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.

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.