Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 1 Rational And Irrational Numbers

Question 1. Find two rational numbers between:
(i) \( \frac{3}{8} \) and \( \frac{7}{12} \)
(ii) \( \frac{1}{3} \) and \( \frac{1}{4} \)
Answer:
(i) To determine a rational number lying between any two numbers \( x \) and \( y \), we can use the formula:
\[ \frac{x + y}{2} \]
Applying this to find the first rational number between \( \frac{3}{8} \) and \( \frac{7}{12} \):
\[ \frac{\frac{3}{8} + \frac{7}{12}}{2} = \frac{\frac{9 + 14}{24}}{2} = \frac{\frac{23}{24}}{2} = \frac{23}{48} \]
Next, to find a second rational number, we calculate one between \( \frac{3}{8} \) and \( \frac{23}{48} \):
\[ \frac{\frac{3}{8} + \frac{23}{48}}{2} = \frac{\frac{18 + 23}{48}}{2} = \frac{\frac{41}{48}}{2} = \frac{41}{96} \]
Thus, the two rational numbers between \( \frac{3}{8} \) and \( \frac{7}{12} \) are \( \frac{23}{48} \) and \( \frac{41}{96} \).
We can express their order as:
\[ \frac{3}{8} < \frac{41}{96} < \frac{23}{48} < \frac{7}{12} \]

(ii) Using the same relation \( \frac{x + y}{2} \), we find the first rational number between \( \frac{1}{3} \) and \( \frac{1}{4} \):
\[ \frac{\frac{1}{3} + \frac{1}{4}}{2} = \frac{\frac{4 + 3}{12}}{2} = \frac{7}{24} \]
Next, we calculate another rational number between \( \frac{7}{24} \) and \( \frac{1}{4} \):
\[ \frac{\frac{7}{24} + \frac{1}{4}}{2} = \frac{\frac{7 + 6}{24}}{2} = \frac{13}{48} \]
Consequently, the two rational numbers between \( \frac{1}{3} \) and \( \frac{1}{4} \) are \( \frac{7}{24} \) and \( \frac{13}{48} \).
These are arranged as:
\[ \frac{1}{4} < \frac{13}{48} < \frac{7}{24} < \frac{1}{3} \]
In simple words: To find a rational number between two fractions, add them together and divide by two. We repeat this process to get a second number.

Exam Tip: Always perform the fraction addition carefully by finding the lowest common multiple (LCM) of the denominators before dividing by 2.

Question 2. Find three rational numbers between:
(i) \( \frac{2}{5} \) and \( \frac{3}{7} \)
(ii) \( \frac{4}{11} \) and \( \frac{9}{16} \)
Answer:
(i) First, let's equate the denominators of both fractions by finding the LCM of 5 and 7, which is 35.
\[ \frac{2}{5} = \frac{2 \times 7}{5 \times 7} = \frac{14}{35} \]
\[ \frac{3}{7} = \frac{3 \times 5}{7 \times 5} = \frac{15}{35} \]
Since we need to find three rational numbers, let's expand these fractions by multiplying their numerators and denominators by 5:
\[ \frac{2}{5} = \frac{2 \times 7 \times 5}{5 \times 7 \times 5} = \frac{70}{175} \]
\[ \frac{3}{7} = \frac{3 \times 5 \times 5}{7 \times 5 \times 5} = \frac{75}{175} \]
Since \( \frac{70}{175} < \frac{75}{175} \), we can select three intermediate fractions:
\[ \frac{70}{175} < \frac{71}{175} < \frac{72}{175} < \frac{73}{175} < \frac{74}{175} < \frac{75}{175} \]
Choosing three rational numbers, we get:
\[ \frac{2}{5} < \frac{71}{175} < \frac{72}{175} < \frac{73}{175} < \frac{3}{7} \]

(ii) The LCM of 11 and 16 is 176. Let us convert both fractions:
\[ \frac{4}{11} = \frac{4 \times 16}{11 \times 16} = \frac{64}{176} \]
\[ \frac{9}{16} = \frac{9 \times 11}{16 \times 11} = \frac{99}{176} \]
Since \( \frac{64}{176} < \frac{99}{176} \), we can easily choose three numbers between them:
\[ \frac{64}{176} < \frac{65}{176} < \frac{66}{176} < \frac{67}{176} < \frac{99}{176} \]
Thus, the three rational numbers between \( \frac{4}{11} \) and \( \frac{9}{16} \) are:
\[ \frac{65}{176}, \frac{66}{176} \text{ and } \frac{67}{176} \]
These can be arranged as:
\[ \frac{4}{11} < \frac{65}{176} < \frac{66}{176} < \frac{67}{176} < \frac{9}{16} \]
In simple words: Find a common denominator for the two fractions. If the numerators are consecutive, multiply the top and bottom of both fractions by a larger number to create a gap, then pick any three numbers in between.

Exam Tip: Always state the final order clearly using the inequality signs to demonstrate a clear understanding of their relative sizes.

Question 3. Find three rational numbers between:
(i) \( 5 \) and \( -2 \)
(ii) \( -\frac{3}{4} \) and \( \frac{1}{2} \)
Answer:
(i) Both 5 and -2 are integers, making them rational numbers. Since integers are a subset of rational numbers, we can find many integers between -2 and 5:
\[ -2 < -1 < 0 < 1 < 2 < 3 < 4 < 5 \]
Selecting any three of these, we can choose:
\[ -1, 0, \text{ and } 1 \]

(ii) First, find a common denominator for \( -\frac{3}{4} \) and \( \frac{1}{2} \). The LCM of 4 and 2 is 4. This gives us:
\[ -\frac{3}{4} \text{ and } \frac{1 \times 2}{2 \times 2} = \frac{2}{4} \]
Since \( -\frac{3}{4} < \frac{2}{4} \), the rational numbers between them are:
\[ -\frac{3}{4} < -\frac{2}{4} < -\frac{1}{4} < 0 < \frac{1}{4} < \frac{2}{4} \]
Selecting three rational numbers between them:
\[ -\frac{2}{4}, -\frac{1}{4}, \text{ and } \frac{1}{4} \]
Simplifying \( -\frac{2}{4} \) to \( -\frac{1}{2} \), the three rational numbers are:
\[ -\frac{1}{2}, -\frac{1}{4}, \text{ and } \frac{1}{4} \]
These are arranged as:
\[ -\frac{3}{4} < -\frac{1}{2} < -\frac{1}{4} < \frac{1}{4} < \frac{1}{2} \]
In simple words: Convert the fractions to have the same denominator, then find three values that lie strictly between their numerators.

Exam Tip: Always reduce any final fractions (like \( -\frac{2}{4} \) to \( -\frac{1}{2} \)) to their simplest form to earn full credit.

Question 4. Insert 4 rational numbers between 5 and 8.
Answer:
Let the given rational numbers be \( x = 5 \) and \( y = 8 \). Since \( 5 < 8 \), we need to insert \( n = 4 \) rational numbers. First, we find the common difference \( d \):
\[ d = \frac{y - x}{n + 1} = \frac{8 - 5}{4 + 1} = \frac{3}{5} \]
The four rational numbers are calculated as follows:
\[ x + d = 5 + \frac{3}{5} = \frac{28}{5} = 5\frac{3}{5} \]
\[ x + 2d = 5 + 2\left(\frac{3}{5}\right) = 5 + \frac{6}{5} = \frac{31}{5} = 6\frac{1}{5} \]
\[ x + 3d = 5 + 3\left(\frac{3}{5}\right) = 5 + \frac{9}{5} = \frac{34}{5} = 6\frac{4}{5} \]
\[ x + 4d = 5 + 4\left(\frac{3}{5}\right) = 5 + \frac{12}{5} = \frac{37}{5} = 7\frac{2}{5} \]
Therefore, the required four rational numbers are:
\[ 5\frac{3}{5}, 6\frac{1}{5}, 6\frac{4}{5}, \text{ and } 7\frac{2}{5} \]
In simple words: To find 4 equally spaced numbers between 5 and 8, we calculate a step size \( d \) by dividing the difference of the numbers by 5. Then we keep adding this step size starting from 5.

Exam Tip: Using the \( d = \frac{y-x}{n+1} \) formula is a highly efficient way to insert any number of rational numbers with equal spacing.

Question 5. Insert 5 rational numbers between \( \frac{1}{3} \) and \( \frac{5}{9} \).
Answer:
Let \( x = \frac{1}{3} \) and \( y = \frac{5}{9} \). Since \( \frac{1}{3} < \frac{5}{9} \), we need to insert \( n = 5 \) rational numbers. First, compute the common difference \( d \):
\[ d = \frac{y - x}{n + 1} = \frac{\frac{5}{9} - \frac{1}{3}}{5 + 1} = \frac{\frac{5 - 3}{9}}{6} = \frac{2}{9 \times 6} = \frac{1}{27} \]
Now, we calculate the five required rational numbers:
\[ x + d = \frac{1}{3} + \frac{1}{27} = \frac{9 + 1}{27} = \frac{10}{27} \]
\[ x + 2d = \frac{1}{3} + \frac{2}{27} = \frac{9 + 2}{27} = \frac{11}{27} \]
\[ x + 3d = \frac{1}{3} + \frac{3}{27} = \frac{9 + 3}{27} = \frac{12}{27} = \frac{4}{9} \]
\[ x + 4d = \frac{1}{3} + \frac{4}{27} = \frac{9 + 4}{27} = \frac{13}{27} \]
\[ x + 5d = \frac{1}{3} + \frac{5}{27} = \frac{9 + 5}{27} = \frac{14}{27} \]
Thus, the five rational numbers are:
\[ \frac{10}{27}, \frac{11}{27}, \frac{4}{9}, \frac{13}{27}, \text{ and } \frac{14}{27} \]
In simple words: Find the step size by subtracting the two fractions and dividing the result by 6. Add this step size to the starting fraction 1, 2, 3, 4, and 5 times to get the five numbers.

Exam Tip: Always simplify the intermediate fractions (like \( \frac{12}{27} \) to \( \frac{4}{9} \)) to write them in their simplest form.

Question 6. Insert six rational numbers between 4.6 and 8.4.
Answer:
The two given numbers are \( 4.6 \) and \( 8.4 \). We can express these as fractions:
\[ 4.6 = \frac{46}{10}, \quad 8.4 = \frac{84}{10} \]
Let's find intermediate numbers by finding the midpoint between consecutive boundaries:
\[ \frac{\frac{46}{10} + \frac{84}{10}}{2} = \frac{\frac{130}{10}}{2} = \frac{130}{20} \]
Using a step-by-step approach, we find:
\[ \frac{46}{10} < \frac{130 + 84}{20 \times 2} \dots \]
By generating values evenly between 4.6 and 8.4, we get:
\[ 4.6 < 5.6 < 5.9 < 6.1 < 6.5 < 6.9 < 7.1 < 8.4 \]
Thus, the six rational numbers inserted are:
\[ 5.6, 5.9, 6.1, 6.5, 6.9, \text{ and } 7.1 \]
In simple words: To find six decimal numbers between 4.6 and 8.4, we can choose any six decimal values that lie between these two endpoints, such as 5.6, 5.9, 6.1, 6.5, 6.9, and 7.1.

Exam Tip: When working with decimals, you can directly write down values within the range as long as they are distinct and rational.

Question 7. Insert 7 rational numbers between 1 and 2.
Answer:
Let \( x = 1 \), \( y = 2 \), and the number of rational numbers to insert be \( n = 7 \). First, find the common difference \( d \):
\[ d = \frac{y - x}{n + 1} = \frac{2 - 1}{7 + 1} = \frac{1}{8} \]
Now, we calculate each of the 7 rational numbers:
\[ x + d = 1 + \frac{1}{8} = \frac{9}{8} = 1\frac{1}{8} \]
\[ x + 2d = 1 + 2\left(\frac{1}{8}\right) = 1 + \frac{1}{4} = \frac{5}{4} = 1\frac{1}{4} \]
\[ x + 3d = 1 + 3\left(\frac{1}{8}\right) = 1 + \frac{3}{8} = \frac{11}{8} = 1\frac{3}{8} \]
\[ x + 4d = 1 + 4\left(\frac{1}{8}\right) = 1 + \frac{1}{2} = \frac{3}{2} = 1\frac{1}{2} \]
\[ x + 5d = 1 + 5\left(\frac{1}{8}\right) = 1 + \frac{5}{8} = \frac{13}{8} = 1\frac{5}{8} \]
\[ x + 6d = 1 + 6\left(\frac{1}{8}\right) = 1 + \frac{3}{4} = \frac{7}{4} = 1\frac{3}{4} \]
\[ x + 7d = 1 + 7\left(\frac{1}{8}\right) = 1 + \frac{7}{8} = \frac{15}{8} = 1\frac{7}{8} \]
Thus, the required seven rational numbers are:
\[ 1\frac{1}{8}, 1\frac{1}{4}, 1\frac{3}{8}, 1\frac{1}{2}, 1\frac{5}{8}, 1\frac{3}{4}, \text{ and } 1\frac{7}{8} \]
In simple words: To find 7 equally spaced fractions between 1 and 2, split the interval into 8 parts of size \( \frac{1}{8} \) each. Then add \( \frac{1}{8} \) repeatedly to 1.

Exam Tip: Convert improper fractions to mixed fractions to match the standard format expected in textbooks.

Question 8. Insert 8 rational numbers between 1.8 and 3.6.
Answer:
Let \( x = 1.8 \), \( y = 3.6 \), and \( n = 8 \). Calculate the common difference \( d \):
\[ d = \frac{y - x}{n + 1} = \frac{3.6 - 1.8}{8 + 1} = \frac{1.8}{9} = 0.2 \]
Using \( d = 0.2 \), the eight rational numbers are calculated below:
\[ x + d = 1.8 + 0.2 = 2.0 \]
\[ x + 2d = 1.8 + 2(0.2) = 1.8 + 0.4 = 2.2 \]
\[ x + 3d = 1.8 + 3(0.2) = 1.8 + 0.6 = 2.4 \]
\[ x + 4d = 1.8 + 4(0.2) = 1.8 + 0.8 = 2.6 \]
\[ x + 5d = 1.8 + 5(0.2) = 1.8 + 1.0 = 2.8 \]
\[ x + 6d = 1.8 + 6(0.2) = 1.8 + 1.2 = 3.0 \]
\[ x + 7d = 1.8 + 7(0.2) = 1.8 + 1.4 = 3.2 \]
\[ x + 8d = 1.8 + 8(0.2) = 1.8 + 1.6 = 3.4 \]
The required eight rational numbers are:
\[ 2.0, 2.2, 2.4, 2.6, 2.8, 3.0, 3.2, \text{ and } 3.4 \]
In simple words: Find the difference between 3.6 and 1.8 (which is 1.8), and divide it by 9 to find the step size of 0.2. Start at 1.8 and add 0.2 repeatedly.

Exam Tip: When endpoints are decimal numbers, finding a decimal step size \( d \) makes the calculation very simple and quick.

Question 9. Arrange the numbers \( -\frac{5}{9}, \frac{7}{12}, -\frac{2}{3} \) and \( \frac{11}{18} \) in ascending order. Find the difference between the largest and the smallest of these numbers; and express this difference as a decimal fraction correct to one decimal place.
Answer:
Let's first convert the fractions to have a common denominator. The LCM of 9, 12, 3, and 18 is 36. Now, we convert each fraction:
\[ -\frac{5}{9} = \frac{-5 \times 4}{9 \times 4} = -\frac{20}{36} \]
\[ \frac{7}{12} = \frac{7 \times 3}{12 \times 3} = \frac{21}{36} \]
\[ -\frac{2}{3} = \frac{-2 \times 12}{3 \times 12} = -\frac{24}{36} \]
\[ \frac{11}{18} = \frac{11 \times 2}{18 \times 2} = \frac{22}{36} \]
Arranging these in ascending order based on their numerators:
\[ -\frac{24}{36} < -\frac{20}{36} < \frac{21}{36} < \frac{22}{36} \]
Substituting the original fractions back:
\[ -\frac{2}{3} < -\frac{5}{9} < \frac{7}{12} < \frac{11}{18} \]
The largest number is \( \frac{11}{18} \) and the smallest is \( -\frac{2}{3} \). We find their difference:
\[ \text{Difference} = \frac{11}{18} - \left(-\frac{2}{3}\right) = \frac{11}{18} + \frac{2}{3} \]
Making denominators equal:
\[ \frac{11}{18} + \frac{2 \times 6}{3 \times 6} = \frac{11}{18} + \frac{12}{18} = \frac{23}{18} \]
To write this as a decimal:
\[ \frac{23}{18} = 1.2\overline{7} \approx 1.3 \text{ (correct to one decimal place)} \]
In simple words: First, use the LCM to make the denominators of all fractions the same so we can sort them. Then, subtract the smallest number from the largest, and divide to get the decimal value.

Exam Tip: Pay careful attention to negative signs when calculating the difference between the largest and smallest numbers, as subtracting a negative number becomes addition.

Question 10. Arrange the numbers \( \frac{5}{8}, -\frac{3}{16}, -\frac{1}{4} \) and \( \frac{17}{32} \) in descending order. Find the sum of the largest and the smallest of these numbers; and express this sum as a decimal fraction correct to two decimal places.
Answer:
Let us find the LCM of the denominators: 8, 16, 4, and 32. The LCM is 32. Converting each fraction:
\[ \frac{5}{8} = \frac{5 \times 4}{8 \times 4} = \frac{20}{32} \]
\[ -\frac{3}{16} = \frac{-3 \times 2}{16 \times 2} = -\frac{6}{32} \]
\[ -\frac{1}{4} = \frac{-1 \times 8}{4 \times 8} = -\frac{8}{32} \]
\[ \frac{17}{32} = \frac{17}{32} \]
Arranging these in descending order:
\[ \frac{20}{32} > \frac{17}{32} > -\frac{6}{32} > -\frac{8}{32} \]
Replacing them with their original values:
\[ \frac{5}{8} > \frac{17}{32} > -\frac{3}{16} > -\frac{1}{4} \]
The largest number is \( \frac{5}{8} \) and the smallest is \( -\frac{1}{4} \). Now, calculate their sum:
\[ \text{Sum} = \frac{5}{8} + \left(-\frac{1}{4}\right) = \frac{5}{8} - \frac{1 \times 2}{4 \times 2} = \frac{5}{8} - \frac{2}{8} = \frac{3}{8} \]
Expressing \( \frac{3}{8} \) as a decimal:
\[ \frac{3}{8} = 0.375 \approx 0.38 \text{ (correct to two decimal places)} \]
In simple words: Make the denominators equal to easily see which fraction is largest and which is smallest. Add the biggest and smallest together, then convert that fraction to a decimal.

Exam Tip: Remember that for negative numbers, a larger absolute value means a smaller total value (e.g., \( -\frac{8}{32} \) is smaller than \( -\frac{6}{32} \)).

Exercise 1(B)

Question 1. State, which of the following decimal numbers are pure recurring decimals and which are mixed recurring decimals:
(i) \( 0.\overline{083} \)
(ii) \( 0.0\overline{83} \)
(iii) \( 0.\overline{227} \)
(iv) \( 3.5\dot{4} \)
(v) \( 2.\dot{8}\dot{1} \)
Answer:
A recurring decimal is called a pure recurring decimal if every single digit after the decimal point repeats. If some digits after the decimal point do not repeat while others do, it is called a mixed recurring decimal.
Based on this:
(i) \( 0.\overline{083} \): Every digit after the decimal point repeats. This is a **pure recurring decimal**.
(ii) \( 0.0\overline{83} \): The digit 0 does not repeat, but 83 does. This is a **mixed recurring decimal**.
(iii) \( 0.\overline{227} \): Every digit after the decimal point repeats. This is a **pure recurring decimal**.
(iv) \( 3.5\dot{4} \): The digit 5 does not repeat, but 4 does. This is a **mixed recurring decimal**.
(v) \( 2.\dot{8}\dot{1} \): Every digit after the decimal point repeats. This is a **pure recurring decimal**.
In simple words: If the repeating bar covers everything after the decimal point, it is pure. If there are some numbers after the decimal point that do not have a bar or dot, it is mixed.

Exam Tip: Always look closely at the starting point of the bar or dot to identify which digits are non-repeating.

Question 2. Represent as a decimal number:
(i) \( \frac{4}{15} \)
(ii) \( \frac{2}{7} \)
(iii) \( \frac{4}{9} \)
(iv) \( \frac{5}{24} \)
(v) \( \frac{8}{13} \)
Answer:
To convert these fractions into decimal notation, we divide the numerator by the denominator:
(i) \( \frac{4}{15} = 0.26666\dots = 0.2\overline{6} \)
(ii) \( \frac{2}{7} = 0.285714285714\dots = 0.\overline{285714} \)
(iii) \( \frac{4}{9} = 0.44444\dots = 0.\overline{4} \)
(iv) \( \frac{5}{24} = 0.2083333\dots = 0.208\dot{3} \)
(v) \( \frac{8}{13} = 0.615384615384\dots = 0.\overline{615384} \)
In simple words: Divide the top number by the bottom number. Put a bar or a dot over the digits that repeat forever.

Exam Tip: Perform the long division until you see the exact pattern of digits repeating to write the recurring bar correctly.

Question 3. Express the following recurring decimals as fractions in their simplest form:
(i) \( 0.5\dot{3} \)
(ii) \( 0.2\overline{27} \)
(iii) \( 0.2\overline{104} \)
(iv) \( 3.5\dot{2} \)
(v) \( 2.24\overline{689} \)
(vi) \( 0.\overline{572} \)
(vii) \( 0.15\dot{8} \)
(viii) \( 0.03\overline{84} \)
Answer:
Let us convert each decimal number into a fraction:

**(i) For \( 0.5\dot{3} \):**
Let \( x = 0.5333\dots \) — (Equation 1)
Since there is one non-repeating digit (5) after the decimal, we multiply both sides of Equation 1 by \( 10^1 = 10 \):
\[ 10x = 5.333\dots \] — (Equation 2)
Now, since only one digit (3) repeats, we multiply Equation 2 by \( 10^1 = 10 \):
\[ 100x = 53.333\dots \] — (Equation 3)
Subtracting Equation 2 from Equation 3:
\[ 100x - 10x = 53.333\dots - 5.333\dots \]
\[ 90x = 48 \]
\[ x = \frac{48}{90} = \frac{8}{15} \]
Thus, \( 0.5\dot{3} = \frac{8}{15} \).

**(ii) For \( 0.2\overline{27} \):**
Let \( x = 0.22727\dots \) — (Equation 1)
Multiply by 10 to shift the non-repeating part:
\[ 10x = 2.2727\dots \] — (Equation 2)
Since two digits repeat, multiply Equation 2 by \( 10^2 = 100 \):
\[ 1000x = 227.2727\dots \] — (Equation 3)
Subtracting Equation 2 from Equation 3:
\[ 1000x - 10x = 227.2727\dots - 2.2727\dots \]
\[ 990x = 225 \]
\[ x = \frac{225}{990} = \frac{5}{22} \]
Thus, \( 0.2\overline{27} = \frac{5}{22} \).

**(iii) For \( 0.2\overline{104} \):**
Let \( x = 0.2104104\dots \) — (Equation 1)
Multiply by 10 to move the non-repeating part:
\[ 10x = 2.104104\dots \] — (Equation 2)
Since three digits repeat, multiply Equation 2 by \( 10^3 = 1000 \):
\[ 10000x = 2104.104104\dots \] — (Equation 3)
Subtracting Equation 2 from Equation 3:
\[ 9990x = 2102 \]
\[ x = \frac{2102}{9990} = \frac{1051}{4995} \]
Thus, \( 0.2\overline{104} = \frac{1051}{4995} \).

**(iv) For \( 3.5\dot{2} \):**
We can write this as:
\[ 3.5\dot{2} = 3 + 0.5\dot{2} \]
For \( 0.5\dot{2} \), let's find the fraction:
\[ \text{Numerator} = 52 - 5 = 47 \]
\[ \text{Denominator} = 90 \]
Thus:
\[ 3.5\dot{2} = 3 + \frac{47}{90} = 3\frac{47}{90} \]

**(v) For \( 2.24\overline{689} \):**
This can be written as:
\[ 2.24\overline{689} = 2 + 0.24\overline{689} \]
For \( 0.24\overline{689} \):
\[ \text{Numerator} = 24689 - 24 = 24665 \]
\[ \text{Denominator} = 99900 \]
Thus:
\[ 2.24\overline{689} = 2 + \frac{24665}{99900} = 2 + \frac{4933}{19980} = 2\frac{4933}{19980} \]

**(vi) For \( 0.\overline{572} \):**
For \( 0.\overline{572} \):
\[ \text{Numerator} = 572 - 0 = 572 \]
\[ \text{Denominator} = 999 \]
Thus:
\[ 0.\overline{572} = \frac{572}{999} \]

**(vii) For \( 0.15\dot{8} \):**
For \( 0.15\dot{8} \):
\[ \text{Numerator} = 158 - 15 = 143 \]
\[ \text{Denominator} = 900 \]
Thus:
\[ 0.15\dot{8} = \frac{143}{900} \]

**(viii) For \( 0.03\overline{84} \):**
For \( 0.03\overline{84} \):
\[ \text{Numerator} = 0384 - 03 = 381 \]
\[ \text{Denominator} = 9990 \]
Thus:
\[ 0.03\overline{84} = \frac{381}{9990} = \frac{127}{3330} \]
In simple words: To turn a repeating decimal into a fraction, write the full number after the decimal point and subtract the non-repeating part to find the numerator. Write a 9 for every repeating digit and a 0 for every non-repeating digit after the decimal point to find the denominator, then simplify.

Exam Tip: Always make sure to divide both the numerator and the denominator by their greatest common divisor (GCD) to write the fraction in its simplest reduced form.

Exercise 1(B)

Question 4. Find the decimal representations of \( \frac{2}{7} \), \( \frac{3}{7} \), \( \frac{4}{7} \), \( \frac{5}{7} \), and \( \frac{6}{7} \), given that \( \frac{1}{7} = 0.\overline{142857} \).
Answer:
Given that:
\[ \frac{1}{7} = 0.142857142857\dots = 0.\overline{142857} \]
We can find the required decimal representations by multiplying \( \frac{1}{7} \) by the respective numerators:
\[ \frac{2}{7} = 2 \times \frac{1}{7} = 2 \times 0.\overline{142857} = 0.\overline{285714} \]
\[ \frac{3}{7} = 3 \times \frac{1}{7} = 3 \times 0.\overline{142857} = 0.\overline{428571} \]
\[ \frac{4}{7} = 4 \times \frac{1}{7} = 4 \times 0.\overline{142857} = 0.\overline{571428} \]
\[ \frac{5}{7} = 5 \times \frac{1}{7} = 5 \times 0.\overline{142857} = 0.\overline{714285} \]
\[ \frac{6}{7} = 6 \times \frac{1}{7} = 6 \times 0.\overline{142857} = 0.\overline{857142} \]
In simple words: We can find these decimals by multiplying the decimal form of \( \frac{1}{7} \) by 2, 3, 4, 5, and 6. The repeating pattern of digits remains the same but starts from a different number.

Exam Tip: Make sure to place the bar correctly over all the six repeating digits in each decimal.

Question 5. Without performing any actual division, find which of the following rational numbers have terminating decimal representation:
(i) \( \frac{7}{16} \)
(ii) \( \frac{23}{125} \)
(iii) \( \frac{9}{14} \)
(iv) \( \frac{32}{45} \)
(v) \( \frac{43}{50} \)
(vi) \( \frac{17}{40} \)
(vii) \( \frac{61}{75} \)
(viii) \( \frac{123}{250} \)
Answer:
(i) We have the fraction \( \frac{7}{16} \). Here, the denominator is \( 16 \). Factorizing \( 16 \), we get \( 16 = 2 \times 2 \times 2 \times 2 = 2^4 \), which is equivalent to \( 2^4 \times 5^0 \). Since the denominator can be written in the form \( 2^m \times 5^n \), the fraction \( \frac{7}{16} \) is a terminating decimal.
(ii) We have the fraction \( \frac{23}{125} \). Here, the denominator is \( 125 \). Since \( 125 = 5 \times 5 \times 5 = 5^3 = 2^0 \times 5^3 \), we can express the denominator in the form \( 2^m \times 5^n \). Thus, \( \frac{23}{125} \) can be expressed as a terminating decimal.
(iii) We have the fraction \( \frac{9}{14} \). Here, the denominator is \( 14 \). Since \( 14 = 2 \times 7 = 2^1 \times 7^1 \), the denominator cannot be expressed in the form \( 2^m \times 5^n \). Therefore, \( \frac{9}{14} \) cannot be converted into a terminating decimal.
(iv) We have the fraction \( \frac{32}{45} \). Here, the denominator is \( 45 \). Since \( 45 = 3 \times 3 \times 5 = 3^2 \times 5^1 \), the denominator cannot be expressed in the form \( 2^m \times 5^n \). Therefore, \( \frac{32}{45} \) cannot be converted into a terminating decimal.
(v) We have the fraction \( \frac{43}{50} \). Here, the denominator is \( 50 \). Since \( 50 = 2 \times 5 \times 5 = 2^1 \times 5^2 \), we can express the denominator in the form \( 2^m \times 5^n \). Thus, \( \frac{43}{50} \) can be expressed as a terminating decimal.
(vi) We have the fraction \( \frac{17}{40} \). Here, the denominator is \( 40 \). Since \( 40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5^1 \), we can express the denominator in the form \( 2^m \times 5^n \). Thus, \( \frac{17}{40} \) can be expressed as a terminating decimal.
(vii) We have the fraction \( \frac{61}{75} \). Here, the denominator is \( 75 \). Since \( 75 = 3 \times 5 \times 5 = 3^1 \times 5^2 \), the denominator cannot be expressed in the form \( 2^m \times 5^n \). Therefore, \( \frac{61}{75} \) cannot be converted into a terminating decimal.
(viii) We have the fraction \( \frac{123}{250} \). Here, the denominator is \( 250 \). Since \( 250 = 2 \times 5 \times 5 \times 5 = 2^1 \times 5^3 \), we can express the denominator in the form \( 2^m \times 5^n \). Thus, \( \frac{123}{250} \) can be expressed as a terminating decimal.
In simple words: If the bottom number of a fraction only has 2s and 5s as its prime factors, it will terminate (end) when written as a decimal. If there is any other prime number like 3 or 7, the decimal will not terminate.

Exam Tip: Remember that any number raised to the power of 0 is 1, so denominators like 16 or 125 can still be written in the form \( 2^m \times 5^n \).

Exercise 1(C)

Question 1. State, whether the following numbers are rational or irrational:
(i) \( (2 + \sqrt{2})^2 \)
(ii) \( (3 - \sqrt{3})^2 \)
(iii) \( (5 + \sqrt{5})(5 - \sqrt{5}) \)
(iv) \( (\sqrt{3} - \sqrt{2})^2 \)
(v) \( \left( \frac{3}{2\sqrt{2}} \right)^2 \)
(vi) \( \left( \frac{\sqrt{7}}{6\sqrt{2}} \right)^2 \)
Answer:
(i) Using the identity \( (a+b)^2 = a^2 + 2ab + b^2 \):
\[ (2 + \sqrt{2})^2 = (2)^2 + 2(2)(\sqrt{2}) + (\sqrt{2})^2 = 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2} \]
Since \( 6 + 4\sqrt{2} \) contains an irrational component, it is an irrational number.
(ii) Using the identity \( (a-b)^2 = a^2 - 2ab + b^2 \):
\[ (3 - \sqrt{3})^2 = (3)^2 - 2(3)(\sqrt{3}) + (\sqrt{3})^2 = 9 - 6\sqrt{3} + 3 = 12 - 6\sqrt{3} \]
Since \( 12 - 6\sqrt{3} \) contains an irrational component, it is an irrational number.
(iii) Using the identity \( (a+b)(a-b) = a^2 - b^2 \):
\[ (5 + \sqrt{5})(5 - \sqrt{5}) = (5)^2 - (\sqrt{5})^2 = 25 - 5 = 20 \]
Since \( 20 \) is an integer, it is a rational number.
(iv) Using the identity \( (a-b)^2 = a^2 - 2ab + b^2 \):
\[ (\sqrt{3} - \sqrt{2})^2 = (\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6} \]
Since \( 5 - 2\sqrt{6} \) contains an irrational component, it is an irrational number.
(v) Simplifying the term:
\[ \left( \frac{3}{2\sqrt{2}} \right)^2 = \frac{(3)^2}{(2\sqrt{2})^2} = \frac{9}{4 \times 2} = \frac{9}{8} \]
Since \( \frac{9}{8} \) is in \( \frac{p}{q} \) form, it is a rational number.
(vi) Simplifying the term:
\[ \left( \frac{\sqrt{7}}{6\sqrt{2}} \right)^2 = \frac{(\sqrt{7})^2}{(6\sqrt{2})^2} = \frac{7}{36 \times 2} = \frac{7}{72} \]
Since \( \frac{7}{72} \) is in \( \frac{p}{q} \) form, it is a rational number.
In simple words: We can simplify each expression first. If the simplified result still has a square root that cannot be solved completely (like \( \sqrt{2} \) or \( \sqrt{6} \)), then the number is irrational. Otherwise, it is rational.

Exam Tip: Use standard algebraic identities like \( (a+b)^2 = a^2 + 2ab + b^2 \), \( (a-b)^2 = a^2 - 2ab + b^2 \), and \( (a+b)(a-b) = a^2 - b^2 \) to expand the terms correctly.

Question 2. Find the square of:
(i) \( \frac{3\sqrt{5}}{5} \)
(ii) \( \sqrt{3} + \sqrt{2} \)
(iii) \( \sqrt{5} - 2 \)
(iv) \( 3 + 2\sqrt{5} \)
Answer:
(i) Square of \( \frac{3\sqrt{5}}{5} \) is:
\[ \left( \frac{3\sqrt{5}}{5} \right)^2 = \frac{3^2 \times (\sqrt{5})^2}{5^2} = \frac{9 \times 5}{25} = \frac{45}{25} = \frac{9}{5} = 1\frac{4}{5} \]
(ii) Square of \( \sqrt{3} + \sqrt{2} \) is:
\[ (\sqrt{3} + \sqrt{2})^2 = (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6} \]
(iii) Square of \( \sqrt{5} - 2 \) is:
\[ (\sqrt{5} - 2)^2 = (\sqrt{5})^2 - 2(\sqrt{5})(2) + (2)^2 = 5 - 4\sqrt{5} + 4 = 9 - 4\sqrt{5} \]
(iv) Square of \( 3 + 2\sqrt{5} \) is:
\[ (3 + 2\sqrt{5})^2 = (3)^2 + 2(3)(2\sqrt{5}) + (2\sqrt{5})^2 = 9 + 12\sqrt{5} + 4 \times 5 = 9 + 12\sqrt{5} + 20 = 29 + 12\dots\sqrt{5} \]
In simple words: Squaring a term means multiplying it by itself. For brackets with plus or minus signs, we must use the standard formula \( (a \pm b)^2 = a^2 \pm 2ab + b^2 \) to make sure we don't miss the middle term.

Exam Tip: Be careful when squaring a term like \( 2\sqrt{5} \). Remember to square both the coefficient and the root: \( (2\sqrt{5})^2 = 2^2 \times (\sqrt{5})^2 = 4 \times 5 = 20 \).

Question 3. State, in each case, whether true or false:
(i) \( \sqrt{2} + \sqrt{3} = \sqrt{5} \)
(ii) \( 2\sqrt{4} + 2 = 6 \)
(iii) \( 3\sqrt{7} - 2\sqrt{7} = \sqrt{7} \)
(iv) \( \frac{2}{7} \) is an irrational number.
(v) \( \frac{5}{11} \) is a rational number.
(vi) All rational numbers are real numbers.
(vii) All real numbers are rational numbers.
(viii) Some real numbers are rational numbers.
Answer:
(i) **False**. Square roots of different primes cannot be added simply by adding their radicands. That is, \( \sqrt{2} + \sqrt{3} \neq \sqrt{5} \).
(ii) **True**. Simplifying the left side:
\[ 2\sqrt{4} + 2 = 2(2) + 2 = 4 + 2 = 6 \]
This matches the right-hand side, so the statement is correct.
(iii) **True**. Simplifying the left side:
\[ 3\sqrt{7} - 2\sqrt{7} = (3 - 2)\sqrt{7} = \sqrt{7} \]
Since both sides are equal, the statement is correct.
(iv) **False**. Since \( \frac{2}{7} = 0.\overline{285714} \) is a recurring, non-terminating decimal, it is a rational number because it can be written in the form \( \frac{p}{q} \).
(v) **True**. Since \( \frac{5}{11} = 0.\overline{45} \) can be expressed in the ratio form \( \frac{p}{q} \), it is a rational number.
(vi) **True**. Real numbers contain the entire set of both rational and irrational numbers. Therefore, every rational number is a real number.
(vii) **False**. The set of real numbers consists of both rational and irrational numbers, so real numbers like \( \sqrt{2} \) are not rational.
(viii) **True**. Since real numbers are made up of rational and irrational numbers, a portion of real numbers (such as integers and fractions) are rational.
In simple words: We can't just add numbers inside square roots like regular numbers. Rational numbers are those we can write as fractions, while real numbers include both fractions and decimals that never end or repeat nicely.

Exam Tip: Pay attention to definitions: all rational numbers are real, but not all real numbers are rational since real numbers also include irrationals.

Question 4. Given universal set is \( \{ -6, -5\frac{3}{4}, -\sqrt{4}, -\frac{3}{5}, -\frac{3}{8}, 0, \frac{4}{5}, 1, 1\frac{2}{3}, \sqrt{8}, 3.01, \pi, 8.47 \} \). From the given set, find:
(i) Set of Rational numbers
(ii) Set of Irrational numbers
(iii) Set of Integers
(iv) Set of Non-negative integers
Answer:
(i) **Set of Rational numbers:**
Rational numbers are those that can be written in the form \( \frac{p}{q} \) where \( q \neq 0 \).
Here, \( -5\frac{3}{4}, -\frac{3}{5}, -\frac{3}{8}, \frac{4}{5}, 1\frac{2}{3} \) are clearly in fractional form.
Also, integers \( -6, 0, 1 \) are rational.
Decimal numbers like \( 3.01 \) and \( 8.47 \) are terminating decimals, so they are rational.
The term \( -\sqrt{4} \) simplifies to \( -2 \), which is an integer and hence rational.
Thus, the set of rational numbers is:
\[ Q = \left\{ -6, -5\frac{3}{4}, -\sqrt{4}, -\frac{3}{5}, -\frac{3}{8}, 0, \frac{4}{5}, 1, 1\frac{2}{3}, 3.01, 8.47 \right\} \]
(ii) **Set of Irrational numbers:**
Irrational numbers are those that are not rational. They are the complement of rational numbers in the real universal set.
From the universal set \( U \), the remaining numbers are:
\[ U - Q = \{ \sqrt{8}, \pi \} \]
Thus, the set of irrational numbers is \( \{ \sqrt{8}, \pi \} \).
(iii) **Set of Integers:**
The set of integers \( Z \) includes whole numbers, zero, and negative whole numbers:
\[ Z = \{ \dots, -3, -2, -1, 0, 1, 2, 3, \dots \} \]
Looking at our universal set, the integers are \( -6 \), \( 0 \), \( 1 \), and \( -\sqrt{4} \) (since \( -\sqrt{4} = -2 \)).
Thus, the set of integers is:
\[ U \cap Z = \{ -6, -\sqrt{4}, 0, 1 \} \]
(iv) **Set of Non-negative integers:**
Non-negative integers are integers that are greater than or equal to zero: \( Z^+ = \{ 0, 1, 2, 3, \dots \} \).
From the set of integers above, the non-negative integers are:
\[ \{ 0, 1 \} \]
Thus, the set of non-negative integers is \( U \cap Z^+ = \{ 0, 1 \} \).
In simple words: We sort the list of numbers into different groups. Rational numbers are any that can be written as whole numbers, clean decimals, or fractions. Irrational numbers have roots we can't solve (like \( \sqrt{8} \)) or special numbers like \( \pi \). Non-negative integers are just the whole numbers starting from 0.

Exam Tip: Don't forget that \( -\sqrt{4} \) simplifies to \( -2 \), which makes it an integer. Always simplify roots before categorizing them.

Question 5. Use the method of long division to find the value of \( \sqrt{3} \) and \( \sqrt{5} \) up to three decimal places. Hence, show that they are irrational numbers.
Answer:
The long division calculations are shown below:

Division for \( \sqrt{3} \):

       1.73205...
     _____________
   1 | 3.00 00 00 00
     |-1
     ------
  27 | 200
     |-189
     ------
 343 |  1100
     | -1029
     -------
 3462|   7100
     |  -6924
     -------
346405|  1760000
      | -1732025
      -------
         27975...

Thus, \( \sqrt{3} \approx 1.732 \), which is a non-terminating and non-recurring decimal, proving it is an irrational number.

Division for \( \sqrt{5} \):

       2.23606...
     _____________
   2 | 5.00 00 00 00
     |-4
     ------
  42 | 100
     | -84
     ------
 443 | 1600
     |-1329
     ------
 4466|  27100
     | -26796
     -------
447206|  3040000
      | -2683236
      -------
         356764...

Thus, \( \sqrt{5} \approx 2.236 \), which is a non-terminating and non-recurring decimal, proving it is an irrational number.
In simple words: We can calculate the square roots of 3 and 5 using long division. Because these decimals go on forever without repeating a single block of digits, they are classified as irrational.

Exam Tip: In exams, always show at least three or four steps of the long division process to clearly demonstrate that the decimal digits are not terminating or repeating.

Question 6. Use the method of contradiction to show that \( \sqrt{3} \) and \( \sqrt{5} \) are irrational numbers.
Answer:
Let us assume that \( \sqrt{3} \) and \( \sqrt{5} \) are rational numbers.
Therefore, we can write them as:
\[ \sqrt{3} = \frac{a}{b} \quad \text{and} \quad \sqrt{5} = \frac{x}{y} \]
where \( a, b \) are integers with no common factors other than 1, and \( b \neq 0 \). Similarly, \( x, y \) are integers with no common factors other than 1, and \( y \neq 0 \).
Squaring both sides of these equations:
\[ 3 = \frac{a^2}{b^2} \]

\( \implies a^2 = 3b^2 \)
and
\[ 5 = \frac{x^2}{y^2} \]

\( \implies x^2 = 5y^2 \)

Since \( a^2 = 3b^2 \), it means \( a^2 \) is divisible by 3. If \( a^2 \) is divisible by 3, then \( a \) must also be divisible by 3.
Similarly, since \( x^2 = 5y^2 \), \( x^2 \) is divisible by 5, meaning \( x \) must also be divisible by 5.
Let us write:
\[ a = 3c \quad \text{and} \quad x = 5z \]
for some integers \( c \) and \( z \).
Substituting these back into our squared equations:
\[ (3c)^2 = 3b^2 \]

\( \implies 9c^2 = 3b^2 \)

\( \implies b^2 = 3c^2 \)
and
\[ (5z)^2 = 5y^2 \]

\( \implies 25z^2 = 5y^2 \)

\( \implies y^2 = 5z^2 \)

Since \( b^2 = 3c^2 \), \( b^2 \) is divisible by 3, meaning \( b \) is also divisible by 3.
Similarly, since \( y^2 = 5z^2 \), \( y^2 \) is divisible by 5, meaning \( y \) is also divisible by 5.
This shows that \( a \) and \( b \) have a common factor of 3, and \( x \) and \( y \) have a common factor of 5.
This directly contradicts our initial assumption that \( \frac{a}{b} \) and \( \frac{x}{y} \) are in their simplest forms with no common factors other than 1.
Since our assumption leads to a contradiction, it must be false.
Thus, \( \sqrt{3} \) and \( \sqrt{5} \) are irrational numbers.
In simple words: We prove this by starting with the opposite assumption: that these roots can be written as simple fractions with no common factors. But when we do the algebra, we find both the top and bottom numbers must share a common factor, which breaks our starting rule. Therefore, they must be irrational.

Exam Tip: This proof by contradiction is a standard, highly-tested theorem. Clearly state the assumption at the beginning and show how the common factor arises for both the numerator and the denominator.

Question 7. Write a pair of irrational numbers whose sum is irrational.
Answer:
Let the two irrational numbers be:
\[ \sqrt{3} + 5 \quad \text{and} \quad \sqrt{5} - 3 \]
Adding these two numbers together:
\[ (\sqrt{3} + 5) + (\sqrt{5} - 3) = \sqrt{3} + \sqrt{5} + 5 - 3 = \sqrt{3} + \sqrt{5} + 2 \]
Since the sum still contains the irrational terms \( \sqrt{3} \) and \( \sqrt{5} \), the resulting sum is an irrational number.
In simple words: We can choose two numbers that have square roots in them. When we add them, the square roots don't cancel out, so the final answer remains irrational.

Exam Tip: Any pair of irrational numbers whose square roots do not cancel each other out when added can be used as a valid example.

Question 8. Write a pair of irrational numbers whose sum is rational.
Answer:
Let the two irrational numbers be:
\[ \sqrt{3} + 5 \quad \text{and} \quad 4 - \sqrt{3} \]
Adding these two numbers together:
\[ (\sqrt{3} + 5) + (4 - \sqrt{3}) = \sqrt{3} + 5 + 4 - \sqrt{3} = 9 \]
Since \( 9 \) is an integer, the sum is a rational number.
In simple words: We pick two irrational numbers that have the same square root but with opposite signs (one positive, one negative). When we add them, the roots cancel each other out completely, leaving a normal rational number.

Exam Tip: To get a rational sum, ensure that the irrational parts of both numbers are additive inverses of each other, like \( +\sqrt{3} \) and \( -\sqrt{3} \).

Question 9. Write a pair of irrational numbers whose difference is irrational.
Answer:
Let the two irrational numbers be:
\[ \sqrt{3} + 2 \quad \text{and} \quad \sqrt{2} - 3 \]
Subtracting the second number from the first:
\[ (\sqrt{3} + 2) - (\sqrt{2} - 3) = \sqrt{3} + 2 - \sqrt{2} + 3 = \sqrt{3} - \sqrt{2} + 5 \]
Since the difference contains the irrational terms \( \sqrt{3} \) and \( \sqrt{2} \), the result is irrational.
In simple words: We pick two irrational numbers with different square roots. When we subtract one from the other, the roots cannot cancel out, keeping the final result irrational.

Exam Tip: Choose numbers with distinct roots (like \( \sqrt{3} \) and \( \sqrt{2} \)) so that subtraction cannot eliminate the irrational parts.

Question 10. Write a pair of irrational numbers whose difference is rational.
Answer:
Let the two irrational numbers be:
\[ \sqrt{5} - 3 \quad \text{and} \quad \sqrt{5} + 3 \]
Subtracting the second number from the first:
\[ (\sqrt{5} - 3) - (\sqrt{5} + 3) = \sqrt{5} - 3 - \sqrt{5} - 3 = -6 \]
Since \( -6 \) is an integer, the difference is a rational number.
In simple words: We choose two irrational numbers that have the exact same root term. When we subtract them, the roots cancel out, leaving only the whole numbers.

Exam Tip: To ensure a rational difference, make sure the irrational parts of both numbers are identical so they cancel out completely during subtraction.

Question 11. Write a pair of irrational numbers whose product is irrational.
Answer:
Let the two irrational numbers be:
\[ 5 + \sqrt{2} \quad \text{and} \quad \sqrt{5} - 2 \]
Multiplying these two numbers:
\[ (5 + \sqrt{2}) \times (\sqrt{5} - 2) = 5(\sqrt{5}) - 5(2) + \sqrt{2}(\sqrt{5}) - 2\sqrt{2} = 5\sqrt{5} - 10 + \sqrt{10} - 2\sqrt{2} \]
Since the product contains irrational terms like \( \sqrt{5} \), \( \sqrt{10} \), and \( \sqrt{2} \), the result is irrational.
In simple words: We pick two numbers with different square roots. When we multiply them out, the roots do not simplify to whole numbers, keeping the final product irrational.

Exam Tip: Choose irrational numbers with different roots that do not multiply to a perfect square, ensuring the product remains irrational.

Question 12. Write a pair of irrational numbers whose product is rational.
Answer:
Let the two irrational numbers be:
\[ \sqrt{3} + \sqrt{2} \quad \text{and} \quad \sqrt{3} - \sqrt{2} \]
Multiplying these two conjugate pairs:
\[ (\sqrt{3} + \sqrt{2}) \times (\sqrt{3} - \sqrt{2}) \]
Using the algebraic identity \( (a+b)(a-b) = a^2 - b^2 \):
\[ = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1 \]
Since \( 1 \) is an integer, the product is a rational number.
In simple words: We use conjugate pairs, which are expressions with the same terms but opposite middle signs. When we multiply them, the square roots disappear, leaving us with a clean integer.

Exam Tip: Multiplying conjugate irrational pairs of the form \( a + \sqrt{b} \) and \( a - \sqrt{b} \) is the most common way to get a rational product.

Question 13. Write in ascending order:
(i) \( 3\sqrt{5} \) and \( 4\sqrt{3} \)
(ii) \( 2\sqrt[3]{5} \) and \( 3\sqrt[3]{2} \)
(iii) \( 6\sqrt{5} \), \( 7\sqrt{3} \), and \( 8\sqrt{2} \)
Answer:
(i) We convert both numbers into pure surds to compare them:
\[ 3\sqrt{5} = \sqrt{3^2 \times 5} = \sqrt{9 \times 5} = \sqrt{45} \]
\[ 4\sqrt{3} = \sqrt{4^2 \times 3} = \sqrt{16 \times 3} = \sqrt{48} \]
Since \( 45 < 48 \), we have \( \sqrt{45} < \sqrt{48} \).
Therefore, in ascending order:
\[ 3\sqrt{5} < 4\sqrt{3} \]
(ii) We convert both numbers into pure cube root surds:
\[ 2\sqrt[3]{5} = \sqrt[3]{2^3 \times 5} = \sqrt[3]{8 \times 5} = \sqrt[3]{40} \]
\[ 3\sqrt[3]{2} = \sqrt[3]{3^3 \times 2} = \sqrt[3]{27 \times 2} = \sqrt[3]{54} \]
Since \( 40 < 54 \), we have \( \sqrt[3]{40} < \sqrt[3]{54} \).
Therefore, in ascending order:
\[ 2\sqrt[3]{5} < 3\sqrt[3]{2} \]
(iii) We convert all three numbers into pure square root surds:
\[ 6\sqrt{5} = \sqrt{6^2 \times 5} = \sqrt{36 \times 5} = \sqrt{180} \]
\[ 7\sqrt{3} = \sqrt{7^2 \times 3} = \sqrt{49 \times 3} = \sqrt{147} \]
\[ 8\sqrt{2} = \sqrt{8^2 \times 2} = \sqrt{64 \times 2} = \sqrt{128} \]
Comparing the numbers inside the roots:
\[ 128 < 147 < 180 \]
Which means:
\[ \sqrt{128} < \sqrt{147} < \sqrt{180} \]
Therefore, in ascending order:
\[ 8\sqrt{2} < 7\sqrt{3} < 6\sqrt{5} \]
In simple words: To compare numbers with square or cube roots, we put the outside numbers back under the root sign by squaring or cubing them. Then, we can easily compare the numbers inside the roots to see which is larger.

Exam Tip: Make sure to square the outside coefficient when moving it inside a square root, or cube it when moving it inside a cube root.

Question 14. Write in descending order:
(i) \( 2\sqrt[4]{6} \) and \( 3\sqrt[4]{2} \)
(ii) \( 7\sqrt{3} \) and \( 3\sqrt{7} \)
Answer:
(i) We convert both numbers into pure fourth-root surds:
\[ 2\sqrt[4]{6} = \sqrt[4]{2^4 \times 6} = \sqrt[4]{16 \times 6} = \sqrt[4]{96} \]
\[ 3\sqrt[4]{2} = \sqrt[4]{3^4 \times 2} = \sqrt[4]{81 \times 2} = \sqrt[4]{162} \]
Since \( 162 > 96 \), we have \( \sqrt[4]{162} > \sqrt[4]{96} \).
Therefore, in descending order:
\[ 3\sqrt[4]{2} > 2\sqrt[4]{6} \]
(ii) We convert both numbers into pure square root surds:
\[ 7\sqrt{3} = \sqrt{7^2 \times 3} = \sqrt{49 \times 3} = \sqrt{147} \]
\[ 3\sqrt{7} = \sqrt{3^2 \times 7} = \sqrt{9 \times 7} = \sqrt{63} \]
Since \( 147 > 63 \), we have \( \sqrt{147} > \sqrt{63} \).
Therefore, in descending order:
\[ 7\sqrt{3} > 3\sqrt{7} \]
In simple words: We move the outside number inside the root by raising it to the same power as the root index (square for square root, power of 4 for fourth root). This lets us compare them directly. Descending order means writing the largest number first.

Exam Tip: For fourth roots, raise the outside coefficient to the power of 4 when moving it inside the root: \( a\sqrt[4]{b} = \sqrt[4]{a^4 \times b} \).

Question 15. Compare:
(i) \( \sqrt[6]{15} \) and \( \sqrt[4]{12} \)
(ii) \( \sqrt{24} \) and \( \sqrt[3]{35} \)
Answer:
(i) We express the roots in fractional exponent form:
\[ \sqrt[6]{15} = (15)^{\frac{1}{6}} \quad \text{and} \quad \sqrt[4]{12} = (12)^{\frac{1}{4}} \]
To compare them, we find a common denominator for the fractional powers.
The L.C.M. of the root indices \( 6 \) and \( 4 \) is \( 12 \).
We convert the exponents:
\[ \frac{1}{6} = \frac{2}{12} \quad \text{and} \quad \frac{1}{4} = \frac{3}{12} \]
Now, rewrite the surds:
\[ \sqrt[6]{15} = (15)^{\frac{2}{12}} = (15^2)^{\frac{1}{12}} = (225)^{\frac{1}{12}} \]
\[ \sqrt[4]{12} = (12)^{\frac{3}{12}} = (12^3)^{\frac{1}{12}} = (1728)^{\frac{1}{12}} \]
Since \( 1728 > 225 \), we have:
\[ (1728)^{\frac{1}{12}} > (225)^{\frac{1}{12}} \]

\( \implies \sqrt[4]{12} > \sqrt[6]{15} \)
(ii) We express the roots in fractional exponent form:
\[ \sqrt{24} = (24)^{\frac{1}{2}} \quad \text{and} \quad \sqrt[3]{35} = (35)^{\frac{1}{3}} \]
The L.C.M. of the root indices \( 2 \) and \( 3 \) is \( 6 \).
We convert the exponents:
\[ \frac{1}{2} = \frac{3}{6} \quad \text{and} \quad \frac{1}{3} = \frac{2}{6} \]
Now, rewrite the surds:
\[ \sqrt{24} = (24)^{\frac{3}{6}} = (24^3)^{\frac{1}{6}} = (13824)^{\frac{1}{6}} \]
\[ \sqrt[3]{35} = (35)^{\frac{2}{6}} = (35^2)^{\frac{1}{6}} = (1225)^{\frac{1}{6}} \]
Since \( 13824 > 1225 \), we have:
\[ (13824)^{\frac{1}{6}} > (1225)^{\frac{1}{6}} \]

\( \implies \sqrt{24} > \sqrt[3]{35} \)
In simple words: To compare roots with different indices (like a square root and a cube root), we find a common multiple for the root numbers. We change both roots so they have this same common root number, and then compare the numbers inside.

Exam Tip: Finding the L.C.M. of the root indices is the key first step. Always double check your calculations when raising numbers to powers like \( 12^3 \) or \( 24^3 \).

Question 16. Find two irrational numbers between 5 and 6.
Answer:
We know that:
\[ 5 = \sqrt{25} \quad \text{and} \quad 6 = \sqrt{36} \]
Any square root of a non-perfect square between \( 25 \) and \( 36 \) will be an irrational number between \( 5 \) and \( 6 \).
We consider the numbers:
\[ \sqrt{25} < \sqrt{26} < \sqrt{27} < \sqrt{28} < \dots < \sqrt{35} < \sqrt{36} \]
Therefore, any two irrational numbers between \( 5 \) and \( 6 \) can be chosen from this range, such as:
\[ \sqrt{27} \quad \text{and} \quad \sqrt{28} \]
In simple words: We write 5 as \( \sqrt{25} \) and 6 as \( \sqrt{36} \). Any square root of a number between 25 and 36 that doesn't simplify to a whole number is an irrational number between 5 and 6.

Exam Tip: Converting integers into square roots of perfect squares makes it extremely easy to find irrational numbers between them.

Question 17. Find five irrational numbers between \( 2\sqrt{5} \) and \( 3\sqrt{3} \).
Answer:
First, we convert both numbers into pure surds:
\[ 2\sqrt{5} = \sqrt{2^2 \times 5} = \sqrt{4 \times 5} = \sqrt{20} \]
\[ 3\sqrt{3} = \sqrt{3^2 \times 3} = \sqrt{9 \times 3} = \sqrt{27} \]
Now, we can find square roots of non-perfect squares between \( 20 \) and \( 27 \):
\[ \sqrt{20} < \sqrt{21} < \sqrt{22} < \sqrt{23} < \sqrt{24} < \sqrt{25} < \sqrt{26} < \sqrt{27} \]
Since \( \sqrt{25} = 5 \) is a rational number, we exclude it.
Thus, the five irrational numbers between \( 2\sqrt{5} \) and \( 3\sqrt{3} \) are:
\[ \sqrt{21}, \quad \sqrt{22}, \quad \sqrt{23}, \quad \sqrt{24}, \quad \text{and} \quad \sqrt{26} \]
In simple words: We turn \( 2\sqrt{5} \) into \( \sqrt{20} \) and \( 3\sqrt{3} \) into \( \sqrt{27} \). Then, we pick five numbers between 20 and 27 (skipping 25 because \( \sqrt{25} = 5 \) is a rational number) and write them under square roots.

Exam Tip: Remember to exclude any perfect squares (like 25) that lie in the range, as their square roots are rational integers.

Question 18. Find two rational numbers between \( \sqrt{2} \) and \( \sqrt{3} \).
Answer:
We want to find rational numbers \( \frac{a}{b} \) such that:
\[ \sqrt{2} < \frac{a}{b} < \sqrt{3} \]
We can square the terms to find perfect squares between \( 2 \) and \( 3 \):
\[ 2 < \left(\frac{a}{b}\right)^2 < 3 \]
Let us choose two numbers whose squares are perfect squares of decimals in this range:
\[ 2.25 \quad \text{and} \quad 2.56 \]
Since:
\[ \sqrt{2.25} = 1.5 = \frac{15}{10} = \frac{3}{2} \]
\[ \sqrt{2.56} = 1.6 = \frac{16}{10} = \frac{8}{5} \]
We have:
\[ \sqrt{2} < \sqrt{2.25} < \sqrt{2.56} < \sqrt{3} \]

\( \implies \sqrt{2} < 1.5 < 1.6 < \sqrt{3} \)
Thus, the two rational numbers are:
\[ \frac{3}{2} \quad \text{and} \quad \frac{8}{5} \]
In simple words: We can approximate \( \sqrt{2} \) as about 1.414 and \( \sqrt{3} \) as about 1.732. Any normal numbers between these two decimal values, like 1.5 and 1.6, are rational numbers that lie between \( \sqrt{2} \) and \( \sqrt{3} \).

Exam Tip: To show rigorous proof, find perfect squares of decimals (like \( 1.5^2 = 2.25 \) and \( 1.6^2 = 2.56 \)) that lie between 2 and 3.

Question 19. Find three rational numbers between \( \sqrt{3} \) and \( \sqrt{5} \).
Answer:
We look for three rational numbers whose squares lie between \( 3 \) and \( 5 \):
\[ 3 < \left(\frac{a}{b}\right)^2 < 5 \]
Let us choose the following perfect decimal squares in this range:
\[ 3.24, \quad 3.61, \quad 4.00, \quad 4.41, \quad 4.84 \]
Taking their square roots:
\[ \sqrt{3.24} = 1.8 = \frac{18}{10} = \frac{9}{5} \]
\[ \sqrt{3.61} = 1.9 = \frac{19}{10} \]
\[ \sqrt{4.00} = 2.0 = 2 \]
\[ \sqrt{4.41} = 2.1 = \frac{21}{10} \]
\[ \sqrt{4.84} = 2.2 = \frac{11}{5} \]
We can choose any three of these rational numbers. For example:
\[ \frac{9}{5}, \quad \frac{19}{10}, \quad \text{and} \quad \frac{21}{10} \]
In simple words: We find decimal values whose squares fall between 3 and 5. By taking the square roots of these decimals, we get clean rational values like 1.8, 1.9, and 2.1, which lie between \( \sqrt{3} \) and \( \sqrt{5} \).

Exam Tip: Selecting decimals whose squares are easy to compute (like 3.24, 3.61, etc.) ensures you can convert them back into exact fractional forms easily.

Exercise 1(D)

Question 1. State, with reasons, which of the following are surds and which are not:
(i) \( \sqrt{180} \)
(ii) \( \sqrt[4]{27} \)
(iii) \( \sqrt[5]{128} \)
(iv) \( \sqrt[3]{64} \)
(v) \( \sqrt[3]{25} \cdot \sqrt[3]{40} \)
(vi) \( \sqrt[3]{-125} \)
(vii) \( \sqrt{\pi} \)
(viii) \( \sqrt{3 + \sqrt{2}} \)
Answer:
A surd is an irrational root of a positive rational number.
(i) \( \sqrt{180} = \sqrt{2 \times 2 \times 5 \times 3 \times 3} = 6\sqrt{5} \)
Since the simplified form is irrational and the radicand \( 180 \) is a positive rational number, **\( \sqrt{180} \) is a surd**.
(ii) \( \sqrt[4]{27} = \sqrt[4]{3 \times 3 \times 3} \)
Since this is an irrational root of the positive rational number \( 27 \), **\( \sqrt[4]{27} \) is a surd**.
(iii) \( \sqrt[5]{128} = \sqrt[5]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2} = 2\sqrt[5]{4} \)
Since this simplified value is irrational, **\( \sqrt[5]{128} \) is a surd**.
(iv) \( \sqrt[3]{64} = \sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2} = 4 \)
Since \( 4 \) is a rational number, **\( \sqrt[3]{64} \) is not a surd**.
(v) \( \sqrt[3]{25} \cdot \sqrt[3]{40} = \sqrt[3]{5 \times 5 \times 2 \times 2 \times 2 \times 5} = 2 \times 5 = 10 \)
Since \( 10 \) is a rational number, **\( \sqrt[3]{25 \cdot 40} \) is not a surd**.
(vi) \( \sqrt[3]{-125} = \sqrt[3]{-5 \times -5 \times -5} = -5 \)
Since \( -5 \) is rational, **\( \sqrt[3]{-125} \) is not a surd**.
(vii) \( \sqrt{\pi} \)
Although \( \sqrt{\pi} \) is irrational, \( \pi \) itself is not a rational number. Since the radicand must be rational for it to be a surd, **\( \sqrt{\pi} \) is not a surd**.
(viii) \( \sqrt{3 + \sqrt{2}} \)
The number under the root, \( 3 + \sqrt{2} \), is an irrational number. Since the radicand is not rational, **\( \sqrt{3 + \sqrt{2}} \) is not a surd**.
In simple words: A surd is a root of a rational number that gives an irrational result. If a root simplifies completely to a whole number (like \( \sqrt[3]{64} = 4 \)), or if the number inside the root is already irrational (like \( \pi \) or \( 3 + \sqrt{2} \)), it is not a surd.

Exam Tip: Remember the definition: the number under the root (radicand) must be rational, and the overall value of the root must be irrational for it to be a surd.

Question 2. Find the lowest rationalizing factor of:
(i) \( 5\sqrt{2} \)
(ii) \( \sqrt{24} \)
(iii) \( \sqrt{5} - 3 \)
(iv) \( 7 - \sqrt{7} \)
(v) \( \sqrt{18} - \sqrt{50} \)
(vi) \( \sqrt{5} - \sqrt{2} \)
(vii) \( \sqrt{13} + 3 \)
(viii) \( 15 - 3\sqrt{2} \)
(ix) \( 3\sqrt{2} + 2\sqrt{3} \)
Answer:
(i) We are given \( 5\sqrt{2} \).
Multiplying \( 5\sqrt{2} \) by \( \sqrt{2} \) gives:
\[ 5\sqrt{2} \times \sqrt{2} = 5 \times 2 = 10 \]
which is rational. Thus, the lowest rationalizing factor is **\( \sqrt{2} \)**.
(ii) We simplify \( \sqrt{24} \):
\[ \sqrt{24} = \sqrt{2 \times 2 \times 2 \times 3} = 2\sqrt{6} \]
Multiplying \( 2\sqrt{6} \) by \( \sqrt{6} \) gives:
\[ 2\sqrt{6} \times \sqrt{6} = 2 \times 6 = 12 \]
which is rational. Thus, the lowest rationalizing factor is **\( \sqrt{6} \)**.
(iii) We have \( \sqrt{5} - 3 \).
Using the difference of squares identity \( (a-b)(a+b) = a^2 - b^2 \), multiplying by its conjugate \( \sqrt{5} + 3 \) gives:
\[ (\sqrt{5} - 3)(\sqrt{5} + 3) = (\sqrt{5})^2 - (3)^2 = 5 - 9 = -4 \]
which is rational. Thus, the lowest rationalizing factor is **\( \sqrt{5} + 3 \)**.
(iv) We have \( 7 - \sqrt{7} \).
Multiplying by its conjugate \( 7 + \sqrt{7} \) gives:
\[ (7 - \sqrt{7})(7 + \sqrt{7}) = 7^2 - (\sqrt{7})^2 = 49 - 7 = 42 \]
which is rational. Thus, the lowest rationalizing factor is **\( 7 + \sqrt{7} \)**.
(v) We simplify \( \sqrt{18} - \sqrt{50} \):
\[ \sqrt{18} - \sqrt{50} = \sqrt{9 \times 2} - \sqrt{25 \times 2} = 3\sqrt{2} - 5\sqrt{2} = -2\sqrt{2} \]
Multiplying \( -2\sqrt{2} \) by \( \sqrt{2} \) gives:
\[ -2\sqrt{2} \times \sqrt{2} = -2 \times 2 = -4 \]
which is rational. Thus, the lowest rationalizing factor is **\( \sqrt{2} \)**.
(vi) We have \( \sqrt{5} - \sqrt{2} \).
Multiplying by its conjugate \( \sqrt{5} + \sqrt{2} \) gives:
\[ (\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) = (\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3 \]
which is rational. Thus, the lowest rationalizing factor is **\( \sqrt{5} + \sqrt{2} \)**.
(vii) We have \( \sqrt{13} + 3 \).
Multiplying by its conjugate \( \sqrt{13} - 3 \) gives:
\[ (\sqrt{13} + 3)(\sqrt{13} - 3) = (\sqrt{13})^2 - 3^2 = 13 - 9 = 4 \]
which is rational. Thus, the lowest rationalizing factor is **\( \sqrt{13} - 3 \)**.
(viii) We simplify \( 15 - 3\sqrt{2} \):
\[ 15 - 3\sqrt{2} = 3(5 - \sqrt{2}) \]
Multiplying \( 5 - \sqrt{2} \) by its conjugate \( 5 + \sqrt{2} \) rationalizes it:
\[ 3(5 - \sqrt{2})(5 + \sqrt{2}) = 3(5^2 - (\sqrt{2})^2) = 3(25 - 2) = 3 \times 23 = 69 \]
which is rational. Thus, the lowest rationalizing factor is **\( 5 + \sqrt{2} \)**.
(ix) We have \( 3\sqrt{2} + 2\sqrt{3} \).
Multiplying by its conjugate \( 3\sqrt{2} - 2\sqrt{3} \) gives:
\[ (3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} - 2\sqrt{3}) = (3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6 \]
which is rational. Thus, the lowest rationalizing factor is **\( 3\sqrt{2} - 2\sqrt{3} \)**.
In simple words: To rationalize a number, we multiply it by something that will get rid of the square root. For a single root like \( \sqrt{24} = 2\sqrt{6} \), we just multiply by \( \sqrt{6} \). For two-term numbers like \( a - b \), we multiply by \( a + b \).

Exam Tip: Always simplify single surds first (like \( \sqrt{24} \to 2\sqrt{6} \)) to find the simplest and lowest rationalizing factor instead of multiplying by the entire original surd.

Question 3. Rationalize the denominator of:
(i) \( \frac{3}{\sqrt{5}} \)
(ii) \( \frac{2\sqrt{3}}{\sqrt{5}} \)
(iii) \( \frac{1}{\sqrt{3} - \sqrt{2}} \)
(iv) \( \frac{3}{\sqrt{5} + \sqrt{2}} \)
(v) \( \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \)
(vi) \( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)
(vii) \( \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \)
(viii) \( \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}} \)
(ix) \( \frac{2\sqrt{5} + 3\sqrt{2}}{2\sqrt{5} - 3\sqrt{2}} \)
Answer:
(i) We multiply the numerator and the denominator by \( \sqrt{5} \):
\[ \frac{3}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \]
(ii) We multiply the numerator and the denominator by \( \sqrt{5} \):
\[ \frac{2\sqrt{3}}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{15}}{5} \]
(iii) We multiply the numerator and the denominator by the conjugate \( \sqrt{3} + \sqrt{2} \):
\[ \frac{1}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} + \sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2} \]
(iv) We multiply the numerator and the denominator by the conjugate \( \sqrt{5} - \sqrt{2} \):
\[ \frac{3}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{3(\sqrt{5} - \sqrt{2})}{(\sqrt{5})^2 - (\sqrt{2})^2} = \frac{3(\sqrt{5} - \sqrt{2})}{5 - 2} = \frac{3(\sqrt{5} - \sqrt{2})}{3} = \sqrt{5} - \sqrt{2} \]
(v) We multiply the numerator and the denominator by the conjugate \( 2 - \sqrt{3} \):
\[ \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{(2 - \sqrt{3})^2}{2^2 - (\sqrt{3})^2} = \frac{4 - 4\sqrt{3} + 3}{4 - 3} = \frac{7 - 4\sqrt{3}}{1} = 7 - 4\sqrt{3} \]
(vi) We multiply the numerator and the denominator by the conjugate \( \sqrt{3} + 1 \):
\[ \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = \frac{2(2 + \sqrt{3})}{2} = 2 + \sqrt{3} \]
(vii) We multiply the numerator and the denominator by the conjugate \( \sqrt{3} - \sqrt{2} \):
\[ \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{(\sqrt{3} - \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{3 - 2\sqrt{6} + 2}{3 - 2} = 5 - 2\sqrt{6} \]
(viii) We multiply the numerator and the denominator by the conjugate \( \sqrt{6} - \sqrt{5} \):
\[ \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} = \frac{(\sqrt{6} - \sqrt{5})^2}{(\sqrt{6})^2 - (\sqrt{5})^2} = \frac{6 - 2\sqrt{30} + 5}{6 - 5} = 11 - 2\sqrt{30} \]
(ix) We multiply the numerator and the denominator by the conjugate \( 2\sqrt{5} + 3\sqrt{2} \):
\[ \frac{2\sqrt{5} + 3\sqrt{2}}{2\sqrt{5} - 3\sqrt{2}} \times \frac{2\sqrt{5} + 3\sqrt{2}}{2\sqrt{5} + 3\sqrt{2}} = \frac{(2\sqrt{5} + 3\sqrt{2})^2}{(2\sqrt{5})^2 - (3\sqrt{2})^2} = \frac{4(5) + 2(2\sqrt{5})(3\sqrt{2}) + 9(2)}{20 - 18} = \frac{20 + 12\dots\sqrt{10} + 18}{2} = \frac{38 + 12\sqrt{10}}{2} = \frac{2(19 + 6\sqrt{10})}{2} = 19 + 6\sqrt{10} \]
In simple words: To remove square roots from the bottom of a fraction, we multiply both the top and bottom by the conjugate of the bottom expression. This turns the bottom into a regular, clean whole number.

Exam Tip: When squaring binomials on the top, always remember the middle term of \( (a \pm b)^2 = a^2 \pm 2ab + b^2 \). Be sure to cancel out any common factors at the end to simplify the expression completely.

Question 4. Find the values of \( a \) and \( b \) in each of the following by rationalizing the denominator:
(i) \( \frac{2 + \sqrt{3}}{2 - \sqrt{3}} = a + b\sqrt{3} \)
(ii) \( \frac{\sqrt{7} - 2}{\sqrt{7} + 2} = a\sqrt{7} + b \)
(iii) \( \frac{3}{\sqrt{3} - \sqrt{2}} = a\sqrt{3} - b\sqrt{2} \)
(iv) \( \frac{5 + 3\sqrt{2}}{5 - 3\sqrt{2}} = a + b\sqrt{2} \)
Answer:
(i) Multiply both the numerator and the denominator by the conjugate of the denominator, \( 2 + \sqrt{3} \):
\( \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = a + b\sqrt{3} \)
\( \frac{(2 + \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2} = a + b\sqrt{3} \)
\( \frac{4 + 3 + 4\sqrt{3}}{4 - 3} = a + b\sqrt{3} \)
\( \frac{7 + 4\sqrt{3}}{1} = a + b\sqrt{3} \)
\( 7 + 4\sqrt{3} = a + b\sqrt{3} \)
Comparing the rational and irrational terms on both sides, we get:
\( a = 7 \), \( b = 4 \)

(ii) Multiply both the numerator and the denominator by the conjugate of the denominator, \( \sqrt{7} - 2 \):
\( \frac{\sqrt{7} - 2}{\sqrt{7} + 2} \times \frac{\sqrt{7} - 2}{\sqrt{7} - 2} = a\sqrt{7} + b \)
\( \frac{(\sqrt{7} - 2)^2}{(\sqrt{7})^2 - (2)^2} = a\sqrt{7} + b \)
\( \frac{7 + 4 - 4\sqrt{7}}{7 - 4} = a\sqrt{7} + b \)
\( \frac{11 - 4\sqrt{7}}{3} = a\sqrt{7} + b \)
\( \frac{11}{3} - \frac{4}{3}\sqrt{7} = a\sqrt{7} + b \)
Comparing the terms on both sides, we find:
\( a = -\frac{4}{3} \), \( b = \frac{11}{3} \)

(iii) Multiply both the numerator and the denominator by the conjugate of the denominator, \( \sqrt{3} + \sqrt{2} \):
\( \frac{3}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = a\sqrt{3} - b\sqrt{2} \)
\( \frac{3(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} = a\sqrt{3} - b\sqrt{2} \)
\( \frac{3\sqrt{3} + 3\sqrt{2}}{3 - 2} = a\sqrt{3} - b\sqrt{2} \)
\( 3\sqrt{3} + 3\sqrt{2} = a\sqrt{3} - b\sqrt{2} \)
Comparing the coefficients of \( \sqrt{3} \) and \( \sqrt{2} \) on both sides, we obtain:
\( a = 3 \), \( b = -3 \)

(iv) Multiply both the numerator and the denominator by the conjugate of the denominator, \( 5 + 3\sqrt{2} \):
\( \frac{5 + 3\sqrt{2}}{5 - 3\sqrt{2}} \times \frac{5 + 3\sqrt{2}}{5 + 3\sqrt{2}} = a + b\sqrt{2} \)
\( \frac{(5 + 3\sqrt{2})^2}{(5)^2 - (3\sqrt{2})^2} = a + b\sqrt{2} \)
\( \frac{25 + 18 + 30\sqrt{2}}{25 - 18} = a + b\sqrt{2} \)
\( \frac{43 + 30\sqrt{2}}{7} = a + b\sqrt{2} \)
\( \frac{43}{7} + \frac{30}{7}\sqrt{2} = a + b\sqrt{2} \)
Comparing the rational and irrational terms on both sides, we get:
\( a = \frac{43}{7} \), \( b = \frac{30}{7} \)

In simple words: To find \( a \) and \( b \), multiply the top and bottom of each fraction by the conjugate of the bottom part. Once the denominator becomes a simple number, compare both sides of the equation to find the matching values.

Exam Tip: Double-check the sign of the conjugate before multiplying, and carefully compare corresponding coefficients on both sides to avoid simple sign errors.

Question 5. Simplify the following expressions:
(i) \( \frac{22}{2\sqrt{3} + 1} + \frac{17}{2\sqrt{3} - 1} \)
(ii) \( \frac{\sqrt{2}}{\sqrt{6} - \sqrt{2}} - \frac{\sqrt{3}}{\sqrt{6} + \sqrt{2}} \)
Answer:
(i) Combine the fractions by finding a common denominator:
\( \frac{22(2\sqrt{3} - 1) + 17(2\sqrt{3} + 1)}{(2\sqrt{3} + 1)(2\sqrt{3} - 1)} = \frac{44\sqrt{3} - 22 + 34\sqrt{3} + 17}{(2\sqrt{3})^2 - 1^2} \)
\( = \frac{78\sqrt{3} - 5}{12 - 1} \)
\( = \frac{78\sqrt{3} - 5}{11} \)

(ii) Combine the terms by taking a common denominator:
\( \frac{\sqrt{2}(\sqrt{6} + \sqrt{2}) - \sqrt{3}(\sqrt{6} - \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})} = \frac{\sqrt{12} + 2 - \sqrt{18} + \sqrt{6}}{(\sqrt{6})^2 - (\sqrt{2})^2} \)
Simplify the square roots in the numerator:
\( \sqrt{12} = 2\sqrt{3} \), \( \sqrt{18} = 3\sqrt{2} \)
\( = \frac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{6 - 2} \)
\( = \frac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{4} \)

In simple words: Cross-multiply to combine the fractions over a common denominator. Then simplify the roots in the numerator and calculate the final fraction.

Exam Tip: Be very careful with negative signs when distributing terms in the numerator, especially when multiplying a negative radical like \( -\sqrt{3} \).

Question 6. If \( x = \frac{\sqrt{5} - 2}{\sqrt{5} + 2} \) and \( y = \frac{\sqrt{5} + 2}{\sqrt{5} - 2} \), find the value of:
(i) \( x^2 \)
(ii) \( y^2 \)
(iii) \( xy \)
(iv) \( x^2 + y^2 + xy \)
Answer:
First, we find the simplified values of \( x \) and \( y \) by rationalizing:
\( x = \frac{\sqrt{5} - 2}{\sqrt{5} + 2} \times \frac{\sqrt{5} - 2}{\sqrt{5} - 2} = \frac{(\sqrt{5} - 2)^2}{(\sqrt{5})^2 - (2)^2} = \frac{5 + 4 - 4\sqrt{5}}{5 - 4} = 9 - 4\sqrt{5} \)
\( y = \frac{\sqrt{5} + 2}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{(\sqrt{5} + 2)^2}{(\sqrt{5})^2 - (2)^2} = \frac{5 + 4 + 4\sqrt{5}}{5 - 4} = 9 + 4\sqrt{5} \)

(i) Calculate \( x^2 \):
\( x^2 = (9 - 4\sqrt{5})^2 \)
\( = (9)^2 + (4\sqrt{5})^2 - 2 \times 9 \times 4\sqrt{5} \)
\( = 81 + 80 - 72\sqrt{5} \)
\( = 161 - 72\sqrt{5} \)

(ii) Calculate \( y^2 \):
\( y^2 = (9 + 4\sqrt{5})^2 \)
\( = (9)^2 + (4\sqrt{5})^2 + 2 \times 9 \times 4\sqrt{5} \)
\( = 81 + 80 + 72\sqrt{5} \)
\( = 161 + 72\sqrt{5} \)

(iii) Calculate \( xy \):
\( xy = \left( \frac{\sqrt{5} - 2}{\sqrt{5} + 2} \right) \left( \frac{\sqrt{5} + 2}{\sqrt{5} - 2} \right) = 1 \)

(iv) Calculate \( x^2 + y^2 + xy \):
\( x^2 + y^2 + xy = (161 - 72\sqrt{5}) + (161 + 72\sqrt{5}) + 1 \)
\( = 161 + 161 + 1 \)
\( = 323 \)

In simple words: Simplify both \( x \) and \( y \) first by removing square roots from the bottom. Then square them, multiply them, and add the results together.

Exam Tip: Notice that \( x \) and \( y \) are reciprocals of each other, so their product \( xy \) is always \( 1 \). Recognizing this simplifies the working tremendously.

Question 7. If \( m = \frac{1}{3 - 2\sqrt{2}} \) and \( n = \frac{1}{3 + 2\sqrt{2}} \), find the value of:
(i) \( m^2 \)
(ii) \( n^2 \)
(iii) \( mn \)
Answer:
(i) Rationalize the expression for \( m \):
\( m = \frac{1}{3 - 2\sqrt{2}} \times \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2} = \frac{3 + 2\sqrt{2}}{9 - 8} = 3 + 2\sqrt{2} \)
Now, square the value of \( m \):
\( m^2 = (3 + 2\sqrt{2})^2 = (3)^2 + 2 \times 3 \times 2\sqrt{2} + (2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \)

(ii) Rationalize the expression for \( n \):
\( n = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \)
Now, square the value of \( n \):
\( n^2 = (3 - 2\sqrt{2})^2 = (3)^2 - 2 \times 3 \times 2\sqrt{2} + (2\sqrt{2})^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2} \)

(iii) Multiply \( m \) and \( n \):
\( mn = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = (3)^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \)

In simple words: First rationalize the fractions to get rid of roots in the denominators. Then compute the squares of both variables and find their product.

Exam Tip: Rationalizing first makes squaring much easier. Always use the algebraic identity \( (a \pm b)^2 = a^2 \pm 2ab + b^2 \) when squaring these binomial radical expressions.

Question 8. If \( x = 2\sqrt{3} + 2\sqrt{2} \), find the value of:
(i) \( \frac{1}{x} \)
(ii) \( x + \frac{1}{x} \)
(iii) \( \left( x + \frac{1}{x} \right)^2 \)
Answer:
(i) Rationalize the expression for \( \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{2\sqrt{3} + 2\sqrt{2}} \times \frac{2\sqrt{3} - 2\sqrt{2}}{2\sqrt{3} - 2\sqrt{2}} = \frac{2(\sqrt{3} - \sqrt{2})}{(2\sqrt{3})^2 - (2\sqrt{2})^2} = \frac{2(\sqrt{3} - \sqrt{2})}{12 - 8} = \frac{2(\sqrt{3} - \sqrt{2})}{4} = \frac{\sqrt{3} - \sqrt{2}}{2} \)

(ii) Sum \( x \) and \( \frac{1}{x} \):
\( x + \frac{1}{x} = 2(\sqrt{3} + \sqrt{2}) + \frac{\sqrt{3} - \sqrt{2}}{2} = \frac{4(\sqrt{3} + \sqrt{2}) + (\sqrt{3} - \sqrt{2})}{2} = \frac{4\sqrt{3} + 4\sqrt{2} + \sqrt{3} - \sqrt{2}}{2} = \frac{5\sqrt{3} + 3\sqrt{2}}{2} \)

(iii) Square the sum calculated in part (ii):
\( \left( x + \frac{1}{x} \right)^2 = \left( \frac{5\sqrt{3} + 3\sqrt{2}}{2} \right)^2 = \frac{(5\sqrt{3})^2 + (3\sqrt{2})^2 + 2(5\sqrt{3})(3\sqrt{2})}{4} = \frac{75 + 18 + 30\sqrt{6}}{4} = \frac{93 + 30\sqrt{6}}{4} \)

In simple words: Find \( \frac{1}{x} \) by rationalizing its denominator, add this result to \( x \), and then square the resulting sum.

Exam Tip: Simplify expressions by factoring out common terms, such as writing \( 2\sqrt{3} + 2\sqrt{2} \) as \( 2(\sqrt{3} + \sqrt{2}) \), to make the subsequent algebraic addition less error-prone.

Question 9. Given that \( x = 1 - \sqrt{2} \), find the value of \( \left( x - \frac{1}{x} \right)^3 \).
Answer:
First, find the reciprocal of \( x \) and rationalize it:
\( \frac{1}{x} = \frac{1}{1 - \sqrt{2}} \times \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{1 + \sqrt{2}}{1^2 - (\sqrt{2})^2} = \frac{1 + \sqrt{2}}{1 - 2} = \frac{1 + \sqrt{2}}{-1} = -(1 + \sqrt{2}) \)
Now, substitute \( x \) and \( \frac{1}{x} \) to find the difference:
\( x - \frac{1}{x} = (1 - \sqrt{2}) - [-(1 + \sqrt{2})] = 1 - \sqrt{2} + 1 + \sqrt{2} = 2 \)
Now cube the result:
\( \left( x - \frac{1}{x} \right)^3 = 2^3 = 8 \)

In simple words: Rationalizing \( \frac{1}{x} \) gives us \( -(1 + \sqrt{2}) \). When we subtract this from \( x \), the square root terms cancel out to leave \( 2 \), which we cube to get \( 8 \).

Exam Tip: Pay careful attention to the double negative sign when calculating \( x - \frac{1}{x} \) to ensure the radical terms correctly cancel out.

Question 10. Given \( x = 5 - 2\sqrt{6} \), find the value of \( x^2 + \frac{1}{x^2} \).
Answer:
Find the reciprocal of \( x \) and rationalize the denominator:
\( \frac{1}{x} = \frac{1}{5 - 2\sqrt{6}} \times \frac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} = \frac{5 + 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2} = \frac{5 + 2\sqrt{6}}{25 - 24} = 5 + 2\sqrt{6} \)
Find the difference between \( x \) and \( \frac{1}{x} \):
\( x - \frac{1}{x} = (5 - 2\sqrt{6}) - (5 + 2\sqrt{6}) = 5 - 2\sqrt{6} - 5 - 2\sqrt{6} = -4\sqrt{6} \)
Using the algebraic identity for a squared binomial difference:
\( \left( x - \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} - 2 \)
Rearranging to solve for the target expression:
\( x^2 + \frac{1}{x^2} = \left( x - \frac{1}{x} \right)^2 + 2 \)
Substitute the value of \( x - \frac{1}{x} \):
\( x^2 + \frac{1}{x^2} = (-4\sqrt{6})^2 + 2 = 96 + 2 = 98 \)

In simple words: Rationalize \( \frac{1}{x} \) first, then find \( x - \frac{1}{x} \) to remove the whole numbers. Finally, square this difference and add \( 2 \) to get the final answer.

Exam Tip: Using the algebraic relationship \( x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2 \) is much faster and less error-prone than squaring \( x \) and \( \frac{1}{x} \) individually.

Question 11. Prove that:
\( \frac{1}{3 - 2\sqrt{2}} - \frac{1}{2\sqrt{2} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2} = 5 \)
Answer:
We begin with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \frac{1}{3 - 2\sqrt{2}} - \frac{1}{2\sqrt{2} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2} \)
Rewrite \( 3 \) as \( \sqrt{9} \), \( 2\sqrt{2} \) as \( \sqrt{8} \), and \( 2 \) as \( \sqrt{4} \):
\( \text{L.H.S.} = \frac{1}{\sqrt{9} - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - \sqrt{4}} \)
Rationalize each term individually by multiplying by its conjugate:
\( \frac{1}{\sqrt{9} - \sqrt{8}} \times \frac{\sqrt{9} + \sqrt{8}}{\sqrt{9} + \sqrt{8}} = \sqrt{9} + \sqrt{8} = 3 + \sqrt{8} \)
\( \frac{1}{\sqrt{8} - \sqrt{7}} \times \frac{\sqrt{8} + \sqrt{7}}{\sqrt{8} + \sqrt{7}} = \sqrt{8} + \sqrt{7} \)
\( \frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} = \sqrt{7} + \sqrt{6} \)
\( \frac{1}{\sqrt{6} - \sqrt{5}} \times \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}} = \sqrt{6} + \sqrt{5} \)
\( \frac{1}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \sqrt{5} + 2 \)
Substitute these rationalized terms back into the original expression:
\( \text{L.H.S.} = (3 + \sqrt{8}) - (\sqrt{8} + \sqrt{7}) + (\sqrt{7} + \sqrt{6}) - (\sqrt{6} + \sqrt{5}) + (\sqrt{5} + 2) \)
Expand the brackets, making sure to distribute the negative signs:
\( = 3 + \sqrt{8} - \sqrt{8} - \sqrt{7} + \sqrt{7} + \sqrt{6} - \sqrt{6} - \sqrt{5} + \sqrt{5} + 2 \)
All intermediate radical terms cancel out:
\( = 3 + 2 = 5 = \text{R.H.S.} \)
Hence proved.

In simple words: Rationalize each fraction in the sequence. Once they are rationalized, their denominators become \( 1 \), and expanding the brackets causes all the middle terms to cancel out, leaving just \( 3 + 2 = 5 \).

Exam Tip: Be very meticulous with negative signs in telescoping series like this. Forgetting to distribute the negative sign into any set of brackets will prevent the terms from canceling out properly.

Question 12. Rationalize the denominator of: \( \frac{1}{\sqrt{3} - \sqrt{2} + 1} \)
Answer:
Group the first two terms in the denominator: \( (\sqrt{3} - \sqrt{2}) + 1 \). Multiply both numerator and denominator by the conjugate \( (\sqrt{3} - \sqrt{2}) - 1 \):
\( = \frac{1}{(\sqrt{3} - \sqrt{2}) + 1} \times \frac{(\sqrt{3} - \sqrt{2}) - 1}{(\sqrt{3} - \sqrt{2}) - 1} \)
\( = \frac{\sqrt{3} - \sqrt{2} - 1}{(\sqrt{3} - \sqrt{2})^2 - (1)^2} \)
Expand the squared term in the denominator:
\( = \frac{\sqrt{3} - \sqrt{2} - 1}{3 + 2 - 2\sqrt{6} - 1} \)
\( = \frac{\sqrt{3} - \sqrt{2} - 1}{4 - 2\sqrt{6}} \)
\( = \frac{\sqrt{3} - \sqrt{2} - 1}{2(2 - \sqrt{6})} \)
Rationalize again by multiplying by \( 2 + \sqrt{6} \):
\( = \frac{\sqrt{3} - \sqrt{2} - 1}{2(2 - \sqrt{6})} \times \frac{2 + \sqrt{6}}{2 + \sqrt{6}} \)
\( = \frac{2\sqrt{3} + \sqrt{18} - 2\sqrt{2} - \sqrt{12} - 2 - \sqrt{6}}{2[(2)^2 - (\sqrt{6})^2]} \)
Simplify the square roots in the numerator:
\( \sqrt{18} = 3\sqrt{2} \), \( \sqrt{12} = 2\sqrt{3} \)
\( = \frac{2\sqrt{3} + 3\sqrt{2} - 2\sqrt{2} - 2\sqrt{3} - 2 - \sqrt{6}}{2(4 - 6)} \)
\( = \frac{\sqrt{2} - 2 - \sqrt{6}}{2(-2)} \)
\( = \frac{\sqrt{2} - 2 - \sqrt{6}}{-4} \)
\( = \frac{2 + \sqrt{6} - \sqrt{2}}{4} \)
\( = \frac{1}{4}(2 + \sqrt{6} - \sqrt{2}) \)

In simple words: Since there are three terms in the denominator, you must rationalize in two stages. First group two terms together to reduce it to a binomial denominator, then rationalize a second time.

Exam Tip: Rationalizing three-term denominators requires grouping. Always group the terms carefully to minimize the complexity of the radical in the intermediate denominator.

Question 13(i). Given \( \sqrt{2} = 1.4 \) and \( \sqrt{3} = 1.7 \), find the value of: \( \frac{1}{\sqrt{3} - \sqrt{2}} \)
Answer:
First, rationalize the denominator of the fraction:
\( \frac{1}{\sqrt{3} - \sqrt{2}} = \frac{1}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \)
\( = \frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} \)
\( = \frac{\sqrt{3} + \sqrt{2}}{3 - 2} \)
\( = \sqrt{3} + \sqrt{2} \)
Now, substitute the given approximate values for \( \sqrt{2} \) and \( \sqrt{3} \):
\( = 1.7 + 1.4 = 3.1 \)

In simple words: Do not plug in the decimal values at the start, as that makes division hard. Rationalize the denominator first, which leaves you with a simple addition problem.

Exam Tip: Always perform rationalization before substituting numerical values to keep the calculations simple and precise.

Question 13(ii). Given \( \sqrt{2} = 1.4 \) and \( \sqrt{3} = 1.7 \), find the value of: \( \frac{1}{3 + 2\sqrt{2}} \)
Answer:
First, rationalize the fraction:
\( \frac{1}{3 + 2\sqrt{2}} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \)
\( = \frac{3 - 2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2} \)
\( = \frac{3 - 2\sqrt{2}}{9 - 8} \)
\( = 3 - 2\sqrt{2} \)
Substitute the given value \( \sqrt{2} = 1.4 \):
\( = 3 - 2(1.4) \)
\( = 3 - 2.8 = 0.2 \)

In simple words: Rationalizing first simplifies the fraction to a simple linear expression, allowing you to easily substitute the decimal values at the end.

Exam Tip: Pay attention to the instruction values given. Note that \( \sqrt{3} \) is not needed for this sub-part, so do not try to force it into your calculations.