Exercise 1.1 — Class 9 Mathematics Solutions

Solution

1. Analyze the decimal representation.

   The given number $2.353535$ has a finite number of decimal places (it terminates after 6 decimal digits).

2. Convert to fraction form $\frac{p}{q}$.

   $$2.353535 = \frac{2353535}{1000000}$$

3. Apply definition of rational numbers.

   Since it can be written as $\frac{p}{q}$ where $p, q \in \mathbb{Z}$ and $q \neq 0$, it is a rational number.

Solution

1. Understand the notation.

   The bar over $6$ denotes that $6$ repeats indefinitely: $0.\overline{6} = 0.6666...$

2. Express as a fraction $\frac{p}{q}$.

   Let:

   $$x = 0.6666... \quad \text{--- (1)}$$

   Multiply equation (1) by $10$:

   $$10x = 6.6666... \quad \text{--- (2)}$$

   Subtract equation (1) from equation (2):

   $$10x - x = 6.6666... - 0.6666...$$ $$9x = 6 \implies x = \frac{6}{9} = \frac{2}{3}$$

3. Conclusion.

   Since $0.\overline{6} = \frac{2}{3}$, it is a recurring decimal that can be written in $\frac{p}{q}$ form.

Solution

1. Examine the decimal digits.

   The dots $(...)$ indicate that the decimal continues infinitely without terminating.

2. Check for periodicity / repeating patterns.

   There is no repeating block or repeating single digit in $2.236067...$

3. Apply definition of irrational numbers.

   A number whose decimal expansion is non-terminating and non-recurring cannot be expressed in the form $\frac{p}{q}$. Therefore, it is an irrational number.

Solution

1. Analyze the radicand.

   The number under the radical is $7$, which is a prime number and not a perfect square.

2. Property of radicals.

   The square root of any positive integer that is not a perfect square is an irrational number.

3. Decimal approximation.

   $\sqrt{7} \approx 2.645751311...$ which is non-terminating and non-recurring.

Solution

1. Identify the constant.

   $e$ is Euler's number (the base of natural logarithms).

2. Decimal value.

   $e \approx 2.718281828459045...$

3. Property.

   $e$ is a well-known transcendental number. Its decimal representation continues infinitely without repeating any pattern, so it cannot be written as $\frac{p}{q}$.

Solution

1. Identify the constant.

   $\pi$ (Pi) is the ratio of a circle's circumference to its diameter.

2. Decimal value.

   $\pi \approx 3.14159265358979...$

3. Common misconception.

   $\frac{22}{7}$ and $3.14$ are only approximate rational values used for practical calculations; the exact value of $\pi$ is an irrational number because its decimal representation is non-terminating and non-recurring.

Solution

1. Analyze the individual terms.

   • $5$ is a rational number ($\frac{5}{1}$).
   • $\sqrt{11}$ is an irrational number because $11$ is not a perfect square.

2. Property of real numbers.

   The sum of a non-zero rational number and an irrational number is always an irrational number.

3. Verification by contradiction.

   If $5 + \sqrt{11} = r$ (where $r$ is rational), then $\sqrt{11} = r - 5$. The difference of two rational numbers $(r - 5)$ must be rational, which would mean $\sqrt{11}$ is rational — a contradiction. Hence, $5 + \sqrt{11}$ is irrational.

Solution

1. Analyze the terms.

   Both $3$ and $13$ are prime numbers (not perfect squares), so $\sqrt{3}$ and $\sqrt{13}$ are both irrational numbers.

2. Sum of distinct square roots.

   The sum of square roots of two distinct prime numbers is always an irrational number.

3. Algebraic proof.

   Suppose $x = \sqrt{3} + \sqrt{13}$ is rational. Squaring both sides:

   $$x^2 = (\sqrt{3} + \sqrt{13})^2 = 3 + 13 + 2\sqrt{39} = 16 + 2\sqrt{39}$$

   Rearranging:

   $$\sqrt{39} = \frac{x^2 - 16}{2}$$

   If $x$ were rational, the right-hand side would be rational, implying $\sqrt{39}$ is rational. But $39 = 3 \times 13$ is not a perfect square, so $\sqrt{39}$ is irrational. Thus, our assumption was false and $\sqrt{3} + \sqrt{13}$ is irrational.

Solution

1. Check the definition of rational numbers.

   A number is rational if it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

2. Compare with definition.

   Here $p = 15 \in \mathbb{Z}$ and $q = 4 \in \mathbb{Z}$ with $q \neq 0$.

3. Decimal form.

   $$\frac{15}{4} = 3.75 \quad (\text{Terminating decimal})$$

4. Conclusion.

   Since it is an exact ratio of two integers with a terminating decimal, it is a rational number.

Solution

1. Apply algebraic identity.

   Recall the difference of squares formula:

   $$(a - b)(a + b) = a^2 - b^2$$

2. Substitute values $a = 2$ and $b = \sqrt{2}$.

   $$(2 - \sqrt{2})(2 + \sqrt{2}) = (2)^2 - (\sqrt{2})^2$$

3. Simplify.

   $$= 4 - 2 = 2$$

4. Conclusion.

   The result is $2 = \frac{2}{1}$, which is an integer. All integers are rational numbers.

Solution

1. Understand the construction principle.

   $\sqrt{2} \approx 1.414$ is an irrational number. We construct its exact location on the real number line using the Pythagorean Theorem with a right-angled triangle.

2. Construct right-angled triangle $\triangle OAB$.

   • Let the origin on the number line be point $O(0)$.
   • Take base $OA = 1$ unit along the positive horizontal axis (from $0$ to $1$).
   • At point $A(1)$, erect a perpendicular line segment $AB = 1$ unit (perpendicular to the number line).
   • Join origin $O$ to point $B$.

3. Calculate the hypotenuse length $OB$.

   Applying the Pythagorean theorem in right-angled $\triangle OAB$:

   $$OB^2 = OA^2 + AB^2 = 1^2 + 1^2 = 1 + 1 = 2$$ $$OB = \sqrt{2} \text{ units}$$

4. Transfer length to the number line.

   • With origin $O(0)$ as the center and radius equal to the hypotenuse $OB = \sqrt{2}$, draw an arc using a compass cutting the positive number line at point $P$.
   • Point $P$ represents $\sqrt{2} \approx 1.414$, situated between $1$ and $2$.

   Geometric Construction of √2 on Real Number Linescale: 1 unit

Solution

1. Understand the construction principle.

   We use the Pythagorean theorem relation:

   $$(\sqrt{3})^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3$$

   Hence, a right-angled triangle with base of length $\sqrt{2}$ and perpendicular altitude of $1$ unit has a hypotenuse of length $\sqrt{3}$.

2. Base construction using $\sqrt{2}$.

   • As constructed in Part (i), locate point $P(\sqrt{2} \approx 1.414)$ on the positive real number line using a right-angled triangle with base $1$ and height $1$.
   • The line segment from origin $O(0)$ to point $P$ forms the base of length $OP = \sqrt{2}$ units.

3. Construct right-angled triangle $\triangle OPQ$.

   • At point $P(\sqrt{2})$, erect a perpendicular line segment $PQ = 1$ unit (perpendicular to the number line).
   • Join the origin $O(0)$ to point $Q$.

4. Calculate the hypotenuse length $OQ$.

   Applying the Pythagorean theorem in right-angled $\triangle OPQ$:

   $$OQ^2 = OP^2 + PQ^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3$$ $$OQ = \sqrt{3} \approx 1.732 \text{ units}$$

5. Transfer length to the number line with a compass.

   • With origin $O(0)$ as center and compass radius equal to hypotenuse $OQ = \sqrt{3}$, draw an arc intersecting the positive real number line at point $R$.
   • Point $R$ represents $\sqrt{3} \approx 1.732$, located between $1$ and $2$ (closer to $2$).

   Geometric Construction of √3 on Real Number Linescale: 1 unit

Solution

1. Identify the interval on the number line.

   $4\frac{1}{3} = 4 + \frac{1}{3} = \frac{13}{3} \approx 4.333...$ This is a positive rational number located strictly between the integers $4$ and $5$.

2. Subdivide the unit interval into equal parts.

   • Locate integer marks $4$ and $5$ on the positive side of the number line.
   • The denominator is $3$, so divide the unit interval from $4$ to $5$ into $3$ equal parts using $2$ equidistant division tick marks.

3. Plot the fractional point.

   • The numerator is $1$. Starting from $4$, move $1$ part to the right.
   • The 1st division mark after $4$ represents $4 + \frac{1}{3} = 4\frac{1}{3} = \frac{13}{3}$.

   Representation of 4 1/3 on Number Linescale: 1 unit

Solution

1. Identify the negative interval.

   $-2\frac{1}{7} = -\left(2 + \frac{1}{7}\right) = -\frac{15}{7} \approx -2.1428...$ Since it is negative, it lies to the left of the origin $0$, specifically between the integers $-2$ and $-3$.

2. Subdivide the unit segment.

   • Locate integers $-2$ and $-3$ on the negative axis of the number line.
   • The denominator is $7$, so divide the distance between $-2$ and $-3$ into $7$ equal parts using $6$ equidistant marks.

3. Plot the target point.

   • The numerator is $1$. Moving to the left from $-2$ towards $-3$, take $1$ step (the first division tick mark to the left of $-2$).
   • This mark represents $-2\frac{1}{7}$.

   Representation of -2 1/7 on Number Linescale: 1 unit

Solution

1. Identify the interval.

   $\frac{5}{8} = 0.625$ is a proper positive fraction with $0 < \frac{5}{8} < 1$. It lies strictly in the unit interval between $0$ and $1$.

2. Subdivide the unit interval.

   • The denominator is $8$, so divide the segment from origin $0$ to $1$ into $8$ equal parts using $7$ equidistant division points.

3. Plot the fraction.

   • The numerator is $5$. Starting from $0$, count $5$ parts to the right.
   • The 5th division mark corresponds exactly to $\frac{5}{8} = 0.625$.

   Representation of 5/8 on Number Linescale: 1 unit

Solution

1. Identify the interval.

   $2\frac{3}{4} = 2 + \frac{3}{4} = \frac{11}{4} = 2.75$ This is a positive mixed fraction located in the unit interval between integers $2$ and $3$.

2. Subdivide the unit segment.

   • Locate integers $2$ and $3$ on the positive number line.
   • The denominator is $4$, so divide the segment from $2$ to $3$ into $4$ equal parts using $3$ equidistant points (representing $2\frac{1}{4}$, $2\frac{2}{4} = 2\frac{1}{2}$, and $2\frac{3}{4}$).

3. Plot the target point.

   • The numerator is $3$. Starting from $2$, count $3$ subdivisions to the right.
   • The 3rd division mark represents $2\frac{3}{4} = 2.75$.

   Representation of 2 3/4 on Number Linescale: 1 unit

Solution

1. Given:

   $$0.\overline{4} = 0.44444\dots$$

   (The bar over $4$ indicates that the single digit $4$ repeats infinitely.)

   ---

   Step-by-Step Solution:

   Step 1 — Set up Equation (1):

   Let $x$ be the given repeating decimal:

   $$x = 0.4444\dots \quad \text{--- (1)}$$

   Step 2 — Multiply by $10$:

   Since one digit ($4$) is repeating, multiply both sides of Equation (1) by $10$:

   $$10 \times x = 10 \times (0.4444\dots)$$ $$10x = 4.4444\dots \quad \text{--- (2)}$$

   Step 3 — Subtract Equation (1) from Equation (2):

   $$\begin{aligned} 10x - x &= (4.4444\dots) - (0.4444\dots) \\[6pt] 9x &= 4.0000\dots \\[6pt] 9x &= 4 \end{aligned}$$

   (Notice how the infinite decimal parts $0.4444\dots$ cancel out completely!)

   Step 4 — Solve for $x$:

   Divide both sides by $9$:

   $$x = \frac{4}{9}$$

   ---

   Verification:

   • $p = 4$ and $q = 9$ are integers with $q \neq 0$.
   • Dividing $4 \div 9 = 0.4444\dots = 0.\overline{4}$.

Solution

1. Given:

   $$0.\overline{37} = 0.373737\dots$$

   (The bar over $37$ indicates that the two-digit block $37$ repeats infinitely.)

   ---

   Step-by-Step Solution:

   Step 1 — Set up Equation (1):

   Let $x$ be the given repeating decimal:

   $$x = 0.373737\dots \quad \text{--- (1)}$$

   Step 2 — Multiply by $100$:

   Since two digits ($37$) are repeating, multiply both sides of Equation (1) by $100$:

   $$100 \times x = 100 \times (0.373737\dots)$$ $$100x = 37.373737\dots \quad \text{--- (2)}$$

   Step 3 — Subtract Equation (1) from Equation (2):

   $$\begin{aligned} 100x - x &= (37.373737\dots) - (0.373737\dots) \\[6pt] 99x &= 37.000000\dots \\[6pt] 99x &= 37 \end{aligned}$$

   (The infinite repeating tails cancel out to zero!)

   Step 4 — Solve for $x$:

   Divide both sides by $99$:

   $$x = \frac{37}{99}$$

   Since $37$ is a prime number and does not divide $99$, the fraction $\frac{37}{99}$ is in its simplest irreducible form.

   ---

   Verification:

   • $p = 37$ and $q = 99$ are integers with $q \neq 0$.
   • Dividing $37 \div 99 = 0.373737\dots = 0.\overline{37}$.

Solution

1. Given:

   $$0.\overline{21} = 0.212121\dots$$

   (The bar over $21$ indicates that the two-digit block $21$ repeats infinitely.)

   ---

   Step-by-Step Solution:

   Step 1 — Set up Equation (1):

   Let $x$ be the given repeating decimal:

   $$x = 0.212121\dots \quad \text{--- (1)}$$

   Step 2 — Multiply by $100$:

   Since two digits ($21$) are repeating, multiply both sides of Equation (1) by $100$:

   $$100 \times x = 100 \times (0.212121\dots)$$ $$100x = 21.212121\dots \quad \text{--- (2)}$$

   Step 3 — Subtract Equation (1) from Equation (2):

   $$\begin{aligned} 100x - x &= (21.212121\dots) - (0.212121\dots) \\[6pt] 99x &= 21.000000\dots \\[6pt] 99x &= 21 \end{aligned}$$

   (The decimal tails cancel out completely!)

   Step 4 — Solve for $x$ and Simplify to Lowest Terms:

   Divide both sides by $99$:

   $$x = \frac{21}{99}$$

   Both $21$ and $99$ share a common factor of $3$. Divide the numerator and denominator by $3$:

   $$x = \frac{21 \div 3}{99 \div 3} = \frac{7}{33}$$

   ---

   Verification:

   • $p = 7$ and $q = 33$ are integers with $q \neq 0$.
   • Dividing $7 \div 33 = 0.212121\dots = 0.\overline{21}$.

Solution

1. Given Equation:

   $$(a + 4) + b = a + (4 + b)$$

   ---

   📌 Concept & Formula Recall:

   Associative Property of Addition: For any three real numbers $x, y, z \in \mathbb{R}$:

   $$(x + y) + z = x + (y + z)$$

   (The order of numbers stays the same, but the grouping brackets change.)

   ---

   Step-by-Step Explanation:

   1. On the LHS, the first two terms are grouped: $(a + 4) + b$.
   2. On the RHS, the last two terms are grouped: $a + (4 + b)$.
   3. Since grouping changes under addition without affecting the result, this uses the Associative Property of Addition.

Solution

1. Given Equation:

   $$\sqrt{2} + \sqrt{3} = \sqrt{3} + \sqrt{2}$$

   ---

   📌 Concept & Formula Recall:

   Commutative Property of Addition: For any two real numbers $x, y \in \mathbb{R}$:

   $$x + y = y + x$$

   (Changing the order of the terms being added does not change the sum.)

   ---

   Step-by-Step Explanation:

   1. Both $\sqrt{2}$ and $\sqrt{3}$ are real numbers (irrational numbers $\in \mathbb{R}$).
   2. On the LHS, $\sqrt{2}$ is added to $\sqrt{3}$.
   3. On the RHS, the order is swapped: $\sqrt{3}$ is added to $\sqrt{2}$.
   4. Therefore, this equation illustrates the Commutative Property of Addition.

Solution

1. Given Equation:

   $$x - x = 0 \quad \Longleftrightarrow \quad x + (-x) = 0$$

   ---

   📌 Concept & Formula Recall:

   Additive Inverse Property: For every real number $x \in \mathbb{R}$, there exists a unique real number $-x \in \mathbb{R}$ such that:

   $$x + (-x) = (-x) + x = 0$$

   (Adding a number to its negative opposite yields the Additive Identity $0$.)

   ---

   Step-by-Step Explanation:

   1. The expression $x - x$ is equivalent to adding the additive inverse $(-x)$ to $x$.
   2. The result is $0$ (the additive identity in $\mathbb{R}$).
   3. Therefore, this represents the Additive Inverse Property.

Solution

1. Given Equation:

   $$a(b + c) = ab + ac$$

   ---

   📌 Concept & Formula Recall:

   Distributive Property of Multiplication over Addition: For any three real numbers $a, b, c \in \mathbb{R}$:

   $$a \cdot (b + c) = a \cdot b + a \cdot c \quad \text{(Left Distributive Law)}$$ $$(a + b) \cdot c = a \cdot c + b \cdot c \quad \text{(Right Distributive Law)}$$

   ---

   Step-by-Step Explanation:

   1. The multiplier $a$ on the outside of the parentheses multiplies each term inside $(b + c)$ individually.
   2. Multiplication distributes over the addition operation from the left.
   3. Therefore, this is the Distributive Property of Multiplication over Addition (Left Distributive Property).

Solution

1. Given Equation:

   $$16 + 0 = 16$$

   ---

   📌 Concept & Formula Recall:

   Additive Identity Property: There exists a unique real number $0 \in \mathbb{R}$ such that for every $x \in \mathbb{R}$:

   $$x + 0 = 0 + x = x$$

   (Adding $0$ to any real number preserves its identity.)

   ---

   Step-by-Step Explanation:

   1. $0$ is added to the real number $16$.
   2. The value of $16$ remains unchanged.
   3. Since $0$ is the additive identity in $\mathbb{R}$, this illustrates the Additive Identity Property.

Solution

1. Given Equation:

   $$100 \times 1 = 100$$

   ---

   📌 Concept & Formula Recall:

   Multiplicative Identity Property: There exists a unique real number $1 \in \mathbb{R}$ such that for every $x \in \mathbb{R}$:

   $$x \times 1 = 1 \times x = x$$

   (Multiplying any real number by $1$ preserves its identity.)

   ---

   Step-by-Step Explanation:

   1. The real number $100$ is multiplied by $1$.
   2. The value of $100$ remains unchanged.
   3. Since $1$ is the multiplicative identity in $\mathbb{R}$, this illustrates the Multiplicative Identity Property.

Solution

1. Given Equation:

   $$4 \times (5 \times 8) = (4 \times 5) \times 8$$

   ---

   📌 Concept & Formula Recall:

   Associative Property of Multiplication: For any three real numbers $x, y, z \in \mathbb{R}$:

   $$x \times (y \times z) = (x \times y) \times z$$

   (Changing the grouping of factors being multiplied does not change the product.)

   ---

   Step-by-Step Explanation:

   1. On the LHS, the last two factors are grouped: $4 \times (5 \times 8) = 4 \times 40 = 160$.
   2. On the RHS, the first two factors are grouped: $(4 \times 5) \times 8 = 20 \times 8 = 160$.
   3. The sequence of factors ($4, 5, 8$) is identical, only the multiplication grouping changed.
   4. Therefore, this illustrates the Associative Property of Multiplication.

Solution

1. Given Equation:

   $$ab = ba \quad \Longleftrightarrow \quad a \times b = b \times a$$

   ---

   📌 Concept & Formula Recall:

   Commutative Property of Multiplication: For any two real numbers $a, b \in \mathbb{R}$:

   $$a \cdot b = b \cdot a$$

   (Changing the order of the factors being multiplied does not change the product.)

   ---

   Step-by-Step Explanation:

   1. On the LHS, $a$ is multiplied by $b$.
   2. On the RHS, the positions of the factors are reversed (commuted) to $b \times a$.
   3. Therefore, this illustrates the Commutative Property of Multiplication.

Solution

1. Given Statement:

   $$-3 < -1 \implies 0 < 2$$

   ---

   📌 Concept & Formula Recall:

   Additive Property of Inequality: For any real numbers $a, b, c \in \mathbb{R}$:

   $$a < b \implies a + c < b + c$$

   (Adding the same real number to both sides of an inequality preserves the inequality relation.)

   ---

   Step-by-Step Analysis:

   1. Start with the given true inequality: $$-3 < -1$$
   2. Add $+3$ to both sides of the inequality: $$-3 + 3 < -1 + 3$$ $$0 < 2$$
   3. Since the same number ($3$) was added to both sides, this is the Additive Property of Inequality.

Solution

1. Given Statement:

   $$a < b \implies \frac{1}{a} > \frac{1}{b} \quad (\text{for } a, b > 0)$$

   ---

   📌 Concept & Formula Recall:

   Reciprocal Property of Inequality: For any two positive real numbers $a, b > 0$:

   $$a < b \iff \frac{1}{a} > \frac{1}{b}$$

   (Taking the reciprocal (multiplicative inverse) of both positive sides reverses the inequality sign.)

   ---

   Step-by-Step Analysis:

   1. Consider positive real numbers, for example $2 < 3$.
   2. Taking the reciprocals of both sides: $$\frac{1}{2} = 0.5, \quad \frac{1}{3} \approx 0.333$$ Clearly, $\frac{1}{2} > \frac{1}{3}$.
   3. Therefore, when $a < b$, taking reciprocals inverts the inequality: $\frac{1}{a} > \frac{1}{b}$.
   4. This is the Reciprocal Property of Inequality.

Solution

1. Given Statement:

   $$a < b \implies a + c < b + c$$

   ---

   📌 Concept & Formula Recall:

   Additive Property of Inequality: For any real numbers $a, b, c \in \mathbb{R}$:

   $$a < b \implies a + c < b + c$$

   (Adding any real constant $c$ to both sides preserves the inequality.)

   ---

   Step-by-Step Analysis:

   1. The original inequality $a < b$ has the same term $c$ added to both sides.
   2. The inequality sign remains unchanged ($<$).
   3. This is the direct algebraic definition of the Additive Property of Inequality.

Solution

1. Given Statement:

   $$ac < bc \quad (\text{where } c > 0) \implies a < b$$

   ---

   📌 Concept & Formula Recall:

   Multiplicative / Cancellation Property of Inequality (Positive Multiplier): For any real numbers $a, b, c \in \mathbb{R}$ with $c > 0$:

   $$ac < bc \implies a < b$$

   (Dividing or canceling a positive number $c > 0$ preserves the inequality direction.)

   ---

   Step-by-Step Analysis:

   1. We are given $ac < bc$ where $c$ is strictly positive ($c > 0$).
   2. Dividing both sides by the positive number $c$ leaves the inequality sign $<$ unchanged: $$\frac{ac}{c} < \frac{bc}{c} \implies a < b$$
   3. This is the Cancellation Property of Inequality with respect to Multiplication (for $c > 0$).

Solution

1. Given Statement:

   $$ac < bc \quad (\text{where } c < 0) \implies a > b$$

   ---

   📌 Concept & Formula Recall:

   Multiplicative / Cancellation Property of Inequality (Negative Multiplier): For any real numbers $a, b, c \in \mathbb{R}$ with $c < 0$:

   $$ac < bc \implies a > b$$

   (Dividing or multiplying by a negative number $c < 0$ reverses the inequality sign.)

   ---

   Step-by-Step Analysis:

   1. We are given $ac < bc$ where $c$ is negative ($c < 0$).
   2. Dividing or canceling the negative number $c$ flips the inequality sign from $<$ to $>$: $$\frac{ac}{c} > \frac{bc}{c} \implies a > b$$
   3. This is the Cancellation / Multiplicative Property of Inequality with a Negative Multiplier (for $c < 0$).

Solution

1. Given Statement:

   $$\text{For any } a, b \in \mathbb{R}, \text{ exactly one of the following holds:} \quad a > b, \quad a = b, \quad a < b$$

   ---

   📌 Concept & Formula Recall:

   Trichotomy Property of Real Numbers: For any two real numbers $a, b \in \mathbb{R}$, exactly one of the three relations must be true:

   1. $a > b$ ($a$ is strictly greater than $b$)
   2. $a = b$ ($a$ is equal to $b$)
   3. $a < b$ ($a$ is strictly less than $b$)

   ---

   Step-by-Step Analysis:

   1. This is the fundamental completeness order axiom of the real line.
   2. Any two real numbers can always be compared, and they satisfy one and only one of these three exclusive conditions.
   3. This axiom is called the Trichotomy Property (Law of Trichotomy).

Solution

1. Given Numbers:

   $$a = \frac{1}{4} = 0.25, \quad b = \frac{1}{3} \approx 0.333 \quad (\text{Notice } \frac{1}{4} < \frac{1}{3})$$

   ---

   Step-by-Step Solution:

   Step 1 — Find the First Rational Number ($q_1$):

   Calculate the average (midpoint) of $\frac{1}{4}$ and $\frac{1}{3}$:

   $$\begin{aligned} q_1 &= \frac{\frac{1}{4} + \frac{1}{3}}{2} \\[8pt] &= \frac{\frac{3 + 4}{12}}{2} \\[8pt] &= \frac{7}{12} \div 2 \\[8pt] &= \frac{7}{12} \times \frac{1}{2} = \mathbf{\frac{7}{24}} \end{aligned}$$

   ---

   Step 2 — Find the Second Rational Number ($q_2$):

   Calculate the average of $q_1 = \frac{7}{24}$ and $b = \frac{1}{3}$:

   $$\begin{aligned} q_2 &= \frac{\frac{7}{24} + \frac{1}{3}}{2} \\[8pt] &= \frac{\frac{7 + 8}{24}}{2} \\[8pt] &= \frac{15}{24} \div 2 \\[8pt] &= \frac{15}{24} \times \frac{1}{2} = \frac{15}{48} \end{aligned}$$

   Simplify $\frac{15}{48}$ by dividing numerator and denominator by $3$:

   $$q_2 = \frac{15 \div 3}{48 \div 3} = \mathbf{\frac{5}{16}}$$

   ---

   Verification of Ordering:

   Expressing all numbers with common denominator $48$:

   $$\frac{1}{4} = \frac{12}{48} < \frac{14}{48} \left(=\frac{7}{24}\right) < \frac{15}{48} \left(=\frac{5}{16}\right) < \frac{16}{48} \left(=\frac{1}{3}\right)$$

   Both $\frac{7}{24}$ and $\frac{5}{16}$ lie strictly between $\frac{1}{4}$ and $\frac{1}{3}$.