Ninth Grade Mathematics: Terminating & Repeating Decimals
The decimal system provides a unified language for expressing real numbers in base ten. When examining the decimal representation of any rational number, an extraordinary regularity emerges: every fraction either terminates after a predictable number of decimal places or enters an infinite periodic loop.
The Structure of Decimal Expansions
Every real number can be written as an integer part followed by a decimal separator and a sequence of fractional digits. The nature of these fractional digits depends entirely on the factors of the denominator when the fraction is reduced to lowest terms.
DECIMAL NUMBERS
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+------------------------+------------------------+
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v v
TERMINATING DECIMALS NON-TERMINATING DECIMALS
(Ends after finite digits) |
Example: 0.375, 1.25, 0.04 +-------------+-------------+
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v v
REPEATING (PERIODIC) NON-REPEATING
(Rational Q) (Irrational I)
Example: 0.666... Example: pi, sqrt(2)
0.142857... 1.010010001...
A terminating decimal has a finite number of non-zero digits after the decimal point, such as 0.75 or 0.128. A repeating decimal, also called a recurring decimal, continues indefinitely but repeats a specific block of one or more digits over and over.
The Denominator Rule for Terminating Decimals
A fraction p/q in simplest form (where p and q share no common integer factors other than 1) terminates if and only if the prime factorization of its denominator q contains no prime factors other than 2 and 5.
Condition for Terminating Decimal:
Denominator q (in lowest terms) = 2^m * 5^n
where m and n are non-negative integers.
To understand why this rule holds, consider that our decimal number system is based on powers of ten: 10 = 2 * 5. Any fraction whose denominator can be multiplied to become a power of 10 will terminate. Because 10 is composed strictly of the prime factors 2 and 5, any fraction containing any other prime factor in its reduced denominator—such as 3, 7, 11, or 13—can never be converted into a base-ten power of ten, and will therefore never terminate.
Example 1: 7 / 40
Factor the denominator: 40 = 2^3 * 5^1.
Prime factors are strictly 2 and 5.
Multiply numerator and denominator by 5^2 to form 10^3:
(7 * 25) / (40 * 25) = 175 / 1000 = 0.175 (Terminates!)
Example 2: 5 / 28
Factor the denominator: 28 = 2^2 * 7.
The presence of prime factor 7 guarantees that 5/28 will REPEAT, not terminate.
Periodicity and Converting Repeating Decimals to Fractions
In a repeating decimal, the repeating sequence of digits is called the period or repetend, and the number of repeating digits is called the length of the period. We indicate a repetend by drawing a horizontal bar (called a vinculum) over the repeating digits. For example, 0.333... is written as 0.(3) with a bar over 3, and 0.181818... is written as 0.(18) with a bar over 18.
Algebraic Conversion Method
Any repeating decimal can be converted into its exact rational fraction p/q using an algebraic technique that cancels out the infinite repeating tail.
Convert x = 0.474747... to a fraction:
Step 1: Assign a variable to the decimal:
x = 0.474747...
Step 2: Identify the period length (here, 2 digits: '47').
Multiply x by 10^2 = 100 to shift one full period to the left of the decimal:
100 * x = 47.474747...
Step 3: Subtract the original equation from the multiplied equation:
100 * x = 47.474747...
- x = 0.474747...
-------------------------
99 * x = 47.000000... (The infinite tail cancels completely!)
Step 4: Solve for x:
x = 47 / 99
Mixed Repeating Decimals
Some decimals have non-repeating digits immediately after the decimal point before the repeating period begins. These are called mixed recurring decimals, such as 0.2353535... where 2 does not repeat but 35 repeats indefinitely.
Convert x = 0.2353535... to a fraction:
Step 1: Multiply by 10 to shift non-repeating digits left of the decimal:
10 * x = 2.353535... (Equation A)
Step 2: Multiply Equation A by 100 (since 2 digits repeat) to shift one period:
1000 * x = 235.353535... (Equation B)
Step 3: Subtract Equation A from Equation B:
1000 * x = 235.353535...
- 10 * x = 2.353535...
---------------------------
990 * x = 233.000000...
Step 4: Solve for x:
x = 233 / 990
Chapter Exercises
Exercise 1: Without performing actual long division, determine whether each of the following fractions produces a terminating or a repeating decimal expansion: 13/80, 21/375, 11/42, and 33/150. Show the prime factorization of each simplified denominator.
Exercise 2: Convert the pure repeating decimal 0.818181... into an exact fraction in simplest form using the algebraic subtraction method.
Exercise 3: Convert the mixed repeating decimal 1.4333... into an improper fraction in lowest terms.
Solutions and Explanations
Solution 1: For 13/80, the fraction is in lowest terms. Factoring the denominator gives 80 = 2^4 * 5. Because the prime factors are exclusively 2 and 5, 13/80 produces a terminating decimal. For 21/375, reduce the fraction first by dividing numerator and denominator by 3: 21/375 = 7/125. The factored denominator is 125 = 5^3. Because only the prime factor 5 appears, 21/375 produces a terminating decimal. For 11/42, the fraction is in simplest form. Factoring 42 gives 42 = 2 * 3 * 7. The presence of the prime factors 3 and 7 guarantees that 11/42 produces a repeating decimal. For 33/150, reduce by dividing by 3: 33/150 = 11/50. Factoring the reduced denominator gives 50 = 2 * 5^2. Because only 2 and 5 are factors, 33/150 produces a terminating decimal.
Solution 2: Let x = 0.818181... The period consists of two repeating digits, '81'. Multiply the equation by 10^2 = 100 to shift one period left of the decimal point, yielding 100x = 81.818181... Subtracting the original equation x = 0.818181... from this new equation eliminates the infinite repeating decimal part: 100x - x = 81.818181... - 0.818181..., which simplifies to 99*x = 81. Dividing both sides by 99 gives x = 81/99. Dividing both numerator and denominator by their greatest common divisor, 9, gives the simplified fraction x = 9/11.
Solution 3: Let x = 1.4333... Multiply by 10 to move the single non-repeating decimal digit (4) to the left of the decimal point: 10x = 14.333... (Equation 1). Because one digit (3) repeats, multiply Equation 1 by 10 to shift one repeating digit: 100x = 143.333... (Equation 2). Subtracting Equation 1 from Equation 2 eliminates the repeating tail: 100x - 10x = 143.333... - 14.333..., yielding 90*x = 129. Solving for x gives x = 129/90. Dividing both numerator and denominator by 3 simplifies the fraction to x = 43/30.