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Q1(iii):
Write the following in decimal form and say what kind of decimal expansion each has : (iii) $4\frac{1}{8}$

Solution :

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.]

8 33.000 4.125 -32 10 - 8 20 -16 40 - 40 0

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.


More Questions from Class 9 Mathematics Number Systems EXERCISE 1.3


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