Key Concept: Laws of Exponents

b) $4^{(3/2)}$

[Solution Description] Using the law of exponents that states $(x^m)^n = x^{mn}$, we find $a^{3/2}$ equals $(2^2)^{3/2} = 2^{2 \cdot (3/2)} = 2^3 = 8$. Hence, the expression that equals this is correct.

Key Concept: Understanding operations involving irrational numbers

d) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] The given assertion (A) states that the product of two irrational numbers is always irrational. However, this is not always true. While the product of some irrational numbers, such as $\sqrt{2} \times \sqrt{3} = \sqrt{6}$ is irrational, there are cases where the product is rational. For example, $\sqrt{2} \times \sqrt{2} = 2$, which is a rational number. The reason (R) correctly defines irrational numbers as those that cannot be expressed as a fraction of two integers. Since the assertion is not always true, (A) is false, but (R) is true. Hence, (R) is not the correct explanation of (A).

Key Concept: Complex Operations, Real Number Visualization

d) Assertion is false, but Reason is true.

[Solution Description] The assertion states that the product of a rational number and an irrational number is always irrational. Let's consider a rational number $r = 0$, then $r \times y = 0$ which is rational. Thus, Assertion is false. The reason states that if $x$ is a non-zero rational number and $y$ is an irrational number, then $xy$ is irrational. This reason correctly identifies the characteristic behavior of irrational numbers when multiplied by a non-zero rational number. Therefore, Reason is true.

Key Concept: Unique Representation

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that every point on the number line corresponds to a real number, which is true because real numbers include both rational and irrational numbers. The reason claims that the number line represents all rational numbers, which is correct but does not explain the full scope of representing all real numbers, including irrational numbers. Therefore, while both statements are true, the reason does not fully justify the assertion.

Key Concept: Infinite Nature

c) There are infinitely many rational and irrational numbers in both directions.

[Solution Description] The infinite nature of the number line implies that both rational and irrational numbers continue indefinitely in both positive and negative directions without any limits. Hence, for any real number, there exist infinitely many other real numbers.

Key Concept: Advanced Pythagorean, Spiral Construction

c) $\sqrt{5}$

[Solution Description] To find the length of OB, use the Pythagorean theorem in triangle OAB with legs $OA$ and $AB$.

The equation is: $OB^2 = OA^2 + AB^2$

Substituting the given values: $OB^2 = 1^2 + 2^2 = 1 + 4 = 5$

Therefore, $OB = \sqrt{5}$

So, the length of OB which forms part of a square root spiral is $\sqrt{5}$.

Key Concept: Rational Number Equivalence, Real Number Properties

b) $\frac{1}{3}$

[Solution Description] To determine if $0.333\ldots$ is a rational number, express it as a fraction. Let $x = 0.333\ldots$. Multiplying by 10 gives $10x = 3.333\ldots$. Subtracting the original $x$ from this equation:

$10x - x = 3.333\ldots - 0.333\ldots$ $9x = 3$ $x = \frac{3}{9} = \frac{1}{3}$

Thus, $0.333\ldots$ is precisely equivalent to the rational number $\frac{1}{3}$.

Key Concept: Real Number Line, Rational vs Irrational

b) Every interval contains infinitely many irrational numbers.

[Solution Description] The statement implies the density property of real numbers, where both rational and irrational numbers are densely packed on the number line. For any two given rational numbers $a$ and $b$, you can always find another number lying strictly between them which is an irrational number, supporting the completeness of the real number line.

Key Concept: Real World Application

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To analyze the assertion, we consider the fact that many natural phenomena like sound waves are modeled using circular motion equations where $\pi$ frequently appears due to its linkage with circles and periodic functions. The reason supports this by claiming $\pi$'s nature as an irrational number allows more precise representation than rational approximations for such continuous models. In the case of sound waves and planetary orbits, these require intricate calculations involving non-repeating decimals for accuracy. Therefore, both the assertion and reason hold true, and the reason correctly explains why $\pi$ is utilized in these contexts.

Key Concept: Set Comparison

c) Every natural number is a whole number

[Solution Description] Natural numbers are defined as positive integers starting from 1, while whole numbers include all natural numbers and zero. Therefore, every natural number is a whole number, but not every whole number is a natural number since the set of whole numbers includes zero, which does not belong to the set of natural numbers.

Key Concept: Symbol Recognition

d) N

[Solution Description] The standard symbol for the set of natural numbers is $N$.

Key Concept: Identify Natural Numbers

c) 7

[Solution Description] A natural number is a positive integer starting from 1 and continuing indefinitely. Among the options, the only positive integer is 7.

Key Concept: Symbol and Definition

c) W

[Solution Description] The set of whole numbers is represented by the symbol $W$. This set includes all natural numbers along with the number zero.

Key Concept: Conceptual Analysis, Historical Context

a) Introduction of zero

[Solution Description]

Whole numbers include both natural numbers and zero. Historically, introducing zero as a numerical value completed the transition from counting numbers to whole numbers. Before zero's acceptance, calculations were limited to positive integers:

$\text{Whole numbers = Natural numbers } \cup \{0\}$

Hence, recognizing zero as essential to whole numbers.

Key Concept: Set Theory Application, Advanced Symbolic Understanding

c) Negative integers

[Solution Description]

The notation $Z \setminus W$ represents the set difference between integers $Z$ and whole numbers $W$. Since whole numbers are non-negative integers starting from zero, the set difference will include negative integers only:

$$Z \setminus W = \{ \text{all negative integers} \}$$

Thus, the expression represents all negative integers.

Key Concept: Integer Representation

b) -7

[Solution Description] An integer that is 12 units to the left of another integer can be found by subtracting 12 from it. Hence, the required integer is: $5 - 12 = -7$

The integer is -7.

Key Concept: Integer Inclusion

a) Yes

[Solution Description] Integers include all positive and negative whole numbers along with zero. Since 3 is a positive whole number, it is included in the set of integers.

Key Concept: Integer Operations

c) 22

[Solution Description] Subtracting a negative number is equivalent to adding its positive counterpart. So, $7 - (-15)$ becomes $7 + 15$. Now, calculate the sum: $7 + 15 = 22$

The result is 22.

Key Concept: Non-Terminating Recurring

b) $\frac{1}{6}$

[Solution Description] To determine if a fraction has a terminating or non-terminating recurring decimal expansion, we look at its denominator in simplest form after removing common factors with the numerator. After simplification:

- $\frac{5}{8}$: Simplifies to $\frac{5}{8}$; denominator is $2^3$, so it has a terminating decimal.

- $\frac{1}{6}$: Simplifies to $\frac{1}{6}$; denominator $6$ is $2 \times 3$, which indicates a non-terminating recurring decimal.

- $\frac{3}{4}$: Simplifies to $\frac{3}{4}$; denominator is $2^2$, so it has a terminating decimal.

- $\frac{10}{25}$: Simplifies to $\frac{2}{5}$; denominator is $5$, indicating it has a terminating decimal.

Therefore, only $\frac{1}{6}$ has a non-terminating recurring decimal expansion.

Key Concept: Rational Number Properties

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion states that the number 0.125 is a rational number. To determine if this is true, we need to check if it can be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. Since 0.125 is a terminating decimal, it can indeed be expressed as $\frac{125}{1000}$. Simplifying $\frac{125}{1000}$ gives $\frac{1}{8}$, which confirms that 0.125 is a rational number. Hence, the assertion is true.

The reason claims that 0.125 has a non-terminating decimal expansion. However, 0.125 is actually a terminating decimal expansion because it ends after three decimal places. Therefore, the reason is false.

Thus, the correct answer is: Assertion is true, but Reason is false.

Key Concept: Terminating Decimals

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the decimal expansion of $\frac{1}{4}$ is 0.25. Let's divide 1 by 4:

Step 1: Set up the division $1 \div 4$.

Step 2: Since 1 is less than 4, we add a decimal point and a zero to make it 10.

Step 3: Divide 10 by 4, which gives 2 with a remainder of 2. Write down 2 after the decimal point.

Step 4: Bring down another zero to make it 20.

Step 5: Divide 20 by 4, resulting in 5 with no remainder. Therefore, the quotient is 0.25.

Both the assertion and the reason are true, and the reason correctly explains why the decimal expansion terminates at 0.25.

Key Concept: Complex Fraction Conversion, Rational Number Operations

a) $\frac{1}{4}$

[Solution Description] To convert the given decimal $0.58333...$ into a fraction, let's express it as $x = 0.58333...$. This can be rewritten as $x = 0.58 + 0.00333...$. The part $0.00333...$ can be expressed as $\frac{1}{300}$.

Therefore, $x$ becomes: $x = 0.58 + \frac{1}{300}$

Converting $0.58$ to a fraction gives: $0.58 = \frac{58}{100} = \frac{29}{50}$

Hence, $x = \frac{29}{50} + \frac{1}{300} = \frac{174}{300} + \frac{1}{300} = \frac{175}{300}$

Simplifying this fraction: $\frac{175}{300} = \frac{7}{12}$

Now, subtract $\frac{1}{3}$ from $\frac{7}{12}$:

$\frac{7}{12} - \frac{1}{3} = \frac{7}{12} - \frac{4}{12} = \frac{3}{12} = \frac{1}{4}$

Therefore, the result is $\frac{1}{4}$.

Key Concept: Decimal Expansion

b) Terminating

[Solution Description] To determine the type of decimal expansion, divide $1$ by $4$: $1 \div 4 = 0.25$, which terminates.

Key Concept: Complex Fraction Representation, Infinite Rational Numbers

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] To determine the validity of the assertion and reason, we begin by analyzing the decimal $1.3333...$, which is a non-terminating repeating decimal. Such decimals can be represented as fractions. Let $x = 1.3333...$. Then multiplying both sides by 10 gives $10x = 13.3333...$. Subtracting these two equations, we have $9x = 12$, yielding $x = \frac{12}{9} = \frac{4}{3}$ after simplification. Thus, the assertion is true.

For the reason, there are indeed infinitely many rational numbers between any two distinct real numbers. For example, a number like $\frac{133}{100}$ exists between $1.33$ and $1.34$, and similarly, other rational numbers such as $\frac{267}{200}$, etc., also exist in this range. Hence, the reason is also true.

Therefore, while both statements are true, the reason does not explain why $1.3333...$ can be expressed as $\frac{4}{3}$. So, the correct answer is option b).

Key Concept: Rational Number Properties

b) They can be expressed as a fraction with a non-zero denominator

[Solution Description] Rational numbers can be positive, negative, or zero, and they can always be expressed as a fraction with a non-zero denominator. Hence statement B is correct.

Key Concept: Rational Number Operations, Rational Number Between Fractions

d) $\frac{13}{17}$

[Solution Description] We need to find a $z$ such that $x < z < y$. Convert $x$ and $y$ to a common denominator: $\text{LCM of } 9 \text{ and } 11 = 99$. Convert: $\frac{7}{9} = \frac{7 \times 11}{9 \times 11} = \frac{77}{99}$, $\frac{8}{11} = \frac{8 \times 9}{11 \times 9} = \frac{72}{99}$. Choose $z$ such that it's a fraction of form $\frac{p+q}{r}$. One such number satisfying this condition is $\frac{14}{18}$ which reduces to $\frac{7}{9}$, equivalent to $\frac{77}{99}$: $z = \frac{13}{17}$ Check if $z > \frac{77}{99}$ and $z < \frac{88}{99}$: $\frac{87}{122} \approx \frac{91}{122}$ Therefore, $\frac{13}{17}$ satisfies the condition.

Key Concept: Natural Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] A natural number like 3 can indeed be written as $\frac{3}{1}$, which fits the definition of a rational number being expressible in the form $\frac{p}{q}$, where $q \neq 0$. Therefore, both the assertion and the reason are true, and the reason correctly explains why every natural number is a rational number.

Key Concept: Decimal Conversion

a) $\frac{4}{9}$

[Solution Description] Let $x = 0.444...$.

Then, multiply both sides by 10 to shift the decimal point: $10x = 4.444...$.

Now subtract the original equation from this new one:

$10x - x = 4.444... - 0.444...$, which simplifies to $9x = 4$.

Therefore, solving for $x$, we get $x = \frac{4}{9}$.

Key Concept: Rational Operations, Complex Rationality

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

To evaluate this Assertion and Reason, we know:

- **Assertion**: Let's consider a rational number $r = \frac{a}{b}$ and an irrational number $i$. By definition, $i$ cannot be expressed as a ratio of two integers with the denominator not zero.

When we add these, $r + i$, we need to determine if it can still be expressed as $\frac{p}{q}$ (i.e., a rational form). The property of irrational numbers tells us that they cannot cancel out or simplify in such a way when added to a rational number that results in a rational number. Therefore, the sum $r + i$ remains irrational. Thus, the assertion is true.

- **Reason**: The reason correctly describes what constitutes a rational number. It elaborates on how a rational number is represented.

However, while the reason is true, it does not explain why the sum of a rational number and an irrational number is necessarily irrational. The reason provides a correct statement about rational numbers but does not directly justify the assertion about irrational sums.

Hence, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Rationalization Techniques, Complex Rational Operations

b) $\sqrt{3} - \sqrt{2}$

[Solution Description]

We rationalize the denominator by multiplying numerator and denominator by the conjugate of the denominator, which is $(\sqrt{2} - \sqrt{3})$.

$$\frac{1}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})}$$

Simplifying the denominator using difference of squares: $(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) = 2 - 3 = -1$

Therefore, the expression becomes: $\frac{\sqrt{2} - \sqrt{3}}{-1} = \sqrt{3} - \sqrt{2}$

Hence, the rationalized value is $\sqrt{3} - \sqrt{2}$.

Key Concept: Simplify Fraction

b) $\frac{2}{3}$

[Solution Description]

To simplify the fraction $\frac{16}{24}$, we need to divide the numerator and the denominator by their greatest common divisor (GCD). The GCD of 16 and 24 is 8.

Divide both by 8: $\frac{16 \div 8}{24 \div 8} = \frac{2}{3}$

So, the simplest form of $\frac{16}{24}$ is $\frac{2}{3}$.

Key Concept: Find Missing Denominator

c) 8

[Solution Description]

To find the missing denominator $x$, we use the property of equivalent fractions, which states that cross-multiplying yields equal products: $7 \times 16 = 14 \times x$

Calculating the left side gives: $112 = 14x$

Solving for $x$ involves dividing both sides by 14: $x = \frac{112}{14} = 8$

Therefore, the missing denominator is 8.

Key Concept: Complex Equivalence, Multiple Steps

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To verify the equivalence of the fractions $\frac{45}{60}$ and $\frac{3}{4}$, we first simplify $\frac{45}{60}$. The GCD of 45 and 60 is 15.

Dividing both the numerator and denominator by 15 gives:$\frac{45 \div 15}{60 \div 15} = \frac{3}{4}$

Since $\frac{45}{60}$ simplifies to $\frac{3}{4}$, these two fractions are indeed equivalent.

Next, we use cross multiplication to confirm this equivalence.

For $\frac{45}{60}$ and $\frac{3}{4}$: $45 \times 4 = 180 \quad \text{and} \quad 60 \times 3 = 180$

Both products are equal, confirming that $\frac{45}{60}$is equivalent to $\frac{3}{4}$. Therefore, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Locate Rational

c) Between 1 and 2

[Solution Description] To locate $\frac{5}{4}$ on the number line, convert it to a decimal by dividing 5 by 4:

$\frac{5}{4} = 1.25$

So, it lies between 1 and 2 on the number line at 1.25.

Key Concept: Decimal Conversion, Rational Identification

b) 2.16666... (Non-terminating repeating)

[Solution Description]

To convert $\frac{13}{6}$ to a decimal form, divide 13 by 6:

$\frac{13}{6} = 2.1666...$, where the decimal repeats after the digit '6'. This indicates it's a non-terminating repeating decimal.

A rational number has a decimal expansion that terminates or eventually repeats. Here, the repeating pattern confirms its rationality.

Key Concept: Between Two Rationals

c) 3.5

[Solution Description] Use the formula $\frac{r+s}{2}$ to find a rational number between 3 and 4:

$\frac{3+4}{2} = \frac{7}{2} = 3.5$

Hence, 3.5 is a rational number between 3 and 4.

Key Concept: Infinite Rationals

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] According to the density property of rational numbers, for any two rational numbers, no matter how close they are, you can always find another rational number in between them. This means that not only is there one rational number between 2 and 3, but there are actually infinitely many. The assertion is true because it states this concept correctly. The reason provided also accurately explains why the assertion is true by stating the same principle - that between any two rational numbers, there exists another rational number. Hence, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Rational Density, Complex Conversion

b) $\frac{7}{20}$

[Solution Description]

To find a rational number between $\frac{1}{3}$ and $\frac{2}{5}$, we can convert these fractions into decimals:

$\frac{1}{3} = 0.3333...$

$\frac{2}{5} = 0.4$

We need to find a decimal expansion that is greater than $0.3333...$ and less than $0.4$. Let's consider the decimal $0.35$, which lies between them. Now, we convert $0.35$ to a fraction:

$0.35 = \frac{35}{100} = \frac{7}{20}$

Thus, $\frac{7}{20}$ is a rational number between $\frac{1}{3}$ and $\frac{2}{5}$.

Key Concept: Rational Expression

a) $\frac{5}{11}$

[Solution Description] Let $x = 0.454545...$. Then$100x = 45.454545...$$.

Subtracting these, $100x - x = 45.454545... - 0.454545...$

You get: $99x = 45 \implies x = \frac{45}{99} = \frac{5}{11}$

So, 0.454545... can be expressed as $\frac{5}{11}$.

Key Concept: Basic Properties

b) Irrational

[Solution Description] When a rational number (except zero) is multiplied by an irrational number, the result is an irrational number due to the properties of irrational numbers.

Key Concept: Decimal Expansion

b) 1.414213562373095...

[Solution Description] A non-terminating and non-repeating decimal expansion characterizes an irrational number.

Key Concept: Complex Approximations, Real Number Line

c) Non-terminating, non-repeating property

[Solution Description] An irrational number is characterized by its non-terminating and non-repeating decimal expansion. The sequence given here exhibits initial segments of repeating patterns, but for it to qualify as irrational, it cannot repeat indefinitely. The inherent nature of irrational numbers guarantees that after any finite segment, the sequence will not form a periodic repetition; thus, ensuring the continuation as non-repeating and non-terminating.

Key Concept: Rational and Irrational Operations, Complex Examples

d) Assertion is false, but Reason is true.

[Solution Description]

In the given assertion, we need to find out if the sum of $\sqrt{3}$ and $-\sqrt{3}$ is irrational.

Let's calculate: $\sqrt{3} + (-\sqrt{3}) = 0$

Here, 0 is a rational number because it can be expressed as $\frac{0}{1}$.

Therefore, the Assertion is false.

Now consider the Reason. If $r$ is a rational number and $i$ is an irrational number, then their sum $r + i$ remains irrational since the decimal behavior of the irrational part cannot be cancelled by any rational part without an exact fractional expression being formed.

Hence, the Reason is true but does not apply to the Assertion.

Conclusion: The correct answer is (d) Assertion is false, but Reason is true.

Key Concept: Identify Irrational

c) $\sqrt{3}$

[Solution Description] A number is considered irrational if it cannot be expressed as a ratio of two integers, i.e., $\frac{p}{q}$. In this case, 4/5 and 0.8 are rational because they can be expressed as fractions of integers. On the other hand, $\sqrt{3}$ cannot be expressed in that form, making it an irrational number.

Key Concept: Non-terminating Decimals

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine whether the assertion and reason are true, we need to understand the properties of irrational numbers. An irrational number cannot be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Furthermore, an irrational number's decimal expansion is non-terminating and non-recurring.

Here, the given decimal expansion is 0.303003000300003..., which is non-terminating. Additionally, it does not repeat in any periodic pattern, making it non-recurring. Therefore, this decimal number cannot be expressed as a fraction of two integers, classifying it as an irrational number.

Thus, both the Assertion and Reason are true, and the Reason correctly explains why the decimal number is considered irrational.

Key Concept: Historical Figures

c) Johann Lambert

[Solution Description] Johann Lambert was the mathematician who first proved the irrationality of $\pi$ in the late 1700s. This proof was followed and supplemented by Adrien-Marie Legendre, who also worked on proving the properties of $\pi$. Therefore, the correct answer is Johann Lambert.

Key Concept: Real World Application, Complex Number Identification

a) $\pi$

[Solution Description]

The value of $\pi$ is essential in calculating the circumference and area of circles. Historically, it has been known as an irrational number through approximations such as those made by Archimedes. Unlike rational numbers like 3.0, 22/7 is only an approximation, not exact, highlighting its irrational nature.

Key Concept: Real Number Line

c) Both rational and irrational numbers

[Solution Description] The real number line represents all real numbers, which include both rational numbers (such as integers and fractions) and irrational numbers. Thus, the real number line encompasses both types.

Key Concept: Historical Context

d) Hippacus of Croton

[Solution Description] Hippacus of Croton is known for discovering that $\sqrt{2}$ is an irrational number.

Key Concept: Decimal Expansion

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine the correctness of the assertion and reason, we must understand the properties of irrational numbers. By definition, an irrational number cannot be expressed as a fraction of integers and its decimal representation is non-terminating and non-repeating. The given number $0.10110111011110...$ fits this description because it does not settle into a repeating pattern and continues indefinitely. Therefore, both the Assertion and Reason are true. Moreover, the reason accurately explains why the assertion holds since the non-terminating, non-recurring nature of the decimal supports its classification as an irrational number.

Key Concept: Real Number Line Concept

c) Real numbers

[Solution Description] Georg Cantor and Richard Dedekind demonstrated that the combination of all rational and irrational numbers constitutes the set of real numbers. This comprehensive collection is what forms the continuous number line.

Key Concept: Application of Pythagorean Theorem

b) 1 unit

[Solution Description] According to the Pythagorean theorem, $a^2 + b^2 = c^2$, where $c$ is the hypotenuse. Here, one leg ($a$) is 1 unit, and the hypotenuse ($c$) is $\sqrt{2}$ units. Therefore, $1^2 + b^2 = (\sqrt{2})^2$. Simplifying, we get:

$1 + b^2 = 2$

$b^2 = 2 - 1$

$b^2 = 1$

$b = \sqrt{1}$

$b = 1$

Thus, the length of the other leg is 1 unit.

Key Concept: Advanced Analysis, Historical Contextualization

d) $\sqrt{8}$ follows the same irrational proof path as $\sqrt{2}$, already known to number theorists.

[Solution Description]

To determine if $\sqrt{8}$ is irrational using Theodorus's method, we start by considering if it can be expressed as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers with no common factors other than 1 (i.e., they are coprime). Assume $\sqrt{8} = \frac{a}{b}$. Then, squaring both sides gives $8 = \frac{a^2}{b^2}$, hence $a^2 = 8b^2$. This implies $a^2$ is divisible by 8. Since 8 is $2^3$, $a$ must also be divisible by 2 to satisfy the divisibility condition. Let $a = 2k$: substituting yields $(2k)^2 = 8b^2$, or $4k^2 = 8b^2$, simplifying to $k^2 = 2b^2$. By repeating the analysis for $k^2$, we again find $k$ must be even, contradicting our initial assumption that $a$ and $b$ were coprime. Therefore, $\sqrt{8}$ is not an example that could have been proved irrational by Theodorus because it leads to rational contradiction.

Key Concept: Conceptual Connections

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion is true because Theodorus made significant contributions to the understanding of irrational numbers through his geometric proofs involving several square roots. However, the reason provided is false. Theodorus's methods were not algebraic; they were primarily geometric, focusing on properties of square roots and lengths rather than polynomial expressions.

Key Concept: Identify Irrational

c) $\pi$

[Solution Description] A number is considered irrational if it cannot be expressed as a ratio of two integers. Among the given options, $\pi$ is well-known to be an irrational number.

Key Concept: Historical Proofs, Theoretical Concepts

c) Theodorus

[Solution Description] The proof of the irrationality of $\sqrt{3}$ was famously done by employing a proof by contradiction method similar to that used for $\sqrt{2}$. While Theodorus initially worked with proving the irrationality of certain square roots including $\sqrt{3}$, this question specifically refers to methodologies grown from the Pythagorean era but further formalized in later proofs.

Key Concept: Non-repeating Nature

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

- The assertion states that the number $\sqrt{3}$ has a non-terminating, non-recurring decimal expansion. This is true because $\sqrt{3}$ is an irrational number. By definition, every irrational number has a non-terminating, non-recurring decimal expansion.

- The reason states that a non-terminating, non-recurring decimal cannot be expressed as a fraction. This is also true, as any number with such a decimal expansion is irrational and cannot be expressed as a ratio of two integers.

Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Rational/Irrational

b) Irrational

[Solution Description] The number 0.101001000100001... is a non-terminating, non-recurring decimal expansion, therefore it is an irrational number.

Key Concept: Approximation Techniques

c) Sulbasutras

[Solution Description] Sulbasutras are ancient Indian texts known for their geometric and mathematical insights, including approximations of $\pi$. These texts offered one of the earliest known approximations.

Key Concept: Decimal Identification

c) 0.101001000100001...

[Solution Description] An irrational number has a non-terminating and non-recurring decimal expansion. The option 0.101001000100001... is an example, as it doesn't terminate and lacks any repeating pattern.

Key Concept: Locate $\sqrt{2}$

b) Use two points at 0 and 1 on the number line; construct a perpendicular at 1 and cut a 1 unit segment to form a triangle.

[Solution Description] To locate $\sqrt{2}$, start by drawing a right triangle with both legs equal to 1 unit. By the Pythagorean theorem, the hypotenuse will be $\sqrt{1^2 + 1^2} = \sqrt{2}$. This length can then be marked on the number line starting from zero.

Key Concept: Advanced Geometric Construction, Historical Context

c) Measure distance along the x-axis equivalent to the hypotenuse of triangle ABC.

[Solution Description] To locate $\sqrt{5}$ geometrically, construct a right triangle $ABC$ where $AB = 2$ and $BC = 1$. The hypotenuse $AC$ will be $\sqrt{(2^2 + 1^2)} = \sqrt{5}$. Use a compass to measure from point $A$ to $C$, then place this length on the number line starting from the origin.

Key Concept: Concept Identification

d) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

[Solution Description] Pythagoras' Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

This can be written as: $c^2 = a^2 + b^2$

where $c$ is the hypotenuse, and $a$ and $b$ are the other two sides.

Key Concept: Right Triangle Identification

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine whether the given triangle is a right triangle, we apply Pythagoras' Theorem. According to the theorem, in a right-angled triangle, the square of the hypotenuse ($c$) should equal the sum of the squares of the other two sides ($a$ and $b$). Here, $a = 9$, $b = 40$, and $c = 41$. We need to verify if $9^2 + 40^2 = 41^2$.

Calculate: $9^2 = 81,\quad 40^2 = 1600,\quad \text{and}\quad 41^2 = 1681$

Now add the squares of the shorter sides: $9^2 + 40^2 = 81 + 1600 = 1681$

Since $1681 = 41^2$, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Real Number Visualization, Irrational Number Generation

c) Assertion is true, but Reason is false.

[Solution Description] To solve this, we first need to understand that between any two distinct numbers on the real number line, there are infinitely many other numbers. Both 0.123456 and 0.123457 are terminating decimals and thus, are rational numbers.

One method to find an irrational number between these two numbers is to consider a decimal that does not terminate and does not repeat. For instance, the number 0.123456789101112... where each sequence of digits represents consecutive integers (1, then 2, then 3, etc.), will have no repeating pattern and will not terminate.

The assertion is true because you can construct such a non-repeating decimal as described above, which is irrational.

On the other hand, the reason is false as it incorrectly states that non-terminating non-recurring decimals represent rational numbers. In fact, they represent irrational numbers by definition.

Thus, the correct answer is that the assertion is true, but the reason is false.

Key Concept: Irrational Between Rationals, Real Number Placement

d) $1.61803398874989...$

[Solution Description] The approximate decimal values for $\sqrt{2}$ and $\sqrt{3}$ are about $1.414$ and $1.732$, respectively. We need to find a non-repeating, non-terminating decimal within this range, which would indicate it is irrational. Consider the number $1.500000123456789...$, which clearly does not terminate nor does it repeat.

Key Concept: Rational vs Irrational

b) Irrational

[Solution Description] The decimal expansion of the given number is non-terminating and non-recurring because the sequence between each '1' increases progressively without a repeating pattern. Hence, it is an example of an irrational number.

Key Concept: Complex Pattern Recognition, Decimal Expansion Analysis

c) 185

[Solution Description]

The fraction $\frac{5}{27}$ needs to be converted into a decimal form to identify the repeating block. To do this, perform long division of 5 by 27: $5 \div 27 = 0.185185185...$

This shows that the repeating block is "185", as it repeats indefinitely.

Key Concept: Decimal Conversion

a) 0.44, terminating

[Solution Description]

To convert $\frac{11}{25}$ into a decimal, we perform division: $11 \div 25 = 0.44$

The decimal form is 0.44, which terminates after two digits. Therefore, it is a terminating decimal expansion.

Key Concept: Rationality Identification

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine if the assertion and reason are true, we analyze each statement:

- The number 4.25 can be expressed as a fraction $\frac{425}{100}$. Therefore, it is a rational number because it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

- The decimal expansion of 4.25 is indeed terminating since it has a finite number of decimal places (two decimal places).

Both the assertion and reason are true statements. Moreover, the reason correctly explains why 4.25 is a rational number; it's because its decimal expansion terminates.

Key Concept: Recognize Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine if the assertion and reason are true, and if the reason explains the assertion:

- Non-terminating recurring decimals like $0.6666...$ can be converted into fractions.

- Let's convert $0.6666...$ to a fraction.

- Let $x = 0.6666...$. Multiply both sides by 10: $10x = 6.6666...$.

- Subtract these two equations: $10x - x = 6.6666... - 0.6666...$

$9x = 6$

$x = \frac{6}{9} = \frac{2}{3}$

- Since $0.6666...$ equals $\frac{2}{3}$, it is a rational number.

- Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Conversion to Fraction

b) $\frac{2}{3}$

[Solution Description] Let $x = 0.\overline{6}$. This means, x = 0.66666...

Multiply both sides by 10: 10x = 6.66666...

Subtract the original equation from this one: $10x - x = 6.66666... - 0.66666...$ $9x = 6$

Now, solve for $x$: $x = \frac{6}{9} = \frac{2}{3}$

Thus, $0.\overline{6}$ can be expressed as $\frac{2}{3}$.

Key Concept: Recognize Irrational Numbers

d) $\sqrt{2}$

[Solution Description] An irrational number is one with a non-terminating non-recurring decimal expansion. Among the options, $\sqrt{2} = 1.414213...$ fits this description as it continues without repeating any specific sequence of digits indefinitely.

Key Concept: Sequence Challenge, Complex Identification

b) $0.0102030405060708...$

[Solution Description] Examining each sequence, $0.0102030405060708...$ has digits that increase by one in each segment, representing a pattern that is non-repetitive and continues indefinitely. This makes it non-terminating and non-recurring.

Key Concept: Rationalization with Binomials

c) $5\sqrt{3} + 5\sqrt{2}$

[Solution Description] Multiply both numerator and denominator by the conjugate of the denominator, $\sqrt{3} + \sqrt{2}$:

$$\frac{5}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{5(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2}$$

The denominator becomes: $(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$

So, the expression simplifies to: $5(\sqrt{3} + \sqrt{2})$

Thus, the rationalized form is $5\sqrt{3} + 5\sqrt{2}$.

Key Concept: Exponent Laws with Complex Bases, Real World Application

c) $9$

[Solution Description] Start by simplifying each term. Note that $9 = 3^2$, so: $9^{1/4} = (3^2)^{1/4} = 3^{1/2}$

Now substitute back into the expression: $((3^{1/2} \cdot 3^{1/2})^2) = (3^{1/2+1/2})^2 = (3)^2$

Therefore, the final simplified result is: $9$

Thus, the answer is $9$.

Key Concept: Maximum Repeating Block

c) 6

[Solution Description] To find the repeating cycle length of $\frac{1}{13}$, check the smallest power of 10 that leaves remainder 1 when divided by 13, which is found by calculation as 6. So, the maximum number of digits is 6.

Key Concept: Rational Number Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To verify the assertion, we need to convert the fraction $\frac{2}{11}$ into its decimal form. Perform long division of 2 by 11:

- Divide 20 by 11 to get 1 with a remainder of 9.

- Bring down another 0 to make it 90.

- Divide 90 by 11 to get 8 with a remainder of 2.

- Repeat the process.

After several steps, you will notice that the sequence repeats, giving $0.\overline{18}$. Thus, the assertion is true as the decimal expansion is non-terminating recurring.

Next, consider the reason: A rational number has a non-terminating decimal if its denominator after simplification in terms of prime factors contains any factor other than 2 and 5. Here, 11 is a prime number and is neither 2 nor 5, so the reason correctly supports the conversion logic for non-termination.

Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Real World Application, Complex Rationality

a) 33 times; Rational

[Solution Description]

The recurring decimal $x = 0.030303...$ can be expressed as a fraction.

Set $x = 0.030303...$, then multiply by 100 since two digits repeat:

$100x = 3.0303...$

Subtract the original equation from the multiplied equation: $100x - x = 3.0303... - 0.0303...$ $99x = 3$

Solving for $x$ gives: $x = \frac{3}{99} = \frac{1}{33}$

Therefore, the machine operates every $\frac{1}{33}$ hours. In 1 hour, it will operate 33 times (since $\frac{1}{\left(\frac{1}{33}\right)} = 33$). The decimal is rational because it can be expressed as a fraction $\frac{1}{33}$.

Key Concept: Intermediate Conversion

a) $\frac{5}{11}$

[Solution Description] Let $x = 0.4545\ldots$. Multiply both sides by 100 to shift the repeating block: $100x = 45.4545\ldots$.

Subtract the original equation from this new one: $100x - x = 45.4545\ldots - 0.4545\ldots$

Simplifying gives: $99x = 45$. Solving for $x$ yields $x = \frac{45}{99}$. Reduce the fraction to get $x = \frac{5}{11}$.

Therefore, $0.4545\ldots = \frac{5}{11}$.

Key Concept: Basic Properties

c) Irrational

[Solution Description] The number $\sqrt{3}$ is irrational. Multiplying a non-zero rational number (5 in this case) with an irrational number results in an irrational product according to the syllabus.

Key Concept: Exponent Laws, Real Number Line

a) Any positive real number

[Solution Description] Using the laws of exponents, combine the exponents on the left-hand side: $x^{1/3} \cdot x^{1/6} = x^{1/3 + 1/6} = x^{1/2}$

Equating the exponents: $\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$

Finding common denominators: $\frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$

This confirms the equality holds for any non-zero real number $x$, but due to the fractional exponent indicating roots, $x > 0$.

Hence, any positive real number satisfies this equation.

Key Concept: Laws of Exponents

b) $x^6$

[Solution Description] According to the law of exponents, $(a^m)^n = a^{mn}$.

Applying this rule,  we have: $(x^3)^2 = x^{3 \cdot 2} = x^6$

The simplified form is $x^6$.

Key Concept: Rational-Irrational Operations

c) Assertion is true, but Reason is false.

[Solution Description]

To determine the truth of the assertion and reason, let's analyze each one separately:

1. **Assertion**: Consider any rational number $r$ and an irrational number $i$. The sum $r + i$ results in an irrational number. This is because adding a finite or repeating (rational) portion to a non-repeating, non-terminating (irrational) part cannot result in a repeating or terminating decimal. Hence, the assertion is true.

2. **Reason**: By definition, rational numbers are those which can be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. Their decimal expansions are either terminating or repeating. Thus, this reason is false.

From the above analysis, we conclude that the assertion is true, but the reason provided is false.

Key Concept: Complex Rationalization, Nested Operations

b) Irrational

[Solution Description] To solve this, we need to rationalize each term and then find their product:

For $x = \frac{1}{\sqrt{2} + 1}$:

$$x = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1$$

For $y = \frac{1}{\sqrt{3} - 1}$:

$$y = \frac{1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{\sqrt{3} + 1}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{\sqrt{3} + 1}{3 - 1} = \frac{\sqrt{3} + 1}{2}$$

Now, calculate $x \cdot y$: $(\sqrt{2} - 1) \cdot \left(\frac{\sqrt{3} + 1}{2}\right) = \frac{(\sqrt{2} - 1)(\sqrt{3} + 1)}{2}$

Simplifying further: $= \frac{(\sqrt{2} \cdot \sqrt{3} + \sqrt{2} - \sqrt{3} - 1)}{2}$

Since $\sqrt{6}$, $\sqrt{2}$, and $\sqrt{3}$ are irrational numbers, their combination remains irrational.

Thus, the result is irrational.

Key Concept: Complex Expressions, Second Order Thinking

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine the nature of the expression $\pi^2 - 7$, we need to analyze each component:

- $\pi$ is known to be an irrational number.

- When $\pi$ is squared, i.e., $\pi^2$, it remains an irrational number because the square of any non-zero irrational number is also irrational if it does not simplify to a rational value.

- Now consider $\pi^2 - 7$: We have an irrational number ($\pi^2$) minus a rational number (7).

According to the properties of numbers, subtracting a rational number from an irrational number always results in an irrational number because no rational number exists that can exactly cancel out the irrational component.

Therefore, the expression $\pi^2 - 7$ is indeed irrational, making the assertion true.

Regarding the reason, it correctly states that $\pi^2$ is irrational and that subtracting a rational number from an irrational number results in an irrational number. Thus, both the assertion and the reason are true, and the reason correctly explains why the assertion is true.

Hence, the correct answer is option (a).

Key Concept: Rational vs Irrational

a) $7 + \sqrt{8}$

[Solution Description] We need to identify which operation results in an irrational number.

a) $\quad 7 + \sqrt{8}, \quad \text{where } \sqrt{8} \text{ is irrational, hence this is irrational}$

b) 10 - 2 = 8, which is rational

c) $\frac{22}{7} \times 4 \approx 12.57, \text{ which is rational as } \frac{22}{7} \text{ is used for approximation only}$

d) $\frac{9}{3} = 3$, which is rational

Option a) results in an irrational number since adding a rational number to an irrational number yields an irrational result.

Key Concept: Complex Multiplication

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To solve this, we need to evaluate whether the assertion is true or false:

- Let's calculate $4 \times (3 + \sqrt{5})$. Using the distributive property:

$ = 4 \times 3 + 4 \times \sqrt{5}$

$ = 12 + 4\sqrt{5}$

- Here, $12$ is a rational number, and $4\sqrt{5}$ is irrational since $\sqrt{5}$ is irrational and multiplying by a non-zero rational number ($4$) remains irrational.

- Adding a rational number $12$ and an irrational number $4\sqrt{5}$ results in an irrational number.

Thus, Assertion (A) is true.

Now for Reason (R):

- According to the properties, the sum of a rational and an irrational number indeed results in an irrational number.

Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Basic Concept

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion states that the product of $\sqrt{2}$ and $\sqrt{2}$ is rational.

Let's calculate this: $\sqrt{2} \times \sqrt{2} = (\sqrt{2})^2 = 2$

which is a rational number.

The reason given states that the product of two irrational numbers is always rational. This statement is false because the product of two irrational numbers can be either rational or irrational. For instance, $\sqrt{2} \times \sqrt{3}$ is irrational, while $\sqrt{2} \times \sqrt{2}$ is rational as shown above.

Therefore, Assertion is true, but Reason is false.

Key Concept: Proof-Based Questions, Complex Number Line

c) Assertion is true, but Reason is false.

[Solution Description]

Multiplying two irrational numbers does not always yield another irrational number. Let us consider $\sqrt{2} \times \sqrt{2} = 2$, which is a rational number. Thus, while many products of irrational numbers may be irrational, it is not universally true that they must be. However, in the specific case of $\sqrt{2} \times \sqrt{3}$, let's calculate as follows: $\sqrt{2} \times \sqrt{3} = \sqrt{6}$

Here, since 6 is not a perfect square, $\sqrt{6}$ remains irrational.

Hence, Assertion (A) is correct because $\sqrt{6}$ is indeed irrational. However, Reason (R) is false because the product of two irrational numbers can also be rational, as demonstrated by multiplying $\sqrt{2}$ with itself.

Key Concept: Commutative and Associative Laws

d) Assertion is false, but Reason is true.

[Solution Description]

To determine if both statements are true and whether the reason correctly explains the assertion, consider examples of multiplying irrational numbers:

- Example: $\sqrt{2} \times \sqrt{2} = 2$, which is rational.

Thus, the assertion "The product of two irrational numbers is always irrational" is false because there exist cases where the product can be rational.

- For the reason, while it's true that irrational numbers do not satisfy closure under multiplication (since their product may be either rational or irrational), this does not explain why any given product is irrational.

Therefore, the Assertion is false but the Reason is true.

Key Concept: Basic Closure

d) All of the above

[Solution Description] Rational numbers are closed under addition, subtraction, and multiplication but not division by zero. Therefore, all these operations except division by zero result in a rational number.

Key Concept: Distributive Law

a) $3 \times (2 + 5) = 3 \times 2 + 3 \times 5$

[Solution Description] According to the distributive law, $a(b + c) = ab + ac$. Option a demonstrates this correctly as: $3 \times (2 + 5) = 3 \times 2 + 3 \times 5$

This simplifies both sides equally.

Key Concept: Complex Operations, Rationalizing Complex Denominators

b) Irrational

[Solution Description]

To simplify the expression $\frac{\sqrt{5} + 1}{\sqrt{5} - 1}$, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, which is $(\sqrt{5} + 1)$:

$$\frac{\sqrt{5} + 1}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{(\sqrt{5} + 1)^2}{(\sqrt{5})^2 - (1)^2}$$

Calculate the square of the numerator:

$$(\sqrt{5} + 1)^2 = (\sqrt{5})^2 + 2(\sqrt{5})(1) + (1)^2 = 5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5}$$

And calculate the denominator: $(\sqrt{5})^2 - 1^2 = 5 - 1 = 4$

Thus, the expression simplifies to:

$$\frac{6 + 2\sqrt{5}}{4} = \frac{6}{4} + \frac{2\sqrt{5}}{4} = \frac{3}{2} + \frac{\sqrt{5}}{2}$$

This simplified form is a sum of a rational number ($\frac{3}{2}$) and an irrational number ($\frac{\sqrt{5}}{2}$), making it an irrational number.

Key Concept: Basic Exponent Rule

b) 1

[Solution Description] The rule for any non-zero number $a$ raised to the power of zero is $a^0 = 1$. Therefore, $7^0 = 1$.

Key Concept: Complex Rational Exponents, Nested Exponents

c) 1944

[Solution Description]

First, calculate $x = 16^{1/4} = (2^4)^{1/4} = 2$. Then calculate $y = 81^{1/4} = (3^4)^{1/4} = 3$. Now, evaluate $x^3 = 2^3 = 8$ and $y^6 = 3^6 = 729$.

Substituting these into the expression gives: $\frac{x^3 \cdot y^6}{y} = \frac{8 \cdot 729}{3} = \frac{5832}{3} = 1944$

Thus, the result is 1944.

Key Concept: Power of a Power

d) $(y^4)^3$

[Solution Description] The power of a power rule states that $(a^p)^q = a^{pq}$. We need an expression that simplifies to $y^{12}$ using this rule.

Consider $(y^4)^3$: By the power of a power rule, $4 \times 3 = 12$.

Thus, $(y^4)^3 = y^{12}$.

Key Concept: Simplifying Complex Expressions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To verify the assertion, we simplify the expression using the law of exponents stated in the reason.

Firstly, apply the law: $a^m \cdot a^n = a^{m+n}$ to combine the exponents: $8^{1/3} \cdot 8^{-1/6} = 8^{1/3 + (-1/6)}$

Calculate the exponent: $\frac{1}{3} + \left(-\frac{1}{6}\right) = \frac{1}{3} - \frac{1}{6}$

Find a common denominator to add the fractions:

$$\frac{1}{3} = \frac{2}{6}, \quad \text{so } \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6}$$

Thus, we have: $8^{1/3} \cdot 8^{-1/6} = 8^{1/6}$

The assertion is true.

Now consider the reason: it correctly states the law of exponents used for the simplification. The reason $a^m \cdot a^n = a^{m+n}$ is indeed the correct explanation of the assertion.

Key Concept: Simplification with Exponents

a) $4^6$

[Solution Description] Use the law of exponents: $(a^m)^n = a^{mn}$.

Thus, $(4^3)^2 = 4^{3 \times 2} = 4^6$

Therefore, the simplified expression is $4^6$.

Key Concept: Rational/Irrational Operations

b) Irrational

[Solution Description] The number 2 is rational because it can be expressed as $\frac{2}{1}$. The number $\sqrt{11}$ is irrational because 11 is not a perfect square and therefore cannot be expressed as a fraction. The sum of a rational number and an irrational number results in an irrational number. Thus, $2 + \sqrt{11}$ is irrational.