf{LCM } & = 4 \\[8pt] -\frac{1 \times 2}{2 \times 2} & = -\frac{2}{4} \\[8pt] \frac{1 \times 1}{4 \times 1} & = \frac{1}{4} \\[8pt] -\frac{2}{4} \text{ and } & \frac{1}{4} \\[8pt] \end{aligned} \] Multiply both numerator and denominator by 4 \[ \begin{aligned} -\frac{2 \times 4}{4 \times 4} = \color{magenta} -\frac{8}{16} \\[8pt] \frac{1 \times 4}{4 \times 4} = \color{magenta} \frac{4}{16} \\[8pt] \end{aligned} \] Three rational numbers between \(-\dfrac{1}{2}\) and \(\dfrac{1}{4}\) are: \( \color{red} \dfrac{1}{16}, \ \dfrac{2}{16} , \ \dfrac{3}{16} \)
3. Simplify the expression: \(\left(-\dfrac{1}{4}\right) + \left(\dfrac{5}{12}\right)\).
Solution
\[ \begin{aligned} & = -\frac{1}{4} + \frac{5}{12} \\[8pt] &= \frac{-3 + 5}{12} \\[8pt] &= \frac{\cancel2^1}{\cancel{12}_6} \\[8pt] &= \color{red} \frac{1}{6} \end{aligned} \]
4. A tailor has \(15\dfrac{3}{4}\) metres of fine silk. If making one kurta requires \(2\dfrac{1}{4}\) metres of silk, exactly how many kurtas can he make?
Solution
\[ \begin{aligned} \text{Total Silk} &= 15\frac{3}{4} \implies \frac{63}{4} \text{ metres} \\[8pt] \text{Silk required to make 1 Kurta} &= 2\frac{1}{4} \implies \frac{9}{4} \text{ metres} \\[8pt] \text{Number of Kurtas} &= \frac{63}{4} \div \frac{9}{4} \\[8pt] &= \frac{\cancel{63}^7}{\cancel4} \times \frac{\cancel4}{\cancel9_1} \\[8pt] &= \color{red} 7 \end{aligned} \]
Answer He can make exactly 7 kurtas.
5. Find three rational numbers between 3.1415 and 3.1416.
Solution
There are infinite rational numbers between any two rational numbers.
3.14151, 3.14152, 3.14153.
6. Can you think of other way(s) to find a rational number between any two rational numbers? Give examples.
Answer
(i) The Average (Mean) Method:
You can take the average \(\left(\dfrac{a+b}{2}\right)\) of the two rational numbers, which will always give you a rational number exactly halfway between them.
(ii) Equivalent Fractions (Common Denominator) Method:
Convert both fractions to a common denominator. If there is still no whole number between the numerators, multiply both the numerator and denominator by a larger number (like 10) to create a wider gap.
(iii) Decimal Conversion Method:
Convert the fractions to decimals, pick any decimal value that falls between them, and convert that decimal back into a fraction.
Exercise Set 3.5
1. Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: \(\dfrac{7}{20}, \dfrac{4}{15}\) and \(\dfrac{13}{250}\). Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.
Solution
Rule: If a rational number \(\dfrac{p}{q}\) is in the simplest form and the prime factorization of denominator (q) contains only 2s or 5s or both the rational number will have terminating decimals if not it will be repeating.
\[
\begin{aligned}
\color{magenta} \textbf{(i) } & \color{magenta} \frac{7}{20} \\[8pt]
\textbf{Denominator }(20) &= 2^2 \times 5 \\[8pt]
\therefore \color{red} \frac{7}{20} & \color{red} \textbf{ has terminating decimals.} \\[8pt]
\textbf{Verification:} \\[8pt]
\begin{array}{l}
\hspace{0.7cm}0.35 \\
20 \enclose{longdiv}{7.00}\\
\hspace{0.3cm}-0\\
\hline
\hspace{0.8cm}70\\
\hspace{0.45cm}-60\\
\hline
\hspace{0.9cm}100 \\
\hspace{0.55cm}-100\\
\hline
\hspace{1.3cm}0 \\
\hline
\end{array} \\ \\
\implies \color{red} \frac{7}{20} & = \color{red} 0.35
\end{aligned}
\]
\[
\begin{aligned}
\color{magenta} \textbf{(ii) } & \color{magenta} \frac{4}{15} \\[8pt]
\textbf{Denominator }(15) &= 3 \times 5 \\[8pt]
\therefore \color{red} \frac{4}{15} & \color{red} \textbf{ has a non-terminating repeating decimal.} \\[8pt]
\textbf{Verification:} \\[8pt]
\begin{array}{l}
\hspace{0.7cm}0.266\ldots \\
15 \enclose{longdiv}{4.000}\\
\hspace{0.3cm}-30\\
\hline
\hspace{0.6cm}100 \\
\hspace{0.5cm}-90\\
\hline
\hspace{.8cm}100 \\
\hspace{0.7cm}-90\\
\hline
\hspace{1cm}10 \\
\end{array} \\ \\
\implies \color{red} \frac{4}{15} & = \color{red} 0.2\overline{6}
\end{aligned}
\]
\[
\begin{aligned}
\color{magenta} \textbf{(iii) } & \color{magenta} \frac{13}{250} \\[8pt]
\textbf{Denominator }(250) &= 2 \times 5^3 \\[8pt]
\therefore \color{red} \frac{13}{250} & \color{red} \textbf{ has terminating decimals.} \\[8pt]
\textbf{Verification:} \\[8pt]
\begin{array}{l}
\hspace{0.9cm}0.052 \\
250 \enclose{longdiv}{13.000}\\
\hspace{0.5cm}-1250\\
\hline
\hspace{1.3cm}500 \\
\hspace{0.95cm}-500\\
\hline
\hspace{1.6cm}0 \\
\hline
\end{array} \\ \\
\implies \color{red} \frac{13}{250} & = \color{red} 0.052
\end{aligned}
\]
2. Perform the long division for \(\dfrac{1}{13}\). Identify the repeating block of digits. Does it show cyclic properties if you evaluate \(\dfrac{2}{13}\)? Now compute \(\dfrac{3}{13}, \dfrac{4}{13}\) etc. What do you notice?
Solution
\[ \begin{aligned} \begin{array}{l} \hspace{0.7cm}0.076923\ldots \\ 13 \enclose{longdiv}{1.0000000}\\ \hspace{0.3cm}-0\\ \hline \hspace{0.6cm}10\\ \hspace{0.45cm}-0\\ \hline \hspace{0.6cm}100 \\ \hspace{0.45cm}-91\\ \hline \hspace{0.9cm}90 \\ \hspace{0.75cm}-78\\ \hline \hspace{0.9cm}120 \\ \hspace{0.75cm}-117\\ \hline \hspace{1.2cm}30 \\ \hspace{1.05cm}-26\\ \hline \hspace{1.35cm}40 \\ \hspace{1.2cm}-39\\ \hline \hspace{1.5cm}10 \\ \end{array} \\ \\ \frac{1}{13} &= \color{red} 0.\overline{076923} \\[5pt] \end{aligned} \]
Evaluating other fractions:
\[ \begin{aligned} \frac{2}{13} = 2 \times \frac{1}{13} & \implies 2 \times 0.\overline{076923} = 0.\overline{153846} \\[8pt] \frac{3}{13} = 3 \times \frac{1}{13} & \implies 3 \times 0.\overline{076923} = 0.\overline{230769} \\[8pt] \frac{4}{13} = 4 \times \frac{1}{13} & \implies 4 \times 0.\overline{076923} = 0.\overline{307692} \\[8pt] \frac{5}{13} = 5 \times \frac{1}{13} & \implies 5 \times 0.\overline{076923} = 0.\overline{384615} \\[8pt] \frac{6}{13} = 6 \times \frac{1}{13} & \implies 6 \times 0.\overline{076923} = 0.\overline{461538} \\[8pt] \end{aligned} \]
What do we notice?
Unlike \( \dfrac{1}{7}\) where all multiples share the exact same repeating block, the fractions of \(13\) split into two distinct cyclic groups:
Group 1 (Cycle 076923): The repeating blocks for \(\dfrac{1}{13}, \dfrac{3}{13}\), and \(\dfrac{4}{13}\) show cyclic properties for the digits 0-7-6-9-2-3.
Group 2 (Cycle 153846): The repeating blocks for \(\dfrac{2}{13}, \dfrac{5}{13}\), and \(\dfrac{6}{13}\) show cyclic properties for different set of digits: 1-5-3-8-4-6.
3. Classify the following numbers as rational or irrational. Find the explicit fractions in case they are rational.
(i) \(\sqrt{81}\)
Solution
\[ \begin{aligned} \sqrt{81} & =9 \\[8pt] \therefore \color{red} \sqrt{81} & \color{red} \textbf{ is rational.} \\[8pt] \textbf{Explicit fraction} & =\frac{9}{1} \\[8pt] \end{aligned} \]
(ii) \(\sqrt{12}\)
Solution
\[ \begin{aligned} \sqrt{12} & =\sqrt{4 \times 3} \\ & = 2\sqrt{3} \\[8pt] \sqrt{3} & \textbf{ is irrational. } \\[5pt] \therefore \color{red} \sqrt{12} & \color{red}\textbf{ is also irrational.} \\[8pt] \end{aligned} \]
(iii) 0.33333...
Solution
\[ \begin{aligned} 0.33333... & = 0.\overline{3} \\ \textbf{It has a repeating decimal} & , \textbf{ so its rational.} \\ \therefore \color{red} 0.33333.... & \color{red}\textbf{ is rational.} \\[8pt] \text{Let } x &= 0.\overline{3} \color{magenta} \longrightarrow \enclose{circle}{1} \\[5pt] \text{Multiply both sides by } & 10 \\ 10x &= 3.\overline{3} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \textbf{Subtract } Eq \ & \enclose{circle}{1} \text{ from } Eq \ \enclose{circle}{2} \\[8pt] 10x - x &= 3.\overline{3} - 0.\overline{3} \\ 9x &= 3 \\[8pt] x &= \frac{3}{9} \\[8pt] x &= \frac{1}{3} \\[8pt] \textbf{Explicit fraction} &= \color{red} \frac{1}{3} \\[8pt] \end{aligned} \]
(iv) 0.123451234512345...
Solution
\[ \begin{aligned} 0.123451234512345... & = 0.\overline{12345} \\ \textbf{It has a repeating block } & 12345 , \textbf{ so its rational.} \\ \therefore \color{red} 0.123451234512345... & \color{red}\textbf{ is rational.} \\[8pt] \text{Let } x &= 0.\overline{12345} \color{magenta} \longrightarrow \enclose{circle}{1} \\[5pt] \text{Multiply both sides by } & 100000 \\ 100000x &= 12345.\overline{12345} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \textbf{Subtract } Eq \ & \enclose{circle}{1} \text{ from } Eq \ \enclose{circle}{2} \\[8pt] 100000x - x &= 12345.\overline{12345} - 0.\overline{12345} \\ 99999x &= 12345 \\[8pt] x &= \frac{12345}{99999} \\[8pt] x &= \frac{4115}{33333} \\[8pt] \textbf{Explicit fraction} &= \color{red} \frac{4115}{33333} \\[8pt] \end{aligned} \]
(v) 1.01001000100001...
Solution There is no repeating block, its non-terminating and non-repeating.
\( \therefore \) 1.01001000100001... is irrational.
(vi) 23.560185612239874790120
Solution
\[ \begin{aligned} \textbf{The block is terminating.} \\[8pt] \therefore \color{red} 23.560185612239874790120 & \color{red}\textbf{ is rational.} \\[8pt] \textbf{Explicit fraction} & =\color{red} \frac{23560185612239874790120}{1000000000000000000000} \\[8pt] \end{aligned} \]
4. The number \(0.\overline{9}\) (which means \(0.99999...\)) is a rational number. Using algebra, explain why \(0.\overline{9}\) is exactly equal to 1.
Solution
\[ \begin{aligned} \text{Let } x &= 0.\overline{9} \color{magenta} \longrightarrow \enclose{circle}{1} \\[5pt] \textbf{Multiply } & \textbf{both sides by } 10 \\ 10x &= 9.\overline{9} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \textbf{Subtract } & \ Eq \ \enclose{circle}{1} \text{ from } \ Eq \ \enclose{circle}{2} \\ 10x - x &= 9.\overline{9} - 0.\overline{9} \\[5pt] 9x &= 9 \\[5pt] x &= \frac{9}{9} \\[5pt] \color{red} x &= \color{red} 1 \\ \therefore 0.\overline{9} & = 1 \end{aligned} \]
5. We have seen that the repeating block of \( \dfrac{1}{7} \) is a cyclic number. Try to find more numbers (n) whose reciprocals \( \dfrac{1}{n} \) produce decimals with repeating blocks that are cyclic.
Solution
\[ \begin{aligned} \frac{1}{17} &= 0.\overline{0588235294117647} \\ \\ \textbf{Repeating block is} &= \color{red} 0588235294117647 \\ \\ \frac{1}{19} &= \color{red} 0.\overline{052631578947368421} \\ \\ \textbf{Repeating block is} &= \color{red} 052631578947368421 \\ \\ \frac{1}{23} &= \color{red} 0.\overline{0434782608695652173913} \\ \\ \textbf{Repeating block is} &= \color{red} 0434782608695652173913 \end{aligned} \]
End-of-Chapter Exercises
1.
Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:
(i) \(\dfrac{3}{50} \)
(ii) \(\dfrac{2}{9} \)
Solution
(i) \(\dfrac{3}{50} \)
\[ \begin{aligned} \begin{array}{l} \hspace{0.7cm}0.06 \\ 50 \enclose{longdiv}{3.00}\\ \hspace{0.3cm}-0\\ \hline \hspace{0.8cm}30\\ \hspace{0.45cm}-0\\ \hline \hspace{0.8cm}300 \\ \hspace{0.45cm}-300\\ \hline \hspace{1.1cm}0 \\ \hline \end{array} \\ \\ \implies \color{red} \frac{3}{50} & = \color{red} 0.06 \textbf{ (Terminating decimal)} \end{aligned} \]
(ii) \(\dfrac{2}{9} \)
\[ \begin{aligned} \begin{array}{l} \hspace{0.7cm}0.222\ldots \\ 9 \enclose{longdiv}{2.000}\\ \hspace{0.3cm}-0\\ \hline \hspace{0.6cm}20\\ \hspace{0.45cm}-18\\ \hline \hspace{0.9cm}20 \\ \hspace{0.75cm}-18\\ \hline \hspace{1.2cm}20 \\ \hspace{1.05cm}-18\\ \hline \hspace{1.35cm}2 \\ \end{array} \\ \\ \frac{2}{9} & = 0.222....\\\\ \color{red} \frac{2}{9} & = \color{red} 0.\overline{2} \textbf{ (Non-terminating, repeating decimal)} \end{aligned} \]
2. Prove that \(\sqrt{5}\) is an irrational number.
Solution
\[ \begin{aligned} \text{Assume } \sqrt{5} & \text{ is a rational number.} \\[8pt] \sqrt{5} = \frac{p}{q} {\color{magenta} \longrightarrow \enclose{circle}{1}} & \begin{cases} \color{green} \text{ p and q are integers} \\ \color{green} q \neq 0 \\ \color{green} \text{HCF (p,q)} = 1 \\ \end{cases}\\[8pt] \textbf{ Square } Eq \ & \enclose{circle}{1} \text{ on both sides}\\[8pt] \left(\sqrt{5} \right)^2 & = \left(\frac{p}{q} \right)^2 \\[8pt] 5 & = \frac{p^2}{q^2} \\[6pt] 5q^2 & = p^2 {\color{magenta} \longrightarrow \enclose{circle}{2}} \\[8pt] \text{Since } p^2 & \text{ is divisible by } 5 \\ \implies p & \text{ is also divisible by } 5 \\[8pt] \text{Let } p & =5 k \\ \text{Substitute } p & =5 k \text{ in } Eq \ \enclose{circle}{2}\\[8pt] 5q^2 & = (5k)^2 \\[8pt] 5q^2 & = 25k^2 \\[8pt] q^2 & = \frac{25k^2}{5} \\[8pt] q^2 & = 5k^2 {\color{magenta} \longrightarrow \enclose{circle}{3}} \\[8pt] \text{Since }q^2 & \text{ is divisible by } 5 \\ \implies q & \text{ is also divisible by } 5 \\[8pt] \therefore \text{Both p and} & \text{ q are divisible by 5} \\ \implies & \textbf{HCF (p,q)} \neq 1 \end{aligned} \]
This contradicts our initial assumption that \( \textbf{HCF (p,q)} = 1\).
Our initial assumption is wrong \(\therefore \sqrt{5}\) must be irrational.
3. Convert the following decimal numbers in the form of \(\dfrac{p}{q}\):
(i) 12.6
Solution
\[ \begin{aligned} & = 12.6 \\[8pt] &= \frac{126}{10} \\[8pt] & = \color{red} \frac{63}{5} \\[8pt] \end{aligned} \]
(ii) 0.0120
Solution
\[ \begin{aligned} & = 0.0120 \\[8pt] &= \frac{12}{1000} \\[8pt] & = \color{red} \frac{3}{250} \\[8pt] \end{aligned} \]
(iii) \(3.0\overline{52}\)
Solution
\[ \begin{aligned} \text{Let } x & = 3.0\overline{52} \color{magenta} \longrightarrow \enclose{circle}{1} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{1} \text{ by }10 \text{ on both sides.}\\[5pt] 10 x & = 30.\overline{52} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{2} \text{ by }100 \text{ on both sides.}\\[5pt] 1000 x & = 3052.\overline{52} \color{magenta} \longrightarrow \enclose{circle}{3} \\[8pt] \textbf{Subtract } Eq \ & \enclose{circle}{2} \text{ from } Eq \ \enclose{circle}{3} \\[8pt] 1000x - 10x & = 3052.\overline{52} - 30.\overline{52} \\[8pt] 990x & = 3022 \\[8pt] x & = \frac{3022}{990} \\[8pt] x & = \frac{1511}{495} \\[8pt] \color{red} 3.0\overline{52} & = \color{red} \frac{1511}{495} \\[8pt] \end{aligned} \]
(iv) \(1.2\overline{35}\)
Solution
\[ \begin{aligned} \text{Let } x & = 1.2\overline{35} \color{magenta} \longrightarrow \enclose{circle}{1} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{1} \text{ by }10 \text{ on both sides.}\\[5pt] 10 x & = 12.\overline{35} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{2} \text{ by }100 \text{ on both sides.}\\[5pt] 1000 x & = 1235.\overline{35} \color{magenta} \longrightarrow \enclose{circle}{3} \\[8pt] \textbf{Subtract } Eq \ & \enclose{circle}{2} \text{ from } Eq \ \enclose{circle}{3} \\[8pt] 1000x - 10x & = 1235.\overline{35} - 12.\overline{35} \\[8pt] 990x & = 1223 \\[8pt] x & = \frac{1223}{990} \\[8pt] \color{red} 1.2\overline{35} & = \color{red} \frac{1223}{990} \\[8pt] \end{aligned} \]
(v) \(0.\overline{23}\)
Solution
\[ \begin{aligned} \text{Let } x & = 0.\overline{23} \color{magenta} \longrightarrow \enclose{circle}{1} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{1} \text{ by }100 \text{ on both sides.}\\[5pt] 100 x & = 23.\overline{23} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \textbf{Subtract } Eq \ & \enclose{circle}{1} \text{ from } Eq \ \enclose{circle}{2} \\[8pt] 100x - x & = 23.\overline{23} - 0.\overline{23} \\[8pt] 99x & = 23 \\[8pt] x & = \frac{23}{99} \\[8pt] \color{red} 0.\overline{23} & = \color{red} \frac{23}{99} \\[8pt] \end{aligned} \]
(vi) \(2.0\overline{5}\)
Solution
\[ \begin{aligned} \text{Let } x & = 2.0\overline{5} \color{magenta} \longrightarrow \enclose{circle}{1} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{1} \text{ by }10 \text{ on both sides.}\\[5pt] 10 x & = 20.\overline{5} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{2} \text{ by }10 \text{ on both sides.}\\[5pt] 100 x & = 205.\overline{5} \color{magenta} \longrightarrow \enclose{circle}{3} \\[8pt] \textbf{Subtract } Eq \ & \enclose{circle}{2} \text{ from } Eq \ \enclose{circle}{3} \\[8pt] 100x - 10x & = 205.\overline{5} - 20.\overline{5} \\[8pt] 90x & = 185 \\[8pt] x & = \frac{185}{90} \\[8pt] x & = \frac{37}{18} \\[8pt] \color{red} 2.0\overline{5} & = \color{red} \frac{37}{18} \\[8pt] \end{aligned} \]
(vii) \(2.12\overline{5}\)
Solution
\[ \begin{aligned} \text{Let } x & = 2.12\overline{5} \color{magenta} \longrightarrow \enclose{circle}{1} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{1} \text{ by }100 \text{ on both sides.}\\[5pt] 100 x & = 212.\overline{5} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{2} \text{ by }10 \text{ on both sides.}\\[5pt] 1000 x & = 2125.\overline{5} \color{magenta} \longrightarrow \enclose{circle}{3} \\[8pt] \textbf{Subtract } Eq \ & \enclose{circle}{2} \text{ from } Eq \ \enclose{circle}{3} \\[8pt] 1000x - 100x & = 2125.\overline{5} - 212.\overline{5} \\[8pt] 900x & = 1913 \\[8pt] x & = \frac{1913}{900} \\[8pt] \color{red} 2.12\overline{5} & = \color{red} \frac{1913}{900} \\[8pt] \end{aligned} \]
(viii) \(3.12\overline{5}\)
Solution
\[ \begin{aligned} \text{Let } x & = 3.12\overline{5} \color{magenta} \longrightarrow \enclose{circle}{1} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{1} \text{ by }100 \text{ on both sides.}\\[5pt] 100 x & = 312.\overline{5} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{2} \text{ by }10 \text{ on both sides.}\\[5pt] 1000 x & = 3125.\overline{5} \color{magenta} \longrightarrow \enclose{circle}{3} \\[8pt] \textbf{Subtract } Eq \ & \enclose{circle}{2} \text{ from } Eq \ \enclose{circle}{3} \\[8pt] 1000x - 100x & = 3125.\overline{5} - 312.\overline{5} \\[8pt] 900x & = 2813 \\[8pt] x & = \frac{2813}{900} \\[8pt] \color{red} 3.12\overline{5} & = \color{red} \frac{2813}{900} \\[8pt] \end{aligned} \]
(ix) \(2.\overline{1625}\)
Solution
\[ \begin{aligned} \text{Let } x & = 2.\overline{1625} \color{magenta} \longrightarrow \enclose{circle}{1} \\[8pt] \text{Multiply} & \ Eq \ \enclose{circle}{1} \text{ by }10000 \text{ on both sides.}\\[5pt] 10000 x & = 21625.\overline{1625} \color{magenta} \longrightarrow \enclose{circle}{2} \\[8pt] \textbf{Subtract } Eq \ & \enclose{circle}{1} \text{ from } Eq \ \enclose{circle}{2} \\[8pt] 10000x - x & = 21625.\overline{1625} - 2.\overline{1625} \\[8pt] 9999x & = 21623 \\[8pt] x & = \frac{21623}{9999} \\[8pt] \color{red} 2.\overline{1625} & = \color{red} \frac{21623}{9999} \\[8pt] \end{aligned} \]
4.
Locate the following rational numbers on the number line:
(i) 0.532
(ii) \(1.1\overline{5}\)
Solution
(i) 0.532
0 < 0.532 < 10.53 < 0.532 < 0.54
(ii) \(1.1\overline{5} = 1.15555...\)
1 < 1.15555.. < 21.15 < 1.15555.. < 1.16
5. Find 6 rational numbers between 3 and 4.
Solution
\[ \begin{aligned} \text{Rational numbers} & =\frac{3}{1} \text{ and } \frac{4}{1} \\[8pt] \text{Multiply both numerator} & \text{ and denominator by } 7 \\[8pt] \frac{3 \times 7}{1 \times 7} & =\frac{21}{7} \\[8pt] \frac{4 \times 7}{1 \times 7} & =\frac{28}{7} \\[8pt] \text{Six Rational numbers} & \text{ between } \frac{21}{7} \text{ and } \frac{28}{7} \\[8pt] \implies \color{red} \frac{22}{7}, \frac{23}{7}, \frac{24}{7}, & \color{red} \frac{25}{7}, \frac{26}{7}, \frac{27}{7} \\[8pt] \end{aligned} \]
6. Find 5 rational numbers between \(\dfrac{2}{5}\) and \(\dfrac{3}{5}\).
Solution
\[ \begin{aligned} \frac{2}{5} & \text{ and } \frac{3}{5} \\[8pt] \text{Multiply both numerator} & \text{ and denominator by } 6 \\[8pt] \frac{2 \times 6}{5 \times 6} & =\frac{12}{30} \\[8pt] \frac{3 \times 6}{5 \times 6} & =\frac{18}{30} \\[8pt] \text{5 Rational numbers} & \text{ between } \frac{12}{30} \text{ and } \frac{18}{30} \\[8pt] \implies \color{red} \frac{13}{30}, \frac{14}{30}, & \color{red} \frac{15}{30}, \frac{16}{30}, \frac{17}{30} \\[8pt] \end{aligned} \]
7. Find 5 rational numbers between \(\dfrac{1}{6}\) and \(\dfrac{2}{5}\).
Solution
\[ \begin{aligned} \frac{1}{6} & \text{ and } \frac{2}{5} \\[8pt] \text{Find common} & \text{ denominator} \\[8pt] \frac{1 \times 5}{6 \times 5} & =\frac{5}{30} \\[8pt] \frac{2 \times 6}{5 \times 6} & =\frac{12}{30} \\[8pt] \text{5 Rational numbers} & \text{ between } \frac{1}{6} \text{ and } \frac{2}{5} \\[8pt] \implies \color{red} \frac{6}{30}, \frac{7}{30}, & \color{red} \frac{8}{30}, \frac{9}{30}, \frac{10}{30} \\[8pt] \textbf{Simplest form:} \\[8pt] \implies \color{red} \frac{1}{5}, \frac{7}{30}, & \color{red} \frac{4}{15}, \frac{3}{10}, \frac{1}{3} \\[8pt] \end{aligned} \]
8. If \(\dfrac{x}{3} + \dfrac{x}{5} = \dfrac{16}{15}\), find the rational number \(x\).
Solution
\[ \begin{aligned} \frac{x}{3} + \frac{x}{5} &= \frac{16}{15} \\[5pt] \frac{5x + 3x}{15} &= \frac{16}{15} \\[5pt] \frac{8x}{15} &= \frac{16}{15} \\[5pt] 8x &= 16 \\[5pt] \color{red} x &= \color{red} 2 \end{aligned} \]
9. Let \(a\) and \(b\) be two non-zero rational numbers such that \(a + \dfrac{1}{b} = 0\). Without assigning any numerical values, determine whether \(ab\) is positive or negative. Justify your answer.
Solution
\[ \begin{aligned} a + \frac{1}{b} &= 0 \\[5pt] a &= -\frac{1}{b} \\[5pt] ab &= -1 \end{aligned} \] Answer Since \(ab = -1\), the product \(ab\) is strictly negative.
10. A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form \(\dfrac{p}{10^4}\) where \(p\) is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by \(2^4\) or \(5^4\)? Give reasons.
Solution
Let the rational number be 𝑥.
Its a terminates decimal and has last non-zero digit exactly at the 4th decimal place.
\( \therefore x = \dfrac{p}{10^4}\)
Because the 4th decimal place was non-zero, the integer \(p\) does not end in zero, so \(p\) is not divisible by 10.
\( \dfrac{p}{10^4} = \dfrac{p}{2^4 \times 5^4}\)
Example:
\[
\begin{aligned}
\frac{1234}{10^4} &= \frac{1234}{10000} \\[8pt]
&= \frac{617}{5000} \\[8pt]
&= \frac{617}{2^3 \times 5^4}
\end{aligned}
\]
No it is not necessary that the denominator of this rational number, when written in the lowest form, is divisible by \(2^4\) or \(5^4\), It may or may not be divisible by \(2^4\) or \(5^4\).
However the denominator in lowest form will always be in the form of \(2^m \times 5^n\).
11. Without performing division, determine whether the decimal expansion of \(\dfrac{18}{125}\) is terminating or non-terminating. If it terminates, state the number of decimal places.
Solution
\[
\begin{aligned}
\frac{18}{125} &= \frac{18}{5^3}
\end{aligned}
\]
The prime factorization of the denominator consists only 5, the decimal expansion is terminating.
\[
\begin{aligned}
\frac{18 \times 8}{125 \times 8} &= \frac{144}{1000} \implies = 0.144
\end{aligned}
\]
Number of decimal places = 3 decimal places.
12. A rational number in its lowest form has denominator \(2^3 \times 5\). How many decimal places will its decimal expansion have? Explain your answer.
Solution
Denominator = \(2^3 \times 5^1\). The maximum power is 3.
We need to multiply the numerator and denominator by \(5^2\) to make the denominator of \(10^3 = 1000\).
The decimal expansion will have 3 decimal places.
13. Let \(a = \dfrac{7}{12}\) and \(b = \dfrac{5}{6}\). Express both 𝑎 and 𝑏 in the form \(\dfrac{k_1}{m}\) and \(\dfrac{k_2}{m}\) where 𝑘₁ , 𝑘₂ and 𝓂 are integers and 𝑘₂ - 𝑘₁ > 6. Using the same denominator 𝓂, write exactly five distinct rational numbers lying between \(a\) and \(b\) keeping an integer numerator. Explain why the condition 𝑘₂ - 𝑘₁ > 𝓃 + 1 is necessary to find 𝓃 such rational numbers.
Solution
\[
\begin{aligned}
a = \frac{7}{12} \ &, \ b = \frac{5}{6} \\[8pt]
\text{Making } & \text{denominator same} \\
a & = \frac{7 \times 1}{12 \times 1} \implies \frac{7}{12} \\[8pt]
b & = \frac{5 \times 2}{6 \times 2} \implies \frac{10}{12} \\[8pt]
a & = \frac{7}{12} = \frac{k_1}{m} \\[8pt]
b & = \frac{10}{12} = \frac{k_2}{m} \\[8pt]
k_2 - k_1 & = 10 - 7 \implies 3 \\[8pt]
3 & \cancel>6 \\ \\
& \text{Multiply both by } 3 \\[5pt]
a & = \frac{7 \times 3}{12 \times 3} \implies \color{green} \frac{21}{36} \\[8pt]
b & = \frac{10 \times 3}{12 \times 3} \implies \color{green} \frac{30}{36} \\[8pt]
k_2 - k_1 & = 30 - 21 \implies 3 \\[8pt]
9 > 6 & \textbf{ (Condition satisfied)} \\
m=36 & \ , \ \ k_1=21 \ , \ k_2=30 \\ \\
\textbf{Rational } & \textbf{numbers beween } \frac{21}{36} \text{ and } \frac{30}{36}\\\\
\color{red} \dfrac{22}{36}, & \color{red}\dfrac{23}{36}, \dfrac{24}{36}, \dfrac{25}{36}, \dfrac{26}{36}
\end{aligned}
\]
\( \color{red} k_2 - k_1 > n+1 \) necessary to get 𝑛 rational numbers which is between \( \dfrac{k_1}{m}\) and \(\dfrac{k_2}{m}\). We need 𝑛 integer numerators beween 𝑘₁ and 𝑘₂.
14. Three rational numbers 𝑥 , 𝑦 , 𝓏 satisfy 𝑥 + 𝑦 + 𝓏 = 0 and 𝑥𝑦 + 𝑦𝓏 + 𝓏𝑥 = 0. Show that all the rational numbers 𝑥 , 𝑦 , 𝓏 must be simultaneously zero.
Solution
\[
\begin{aligned}
\textbf{Given: } x+y+z = 0 \ , & \quad xy+yz+zx = 0 \\[8pt]
\color{magenta} (x + y + z)^2 &= \color{magenta} x^2 + y^2 + z^2 + 2(xy + yz + zx) \\[5pt]
(0)^2 &= x^2 + y^2 + z^2 + 2(0) \\[5pt]
0 &= x^2 + y^2 + z^2 \\[5pt]
x^2 & = 0 \implies x = 0 \\[5pt]
y^2 & = 0 \implies y = 0 \\[5pt]
z^2 & = 0 \implies z = 0 \\[5pt]
\therefore \color{red} x \ & \color{red}= y = z \implies 0
\end{aligned}
\]
𝑥 , 𝑦 , 𝓏 are rational numbers, their squares cannot be negative.
Hence all the rational numbers 𝑥 , 𝑦 , 𝓏 must be simultaneously zero.
15. Show that the rational number \(\dfrac{(a+b)}{2}\) lies between the rational numbers 𝑎 and 𝑏.
Solution
𝑎 and 𝑏 are two rational numbers. \[ \begin{array}{c|c} \begin{aligned} & a < b \\ \\ \textbf{Add } {\color{brown}a} & \textbf{ on both sides} \\ {\color{brown}a} + a & < {\color{brown}a} + b \\ 2a & < a + b \\[8pt] \textbf{Divide by } {\color{brown}2} & \textbf{ on both sides} \\[8pt] \frac{2a}{{\color{brown}2}} & < \frac{a + b}{{\color{brown}2}} \\[8pt] a & < \frac{a + b}{2} \color{magenta} \longrightarrow \enclose{circle}{1} \\ \end{aligned} & \begin{aligned} & a < b \\ \\ \textbf{Add } {\color{brown}b} & \textbf{ on both sides} \\ a + {\color{brown}b} & < b + {\color{brown}b} \\ a + b & < 2b \\[8pt] \textbf{Divide by } {\color{brown}2} & \textbf{ on both sides} \\[8pt] \frac{a + b}{{\color{brown}2}} & <\frac{2b}{{\color{brown}2}} \\[8pt] \frac{a + b}{2} & < b \color{magenta} \longrightarrow \enclose{circle}{2} \\ \end{aligned}\\ \end{array}\\ \] \[ \begin{aligned} \textbf{From } & \enclose{circle}{1} \text{ and } \enclose{circle}{2} \\[8pt] a & < \frac{a + b}{2} < b \\[8pt] \end{aligned} \]
\( \therefore \color{red} \dfrac{(a+b)}{2}\) lies between the rational numbers 𝑎 and 𝑏.
16. Find the lengths of the hypotenuses of all the right triangles in below figure which is referred to as the square root spiral.
Solution
The square root spiral is built using right triangles where one side is always 1, and the other leg is the hypotenuse of the previous triangle.
By Pythagorean theorem (\(Hypotenuse = \sqrt{(base)^2 + (perpendicular)^2} \))
Triangle 1 : Hypotenuse = \(\sqrt{1^2 + 1^2} \implies \color{red} \sqrt{2}\)
Triangle 2 : Hypotenuse = \(\sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} \implies \color{red} \sqrt{3}\)
Triangle 3: Hypotenuse = \(\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} \implies \color{red} \sqrt{4} \)
Triangle 4: Hypotenuse = \(\color{red} \sqrt{5}\) ...and so on.
\(\therefore \)The lengths of the hypotenuses of all the right triangles in above figure
\( = \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \sqrt{6} , \sqrt{7} , \sqrt{8} , \sqrt{9} , \sqrt{10} , \sqrt{11}\).