Recurring Decimals & Fraction Conversions for CSAT UPSC

1. Theoretical Foundation & Mathematical Underpinnings

Formal Definition & Scope

A rational number is any number expressible in the form ^p/_q where p ∈ ℤ, q ∈ ℤ⁺, gcd(p, q) = 1.

Every rational number possesses a decimal expansion belonging to exactly one of the following categories:

• Terminating: The decimal expansion terminates after a finite sequence of n digits:
  x = A.a₁a₂…a_n
• Pure Recurring (Pure Repetend): The repeating cycle begins immediately following the decimal point:
  x = A.b₁b₂…b_r, r ≥ 1
  Examples: 0.27, 2.142857
• Mixed Recurring (Mixed Repetend): A non-repeating block of n ≥ 1 digits precedes a recurring block of r ≥ 1 digits:
  x = A.c₁…c_nb₁…b_r
  Examples: 0.16, 0.235

Scope Boundary — The 0.9 = 1 Exception: 0.9 is not strictly less than 1; it is an identical mathematical representation of 1. This represents the only non-unique decimal representation among real numbers and serves as a classic examiner trap. Consequently, every terminating decimal can also be expressed as an expansion ending in 9 (e.g., 0.5 = 0.49).

The “Why” Behind the Math

Why Infinite Tail Cancellation Works:

Any recurring decimal can be formulated as an infinite geometric series. For a 2-digit repetend xy:

x = 0.xy
x = ^xy/_10² + ^xy/_10⁴ + ^xy/_10⁶ + ···

This is a geometric series with first term a = ^xy/₁₀₀ and common ratio r = ¹/₁₀₀ < 1:

S = ^a/_{(1 − r)} = ^{(xy / 100)}/_{(1 − 1/100)} = ^{(xy / 100)}/_{(99 / 100)} = ^xy/₉₉

The algebraic technique of scaling by powers of 10 and subtracting cancels the infinite tails. Multiplying by 10^r (where r is the repetend length) shifts the decimal point by exactly one full repeating period, rendering the non-terminating tails identical.

Proof for Mixed Recurring Decimals: Let x = 0.abc where n = 1 and r = 2:

x = 0.abc
10ⁿx = 10x = a.bc — (Eq. 1) [Repetend starts at decimal point]
10^n+rx = 1000x = abc.bc — (Eq. 2) [Shifted by one full period]

Subtracting Eq. 1 from Eq. 2:

(10^n+r − 10ⁿ)x = abc − a
990x = abc − a
⇒ x = ^{(abc − a)}/₉₉₀

(where abc represents the integer formed by the concatenation of digits a, b, c).

General Master Theorem:
For any decimal of the form x = A.c₁…c_nb₁…b_r:

x = A + ^{[(c₁…c_nb₁…b_r) − (c₁…c_n)]} / _{[99…9 (r times) 00…0 (n times)]}

• If n = 0 (Pure Recurring): x = A + ^(b₁…b_r) / _{[99…9 (r times)]}
• If r = 0 (Terminating): x = A + ^(c₁…c_n) / _10ⁿ

Retention Anchor: 9 for Bar, 0 for Gap — The denominator takes r nines followed by n zeros. The numerator is the entire sequence of digits after the decimal minus the non-repeating prefix.

Key Theorems & Properties

• Termination Criterion: A rational number ^p/_q in irreducible form (gcd(p, q) = 1) is terminating if and only if the prime factorization of q contains no primes other than 2 and 5:
  q = 2^a5^b (a, b ∈ ℕ₀)
  Otherwise, the expansion is recurring.
  • Example: ¹³/₄₀ = ¹³/_{(2³ · 5)} ⇒ Terminating (0.325).
  • Example: ¹³/₃₀ = ¹³/_{(2 · 3 · 5)} ⇒ Mixed recurring (0.43) due to the prime factor 3.
  • Reduction Warning: For ⁶/₉₀, reducing to lowest terms yields ¹/₁₅ = ¹/_{(3 · 5)}. Because the denominator still contains 3, the decimal is recurring (0.06). Always reduce the fraction before factoring the denominator.

x

• The 0.9 = 1 Identity: Let x = 0.9 ⇒ 10x = 9.9 ⇒ 9x = 9 ⇒ x = 1.
• Period Length Theorem: For ^p/_q with gcd(q, 10) = 1, the length r of the repetend divides Euler’s totient φ(q) and is the smallest positive integer r satisfying:
  10^r ≡ 1 (mod q)
  For instance, the period length of ¹/₇ is 6 because 10⁶ ≡ 1 (mod 7) and no smaller positive integer exponent satisfies the congruence.
• Factor Pairing Property: If x = ^N/_99…9, then N plus its 9’s complement equals 99…9. Consequently, ¹/₁₁ = 0.09 and ¹⁰/₁₁ = 0.90.

2. Comparative Matrix & Conceptual Distinctions

Diagnostic Rule: A prime denominator does not ensure a recurring expansion. Denominators q = 2 and q = 5 produce strictly terminating values.

3. High-Yield Data Anchors & Memorization Benchmarks

A. The Instant Conversion Invariants

B. Deceptive Composites & Digit Misalignments

• 0.1666… = 0.16 = ^{(16 − 1)}/₉₀ = ¹/₆. Do not evaluate this as ¹⁶/₉₉. Note that 0.16 and 0.166 are mathematically identical (0.1666…), whereas 0.16 = 0.161616… = ¹⁶/₉₉.
• 0.090909… = 0.09 = ¹/₁₁, which is distinct from 0.099 = 0.1000… = 0.1.
• 0.9 = 1, whereas 0.09 = 0.0999… = ^{(09 − 0)}/₉₀ = ⁹/₉₀ = 0.1.

C. Algebraic Invariant Anchors for Fast Simplification

• 99 = 9 × 11
• 999 = 27 × 37
• 9999 = 9 × 1111 = 9 × 11 × 101
• 99999 = 9 × 11111 = 9 × 41 × 271

Pattern Trigger: If the numerator features values such as 37, 74, or 111, anticipate factorization against denominators containing 999 or 9999.

4. Standard Algorithmic Protocols (Decision Logic)

Fast Decimal-to-Fraction Protocol (≤ 15 Seconds)

To convert any decimal value D into an irreducible rational fraction ^p/_q, follow this 4-step sequence:

1. Step 1: Isolate the Integer Component
   D = A + Decimal Part
   (Separating the integer component A prevents handling unwieldy numerators during factorization).
2. Step 2: Determine Repetend and Non-Repetend Counts
   • Identify n = number of non-repeating digits after the decimal point (Gap).
   • Identify r = number of repeating digits under the repetend bar (Bar).
3. Step 3: Construct the Fractional Quotient
   • Case I: Terminating (r = 0)
     Fraction = ^{(All digits after decimal)} / _10ⁿ
   • Case II: Pure Recurring (n = 0, r ≥ 1)
     Fraction = ^{(Repetend digits)} / _{[99…9 (r nines)]}
   • Case III: Mixed Recurring (n ≥ 1, r ≥ 1)
     Fraction = ^{[(All digits after decimal) − (Non-repeating prefix)]} / _{[99…9 (r nines) 00…0 (n zeros)]}
4. Step 4: Recombine and Simplify
   D = A + ^Numerator/_Denominator → (Reduce by gcd) → ^p/_q

Step-by-Step Worked Demonstration Models

Model 1: Terminating Decimal (r = 0)

Convert D = 0.375 to ^p/_q:

• A = 0, n = 3, r = 0
• Fraction = ³⁷⁵/_10³ = ³⁷⁵/₁₀₀₀
• Divide numerator and denominator by gcd(375, 1000) = 125:
  ^{(375 ÷ 125)} / _{(1000 ÷ 125)} = ³/₈

Model 2: Pure Recurring Decimal (n = 0)

Convert D = 0.45 to ^p/_q:

• A = 0, n = 0, r = 2 (repeating block: 45)
• Denominator = 99 (2 nines)
• Fraction = ⁴⁵/₉₉
• Divide numerator and denominator by gcd(45, 99) = 9:
  ^{(45 ÷ 9)} / _{(99 ÷ 9)} = ⁵/₁₁

Model 3: Mixed Recurring Decimal with Zero Integer (A = 0, n ≥ 1)

Convert D = 0.1635 to ^p/_q:

• A = 0, n = 2 (non-repeating digits 16), r = 2 (repeating block 35)
• Numerator = 1635 − 16 = 1619
• Denominator = 9900 (r = 2 nines, n = 2 zeros)
• D = ¹⁶¹⁹/₉₉₀₀
  (Since gcd(1619, 9900) = 1, the fraction is in simplest form).

Model 4: Mixed Recurring with Non-Zero Integer Part (A > 0)

Convert D = 1.42 to ^p/_q:

• Method A (Integer Separation):
  • A = 1, n = 1 (digit 4), r = 1 (digit 2)
  • Numerator = 42 − 4 = 38
  • Denominator = 90 (1 nine, 1 zero)
  • D = 1 + ³⁸/₉₀ = 1 + ¹⁹/₄₅ = ^{(45 + 19)}/₄₅ = ⁶⁴/₄₅
• Method B (Direct Single-Fraction Conversion):
  • Numerator = (Entire number without decimal) − (Digits before bar) = 142 − 14 = 128
  • Denominator = 90
  • D = ¹²⁸/₉₀ = ⁶⁴/₄₅

5. Standard Question Typologies & Analytical Solutions

Type A: Parity & Multi-Variable Constraints

Archetype: Pure recurring fractions of the form ^m/₉ and ⁿ/₉₉ subject to range constraints.

Problem: Consider the numbers x = 0.a and y = 0.b, where a and b are single non-zero digits (1 ≤ a, b ≤ 9). If x + y = 0.c, where c is also a single digit, find the number of distinct ordered pairs (a, b) that satisfy this equation.

Analytical Solution:

Step 1: Convert to Fractional Form
x = ^a/₉, y = ^b/₉ ⇒ x + y = ^{(a + b)}/₉
The target sum is 0.c = ^c/₉, where 1 ≤ c ≤ 9.

Step 2: Establish Boundary Constraints
For the sum ^{(a + b)}/₉ to remain strictly in the form 0.c, the sum of the digits must satisfy:
a + b = c ≤ 9
If a + b > 9, the sum becomes 1 + ^{(a + b − 9)}/₉ = 1.a + b − 9, which is no longer of the form 0.c.

Step 3: Count Valid Integer Pairs
For each integer k = a + b such that 2 ≤ k ≤ 9, the number of valid pairs (a, b) with a, b ≥ 1 is (k − 1):

Total Pairs = ∑_k=2⁹ (k − 1) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

Step 4: Boundary Evaluation on c = 9

• If c = 9 is permissible (0.9 = 1), there are 36 pairs.
• If the problem strictly defines 0.c < 1 (excluding c = 9), then a + b ≤ 8:
  ∑_k=2⁸ (k − 1) = 1 + 2 + ··· + 7 = 28 pairs

Exam Shortcut: In additions involving pure single-digit repetends 0.a + 0.b, convert directly to denominators of 9. The non-carry condition a + b ≤ 9 governs validity.

Type B: Statement Evaluation & Conceptual Validity

Archetype: Validating termination properties and the 0.9 = 1 identity.

Problem: Consider the following statements:

1. The value of 0.9 is strictly less than 1.
2. The rational equivalent of 1.42 in lowest terms has a denominator divisible by 3.
3. Every fraction with a denominator of 40 yields a terminating decimal expansion.

Which of the above statements are correct?

Analytical Solution:

Statement 1:
Let x = 0.9 ⇒ 10x = 9.9 ⇒ 9x = 9 ⇒ x = 1. Thus, 0.9 = 1 exactly. (Incorrect)

Statement 2:
1.42 = 1 + ^{(42 − 4)}/₉₀ = 1 + ³⁸/₉₀ = ¹²⁸/₉₀ = ⁶⁴/₄₅
The simplified denominator is 45 = 3² × 5, which is divisible by 3. (Correct)

Statement 3:
The prime factorization of the denominator is 40 = 2³ × 5¹. Since the prime factors are exclusively from the set {2, 5}, any fraction ^p/₄₀ will simplify to a denominator of the form 2^a5^b. Consequently, the expansion always terminates. (Correct)

Correct Answer: Statements 2 and 3 are correct.

Type C: Existence & Combinatorial Counting

Archetype: Counting distinct rational outputs generated by digit combinations.

Problem: How many distinct rational numbers of the form 0.pq can be formed where p and q are distinct non-zero single digits (p, q ∈ {1, 2, …, 9} with p ≠ q)?

Analytical Solution:

Step 1: Algebraic Representation
0.pq = ^{(10p + q − p)}/₉₀ = ^{(9p + q)}/₉₀

Step 2: Injectivity & Uniqueness Check
We evaluate whether two distinct digit pairs (p₁, q₁) ≠ (p₂, q₂) can yield the same value:
9p₁ + q₁ = 9p₂ + q₂ ⇒ 9(p₁ − p₂) = q₂ − q₁
Because q₁, q₂ ∈ {1, …, 9}, the difference |q₂ − q₁| ≤ 8.
The only multiple of 9 in this range is 0, meaning p₁ − p₂ = 0 ⇒ p₁ = p₂ and q₁ = q₂.
Thus, every distinct ordered pair (p, q) generates a unique numerator 9p + q.

Step 3: Combinatorial Computation
Selecting 2 distinct digits from 9 without replacement:
P(9, 2) = 9 × 8 = 72 distinct rational values

Type D: Algebraic Simplification with Cyclics

Archetype: Factor cancellation using 999 and prime period properties.

Problem: Evaluate the exact value of:
E = (0.037 × 27) + 0.83

Analytical Solution:

Step 1: Convert Individual Terms
0.037 = ³⁷/₉₉₉
Recall the algebraic identity 999 = 27 × 37:
0.037 × 27 = ^{(37 × 27)}/₉₉₉ = ⁹⁹⁹/₉₉₉ = 1

Step 2: Convert the Mixed Recurring Term
0.83 = ^{(83 − 8)}/₉₀ = ⁷⁵/₉₀ = ⁵/₆

Step 3: Sum the Terms
E = 1 + ⁵/₆ = ¹¹/₆ = 1.83

Type E: Modular Periodicity & Co-Prime Denominators

Archetype: Counting coprime period lengths and irreducible fractions.

Problem: Let N = 0.ab where ab is a two-digit integer (10 ≤ ab ≤ 99). When expressed as an irreducible fraction ^p/_q, the denominator is given by q = ⁹⁹/_{gcd(ab, 99)}. How many two-digit integers ab ∈ [10, 99] are co-prime to 99?

Analytical Solution:

Step 1: Factorization and Totient Function
99 = 3² × 11
The number of integers in [1, 99] co-prime to 99 is given by Euler’s totient function φ(99):
φ(99) = 99 × (1 − ¹/₃) × (1 − ¹/₁₁) = 99 × ²/₃ × ¹⁰/₁₁ = 60

Step 2: Exclude Single-Digit Values
The single-digit values in [1, 9] co-prime to 99 are 1, 2, 4, 5, 7, 8 (a total of 6 values).

Step 3: Compute the Target Count
Count = φ(99) − 6 = 60 − 6 = 54
(For these 54 values, the denominator remains strictly q = 99).

Type F: Composite Arithmetic & Denominator LCM

Archetype: Addition of mixed and pure recurring decimals with fractional reduction.

Problem: Evaluate the sum S = 0.16 + 0.27. Express the result as an irreducible fraction ^p/_q and find (q − p).

Analytical Solution:

Step 1: Fractional Conversion
0.16 = ^{(16 − 1)}/₉₀ = ¹⁵/₉₀ = ¹/₆
0.27 = ²⁷/₉₉ = ³/₁₁

Step 2: Compute the Sum
S = ¹/₆ + ³/₁₁ = ^{(11 + 18)}/₆₆ = ²⁹/₆₆

Step 3: Evaluate Invariant
Since gcd(29, 66) = 1, we have p = 29 and q = 66:
q − p = 66 − 29 = 37

6. The Examiner Trap Matrix

7. Golden Rules & Instant Exam Triggers

Final Retention Formula: “9 for Bar, 0 for Gap, Subtract the Prefix”
Rational Fraction = ^{[(All digits after decimal) − (Non-repeating prefix)]} / _{[99…9 (Length of Bar) 00…0 (Length of Gap)]}