Step 1: Initial Setup & Variable Assignment

Let the given repeating decimal be assigned to a variable $x$. The vinculum (the bar over the digits) indicates the specific block of digits that repeats infinitely.

$x = 0.\overline{001}$

Expanding the repeating block, we can write this as an infinite decimal:

$x = 0.001001001... \quad \text{--- (Equation 1)}$

Step 2: Determining the Multiplier [Base-10 Shift]

To convert a recurring decimal into a rational fraction of the form $\frac{p}{q}$, we must eliminate the infinitely repeating fractional part. [Per the properties of repeating decimals], we achieve this by shifting the decimal point to the right by exactly the number of digits in the repeating block.

The repeating block is "001", which consists of exactly $3$ digits. Therefore, we multiply both sides of Equation 1 by $10^3$ (which is $1000$).

$1000 \cdot x = 1000 \cdot (0.001001001...)$

$1000x = 1.001001001... \quad \text{--- (Equation 2)}$

Step 3: Algebraic Subtraction & Elimination of the Infinite Series

We now have two equations where the fractional parts (the digits after the decimal point) are identical. By subtracting Equation 1 from Equation 2, the infinite repeating sequence will cancel out completely [By the Subtraction Property of Equality].

Mathematically, this is executed as:

$1000x - x = (1.001001001...) - (0.001001001...)$

$999x = 1$

Step 4: Isolation of the Variable

Divide both sides of the equation by $999$ to isolate $x$ [By the Division Property of Equality]:

$x = \frac{1}{999}$

Step 5: Verification of Rational Conditions

The problem requires the final answer to be in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

• $p = 1$, which is an integer ($1 \in \mathbb{Z}$).
• $q = 999$, which is an integer ($999 \in \mathbb{Z}$).
• The denominator $q = 999 \neq 0$.

All conditions for a rational number are strictly satisfied. Furthermore, the greatest common divisor $\text{GCD}(1, 999) = 1$, meaning the fraction is in its simplest form.

Final Solution: The repeating decimal $0.\overline{001}$ expressed in the form $\frac{p}{q}$ is $\frac{1}{999}$.

Initial Setup & Given Variables

We are tasked with evaluating the rational number provided in fractional form:

$x = \frac{36}{100}$

The objective is to convert this fraction into its equivalent decimal representation and mathematically classify the nature of its decimal expansion.

Step 1: Conversion to Decimal Form

To convert a fraction to a decimal, we perform the division of the numerator by the denominator. When the denominator is a perfect power of $10$ (i.e., $10^k$), the division can be executed by shifting the decimal point of the numerator $k$ places to the left.

• Numerator: $36$ (which can be written as $36.0$)
• Denominator: $100 = 10^2$

Since the exponent $k = 2$, we shift the decimal point two places to the left:

$x = \frac{36}{100} = 0.36$

Step 2: Analyzing the Nature of the Decimal Expansion

A decimal expansion is classified based on its termination behavior:

• Terminating Decimal: The division process yields a remainder of zero after a finite number of steps, resulting in a finite number of digits after the decimal point.
• Non-Terminating Repeating Decimal: The division process never yields a zero remainder, but a digit or a block of digits repeats infinitely.

Because the value $0.36$ ends exactly after the hundredths place (two decimal digits), it is classified as a terminating decimal expansion.

Step 3: Theoretical Justification via Prime Factorization

[Per the Fundamental Theorem of Arithmetic and the properties of rational numbers], a rational number expressed in its simplest form $\frac{p}{q}$ will have a terminating decimal expansion if and only if the prime factorization of its denominator $q$ is strictly of the form:

$q = 2^n \times 5^m$

where $n$ and $m$ are non-negative integers.

Let us analyze the denominator $q = 100$:

$100 = 10 \times 10 = (2 \times 5) \times (2 \times 5) = 2^2 \times 5^2$

Here, $n = 2$ and $m = 2$. Because the prime factors of the denominator consist exclusively of $2$ and $5$, the mathematical theorem guarantees that the fraction $\frac{36}{100}$ must resolve into a terminating decimal.

Step 4: Visual Representation of the Decimal

To conceptualize $0.36$ geometrically, we can map the fraction onto a $10 \times 10$ Cartesian grid representing $100$ equal units. Shading $36$ of these units provides a precise visual truth of the decimal's magnitude.

Final Solution: The decimal form of $\frac{36}{100}$ is $0.36$, and it has a terminating decimal expansion.

Initial Setup & Theoretical Foundation

We are tasked with classifying the following real number:

$x = 1.101001000100001...$

To classify a real number as either rational ($\mathbb{Q}$) or irrational ($\mathbb{I}$) based on its decimal representation, we rely on the fundamental properties of real numbers:

• Rational Numbers ($\mathbb{Q}$): Possess decimal expansions that are either terminating (e.g., $0.25$) or non-terminating but repeating/recurring (e.g., $0.333...$). They can be expressed in the form $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \neq 0$.
• Irrational Numbers ($\mathbb{I}$): Possess decimal expansions that are strictly non-terminating and non-repeating. They cannot be expressed as a ratio of two integers.

Step 1: Analyzing the Termination of the Decimal

Observe the given number $x = 1.101001000100001...$. The presence of the ellipsis ($...$) at the end of the number explicitly indicates that the decimal digits continue infinitely. [Per the standard mathematical notation for infinite sequences].

Therefore, the decimal expansion is definitively non-terminating.

Step 2: Analyzing the Recurrence (Repeating Pattern)

Next, we examine the sequence of digits after the decimal point to determine if there is a fixed, finite block of digits that repeats infinitely.

The sequence of fractional digits is: $101001000100001...$

Let us break down the structural pattern of zeros and ones:

• First segment: $10$ (contains exactly one zero)
• Second segment: $100$ (contains exactly two zeros)
• Third segment: $1000$ (contains exactly three zeros)
• Fourth segment: $10000$ (contains exactly four zeros)

Because the number of zeros between consecutive ones increases by exactly one each time, the sequence constantly changes its structure. [By the mathematical definition of periodicity, a sequence is periodic if $a_{n+p} = a_n$ for some constant period $p$. The strictly increasing run-length of zeros mathematically prevents any constant period $p$ from existing].

Therefore, the decimal expansion is non-repeating (non-recurring).

Step 3: Visualizing the Classification Framework

The following flowchart illustrates the logical path used to classify the given number based on its decimal expansion.

Step 4: Final Classification

Since the decimal expansion of $1.101001000100001...$ is proven to be both non-terminating and non-repeating, it cannot be expressed as a simple fraction $\frac{p}{q}$.

[Per the fundamental theorem of rational numbers, only numbers with terminating or repeating decimals belong to the set of rational numbers $\mathbb{Q}$. All other real numbers belong to the set of irrational numbers $\mathbb{I}$].

Final Solution: The number $1.101001000100001...$ is an irrational number.

Theoretical Foundation: Decimal Expansions of Real Numbers

In the real number system, numbers are broadly classified into two categories based on their decimal expansions: rational numbers and irrational numbers. [Per the fundamental classification of Real Numbers], a rational number's decimal expansion either terminates (ends) or is non-terminating but recurring (repeats a specific block of digits infinitely). Conversely, an irrational number cannot be expressed as a simple fraction $\frac{p}{q}$ (where $p$ and $q$ are integers and $q \neq 0$). Consequently, the defining characteristic of an irrational number is that its decimal expansion is strictly non-terminating and non-recurring.

To satisfy the prompt, we must provide three distinct examples of irrational numbers. We will derive these using three different mathematical approaches: pattern construction, algebraic roots, and transcendental constants.

Step 1: Constructing a Pattern-Based Irrational Number

The most direct way to generate a non-terminating, non-recurring decimal is to construct a sequence of digits that follows a predictable, yet non-repeating pattern. We can achieve this by systematically increasing the number of zeros between a repeating non-zero digit.

• Let the first decimal place be $0$.
• Follow it with a $1$, then one $0$ ($0.01\dots$).
• Follow it with a $1$, then two $0$s ($0.01001\dots$).
• Follow it with a $1$, then three $0$s ($0.010010001\dots$).

Number 1: $0.01001000100001\dots$

[Justification: Because the block of zeros expands by one with each iteration, there is no fixed finite block of digits that will ever repeat infinitely. Thus, it is non-recurring. The ellipsis ($\dots$) indicates it is non-terminating.]

Step 2: Utilizing Square Roots of Non-Perfect Squares

In algebra, the square root of any positive integer that is not a perfect square (such as $2, 3, 5, 6, 7$) is inherently an irrational number. [Per the theorem of irrationality of roots: If $p$ is a prime number, then $\sqrt{p}$ is irrational].

Let us take the square root of the first prime number, $2$.

Number 2: $\sqrt{2} = 1.41421356237309504\dots$

[Justification: The decimal expansion of $\sqrt{2}$ has been mathematically proven to extend infinitely without any repeating sequence of digits.]

Step 3: Referencing Transcendental Mathematical Constants

Certain fundamental constants in mathematics cannot be expressed as the root of any non-zero polynomial equation with rational coefficients. These are known as transcendental numbers, which are a subset of irrational numbers. The most famous example is Pi ($\pi$), which represents the ratio of a circle's circumference to its diameter.

Number 3: $\pi = 3.141592653589793238\dots$

[Justification: Johann Lambert proved in 1761 that $\pi$ is irrational. Therefore, its decimal representation goes on forever without settling into a permanently repeating pattern.]

Final Solution: Three distinct numbers whose decimal expansions are non-terminating and non-recurring are:
1) $0.01001000100001\dots$ (A constructed non-repeating pattern)
2) $\sqrt{2} = 1.41421356\dots$ (The square root of a non-perfect square)
3) $\pi = 3.14159265\dots$ (A transcendental mathematical constant)

Initial Setup & Given Number

We are tasked with classifying the given real number as either rational or irrational based on its decimal expansion.

Let the given number be $x$:

$x = 7.478478...$

Step 1: Analyzing the Decimal Expansion

By observing the sequence of digits after the decimal point, we can identify a distinct repeating pattern. The block of digits $478$ repeats infinitely. Therefore, the number can be expressed using bar notation over the repeating block:

$x = 7.\overline{478}$

[Per the fundamental theorem of real number decimal expansions, any number that exhibits a non-terminating but repeating (recurring) decimal expansion is, by definition, a rational number. Conversely, non-terminating and non-repeating decimals are irrational.]

Step 2: Algebraic Proof of Rationality

To rigorously prove that $x$ is rational, we must demonstrate that it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers ($\in \mathbb{Z}$) and $q \neq 0$.

Let our initial equation be:

$x = 7.478478... \quad \text{--- (Equation 1)}$

Since the periodicity (the number of digits in the repeating block) is $3$, we multiply both sides of Equation 1 by $10^3$ (which is $1000$) to shift the decimal point past the first repeating block:

$1000x = 7478.478478... \quad \text{--- (Equation 2)}$

Next, we subtract Equation 1 from Equation 2 to eliminate the infinite repeating decimal part:

$1000x - x = 7478.478478... - 7.478478...$
$999x = 7471.000000...$

Solving for $x$, we isolate the variable:

$x = \frac{7471}{999}$

Step 3: Evaluating the Result

We have successfully expressed $x$ as a fraction $\frac{7471}{999}$.

• $p = 7471$, which is an integer ($p \in \mathbb{Z}$).
• $q = 999$, which is an integer ($q \in \mathbb{Z}$).
• $q \neq 0$ ($999 \neq 0$).

[By the formal definition of rational numbers ($\mathbb{Q}$), any number that satisfies these conditions is strictly rational.]

Visual Classification Flowchart

Final Solution: The number $7.478478...$ is a rational number.

Initial Setup & Mathematical Objective

We are given the rational number $\frac{2}{11}$. Our objective is to convert this fraction into its decimal form using the long division algorithm and to classify the nature of its decimal expansion [Per the fundamental properties of rational numbers].

Step 1: Executing the Long Division Algorithm

To find the decimal expansion, we divide the numerator ($2$) by the denominator ($11$). Since $2 < 11$, the quotient begins with $0$. We place a decimal point in the quotient and append zeros to the dividend to continue the division process.

The step-by-step division proceeds as follows:

• Iteration 1: Bring down a $0$ to make $20$. Divide $20$ by $11$. The quotient is $1$. $11 \times 1 = 11$. Subtracting $11$ from $20$ yields a remainder of $9$.
• Iteration 2: Bring down a $0$ to make $90$. Divide $90$ by $11$. The quotient is $8$. $11 \times 8 = 88$. Subtracting $88$ from $90$ yields a remainder of $2$.
• Iteration 3: Bring down a $0$ to make $20$. Divide $20$ by $11$. The quotient is $1$. $11 \times 1 = 11$. Subtracting $11$ from $20$ yields a remainder of $9$.
• Iteration 4: Bring down a $0$ to make $90$. Divide $90$ by $11$. The quotient is $8$. $11 \times 8 = 88$. Subtracting $88$ from $90$ yields a remainder of $2$.

Step 2: Analyzing the Sequence of Remainders

Observing the division process, the sequence of remainders is $9, 2, 9, 2, \dots$. Because the remainder $2$ (which was our original dividend) reappears, the sequence of calculations will loop infinitely. Consequently, the block of digits in the quotient, $18$, will repeat indefinitely.

Mathematically, this is expressed as:

$\frac{2}{11} = 0.181818\dots$

Using standard mathematical bar notation to denote the repeating block of digits, we write this as $0.\overline{18}$.

Step 3: Theoretical Classification of the Decimal Expansion

According to the fundamental theorem of arithmetic applied to rational numbers, if the prime factorization of the denominator of a fraction (in simplest form) contains prime factors other than $2$ or $5$, the fraction will result in a non-terminating repeating decimal. Here, the denominator is $11$, which is a prime number distinct from $2$ and $5$.

Since the division never yields a remainder of $0$ and a specific block of digits ($18$) repeats infinitely, the decimal expansion is formally classified as non-terminating repeating (or non-terminating recurring).

Final Solution: The decimal form of $\frac{2}{11}$ is $0.\overline{18}$, and its decimal expansion is non-terminating repeating.

Step 1: Decimal Expansion of the Given Rational Numbers

To find irrational numbers between two rational numbers, we must first determine the decimal expansions of the given boundaries. This allows us to establish a clear numerical range.

• First Rational Number ($\frac{5}{7}$):
  Performing long division of $5$ by $7$:
  $5 \div 7 = 0.714285714285\dots$
  Since the block of digits "$714285$" repeats infinitely, we express this as $0.\overline{714285}$.


• Second Rational Number ($\frac{9}{11}$):
  Performing long division of $9$ by $11$:
  $9 \div 11 = 0.818181\dots$
  Since the block of digits "$81$" repeats infinitely, we express this as $0.\overline{81}$.

Step 2: Establishing the Range and Criteria

[Per the fundamental properties of the Real Number System], an irrational number is defined as a number whose decimal expansion is strictly non-terminating and non-repeating. Therefore, our objective is to construct three distinct decimal numbers, say $x$, $y$, and $z$, that satisfy two conditions:

1. Range Condition: $0.\overline{714285} < x, y, z < 0.\overline{81}$
2. Irrationality Condition: The decimals must not terminate and must not settle into a repeating periodic block.

Step 3: Construction of the Irrational Numbers

We can systematically construct non-terminating, non-repeating decimals by choosing starting digits that fall strictly between $0.71$ and $0.81$, and then appending a pattern of digits separated by an increasing number of zeros. This guarantees the sequence never repeats a fixed block.

• First Irrational Number ($x$):
  Choose a prefix greater than $0.714\dots$, such as $0.72$.
  Construct the non-repeating pattern: $x = 0.720720072000720000\dots$


• Second Irrational Number ($y$):
  Choose a prefix greater than $0.72$ but less than $0.81$, such as $0.75$.
  Construct the non-repeating pattern: $y = 0.750750075000750000\dots$


• Third Irrational Number ($z$):
  Choose a prefix greater than $0.75$ but less than $0.818\dots$, such as $0.80$.
  Construct the non-repeating pattern: $z = 0.808008000800008\dots$

Step 4: Visualizing the Real Number Line

The following high-precision number line maps the exact spatial relationship between the rational boundaries and the newly constructed irrational numbers.

Note: There are infinitely many irrational numbers between any two rational numbers. The three numbers constructed above are just one valid set of examples.

Final Solution: Three different irrational numbers between $\frac{5}{7}$ and $\frac{9}{11}$ are $0.720720072000\dots$, $0.750750075000\dots$, and $0.80800800080000\dots$.

Initial Setup & Theoretical Foundation

To classify a given real number as rational or irrational, we must analyze its decimal expansion or its ability to be expressed as a fraction. The classification is governed by the following fundamental definitions in real analysis:

• Rational Number ($\mathbb{Q}$): A number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers ($\mathbb{Z}$) and $q \neq 0$. In decimal form, rational numbers have expansions that are either terminating or non-terminating but repeating.
• Irrational Number ($\mathbb{I}$): A number that cannot be expressed as a simple fraction. Its decimal expansion is strictly non-terminating and non-repeating.

Step 1: Analyzing the Decimal Expansion

The given number is $0.3796$.

By observing the digits after the decimal point, we can see that the expansion ends exactly after the fourth decimal place (the ten-thousandths place). There are no trailing ellipses ($\dots$) or bar notations ($\overline{3796}$) indicating infinite continuation.

[Per the properties of real numbers, any decimal number that comes to an end after a finite number of digits is classified as a terminating decimal].

Step 2: Conversion to Fractional Form ($\frac{p}{q}$)

To rigorously prove that a terminating decimal is a rational number, we must demonstrate that it can be written in the standard fractional form $\frac{p}{q}$.

Since there are four digits after the decimal point, we multiply and divide the number by $10^4$ ($10000$):

$0.3796 = \frac{0.3796 \times 10000}{10000} = \frac{3796}{10000}$

Here, $p = 3796$ and $q = 10000$. Both $3796$ and $10000$ are integers, and the denominator $10000 \neq 0$. This satisfies the strict definition of a rational number.

Step 3: Simplification and Prime Factorization Analysis (Verification)

While $\frac{3796}{10000}$ is sufficient to prove rationality, reducing the fraction to its lowest terms provides complete mathematical rigor. We find the Greatest Common Divisor (GCD) of $3796$ and $10000$.

• Both numbers are divisible by $4$.
• Numerator: $3796 \div 4 = 949$
• Denominator: $10000 \div 4 = 2500$

Thus, the simplest fractional form is $\frac{949}{2500}$.

[By the Rational Number Theorem, a fraction $\frac{p}{q}$ in its simplest form yields a terminating decimal if and only if the prime factorization of $q$ is of the form $2^n \times 5^m$ for non-negative integers $n, m$].

Checking the denominator: $2500 = 25 \times 100 = 5^2 \times (2^2 \times 5^2) = 2^2 \times 5^4$. Because the prime factors of the denominator consist exclusively of $2$ and $5$, the mathematical theorem perfectly corroborates that $0.3796$ is a terminating, rational number.

Final Solution: The number $0.3796$ has a terminating decimal expansion and can be expressed in the form $\frac{p}{q}$ as $\frac{3796}{10000}$ (or $\frac{949}{2500}$). Therefore, it is a Rational Number.

Given Expression & Theoretical Foundation

We are tasked with classifying the number $\sqrt{23}$ as either a rational or an irrational number. To do this rigorously, we must first establish the mathematical definitions of these two sets of numbers:

• Rational Number ($\mathbb{Q}$): Any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers ($p, q \in \mathbb{Z}$) and $q \neq 0$. The decimal expansion of a rational number is always either terminating or non-terminating but repeating.
• Irrational Number ($\mathbb{I}$): Any real number that cannot be expressed as a simple fraction $\frac{p}{q}$. The decimal expansion of an irrational number is strictly non-terminating and non-repeating.

Step 1: Analyzing the Radicand

The given expression is $\sqrt{23}$. The number inside the square root, $23$, is called the radicand. We must first determine if the radicand is a perfect square.

Let us evaluate the squares of integers in the vicinity of $23$:

• $4^2 = 16$
• $5^2 = 25$

Since $16 < 23 < 25$, it follows that $\sqrt{16} < \sqrt{23} < \sqrt{25}$, which simplifies to $4 < \sqrt{23} < 5$. Because $23$ lies strictly between two consecutive perfect squares, it is not a perfect square. Its square root will not be an integer.

Step 2: Prime Factorization & Theorem Application

Next, we analyze the factors of the radicand $23$. The only positive divisors of $23$ are $1$ and $23$ itself. Therefore, $23$ is a prime number.

[Per the properties of real numbers], there is a fundamental theorem regarding the square roots of prime numbers:

Theorem: If $p$ is a prime number, then $\sqrt{p}$ is an irrational number.

Since $23$ is a prime number, applying this theorem directly classifies $\sqrt{23}$ as an irrational number.

Step 3: Rigorous Proof by Contradiction (Analytical Depth)

To provide a masterclass-level justification, we will prove that $\sqrt{23}$ is irrational using a proof by contradiction.

Assumption: Assume the contrary, that $\sqrt{23}$ is a rational number. Therefore, it can be written in the simplest fractional form:

$\sqrt{23} = \frac{a}{b}$

where $a$ and $b$ are coprime integers (meaning their greatest common divisor is $1$, $\gcd(a,b) = 1$), and $b \neq 0$.

Algebraic Manipulation:

1. Square both sides of the equation:
   $23 = \frac{a^2}{b^2}$
2. Multiply both sides by $b^2$:
   $a^2 = 23b^2$     (Equation 1)

Logical Deduction:

Equation 1 states that $a^2$ is a multiple of $23$. [Per Euclid's Lemma and the Fundamental Theorem of Arithmetic], if a prime number divides the square of an integer, it must also divide the integer itself. Therefore, $23$ divides $a$.

Let $a = 23k$, where $k$ is some integer. Substitute this back into Equation 1:

$(23k)^2 = 23b^2$

$529k^2 = 23b^2$

Divide both sides by $23$:

$23k^2 = b^2$

This new equation shows that $b^2$ is also a multiple of $23$. By the same logic applied previously, $23$ must divide $b$.

The Contradiction:

We have deduced that $23$ divides both $a$ and $b$. This means $a$ and $b$ share a common factor of $23$. However, this directly contradicts our initial assumption that $a$ and $b$ are coprime ($\gcd(a,b) = 1$). Because our assumption led to a logical contradiction, the assumption must be false.

Therefore, $\sqrt{23}$ cannot be expressed as a rational fraction.

Step 4: Visualizing on the Real Number Line

To further contextualize this irrational number, we can map its approximate decimal expansion ($\sqrt{23} \approx 4.795831523...$) onto a real number line. The non-terminating, non-repeating nature of this decimal confirms its irrationality.

Final Solution: The number $\sqrt{23}$ is an irrational number.

Initial Setup & Variable Assignment

We are tasked with converting the mixed recurring decimal $0.4\overline{7}$ into a rational number of the form $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \neq 0$.

Let the given number be represented by the variable $x$:

$x = 0.47777\dots$     (Equation 1)

[By definition of the vinculum (bar) notation, only the digit $7$ repeats infinitely, while the digit $4$ is a non-repeating decimal component].

Step 1: Isolating the Repeating Decimal Block

To eliminate the infinite repeating sequence, we must first align the decimal point immediately before the repeating block. Since there is exactly one non-repeating digit ($4$) after the decimal point, we multiply both sides of Equation 1 by $10^1$.

$10 \cdot x = 10 \cdot (0.47777\dots)$

$10x = 4.7777\dots$     (Equation 2)

[Per the properties of base-10 positional notation, multiplying by $10$ shifts the decimal point one place to the right].

Step 2: Shifting the Decimal Past the Repeating Block

Next, we need a second equation where the fractional part is identical to the fractional part of Equation 2 (which is $.7777\dots$). Since the repeating block consists of exactly one digit ($7$), we multiply Equation 2 by $10^1$.

$10 \cdot 10x = 10 \cdot (4.7777\dots)$

$100x = 47.7777\dots$     (Equation 3)

Step 3: Algebraic Elimination of the Infinite Decimal

We now subtract Equation 2 from Equation 3. Because the infinite repeating decimal tails ($.7777\dots$) are perfectly aligned, they subtract to zero, leaving only integers.

[By the Subtraction Property of Equality, if $a = b$ and $c = d$, then $a - c = b - d$].

Step 4: Solving for the Rational Form $\frac{p}{q}$

We now have a simple linear equation with integer coefficients:

$90x = 43$

Isolating $x$ by dividing both sides by $90$:

$x = \frac{43}{90}$

We must verify that this fraction is in its simplest form. The numerator, $43$, is a prime number. The denominator, $90$, is not a multiple of $43$ (since $43 \times 2 = 86$). Therefore, the greatest common divisor $\text{GCD}(43, 90) = 1$. The fraction is irreducible, and both $p = 43$ and $q = 90$ are integers with $q \neq 0$.

Final Solution: The mixed recurring decimal $0.4\overline{7}$ expressed in the rational form $\frac{p}{q}$ is $\frac{43}{90}$.

Step 1: Initial Setup & Conversion to Improper Fraction

We are given the mixed fraction $4\frac{1}{8}$. To analyze its decimal expansion, we first convert it into an improper fraction of the form $\frac{p}{q}$.

The conversion formula is:

$\text{Improper Fraction} = \frac{(\text{Whole Number} \times \text{Denominator}) + \text{Numerator}}{\text{Denominator}}$

Substituting the given values:

$4\frac{1}{8} = \frac{(4 \times 8) + 1}{8} = \frac{32 + 1}{8} = \frac{33}{8}$

Step 2: Decimal Conversion via Long Division

To find the decimal form, we divide the numerator ($33$) by the denominator ($8$).

• Divide 33 by 8: $8 \times 4 = 32$. The quotient is $4$, and the remainder is $33 - 32 = 1$.
• Add a decimal point: Place a decimal point in the quotient and append a $0$ to the remainder, making it $10$.
• Divide 10 by 8: $8 \times 1 = 8$. The quotient digit is $1$, and the remainder is $10 - 8 = 2$.
• Append another 0: The remainder becomes $20$.
• Divide 20 by 8: $8 \times 2 = 16$. The quotient digit is $2$, and the remainder is $20 - 16 = 4$.
• Append another 0: The remainder becomes $40$.
• Divide 40 by 8: $8 \times 5 = 40$. The quotient digit is $5$, and the remainder is $40 - 40 = 0$.

[Because the remainder has reached exactly $0$, the division process terminates.]

Step 3: Theoretical Verification via Prime Factorization

We can verify the nature of the decimal expansion without long division by analyzing the denominator of the rational number $\frac{33}{8}$.

[Per the Rational Number Decimal Expansion Theorem: A rational number $\frac{p}{q}$ (where $p$ and $q$ are co-prime) has a terminating decimal expansion if and only if the prime factorization of $q$ is of the form $2^n \times 5^m$, where $n$ and $m$ are non-negative integers.]

• The denominator is $q = 8$.
• The prime factorization of $8$ is $2 \times 2 \times 2 = 2^3$.
• This can be written in the standard form as $2^3 \times 5^0$.

Since the prime factors of the denominator consist entirely of the digit $2$ (fitting the $2^n \times 5^m$ condition), the fraction is mathematically guaranteed to have a terminating decimal expansion.

Step 4: Alternative Calculation Method (Powers of 10)

To bypass long division, we can force the denominator to become a power of $10$ ($10, 100, 1000, \dots$).

Starting with the fractional part $\frac{1}{8}$:

$\frac{1}{8} = \frac{1}{2^3}$

To make the denominator a power of $10$, we multiply both the numerator and the denominator by $5^3$ ($125$):

$\frac{1 \times 125}{8 \times 125} = \frac{125}{1000} = 0.125$

Adding this back to the whole number part:

$4 + 0.125 = 4.125$

Final Solution: The decimal form of $4\frac{1}{8}$ is $4.125$, and it has a terminating decimal expansion.

Initial Setup & Mathematical Definitions

We are tasked with classifying the real number $\sqrt{225}$ as either a rational or an irrational number. To do this rigorously, we must evaluate the radical expression and apply the formal definitions of the real number system.

• Rational Number: A number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers ($p, q \in \mathbb{Z}$) and $q \neq 0$.
• Irrational Number: A real number that cannot be expressed as a simple fraction of two integers. Its decimal expansion is non-terminating and non-repeating.

Step 1: Prime Factorization of the Radicand

The number inside the square root symbol is called the radicand. Here, the radicand is $225$. To simplify the square root, we first determine the prime factorization of $225$ [Per the Fundamental Theorem of Arithmetic].

We systematically divide $225$ by the smallest possible prime numbers:

• $225$ ends in $5$, but let us check divisibility by $3$. The sum of the digits is $2 + 2 + 5 = 9$, which is divisible by $3$.
• $225 \div 3 = 75$
• $75 \div 3 = 25$
• $25 \div 5 = 5$
• $5 \div 5 = 1$

Thus, the prime factorization of $225$ is:

$225 = 3 \times 3 \times 5 \times 5 = 3^2 \times 5^2$

Step 2: Simplification of the Radical Expression

Substitute the prime factorization back into the radical expression. Using the property of radicals $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ (where $a, b \ge 0$):

$\sqrt{225} = \sqrt{3^2 \times 5^2}$

$\sqrt{225} = \sqrt{3^2} \times \sqrt{5^2}$

Since the square root of a squared positive real number is the number itself ($\sqrt{x^2} = x$ for $x \ge 0$):

$\sqrt{225} = 3 \times 5 = 15$

Step 3: Geometric Interpretation of a Perfect Square

Geometrically, finding the square root of $225$ is equivalent to finding the side length of a square whose total area is $225$ square units. Because the side length evaluates to exactly $15$ (a whole number), $225$ is classified as a perfect square.

Step 4: Classification Based on Number Theory

We have established that $\sqrt{225} = 15$. We must now determine if $15$ fits the definition of a rational number.

Any integer $n$ can be written as a fraction by placing it over $1$ ($n = \frac{n}{1}$). Therefore:

$15 = \frac{15}{1}$

In this fraction:

• $p = 15$, which is an integer ($15 \in \mathbb{Z}$).
• $q = 1$, which is an integer ($1 \in \mathbb{Z}$).
• The denominator $q \neq 0$ ($1 \neq 0$).

Because $\sqrt{225}$ can be expressed exactly as the ratio of two integers, it perfectly satisfies the criteria for being a rational number.

Final Solution: $\sqrt{225}$ simplifies to $15$, which can be written as $\frac{15}{1}$. Therefore, $\sqrt{225}$ is a rational number.

Step 1: Variable Assignment and Expansion

To express the non-terminating, repeating decimal $0.\overline{6}$ as a rational number in the form $\frac{p}{q}$, we begin by assigning it to a variable. Let $x$ represent the given value.

$x = 0.\overline{6}$

Expanding the vinculum (the overline indicating the repeating sequence), we write the number as an infinite decimal series:

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

Step 2: Multiplication by a Power of 10

[Per the fundamental algorithm for converting repeating decimals to fractions], we must shift the decimal point to the right by exactly the number of digits in the repeating block. This aligns the infinite repeating tails so they can be algebraically eliminated.

In the decimal $0.\overline{6}$, there is exactly one repeating digit (the digit 6). Therefore, we multiply both sides of Equation 1 by $10^1$ (which is $10$).

$10 \cdot x = 10 \cdot (0.6666\dots)$

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

Step 3: Algebraic Subtraction to Eliminate the Repeating Tail

Next, we subtract Equation 1 from Equation 2. Because the infinite sequence of $6$s to the right of the decimal point is identical in both equations, subtracting them will perfectly cancel out the fractional part, leaving an integer.

This yields the simplified linear equation:

$9x = 6$

Step 4: Solving for $x$ and Simplification

Isolate $x$ by dividing both sides of the equation by $9$:

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

[By the Fundamental Theorem of Arithmetic], fractions must be expressed in their simplest form. We find the Greatest Common Divisor (GCD) of the numerator ($6$) and the denominator ($9$).

• Prime factorization of $6 = 2 \times 3$
• Prime factorization of $9 = 3 \times 3$
• $\text{GCD}(6, 9) = 3$

Dividing both the numerator and the denominator by $3$:

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

Geometric Verification

The fraction $\frac{2}{3}$ represents two parts out of three equal subdivisions of a whole. When $2$ is divided by $3$ using long division, the quotient is exactly $0.666\dots$, confirming our algebraic derivation.

Final Solution: The repeating decimal $0.\overline{6}$ expressed in the form $\frac{p}{q}$ is $\frac{2}{3}$, where $p=2$ and $q=3$ are integers, and $q \neq 0$.

Initial Setup & Given Expression

We are tasked with converting the rational number $\frac{3}{13}$ into its decimal form and determining the specific nature of its decimal expansion. To achieve this, we will perform the long division algorithm, dividing the numerator ($3$) by the denominator ($13$).

Step 1: Executing the Long Division Algorithm

Since the dividend ($3$) is strictly less than the divisor ($13$), we place a decimal point in the quotient and append zeros to the dividend to proceed with the division.

Let us break down the sequential arithmetic of the division:

• $30 \div 13 = 2$ with a remainder of $4$ $\quad (13 \times 2 = 26)$
• $40 \div 13 = 3$ with a remainder of $1$ $\quad (13 \times 3 = 39)$
• $10 \div 13 = 0$ with a remainder of $10$ $\quad (13 \times 0 = 0)$
• $100 \div 13 = 7$ with a remainder of $9$ $\quad (13 \times 7 = 91)$
• $90 \div 13 = 6$ with a remainder of $12$ $\quad (13 \times 6 = 78)$
• $120 \div 13 = 9$ with a remainder of $3$ $\quad (13 \times 9 = 117)$

Step 2: Analyzing the Sequence of Remainders

The sequence of remainders generated during the division is: $3, 4, 1, 10, 9, 12$. At the 6th step, the remainder is $3$, which is the exact value of our original dividend. [Per the properties of rational numbers], whenever a remainder repeats in long division, the sequence of digits in the quotient will also begin to repeat infinitely in the exact same order.

The block of digits in the quotient corresponding to this cycle is $230769$. Therefore, the decimal expansion can be written using a vinculum (bar) over the repeating block:

$\frac{3}{13} = 0.230769230769... = 0.\overline{230769}$

Step 3: Theoretical Justification & Classification

We can verify the nature of this decimal expansion without division by analyzing the prime factorization of the denominator. [By the Fundamental Theorem of Arithmetic and the properties of rational numbers], a rational number $\frac{p}{q}$ (in simplest form) will have a terminating decimal expansion if and only if the prime factorization of $q$ is of the form $2^n \times 5^m$, where $n$ and $m$ are non-negative integers.

In our given fraction $\frac{3}{13}$:

• The denominator is $q = 13$.
• The prime factorization of $13$ is simply $13^1$.
• Because the prime factorization contains a prime number other than $2$ or $5$, the decimal expansion cannot terminate.

Since it does not terminate but represents a rational number, it must be a non-terminating repeating (or recurring) decimal.

Final Solution: The decimal form of $\frac{3}{13}$ is $0.\overline{230769}$, and it has a non-terminating repeating decimal expansion.

Initial Setup & Mathematical Objective

The objective is to express the rational number $\frac{1}{11}$ in its decimal form and classify the nature of its decimal expansion. To achieve this, we apply the standard long division algorithm, dividing the numerator ($1$) by the denominator ($11$).

Step 1: Execution of the Long Division Algorithm

We set up the division $1 \div 11$. Since the dividend ($1$) is strictly less than the divisor ($11$), we place a decimal point in the quotient and append zeros to the dividend to proceed with the fractional parts.

• Tenths place: Bring down a $0$ to make the dividend $10$. Since $10 < 11$, the quotient digit is $0$. The remainder is $10$.
• Hundredths place: Bring down another $0$ to make the dividend $100$. Divide $100$ by $11$. The maximum multiple is $11 \times 9 = 99$. The quotient digit is $9$, and the remainder is $100 - 99 = 1$.
• Thousandths place: Bring down a $0$ to the remainder $1$, making it $10$. Since $10 < 11$, the quotient digit is $0$. The remainder is $10$.
• Ten-thousandths place: Bring down another $0$ to make it $100$. Divide $100$ by $11$. The quotient digit is $9$, and the remainder is $100 - 99 = 1$.

The visual representation of this algorithmic process is detailed below:

Step 2: Analysis of the Remainder and Quotient Sequences

By tracking the division process, we can observe a strict mathematical pattern in the remainders and the resulting quotient digits.

[Per the properties of rational numbers], because the remainder sequence $(10, 1, 10, 1, \dots)$ begins to repeat, the sequence of digits in the quotient will also repeat infinitely. The repeating block of digits in the quotient is $09$.

Step 3: Classification of the Decimal Expansion

Because the remainder never becomes zero, the decimal expansion is non-terminating. Furthermore, because a specific block of digits ($09$) repeats infinitely, the expansion is repeating (or recurring). Mathematically, this is denoted by placing a vinculum (bar) over the repeating block of digits:

$ \frac{1}{11} = 0.090909\dots = 0.\overline{09} $

Final Solution: The decimal form of $\frac{1}{11}$ is $0.\overline{09}$, and it has a non-terminating repeating (recurring) decimal expansion.

Given Variables & Initial Setup

We are given the decimal expansion of the rational number $\frac{1}{7}$:

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

The bar over the digits $142857$ indicates that this block of six digits repeats infinitely. We are tasked with predicting the decimal expansions of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7},$ and $\frac{6}{7}$ without performing long division, utilizing the properties of the remainders generated during the division of $1$ by $7$.

Step 1: Understanding the Cyclic Property of $\frac{1}{7}$

The number $142857$ is a well-known cyclic number. [Per number theory, a cyclic number is an integer in which cyclic permutations of the digits are produced when multiplied by successive integers]. Because $7$ is a prime number and the length of the repeating decimal block is $7 - 1 = 6$, the decimal expansions for any fraction $\frac{k}{7}$ (where $1 \le k \le 6$) will consist of the exact same six digits in the exact same cyclic order. The only difference is the starting digit.

Step 2: Analyzing the Remainders (The Hint)

When we perform the long division of $1 \div 7$, we append a zero to the remainder at each step to continue the division. The sequence of dividends and their corresponding quotients dictates the repeating block:

• $10 \div 7 = 1$ (Remainder $3$) → First digit is $1$
• $30 \div 7 = 4$ (Remainder $2$) → Second digit is $4$
• $20 \div 7 = 2$ (Remainder $6$) → Third digit is $2$
• $60 \div 7 = 8$ (Remainder $4$) → Fourth digit is $8$
• $40 \div 7 = 5$ (Remainder $5$) → Fifth digit is $5$
• $50 \div 7 = 7$ (Remainder $1$) → Sixth digit is $7$

[By the Division Algorithm], to find the decimal expansion of $\frac{k}{7}$, we simply look at the first step of dividing $k$ by $7$ (i.e., $10k \div 7$). The quotient gives us the starting digit of the cycle, and the rest of the digits follow the established cyclic order: $1 \rightarrow 4 \rightarrow 2 \rightarrow 8 \rightarrow 5 \rightarrow 7 \rightarrow 1$.

Step 3: Predicting the Decimal Expansions

Using the logic established in Step 2, we can predict the expansions through simple scalar multiplication or by identifying the starting digit:

Alternatively, this can be verified algebraically by multiplying the original repeating decimal by the respective numerators. For example:
$\frac{2}{7} = 2 \times \frac{1}{7} = 2 \times 0.\overline{142857} = 0.\overline{285714}$.
Because there are no carry-over digits that disrupt the cycle, the permutation holds perfectly true for all scalar multiples up to $6$.

Final Solution: Yes, the decimal expansions can be predicted without long division by utilizing the cyclic permutation of the digits $142857$. The predicted expansions are:
$\frac{2}{7} = 0.\overline{285714}$
$\frac{3}{7} = 0.\overline{428571}$
$\frac{4}{7} = 0.\overline{571428}$
$\frac{5}{7} = 0.\overline{714285}$
$\frac{6}{7} = 0.\overline{857142}$

Initial Setup & Theoretical Foundation

We are tasked with converting the rational number $\frac{329}{400}$ into its decimal form and determining the nature of its decimal expansion.

[Per the Rational Number Decimal Expansion Theorem]: Let $x = \frac{p}{q}$ be a rational number, such that the prime factorization of $q$ is of the form $2^n \times 5^m$, where $n$ and $m$ are non-negative integers. Then $x$ has a decimal expansion which terminates. If the prime factorization of $q$ contains any prime other than $2$ or $5$, the decimal expansion is non-terminating repeating.

Step 1: Prime Factorization of the Denominator

To predict the type of decimal expansion, we first analyze the denominator, $q = 400$.

• We decompose $400$ into its prime factors:
• $400 = 4 \times 100$
• $400 = 2^2 \times (10^2)$
• $400 = 2^2 \times (2 \times 5)^2$
• $400 = 2^2 \times 2^2 \times 5^2 = 2^4 \times 5^2$

Since the prime factors of the denominator consist exclusively of the primes $2$ and $5$ (where $n=4$ and $m=2$), we can definitively conclude that the fraction will yield a terminating decimal expansion.

Step 2: Algebraic Manipulation to Base 10

To convert the fraction to a decimal without long division, we can manipulate the denominator to become a power of $10$. A power of $10$ requires equal powers of $2$ and $5$ ($10^k = 2^k \times 5^k$).

Currently, the denominator is $2^4 \times 5^2$. To make the powers of $2$ and $5$ equal, we must multiply both the numerator and the denominator by $5^2$ (which is $25$).

$\frac{329}{400} = \frac{329}{2^4 \times 5^2}$

Multiplying numerator and denominator by $5^2$:

$= \frac{329 \times 5^2}{2^4 \times 5^2 \times 5^2}$

$= \frac{329 \times 25}{2^4 \times 5^4}$

$= \frac{8225}{(2 \times 5)^4}$

$= \frac{8225}{10^4}$

$= \frac{8225}{10000}$

Step 3: Decimal Conversion

Dividing by $10000$ (which has four zeros) requires shifting the decimal point of the numerator four places to the left.

$\frac{8225}{10000} = 0.8225$

Visual Representation of the Transformation

Step 4: Verification via Long Division

Alternatively, we can verify this result using the standard long division algorithm:

• $329 \div 400 = 0$ with a remainder of $329$. (Add a decimal point and a zero).
• $3290 \div 400 = 8$ ($400 \times 8 = 3200$), remainder is $90$. (Bring down a zero).
• $900 \div 400 = 2$ ($400 \times 2 = 800$), remainder is $100$. (Bring down a zero).
• $1000 \div 400 = 2$ ($400 \times 2 = 800$), remainder is $200$. (Bring down a zero).
• $2000 \div 400 = 5$ ($400 \times 5 = 2000$), remainder is $0$.

The division process ends with a remainder of $0$, confirming that the decimal expansion terminates.

Final Solution: The decimal form of $\frac{329}{400}$ is $0.8225$, and it has a terminating decimal expansion.

Step 1: Theoretical Foundation & Initial Setup

We are tasked with analyzing rational numbers of the form $\frac{p}{q}$, where:

• $p$ and $q$ are integers.
• $q \neq 0$.
• $\gcd(p, q) = 1$ [This indicates that $p$ and $q$ are coprime; they share no common factors other than 1, meaning the fraction is in its simplest form].
• The decimal representation of $\frac{p}{q}$ is terminating (it has a finite number of digits after the decimal point).

Our objective is to deduce the specific mathematical property that the denominator $q$ must satisfy to guarantee a terminating decimal expansion.

Step 2: Empirical Observation & Data Collection

To identify the pattern, we will generate several examples of rational numbers that yield terminating decimals, ensure they are in their simplest form, and examine the prime factorization of their denominators.

Step 3: Analyzing the Denominators ($q$)

By observing the "Prime Factorization of $q$" column, a distinct pattern emerges. The prime factors of every denominator $q$ in these terminating examples consist exclusively of the numbers $2$ and $5$.

To verify this exclusivity, let us briefly look at fractions where the denominator contains prime factors other than $2$ or $5$:

• $\frac{1}{3} = 0.3333...$ [Non-terminating, repeating; prime factor is $3$]
• $\frac{1}{7} = 0.142857...$ [Non-terminating, repeating; prime factor is $7$]
• $\frac{5}{6} = \frac{5}{2 \times 3} = 0.8333...$ [Non-terminating, repeating; prime factors are $2$ and $3$]

Step 4: Theoretical Justification

Any terminating decimal can be expressed as a fraction where the denominator is a power of $10$ (i.e., $10^k$ for some integer $k \geq 1$).

[Per the Fundamental Theorem of Arithmetic], the prime factorization of $10$ is strictly $2 \times 5$. Therefore, the prime factorization of any power of $10$ is:

$10^k = (2 \times 5)^k = 2^k \times 5^k$

When a fraction $\frac{x}{10^k}$ is reduced to its simplest form $\frac{p}{q}$ by canceling common factors, the only prime factors that can possibly remain in the denominator $q$ are $2$ and $5$. No new prime factors can be introduced during simplification.

Step 5: Visualizing the Logical Flow

The following diagram illustrates the structural transformation from a terminating decimal to its simplest fractional form, proving why $q$ is restricted to powers of $2$ and $5$.

Step 6: Formulating the General Property

Based on the empirical data and theoretical proof, for any rational number $\frac{p}{q}$ (where $p$ and $q$ are coprime) to have a terminating decimal expansion, the prime factorization of the denominator $q$ must not contain any prime numbers other than $2$ and $5$.

Mathematically, this is expressed as:

$q = 2^m \times 5^n$

where $m$ and $n$ are non-negative integers ($m, n \geq 0$).

Final Solution: For a rational number $\frac{p}{q}$ (where $p$ and $q$ are coprime integers) to have a terminating decimal representation, the prime factorization of the denominator $q$ must consist exclusively of powers of $2$ and/or $5$. That is, $q$ must satisfy the property $q = 2^m \times 5^n$, where $m$ and $n$ are non-negative integers.

Theoretical Foundation: Maximum Digits in a Repeating Block

To determine the maximum number of digits in the repeating block of the decimal expansion of a rational number $\frac{p}{q}$ (where $p$ and $q$ are coprime integers), we analyze the possible remainders generated during the long division process.

[Per the Division Algorithm], when dividing by a divisor $q$, the only possible non-zero remainders are the integers from $1$ to $q - 1$. If a remainder of $0$ is reached, the decimal terminates. If the decimal is non-terminating and repeating, the sequence of remainders must eventually repeat. Because there are exactly $q - 1$ possible non-zero remainders, the division process can generate at most $q - 1$ distinct remainders before a previous remainder reappears, forcing the sequence of quotient digits to cycle.

For the fraction $\frac{1}{17}$:

• The divisor is $q = 17$.
• The maximum possible number of distinct non-zero remainders is $17 - 1 = 16$.

Therefore, the maximum number of digits in the repeating block of $\frac{1}{17}$ is strictly bounded by 16.

Step-by-Step Long Division Execution

To verify this theoretical maximum, we perform the long division of $1 \div 17$. We append zeros to the dividend and track the quotient digits and remainders at each step [applying $a = bq + r$].

At Step 16, we obtain a remainder of $1$. This is the exact same value we started with (the numerator of $\frac{1}{17}$). Because the remainder has repeated, the sequence of quotient digits will now repeat infinitely in the exact same order.

Visualizing the Cyclic Nature of Remainders

The following diagram illustrates the complete cycle of the 16 distinct remainders generated during the division process. The cycle proves that the repeating block has reached its maximum theoretical length.

Final Decimal Expansion

Collecting the quotient digits from Step 1 through Step 16, we construct the repeating decimal block:

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

Counting the digits under the vinculum (overline), we find exactly 16 digits.

Final Solution: The maximum number of digits in the repeating block of the decimal expansion of $\frac{1}{17}$ is 16. Performing the long division confirms that the repeating block is $0588235294117647$, which contains exactly 16 digits.

Step 1: Variable Assignment and Equation Formulation

To express the non-terminating, repeating decimal $0.99999...$ (which can also be written as $0.\overline{9}$) in the rational form $\frac{p}{q}$, we begin by assigning it to a variable $x$.

Let $x = 0.99999... \quad \text{--- (Equation 1)}$

Step 2: Algebraic Multiplication

We observe the periodicity of the repeating decimal. Since exactly one digit (the digit $9$) repeats infinitely, we multiply both sides of Equation 1 by $10^1$ (or $10$). [Per the fundamental properties of equality and base-10 positional notation, multiplying by 10 shifts the decimal point one place to the right].

$10x = 9.99999... \quad \text{--- (Equation 2)}$

Step 3: Subtraction and Isolation of the Variable

To eliminate the infinitely repeating fractional part, we subtract Equation 1 from Equation 2. Because the infinite tail of $9$s is identical in both equations, their difference is exactly zero.

$10x = 9.99999...$
$- \quad x = 0.99999...$

Now, solve for $x$ by dividing both sides by $9$:

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

$x = 1$

Step 4: Expressing in $\frac{p}{q}$ Form

The integer $1$ can be expressed as a rational number in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

$x = \frac{1}{1}$

Step 5: Analytical Discussion & Theoretical Justification

The result $0.99999... = 1$ often causes initial surprise because they appear to be distinct numbers. However, in the real number system, they are two different representations of the exact same mathematical value. This makes perfect logical sense when analyzed through multiple mathematical frameworks:

• Argument 1: Fractional Equivalency
  Consider the fraction $\frac{1}{3}$. In decimal form, $\frac{1}{3} = 0.33333...$
  If we multiply both sides of this equation by $3$, we get:
  $3 \times \frac{1}{3} = 3 \times 0.33333...$
  $1 = 0.99999...$


• Argument 2: Infinite Geometric Series (Real Analysis)
  The decimal $0.999...$ can be expanded as a sum of fractions:
  $0.9 + 0.09 + 0.009 + \dots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \dots$
  This is an infinite geometric series where the first term $a = 0.9$ and the common ratio $r = 0.1$.
  [Per the formula for the sum of an infinite geometric series, $S = \frac{a}{1 - r}$ for $|r| < 1$]:
  $S = \frac{0.9}{1 - 0.1} = \frac{0.9}{0.9} = 1$.


• Argument 3: The Archimedean Property
  In the real number system, if two numbers are strictly different, there must exist a non-zero distance between them. The difference between $1$ and $0.999...$ is $0.000...$ with an infinite string of zeros. Since there is no "final" position to place a $1$, the difference is exactly $0$. Therefore, the numbers are identical.

Final Solution: The repeating decimal $0.99999...$ expressed in the rational form $\frac{p}{q}$ is $\frac{1}{1}$. The value is exactly equal to $1$, which is mathematically sound and justified by the properties of infinite geometric series, fractional equivalency, and the definition of limits in the real number system.