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Q1(iv):
Write the following in decimal form and say what kind of decimal expansion each has : (iv) $\frac{3}{13}$

Solution :

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.

0 . 2 3 0 7 6 9 13 3 . 0 0 0 0 0 0 2 6 - 4 0 3 9 - 1 0 0 9 1 - 9 0 7 8 - 1 2 0 1 1 7 - 3 ← Remainder is 3 (Initial dividend repeats)

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.


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