Theorems on Area

Mathematics began with the fundamental human necessity of counting concrete objects—herds of cattle, measures of grain, and cycles of the moon. This practical intuition gave birth to the Natural Numbers ($\mathbb{N}$). As ancient civilizations developed trade, commerce, and accounting, the Indian mathematicians Brahmagupta (7th century CE) and Aryabhata formalized the concept of zero (śūnya) as both a place-holder and an independent numerical quantity, expanding natural numbers into the set of Whole Numbers ($\mathbb{W}$).

Commercial transactions involving debts and financial deficits, alongside the physical requirement of measuring opposite directions and temperatures below freezing, mandated negative quantities. The integration of negative integers with whole numbers produced the Integers ($\mathbb{Z}$, from the German Zahlen, meaning 'numbers'). However, dividing a continuous quantity (such as dividing a harvest among heirs or measuring fractional lengths) showed that integers were not closed under division. This led to the creation of Rational Numbers ($\mathbb{Q}$, from the Italian quoziente, meaning 'quotient', formalized by Giuseppe Peano).

The mathematical hierarchy of discrete and fractional numbers is formally defined as follows:

• Natural Numbers ($\mathbb{N}$): The set of positive counting numbers:
  $$\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}$$ The smallest natural number is $1$. There is no greatest natural number (the set is countably infinite). $\mathbb{N}$ is closed under addition ($a + b \in \mathbb{N}$) and multiplication ($a \cdot b \in \mathbb{N}$), but not closed under subtraction ($3 - 7 = -4 otin \mathbb{N}$) or division ($2 \div 5 otin \mathbb{N}$).
• Whole Numbers ($\mathbb{W}$): The union of zero with the set of natural numbers:
  $$\mathbb{W} = \{0\} \cup \mathbb{N} = \{0, 1, 2, 3, 4, \dots\}$$ The smallest whole number is $0$. The addition of $0$ provides the additive identity ($a + 0 = a$).
• Integers ($\mathbb{Z}$): The set encompassing all positive whole numbers, zero, and negative whole numbers:
  $$\mathbb{Z} = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$$ $\mathbb{Z}$ is closed under addition, subtraction, and multiplication. For every integer $a$, there exists an additive inverse $-a$ such that $a + (-a) = 0$. However, $\mathbb{Z}$ is not closed under division (e.g., $1 \div 2 = 0.5 otin \mathbb{Z}$).
• Rational Numbers ($\mathbb{Q}$): Any number that can be expressed as the ratio or quotient of two integers:
  $$\mathbb{Q} = \left\{rac{p}{q} : p, q \in \mathbb{Z}, \, q eq 0, \, \gcd(p, q) = 1 ight\}$$ Here, $p$ is the numerator and $q$ is the non-zero denominator. We stipulate $\gcd(p, q) = 1$ to ensure that the fraction is in its simplest (irreducible) form.

In the integer set $\mathbb{Z}$, each number has an immediate successor and predecessor: between $2$ and $3$, there is no other integer. The integers are discrete. In sharp contrast, the set of rational numbers $\mathbb{Q}$ is dense.

Density Property: Between any two distinct rational numbers $a$ and $b$ (with $a < b$), there exist infinitely many rational numbers. Consequently, there is no such concept as the 'next' rational number after any given rational number.

To find rational numbers between two distinct rationals $a$ and $b$ ($a < b$), we deploy two standard mathematical techniques:

1. The Arithmetic Mean Method:
   For any two distinct rational numbers $a$ and $b$ with $a < b$: $$q_1 = rac{a + b}{2}$$ Because $a < b$, adding $a$ to both sides gives $2a < a + b \implies a < rac{a+b}{2}$. Adding $b$ to both sides gives $a + b < 2b \implies rac{a+b}{2} < b$. Hence: $$a < q_1 < b$$ By iteratively taking the arithmetic mean of $a$ and $q_1$, and of $q_1$ and $b$, infinitely many distinct rational numbers can be generated.
2. The Equidistant Common Difference Step Formula:
   When we need to insert exactly $n$ equidistant rational numbers between $a$ and $b$ ($a < b$), we partition the interval $[a, b]$ into $(n + 1)$ equal sub-intervals. The common step length is:
   $$d = rac{b - a}{n + 1}$$
   The $n$ rational numbers inserted in strictly ascending order are: $$a + d, \quad a + 2d, \quad a + 3d, \quad \dots, \quad a + nd$$ Each of these numbers is strictly rational and satisfies $a < a+d < a+2d < \dots < a+nd < b$.

When a rational number $rac{p}{q}$ in irreducible form is converted to a decimal via long division, two mutually exclusive possibilities arise:

1. Terminating Decimal Expansions: The division terminates after a finite number of decimal places because the remainder becomes zero at some stage (e.g., $rac{1}{4} = 0.25$, $rac{3}{8} = 0.375$, $rac{7}{20} = 0.35$).
2. Non-Terminating Repeating (Recurring / Periodic) Decimal Expansions: The remainder never becomes zero. Instead, after a certain stage, the remainders begin repeating in a cyclic loop. The quotient exhibits a repeating sequence of digits indefinitely (e.g., $rac{1}{3} = 0.333\dots = 0.\overline{3}$, $rac{1}{7} = 0.142857142857\dots = 0.\overline{142857}$).

We do not need to execute long division to determine whether a given rational number terminates. The decimal structure is governed entirely by the prime factorization of its denominator.

Why only prime factors 2 and 5? The decimal number system is constructed on base $10$. Any terminating decimal with $k$ decimal digits can be written as an integer over $10^k$: $$0.d_1 d_2 \dots d_k = rac{d_1 d_2 \dots d_k}{10^k} = rac{N}{(2 \cdot 5)^k} = rac{N}{2^k \cdot 5^k}$$ The only prime divisors of the base $10$ are $2$ and $5$. Therefore, the reduced denominator $q$ cannot contain any prime factors other than $2$ or $5$. If $q$ contains any other prime factor (such as $3, 7, 11, 13, \dots$), no power of $10$ can clear the denominator, forcing the decimal expansion to be non-terminating and repeating.

Non-terminating repeating decimals are categorized based on where the repeating cycle begins:

• Pure Recurring Decimal: A recurring decimal in which the repeating cycle of digits begins immediately after the decimal point.
  Examples: $0.\overline{3} = 0.333\dots$, $0.\overline{27} = 0.272727\dots$, $3.\overline{142857} = 3.142857142857\dots$.
  Period: The repeating block of digits (e.g., in $0.\overline{27}$, the period is $27$).
  Periodicity: The number of digits in the repeating block (e.g., periodicity of $0.\overline{3}$ is $1$; periodicity of $0.\overline{27}$ is $2$; periodicity of $0.\overline{142857}$ is $6$).
• Mixed Recurring Decimal: A recurring decimal in which there is at least one non-repeating digit between the decimal point and the repeating cycle.
  Examples: $0.1\overline{6} = 0.1666\dots$ (non-repeating digit: $1$; repeating period: $6$), $0.2\overline{35} = 0.2353535\dots$ (non-repeating: $2$; repeating period: $35$).

Every recurring decimal represents a rational number and can be converted into an irreducible fraction $rac{p}{q}$ using algebraic multiplication and subtraction:

1. Let $x = 0.545454\dots$   ... (Equation 1)
2. Identify the periodicity $k = 2$. Multiply Equation 1 by $10^k = 10^2 = 100$:
   $$100x = 54.545454\dots$$   ... (Equation 2)
3. Subtract Equation 1 from Equation 2 to eliminate the infinite recurring tail:
   $$100x - x = (54.545454\dots) - (0.545454\dots)$$
   $$99x = 54 \implies x = rac{54}{99}$$
4. Reduce to simplest form by dividing numerator and denominator by $\gcd(54, 99) = 9$:
   $$x = rac{6}{11}$$

1. Let $x = 0.2353535\dots$   ... (Equation 1)
2. There is $m = 1$ non-repeating digit after the decimal. Multiply Equation 1 by $10^m = 10^1 = 10$ to shift the non-repeating part to the left of the decimal:
   $$10x = 2.353535\dots$$   ... (Equation 2)
3. The periodicity is $n = 2$. Multiply Equation 2 by $10^n = 10^2 = 100$:
   $$1000x = 235.353535\dots$$   ... (Equation 3)
4. Subtract Equation 2 from Equation 3:
   $$1000x - 10x = (235.353535\dots) - (2.353535\dots)$$
   $$990x = 235 - 2 = 233 \implies x = rac{233}{990}$$
5. Check if irreducible: $\gcd(233, 990) = 1$. The fraction in simplest form is $rac{233}{990}$.

In the 5th century BCE, the philosophical school of Pythagoras held as a core theological dogma that "Number rules the universe". By 'number', they meant whole numbers and their ratios (fractions). They believed that any two physical lengths in geometry were commensurable—meaning there existed a common fundamental unit of length such that both could be expressed as exact integer multiples of that unit.

This worldview collapsed when a Pythagorean mathematician, Hippasus of Metapontum, examined the diagonal of a unit square. By the Pythagorean theorem, a square with side length $1$ has a diagonal $d$ given by: $$d^2 = 1^2 + 1^2 = 2 \implies d = \sqrt{2}$$ Hippasus demonstrated that no two integers $p$ and $q$ could ever satisfy $rac{p^2}{q^2} = 2$. The side and diagonal of a square were fundamentally incommensurable. Greek geometry was thrown into crisis, revealing an entirely new realm of numbers: the Irrational Numbers ($\mathbb{Q}'$).

Definition: A real number is defined as an irrational number ($\mathbb{Q}'$) if it cannot be written in the form $rac{p}{q}$, where $p$ and $q$ are integers and $q eq 0$.

Decimal Characterization: A real number is irrational if and only if its decimal expansion is non-terminating and non-recurring.

• Unlike rational numbers, the digits of an irrational number continue infinitely without ever entering a repeating cycle.
• Examples of Irrational Numbers:
  • Square roots of non-perfect squares: $\sqrt{2} pprox 1.41421356\dots$, $\sqrt{3} pprox 1.7320508\dots$, $\sqrt{5} pprox 2.2360679\dots$, $\sqrt{10} pprox 3.1622776\dots$.
  • Cube roots of non-perfect cubes: $\sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{9}$.
  • Transcendental mathematical constants: $\pi pprox 3.14159265\dots$ (the ratio of circumference to diameter) and Euler's number $e pprox 2.7182818\dots$.
  • Synthetic patterned decimals: $0.101001000100001\dots$ (where the number of zeros between successive ones increases by one indefinitely). Because the pattern never repeats periodically, it is strictly irrational.

The classical proof utilizes Reductio ad Absurdum (proof by contradiction) alongside Euclid's Lemma on prime divisibility:

Euclid's Divisibility Lemma: If a prime number $p$ divides $a^2$ (where $a \in \mathbb{Z}$), then $p$ must divide $a$.

To prove that an algebraic combination of a rational and an irrational is irrational, we use proof by contradiction:

1. Assume $x = 3 + 2\sqrt{5}$ is rational. Then $x = rac{p}{q}$ for integers $p, q$ ($q eq 0$).
2. Rearrange to isolate the surd $\sqrt{5}$: $$2\sqrt{5} = rac{p}{q} - 3 = rac{p - 3q}{q} \implies \sqrt{5} = rac{p - 3q}{2q}$$
3. Since $p, q \in \mathbb{Z}$, the numerator $(p - 3q)$ and denominator $2q$ are both integers with $2q eq 0$. Therefore, the right-hand side is a rational number.
4. This implies that $\sqrt{5}$ is rational, which contradicts the known truth that $\sqrt{5}$ is irrational.
5. Hence, our assumption is false, and $3 + 2\sqrt{5}$ is strictly irrational.

To find irrational numbers between two distinct rational numbers $a$ and $b$ (e.g., between $rac{1}{7}$ and $rac{2}{7}$):

• Compute the decimal expansions of both numbers: $$rac{1}{7} = 0.142857\dots \quad ext{and} \quad rac{2}{7} = 0.285714\dots$$
• Construct non-terminating, non-repeating decimals strictly lying between $0.142857\dots$ and $0.285714\dots$:
  $$I_1 = 0.1501001000100001\dots$$
  $$I_2 = 0.201001000100001\dots$$
  $$I_3 = 0.2502002000200002\dots$$
• Alternatively, find a non-perfect square between their squares: between $2$ and $3$, $\sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}$ are all irrational numbers lying strictly between $2 = \sqrt{4}$ and $3 = \sqrt{9}$.

Can irrational numbers be pinpointed on the geometric number line? In the late 19th century, German mathematicians Richard Dedekind and Georg Cantor established the foundational postulate connecting geometry and arithmetic:

By applying the Pythagorean Theorem ($c = \sqrt{a^2 + b^2}$), we construct line segments whose lengths are irrational roots, and then transfer these lengths to the number line using a compass:

1. Geometric Construction of $\sqrt{2}$:
   • Let the origin $O$ represent $0$ on the number line. Mark point $A$ representing $1$ unit, so $OA = 1$.
   • At point $A$, erect a perpendicular line segment $AB$ of length $1$ unit ($AB \perp OA$).
   • Join $OB$. In right-angled triangle $ riangle OAB$, by Pythagoras' theorem: $$OB = \sqrt{OA^2 + AB^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} ext{ units}$$
   • With center $O$ and radius $OB = \sqrt{2}$, draw an arc cutting the positive number line at point $P$.
   • Point $P$ represents the irrational number $\sqrt{2} pprox 1.414$ on the number line.
2. Geometric Construction of $\sqrt{3}$:
   • With $OB$ (length $\sqrt{2}$) as base, erect a perpendicular line segment $BC$ of length $1$ unit ($BC \perp OB$).
   • Join $OC$. In right-angled triangle $ riangle OBC$: $$OC = \sqrt{OB^2 + BC^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3} ext{ units}$$
   • With center $O$ and radius $OC = \sqrt{3}$, draw an arc cutting the positive number line at point $Q$.
   • Point $Q$ represents $\sqrt{3} pprox 1.732$ on the number line.
3. Geometric Construction of $\sqrt{5}$:
   • Mark point $A$ at $2$ units from $O$ ($OA = 2$). Erect perpendicular $AB = 1$ unit.
   • In right-angled triangle $ riangle OAB$: $$OB = \sqrt{OA^2 + AB^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} ext{ units}$$
   • Swing an arc with radius $OB$ to intersect the number line at $R$. Point $R$ represents $\sqrt{5} pprox 2.236$.

This sequential Pythagorean construction can be continued indefinitely, forming the famous Spiral of Theodorus (discovered by Theodorus of Cyrene in the 5th century BCE): $$\sqrt{1} o \sqrt{2} o \sqrt{3} o \sqrt{4} o \sqrt{5} o \dots o \sqrt{17}$$ Each triangle has an outer leg of $1$ unit perpendicular to the hypotenuse of the preceding triangle, yielding the square root of the next positive integer as its new hypotenuse.

While the Pythagorean method is ideal for square roots of whole numbers, representing decimal square roots such as $\sqrt{9.3}$ or $\sqrt{4.5}$ requires the classical semicircle altitude construction:

We prove rigorously that $BD = \sqrt{x}$ using Pythagoras' Theorem on right-angled triangle $ riangle MBD$:

1. The diameter of the semicircle is $AC = AB + BC = x + 1$.
2. Therefore, the radius of the semicircle is: $$MD = MA = MC = rac{x + 1}{2}$$
3. Now determine the distance $MB$: $$MB = MC - BC = rac{x + 1}{2} - 1 = rac{x + 1 - 2}{2} = rac{x - 1}{2}$$
4. Since $BD \perp AC$, triangle $ riangle MBD$ is a right-angled triangle at $B$. By Pythagoras' Theorem: $$MD^2 = MB^2 + BD^2 \implies BD^2 = MD^2 - MB^2$$
5. Substitute the expressions for $MD$ and $MB$: $$BD^2 = \left(rac{x + 1}{2} ight)^2 - \left(rac{x - 1}{2} ight)^2$$ $$BD^2 = rac{(x^2 + 2x + 1) - (x^2 - 2x + 1)}{4} = rac{x^2 + 2x + 1 - x^2 + 2x - 1}{4} = rac{4x}{4} = x$$
6. Taking the positive square root on both sides: $$BD = \sqrt{x}$$ This completes the formal proof. $\quad lacksquare$

The Real Number System ($\mathbb{R}$) is defined as the union of the set of all rational numbers ($\mathbb{Q}$) and the set of all irrational numbers ($\mathbb{Q}'$):

Rational and irrational numbers are strictly disjoint: no number can be both rational and irrational. Every real number is either rational or irrational.

The set of real numbers equipped with addition and multiplication, denoted $(\mathbb{R}, +, \cdot)$, satisfies the 11 algebraic axioms of a mathematical Field:

• Closure Properties: For all $a, b \in \mathbb{R}$:
  $$a + b \in \mathbb{R} \quad ext{and} \quad a \cdot b \in \mathbb{R}$$
• Commutative Properties:
  $$a + b = b + a \quad ext{and} \quad a \cdot b = b \cdot a$$
• Associative Properties:
  $$(a + b) + c = a + (b + c) \quad ext{and} \quad (a \cdot b) \cdot c = a \cdot (b \cdot c)$$
• Distributive Property of Multiplication over Addition:
  $$a \cdot (b + c) = a \cdot b + a \cdot c$$
• Identity Elements:
  Additive Identity: There exists $0 \in \mathbb{R}$ such that $a + 0 = a$.
  Multiplicative Identity: There exists $1 \in \mathbb{R}$ ($1 eq 0$) such that $a \cdot 1 = a$.
• Inverse Elements:
  Additive Inverse: For every $a \in \mathbb{R}$, there exists $-a \in \mathbb{R}$ such that $a + (-a) = 0$.
  Multiplicative Inverse: For every non-zero $a \in \mathbb{R}$ ($a eq 0$), there exists $a^{-1} = rac{1}{a} \in \mathbb{R}$ such that $a \cdot rac{1}{a} = 1$.

When arithmetic operations combine a rational number with an irrational number, the outcome is governed by precise mathematical theorems:

Crucial Concept: The set of irrational numbers $\mathbb{Q}'$ is NOT closed under addition, subtraction, multiplication, or division! The sum, difference, product, or quotient of two irrational numbers can be either rational or irrational:

• Sum of Two Irrationals:
  • Can be rational: $\sqrt{3} + (-\sqrt{3}) = 0 \in \mathbb{Q}$, or $(2 + \sqrt{5}) + (2 - \sqrt{5}) = 4 \in \mathbb{Q}$.
  • Can be irrational: $\sqrt{2} + \sqrt{3} \in \mathbb{Q}'$.
• Difference of Two Irrationals:
  • Can be rational: $(3 + \sqrt{7}) - \sqrt{7} = 3 \in \mathbb{Q}$, or $\sqrt{2} - \sqrt{2} = 0 \in \mathbb{Q}$.
  • Can be irrational: $\sqrt{5} - \sqrt{2} \in \mathbb{Q}'$.
• Product of Two Irrationals:
  • Can be rational: $\sqrt{2} \cdot \sqrt{2} = 2 \in \mathbb{Q}$, $\sqrt{8} \cdot \sqrt{2} = \sqrt{16} = 4 \in \mathbb{Q}$, or $(3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7 \in \mathbb{Q}$.
  • Can be irrational: $\sqrt{2} \cdot \sqrt{3} = \sqrt{6} \in \mathbb{Q}'$.
• Quotient of Two Irrationals:
  • Can be rational: $rac{\sqrt{18}}{\sqrt{2}} = \sqrt{9} = 3 \in \mathbb{Q}$, or $rac{\pi}{\pi} = 1 \in \mathbb{Q}$.
  • Can be irrational: $rac{\sqrt{10}}{\sqrt{2}} = \sqrt{5} \in \mathbb{Q}'$.

Definition: An irrational root of a positive rational number is called a surd (or radical). More formally, if $a$ is a positive rational number and $n$ is a positive integer ($n \ge 2$) such that $a$ is not the $n$-th power of any rational number, then the real root: $$\sqrt[n]{a} \quad ext{or} \quad a^{1/n}$$ is called a surd of order $n$. Here, the symbol $\sqrt{}$ is the radical sign, $n$ is the order (or index) of the surd, and $a$ is the radicand.

• Quadratic Surd: A surd of order $2$ ($n = 2$), written simply as $\sqrt{a}$ (e.g., $\sqrt{2}, \sqrt{7}, \sqrt{15}$).
• Cubic Surd: A surd of order $3$ ($n = 3$), written as $\sqrt[3]{a}$ (e.g., $\sqrt[3]{2}, \sqrt[3]{5}$).
• Surd vs. Irrational Distinction: All surds are irrational numbers, but not all irrational numbers are surds! For example, $\pi$ and $e$ are transcendental irrational numbers; they cannot be expressed as roots of any rational numbers, so they are not surds. Furthermore, $\sqrt{9} = 3$ and $\sqrt[3]{27} = 3$ are not surds because $9$ and $27$ are perfect powers, making their roots rational.

1. Pure Surd: A surd whose only rational factor is $1$ (e.g., $\sqrt{5}, \sqrt{18}, \sqrt[3]{7}$).
2. Mixed Surd: A surd that has a rational coefficient other than $\pm 1$ multiplying a pure surd (e.g., $3\sqrt{2}, 5\sqrt{7}$).
   Conversion: Every pure surd containing a square factor in its radicand can be converted to a mixed surd: $$\sqrt{72} = \sqrt{36 imes 2} = \sqrt{6^2 imes 2} = 6\sqrt{2}$$
3. Similar (Like) Surds: Two or more surds that, when reduced to their simplest forms, have identical surd parts.
   Example: $\sqrt{12} = 2\sqrt{3}$, $\sqrt{27} = 3\sqrt{3}$, and $\sqrt{75} = 5\sqrt{3}$ are similar surds because each has $\sqrt{3}$ as its radical factor.
   Addition/Subtraction: Only similar surds can be directly added or subtracted: $$2\sqrt{3} + 5\sqrt{3} - 3\sqrt{3} = (2 + 5 - 3)\sqrt{3} = 4\sqrt{3}$$
4. Dissimilar (Unlike) Surds: Surds that have different surd parts in their simplest form.
   Example: $\sqrt{8} = 2\sqrt{2}$ and $\sqrt{12} = 2\sqrt{3}$ are dissimilar surds. They cannot be combined into a single term ($2\sqrt{2} + 2\sqrt{3}$ cannot be simplified further).
5. Compound and Binomial Quadratic Surds: The algebraic sum or difference of two or more surds, or of a rational number and a surd (e.g., $a + \sqrt{b}$, $\sqrt{a} + \sqrt{b}$).

Rationalizing Factor (RF): When two surds are multiplied together and their product is a rational number, each surd is called a rationalizing factor of the other.

Conjugate Surds: Two binomial quadratic surds are called conjugate surds (or complementary surds) if both their sum AND product are rational numbers.

When an algebraic fraction contains a surd in its denominator, dividing or computing numerical values is difficult. Rationalization is the algebraic technique of multiplying both the numerator and denominator by a suitable rationalizing factor (the conjugate of the denominator) to transform the denominator into a rational number:

Standard Identity: $$rac{1}{a + \sqrt{b}} = rac{1 \cdot (a - \sqrt{b})}{(a + \sqrt{b})(a - \sqrt{b})} = rac{a - \sqrt{b}}{a^2 - b}$$ $$rac{1}{\sqrt{a} + \sqrt{b}} = rac{\sqrt{a} - \sqrt{b}}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})} = rac{\sqrt{a} - \sqrt{b}}{a - b}$$

To compare two or more surds of different orders (such as $\sqrt[3]{4}$ and $\sqrt[4]{5}$):

1. Identify the orders of the surds: here $m = 3$ and $n = 4$.
2. Find the Lowest Common Multiple (LCM) of the orders: $ ext{LCM}(3, 4) = 12$.
3. Express each surd as an equivalent surd of order $12$: $$\sqrt[3]{4} = 4^{1/3} = 4^{4/12} = \sqrt[12]{4^4} = \sqrt[12]{256}$$ $$\sqrt[4]{5} = 5^{1/4} = 5^{3/12} = \sqrt[12]{5^3} = \sqrt[12]{125}$$
4. Since the orders are now identical ($12$), compare the radicands: $$256 > 125 \implies \sqrt[12]{256} > \sqrt[12]{125} \implies \sqrt[3]{4} > \sqrt[4]{5}$$