Number System

Given: The number zero (0).

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

Working:
Yes, zero is a rational number. We can write:
$$0 = \frac{0}{1} = \frac{0}{2} = \frac{0}{3} = \frac{0}{-1} \text{ etc.}$$
In each case, $p = 0$ (an integer) and $q \neq 0$ (an integer).

Conclusion: Zero is indeed a rational number and can be written in the form $\frac{p}{q}$ in infinitely many ways.

Given: Two rational numbers 3 and 4.

Concept: To find $n$ rational numbers between two numbers $a$ and $b$, multiply numerator and denominator to create a gap. Here we need 6 rational numbers, so we write:
$$3 = \frac{3 \times 7}{7} = \frac{21}{7} \quad \text{and} \quad 4 = \frac{4 \times 7}{7} = \frac{28}{7}$$

Working:
The rational numbers between $\frac{21}{7}$ and $\frac{28}{7}$ are:
$$\frac{22}{7},\ \frac{23}{7},\ \frac{24}{7},\ \frac{25}{7},\ \frac{26}{7},\ \frac{27}{7}$$

Answer: Six rational numbers between 3 and 4 are:
$$\frac{22}{7},\ \frac{23}{7},\ \frac{24}{7},\ \frac{25}{7},\ \frac{26}{7},\ \frac{27}{7}$$
(Note: There are infinitely many such rational numbers; this is one possible set.)

Given: $\frac{3}{5}$ and $\frac{4}{5}$.

Concept: To find 5 rational numbers between them, convert both fractions to equivalent fractions with a larger denominator (multiply by 6):
$$\frac{3}{5} = \frac{3 \times 6}{5 \times 6} = \frac{18}{30} \quad \text{and} \quad \frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}$$

Working:
The rational numbers between $\frac{18}{30}$ and $\frac{24}{30}$ are:
$$\frac{19}{30},\ \frac{20}{30},\ \frac{21}{30},\ \frac{22}{30},\ \frac{23}{30}$$

Answer: Five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ are:
$$\frac{19}{30},\ \frac{20}{30},\ \frac{21}{30},\ \frac{22}{30},\ \frac{23}{30}$$

(i) Every natural number is a whole number.

Answer: TRUE

Reason: The set of natural numbers is $\mathbb{N} = \{1, 2, 3, 4, \ldots\}$ and the set of whole numbers is $\mathbb{W} = \{0, 1, 2, 3, \ldots\}$. Every natural number is present in the set of whole numbers. Hence, every natural number is a whole number.

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(ii) Every integer is a whole number.

Answer: FALSE

Reason: The set of integers is $\mathbb{Z} = \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$. Negative integers such as $-1, -2, -3, \ldots$ are integers but they are NOT whole numbers. Hence, every integer is not a whole number.

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(iii) Every rational number is a whole number.

Answer: FALSE

Reason: Rational numbers include fractions such as $\frac{1}{2}, \frac{3}{4}, \frac{-2}{3}$, etc. These are not whole numbers. For example, $\frac{1}{2}$ is a rational number but not a whole number. Hence, every rational number is not a whole number.

(i) Every irrational number is a real number.

Answer: TRUE

Reason: The set of real numbers consists of all rational numbers and all irrational numbers together. Therefore, every irrational number is a part of the collection of real numbers, making this statement true.

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(ii) Every point on the number line is of the form $\sqrt{m}$, where $m$ is a natural number.

Answer: FALSE

Reason: Points on the number line include negative numbers (e.g., $-1, -2$), zero, and positive numbers. Negative numbers cannot be expressed as $\sqrt{m}$ where $m$ is a natural number (since square roots of natural numbers are non-negative). Also, numbers like $2, 3$ are on the number line but $\sqrt{m}$ for natural number $m$ gives $\sqrt{1}=1, \sqrt{2}, \sqrt{3}, \sqrt{4}=2, \ldots$ — not every point is covered. Hence the statement is false.

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(iii) Every real number is an irrational number.

Answer: FALSE

Reason: Real numbers include both rational and irrational numbers. For example, $2, \frac{3}{4}, 0$ are real numbers but they are rational, not irrational. Hence, not every real number is irrational.

Answer: No, the square roots of all positive integers are not irrational.

Example:
$$\sqrt{4} = 2, \quad \sqrt{9} = 3, \quad \sqrt{16} = 4$$
Here, $\sqrt{4} = 2$ is a rational number (it can be written as $\frac{2}{1}$).

Conclusion: The square roots of perfect squares like $1, 4, 9, 16, 25, \ldots$ are rational numbers. Only the square roots of non-perfect-square positive integers are irrational.

Concept: We use the Pythagorean theorem to construct $\sqrt{5}$.

Steps:

Step 1: Draw a number line and mark the origin $O$ (representing 0) and point $A$ representing 2, so $OA = 2$ units.

Step 2: At point $A$, draw $AB$ perpendicular to the number line such that $AB = 1$ unit.

Step 3: Join $OB$. By the Pythagorean theorem:
$$OB = \sqrt{OA^2 + AB^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$$

Step 4: With $O$ as centre and $OB$ as radius, draw an arc that cuts the number line at point $P$.

Conclusion: The point $P$ on the number line represents $\sqrt{5}$, since $OP = OB = \sqrt{5}$.

This is a classroom activity. Below is the mathematical justification:

Step 1: Start at point $O$. Draw $OP_1 = 1$ unit along the number line.
$$OP_1 = \sqrt{1} = 1$$

Step 2: Draw $P_1P_2 \perp OP_1$, with $P_1P_2 = 1$ unit.
$$OP_2 = \sqrt{OP_1^2 + P_1P_2^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$$

Step 3: Draw $P_2P_3 \perp OP_2$, with $P_2P_3 = 1$ unit.
$$OP_3 = \sqrt{OP_2^2 + P_2P_3^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2+1} = \sqrt{3}$$

Step 4: Draw $P_3P_4 \perp OP_3$, with $P_3P_4 = 1$ unit.
$$OP_4 = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$$

General Pattern: At each step $n$:
$$OP_n = \sqrt{n}$$

Conclusion: By continuing this process, we obtain a spiral (called the square root spiral or Theodorus spiral) where the distance from $O$ to $P_n$ equals $\sqrt{n}$, representing $\sqrt{2}, \sqrt{3}, \sqrt{4}, \ldots$ on the plane.

(i) $\frac{36}{100}$
$$\frac{36}{100} = 0.36$$
Type: Terminating decimal

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(ii) $\frac{1}{11}$
Performing long division: $1 \div 11$:
$$\frac{1}{11} = 0.090909\ldots = 0.\overline{09}$$
Type: Non-terminating recurring (repeating block: 09)

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(iii) $4\frac{1}{8}$
$$4\frac{1}{8} = \frac{33}{8}$$
Performing long division: $33 \div 8$:
$$\frac{33}{8} = 4.125$$
Type: Terminating decimal

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(iv) $\frac{3}{13}$
Performing long division: $3 \div 13$:
$$\frac{3}{13} = 0.230769230769\ldots = 0.\overline{230769}$$
Type: Non-terminating recurring (repeating block: 230769)

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(v) $\frac{2}{11}$
Performing long division: $2 \div 11$:
$$\frac{2}{11} = 0.181818\ldots = 0.\overline{18}$$
Type: Non-terminating recurring (repeating block: 18)

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(vi) $\frac{329}{400}$
Performing long division: $329 \div 400$:
$$\frac{329}{400} = 0.8225$$
Type: Terminating decimal

Given: $\frac{1}{7} = 0.\overline{142857}$

Concept: Since $\frac{2}{7} = 2 \times \frac{1}{7}$, $\frac{3}{7} = 3 \times \frac{1}{7}$, etc., we can multiply the repeating block. Also, the remainders while dividing $1$ by $7$ cycle through $1, 3, 2, 6, 4, 5$ — each remainder corresponds to a cyclic permutation of the block $142857$.

Predictions:
$$\frac{2}{7} = 2 \times 0.\overline{142857} = 0.\overline{285714}$$
$$\frac{3}{7} = 3 \times 0.\overline{142857} = 0.\overline{428571}$$
$$\frac{4}{7} = 4 \times 0.\overline{142857} = 0.\overline{571428}$$
$$\frac{5}{7} = 5 \times 0.\overline{142857} = 0.\overline{714285}$$
$$\frac{6}{7} = 6 \times 0.\overline{142857} = 0.\overline{857142}$$

Observation: Each decimal is a cyclic permutation of the digits $142857$. This happens because the remainders when dividing by 7 cycle through all non-zero residues.

(i) $0.\overline{6}$

Let $x = 0.6666\ldots$ $\quad \cdots (1)$

Multiply both sides by 10:
$$10x = 6.6666\ldots \quad \cdots (2)$$

Subtract (1) from (2):
$$10x - x = 6.6666\ldots - 0.6666\ldots$$
$$9x = 6$$
$$x = \frac{6}{9} = \frac{2}{3}$$

$$\boxed{0.\overline{6} = \frac{2}{3}}$$

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(ii) $0.4\overline{7}$

Let $x = 0.4777\ldots$ $\quad \cdots (1)$

Multiply both sides by 10:
$$10x = 4.777\ldots \quad \cdots (2)$$

Multiply both sides by 100:
$$100x = 47.777\ldots \quad \cdots (3)$$

Subtract (2) from (3):
$$100x - 10x = 47.777\ldots - 4.777\ldots$$
$$90x = 43$$
$$x = \frac{43}{90}$$

$$\boxed{0.4\overline{7} = \frac{43}{90}}$$

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(iii) $0.\overline{001}$

Let $x = 0.001001001\ldots$ $\quad \cdots (1)$

Multiply both sides by 1000:
$$1000x = 1.001001001\ldots \quad \cdots (2)$$

Subtract (1) from (2):
$$1000x - x = 1.001001\ldots - 0.001001\ldots$$
$$999x = 1$$
$$x = \frac{1}{999}$$

$$\boxed{0.\overline{001} = \frac{1}{999}}$$

Let $x = 0.9999\ldots$ $\quad \cdots (1)$

Multiply both sides by 10:
$$10x = 9.9999\ldots \quad \cdots (2)$$

Subtract (1) from (2):
$$10x - x = 9.9999\ldots - 0.9999\ldots$$
$$9x = 9$$
$$x = 1$$

$$\boxed{0.99999\ldots = \frac{1}{1} = 1}$$

Discussion: This result may seem surprising, but it makes sense because $0.9999\ldots$ is a non-terminating recurring decimal and the difference between 1 and $0.9999\ldots$ is $0.0000\ldots = 0$. There is no gap between $0.9999\ldots$ and $1$; they represent the same number. This shows that every non-terminating recurring decimal is a rational number.

Concept: When we divide $1$ by $17$, the remainders at each step can only be $1, 2, 3, \ldots, 16$ (i.e., at most $16$ different non-zero remainders). Once a remainder repeats, the decimal block repeats. Therefore, the maximum length of the repeating block is $\mathbf{16}$.

Verification by long division:

Performing $1 \div 17$:

$$1.000000000000000 \div 17$$

Step-by-step remainders: $10, 100\to 15, 150\to 14, 140\to 4, 40\to 6, 60\to 9, 90\to 5, 50\to 16, 160\to 7, 70\to 2, 20\to 3, 30\to 13, 130\to 11, 110\to 8, 80\to 12, 120\to 1$ (remainder 1 repeats)

$$\frac{1}{17} = 0.\overline{0588235294117647}$$

The repeating block is $0588235294117647$, which has 16 digits.

Conclusion: The maximum number of digits in the repeating block of $\frac{1}{17}$ is $16$, which is confirmed by the division.

Examples of terminating decimals:
$$\frac{1}{2} = 0.5 \quad (q = 2 = 2^1)$$
$$\frac{1}{4} = 0.25 \quad (q = 4 = 2^2)$$
$$\frac{1}{5} = 0.2 \quad (q = 5 = 5^1)$$
$$\frac{1}{8} = 0.125 \quad (q = 8 = 2^3)$$
$$\frac{1}{10} = 0.1 \quad (q = 10 = 2^1 \times 5^1)$$
$$\frac{1}{25} = 0.04 \quad (q = 25 = 5^2)$$
$$\frac{7}{20} = 0.35 \quad (q = 20 = 2^2 \times 5^1)$$

Observation and Conclusion:

In all the above examples, the denominator $q$ (in lowest terms) has only $2$ and $5$ as its prime factors.

Property: A rational number $\frac{p}{q}$ (in lowest terms) has a terminating decimal expansion if and only if $q$ is of the form $2^m \times 5^n$, where $m$ and $n$ are non-negative integers.

Concept: Non-terminating non-recurring decimals represent irrational numbers.

Three such numbers are:

1. $\sqrt{2} = 1.41421356\ldots$ (non-terminating, non-recurring)

2. $\sqrt{3} = 1.73205080\ldots$ (non-terminating, non-recurring)

3. $\pi = 3.14159265\ldots$ (non-terminating, non-recurring)

Alternatively, one can write numbers like:
$$0.101001000100001\ldots$$
$$0.202002000200002\ldots$$
$$0.303003000300003\ldots$$
where the pattern never repeats.

Given: $\frac{5}{7}$ and $\frac{9}{11}$.

Step 1: Convert to decimals.
$$\frac{5}{7} = 0.\overline{714285} \approx 0.7142\ldots$$
$$\frac{9}{11} = 0.\overline{81} = 0.8181\ldots$$

Step 2: We need irrational numbers between $0.7142\ldots$ and $0.8181\ldots$

Three irrational numbers between them:

1. $0.73073007300073000073\ldots$ (non-terminating, non-recurring)

2. $0.75075007500075000075\ldots$ (non-terminating, non-recurring)

3. $0.79079007900079000079\ldots$ (non-terminating, non-recurring)

Conclusion: These three numbers lie between $\frac{5}{7}$ and $\frac{9}{11}$ and are irrational (non-terminating non-recurring decimals).

(i) $\sqrt{23}$

23 is not a perfect square. Its square root cannot be expressed as $\frac{p}{q}$.
$$\sqrt{23} = 4.79583\ldots \text{ (non-terminating, non-recurring)}$$
Classification: Irrational number

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(ii) $\sqrt{225}$
$$\sqrt{225} = \sqrt{15^2} = 15$$
15 is an integer and can be written as $\frac{15}{1}$.
Classification: Rational number

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(iii) $0.3796$

This is a terminating decimal.
$$0.3796 = \frac{3796}{10000} = \frac{949}{2500}$$
Classification: Rational number

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(iv) $7.478478\ldots = 7.\overline{478}$

This is a non-terminating but recurring (repeating) decimal.
Classification: Rational number

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(v) $1.101001000100001\ldots$

The pattern shows one more zero is inserted between consecutive 1s each time. This decimal is non-terminating and non-recurring (the block never repeats).
Classification: Irrational number

(i) $2 - \sqrt{5}$

Here, $2$ is rational and $\sqrt{5}$ is irrational.
By the property: rational $-$ irrational $=$ irrational.
Classification: Irrational number

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(ii) $(3 + \sqrt{23}) - \sqrt{23}$

$$(3 + \sqrt{23}) - \sqrt{23} = 3 + \sqrt{23} - \sqrt{23} = 3$$
3 can be written as $\frac{3}{1}$.
Classification: Rational number

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(iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$

$$\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}$$
$\frac{2}{7}$ is of the form $\frac{p}{q}$ where $p = 2$, $q = 7 \neq 0$.
Classification: Rational number

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(iv) $\frac{1}{\sqrt{2}}$

$\sqrt{2}$ is irrational. The reciprocal of an irrational number is irrational.
$$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.7071\ldots \text{ (non-terminating, non-recurring)}$$
Classification: Irrational number

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(v) $2\pi$

$\pi$ is irrational. A non-zero rational number ($2$) multiplied by an irrational number is irrational.
$$2\pi = 6.28318\ldots \text{ (non-terminating, non-recurring)}$$
Classification: Irrational number

(i) $(3 + \sqrt{3})(2 + \sqrt{2})$

Using distributive property:
$$(3 + \sqrt{3})(2 + \sqrt{2}) = 3 \cdot 2 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{3} \cdot \sqrt{2}$$
$$= 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$$

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(ii) $(3 + \sqrt{3})(3 - \sqrt{3})$

Using identity $(a+b)(a-b) = a^2 - b^2$, with $a = 3$, $b = \sqrt{3}$:
$$(3 + \sqrt{3})(3 - \sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9 - 3 = 6$$

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(iii) $(\sqrt{5} + \sqrt{2})^2$

Using identity $(a + b)^2 = a^2 + 2ab + b^2$, with $a = \sqrt{5}$, $b = \sqrt{2}$:
$$(\sqrt{5} + \sqrt{2})^2 = (\sqrt{5})^2 + 2 \cdot \sqrt{5} \cdot \sqrt{2} + (\sqrt{2})^2$$
$$= 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}$$

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(iv) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$

Using identity $(a - b)(a + b) = a^2 - b^2$, with $a = \sqrt{5}$, $b = \sqrt{2}$:
$$(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) = (\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3$$

Understanding the apparent contradiction:

The definition $\pi = \frac{c}{d}$ (circumference divided by diameter) looks like a ratio of two quantities, which might suggest $\pi$ is rational.

Resolution:

For $\pi$ to be rational, both $c$ and $d$ must be integers (or at least rational numbers with $d \neq 0$). However, when we measure the circumference $c$ and diameter $d$ of any circle, at least one of them will be irrational.

For example:
- If $d = 1$ (rational), then $c = \pi \times 1 = \pi$ (irrational).
- If $d = 2$ (rational), then $c = 2\pi$ (irrational).

So $\pi = \frac{c}{d}$ is a ratio of two real numbers (not necessarily integers), and a ratio of two real numbers need not be rational.

Conclusion: There is no contradiction. The definition $\pi = \frac{c}{d}$ means $\pi$ is the ratio of two lengths (real numbers), not necessarily integers. Since $c$ and $d$ cannot both be rational simultaneously for any circle, $\pi$ remains irrational.

Steps to represent $\sqrt{9.3}$ on the number line:

Step 1: Draw a line segment $AB = 9.3$ units on the number line (from point $A$ at 0 to point $B$ at 9.3).

Step 2: Extend $AB$ by 1 unit to point $C$, so $BC = 1$ unit. Thus $AC = 10.3$ units.

Step 3: Find the midpoint $M$ of $AC$. So $AM = MC = \frac{10.3}{2} = 5.15$ units.

Step 4: With $M$ as centre and $MC$ (= 5.15 units) as radius, draw a semicircle above the line $AC$.

Step 5: At point $B$, draw a perpendicular to $AC$. Let it meet the semicircle at point $D$.

Step 6: By the geometric mean relation:
$$BD = \sqrt{AB \times BC} = \sqrt{9.3 \times 1} = \sqrt{9.3}$$

Step 7: With $B$ as centre and $BD$ as radius, draw an arc to cut the number line at point $P$.

Conclusion: The point $P$ on the number line represents $\sqrt{9.3}$.

(i) $\frac{1}{\sqrt{7}}$

Multiply numerator and denominator by $\sqrt{7}$:
$$\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}$$

$$\boxed{\frac{1}{\sqrt{7}} = \frac{\sqrt{7}}{7}}$$

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(ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$

Multiply numerator and denominator by the conjugate $(\sqrt{7} + \sqrt{6})$:
$$\frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \frac{\sqrt{7} + \sqrt{6}}{1}$$

$$\boxed{\frac{1}{\sqrt{7} - \sqrt{6}} = \sqrt{7} + \sqrt{6}}$$

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(iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$

Multiply numerator and denominator by the conjugate $(\sqrt{5} - \sqrt{2})$:
$$\frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{(\sqrt{5})^2 - (\sqrt{2})^2} = \frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3}$$

$$\boxed{\frac{1}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{3}}$$

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(iv) $\frac{1}{\sqrt{7} - 2}$

Multiply numerator and denominator by the conjugate $(\sqrt{7} + 2)$:
$$\frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{\sqrt{7} + 2}{(\sqrt{7})^2 - (2)^2} = \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}$$

$$\boxed{\frac{1}{\sqrt{7} - 2} = \frac{\sqrt{7} + 2}{3}}$$

(i) $64^{\frac{1}{2}}$

$$64^{\frac{1}{2}} = \sqrt{64} = \sqrt{8^2} = 8$$

$$\boxed{64^{\frac{1}{2}} = 8}$$

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(ii) $32^{\frac{1}{5}}$

$$32 = 2^5$$
$$32^{\frac{1}{5}} = (2^5)^{\frac{1}{5}} = 2^{5 \times \frac{1}{5}} = 2^1 = 2$$

$$\boxed{32^{\frac{1}{5}} = 2}$$

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(iii) $125^{\frac{1}{3}}$

$$125 = 5^3$$
$$125^{\frac{1}{3}} = (5^3)^{\frac{1}{3}} = 5^{3 \times \frac{1}{3}} = 5^1 = 5$$

$$\boxed{125^{\frac{1}{3}} = 5}$$

(i) $9^{\frac{3}{2}}$

$$9^{\frac{3}{2}} = (9^{\frac{1}{2}})^3 = (\sqrt{9})^3 = 3^3 = 27$$

$$\boxed{9^{\frac{3}{2}} = 27}$$

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(ii) $32^{\frac{2}{5}}$

$$32 = 2^5$$
$$32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^{5 \times \frac{2}{5}} = 2^2 = 4$$

$$\boxed{32^{\frac{2}{5}} = 4}$$

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(iii) $16^{\frac{3}{4}}$

$$16 = 2^4$$
$$16^{\frac{3}{4}} = (2^4)^{\frac{3}{4}} = 2^{4 \times \frac{3}{4}} = 2^3 = 8$$

$$\boxed{16^{\frac{3}{4}} = 8}$$

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(iv) $125^{-\frac{1}{3}}$

$$125 = 5^3$$
$$125^{-\frac{1}{3}} = (5^3)^{-\frac{1}{3}} = 5^{3 \times (-\frac{1}{3})} = 5^{-1} = \frac{1}{5}$$

$$\boxed{125^{-\frac{1}{3}} = \frac{1}{5}}$$

(i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$

Using law $a^m \cdot a^n = a^{m+n}$:
$$2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}} = 2^{\frac{2}{3} + \frac{1}{5}} = 2^{\frac{10}{15} + \frac{3}{15}} = 2^{\frac{13}{15}}$$

$$\boxed{2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}} = 2^{\frac{13}{15}}}$$

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(ii) $\left(\frac{1}{3^3}\right)^7$

$$\left(\frac{1}{3^3}\right)^7 = \left(3^{-3}\right)^7 = 3^{-3 \times 7} = 3^{-21} = \frac{1}{3^{21}}$$

$$\boxed{\left(\frac{1}{3^3}\right)^7 = \frac{1}{3^{21}}}$$

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(iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

Using law $\frac{a^m}{a^n} = a^{m-n}$:
$$\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} = 11^{\frac{1}{2} - \frac{1}{4}} = 11^{\frac{2}{4} - \frac{1}{4}} = 11^{\frac{1}{4}}$$

$$\boxed{\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} = 11^{\frac{1}{4}}}$$

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(iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$

Using law $a^m \cdot b^m = (ab)^m$:
$$7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} = (7 \times 8)^{\frac{1}{2}} = 56^{\frac{1}{2}} = \sqrt{56}$$

$$\boxed{7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} = \sqrt{56}}$$