Real Numbers & Integer Properties for CSAT

1. Theoretical Foundation & Mathematical Underpinnings

Formal Definition & Scope

The quantitative framework rests upon the Real Number System (ℝ) — the set of all quantities representable on a one-dimensional continuous number line. It forms a mutually exclusive and exhaustive partition:

ℝ = ℚ ∪ ℚ^c  and  ℚ ∩ ℚ^c = ∅

where ℚ denotes the set of Rational Numbers, and ℚ^c denotes the set of Irrational Numbers (ℝ \ ℚ).

1. Rational Numbers (ℚ):

ℚ = { ^p/_q | p, q ∈ ℤ, q ≠ 0 }

A real number is rational if and only if its decimal expansion is either Terminating or Non-Terminating but Repeating (Recurring).

• Terminating: ¹/₂ = 0.5, ⁷/₈ = 0.875, ¹³/₄₀₀ = 0.0325
• Repeating: ¹/₃ = 0.3, ¹/₇ = 0.142857, ¹/₆ = 0.16

2. Irrational Numbers (ℚ^c):

Numbers that cannot be expressed in the form ^p/_q where p, q ∈ ℤ and q ≠ 0. Defining property: the decimal expansion is Non-Terminating and Non-Repeating. No finite sequence of digits recurs indefinitely.

• Canonical Examples: √2, √3, √5, √7, … (surds of non-perfect squares); transcendental constants such as π, e; and constructed decimals such as 0.101001000100001…

3. Integers (ℤ):

A critical subset of ℚ where the denominator in reduced form is q = 1.

ℤ = {…, -3, -2, -1, 0, 1, 2, 3, …}

ℤ⁺ = {1, 2, 3, …} (Natural Numbers / Positive Integers)

ℤ^– = {-1, -2, -3, …} (Negative Integers)

The “Why” Behind the Math — Proofs for Retention

Why 1: Why does a rational number always terminate or repeat?
When dividing an integer p by a positive integer q, the remainder at each step of standard long division must strictly belong to the set:

R ∈ {0, 1, 2, …, q – 1}

This yields at most q distinct possible remainders:

1. If the remainder becomes 0 at any step, the division process halts, producing a terminating decimal.
2. If the remainder never reaches 0, then by the Pigeonhole Principle, a remainder must recur within at most q subsequent division steps. Once a remainder repeats, the exact sequence of quotient digits generated thereafter repeats identically.

Hence, every rational representation yields a periodic or finite decimal, whereas irrationality corresponds to an infinite, non-repeating sequence of remainders.

Why 2: Why is a rational decimal terminating if and only if its reduced denominator contains only prime factors of 2 and 5?
Any terminating decimal with k decimal places can be expressed as a base-10 fraction:

^N/_10^k = ^N/_{(2 · 5)^k} = ^N/_{(2^k · 5^k)}

for some integer N and non-negative integer k ≥ 0. After cancelling all common factors between numerator and denominator:

^p/_q = ^{(N / gcd(N, 10k))}/_{(10^k / gcd(N, 10^k))}

The simplified denominator q cannot contain any prime factors other than 2 and 5.

Verification Test:

• ⁷/₄₀ = ⁷/_{(2³ · 5)} ⇒ Denominator has only primes 2 and 5 ⇒ Terminating (0.175).
• ⁵/₁₄ = ⁵/_{(2 · 7)} ⇒ Denominator contains the prime factor 7 ⇒ Non-terminating repeating (0.3571428).

Note: This test is valid only after the fraction is reduced to its lowest terms.

Why 3: Why is Rational + Irrational = Irrational always true?

Proof by Contradiction:
Let r ∈ ℚ and s ∈ ℚ^c. Assume, for contradiction, that their sum is rational:

r + s = t  where  t ∈ ℚ

Isolating the irrational term yields:

s = t – r

Because the set of rational numbers ℚ is closed under subtraction, the difference of two rational numbers (t – r) must be rational:

(t – r) ∈ ℚ ⇒ s ∈ ℚ

This directly contradicts the initial premise that s ∈ ℚ^c. Thus, the sum t must be irrational.

Corollary Trap: This closure property does not extend to the sum of two irrational numbers (ℚ^c + ℚ^c). The set of irrational numbers is not closed under addition:

(2 + √3) + (2 – √3) = 4 ∈ ℚ

Why 4: Why is Zero (0) Even? Two Rigorous Proofs

• Algebraic Definition: An integer n is even if there exists an integer k ∈ ℤ such that n = 2k. Setting k = 0:
  2k = 2(0) = 0 ⇒ 0 ∈ 2ℤ (Even)
• Sequential Parity Invariance: Parity strictly alternates across consecutive integers on the number line:
  …, -3 (Odd), -2 (Even), -1 (Odd), 0 (?), 1 (Odd), 2 (Even), …
  To preserve strict alternating parity (Odd ↔ Even), 0 must be Even.

Consequence: Zero (0) is an even integer, but it is neither positive nor negative; it serves as the neutral sign boundary on ℝ.

Key Theorems & Properties

• Closure Properties:
  • ℚ + ℚ = ℚ and ℚ × ℚ = ℚ (Closed under addition and multiplication)
  • If r ∈ ℚ \ {0} and s ∈ ℚ^c, then r · s ∈ ℚ^c
  • ℚ^c + ℚ^c = Indeterminate (ℚ or ℚ^c)
  • ℚ^c × ℚ^c = Indeterminate (ℚ or ℚ^c)

• Parity Arithmetic (0 is Even):
  • Addition / Subtraction:
    Even ± Even = Even,   Odd ± Odd = Even,   Even ± Odd = Odd
  • Multiplication:
    Even × (Any Integer) = Even,   Odd × Odd = Odd
    (An odd product requires every factor in the product to be odd.)
  • Operational Examples:
    4 (Even) + 0 (Even) = 4 (Even)
    (-3) (Odd) × 0 (Even) = 0 (Even)
• Independence of Sign and Parity: Sign and parity are independent mathematical properties. For instance, -4 is Negative and Even, while -3 is Negative and Odd.

2. Comparative Matrix & Conceptual Distinctions

Table 1: Rational vs. Irrational — The Decisive Divide

Table 2: Integer Terminology Spectrum — The Zero Boundary

Key Takeaway: In standardized competitive examinations, problem stems frequently distinguish between Positive and Non-Negative to test boundary-condition awareness.

Table 3: The π versus ²²/₇ Fallacy

3. High-Yield Data Anchors & Memorization Benchmarks

A. Terminating Decimal Test (Denominator Factorization)

Given a rational fraction ^p/_q in irreducible form (gcd(p, q) = 1):

• If q = 2^a · 5^b for non-negative integers a, b ≥ 0 ⇒ Terminating Decimal
• If q contains any prime factor p_k ∉ {2, 5} ⇒ Non-Terminating Repeating Decimal

Examples:

• ³/₈₀ ⇒ 80 = 2⁴ · 5¹ ⇒ Terminates (0.0375)
• ⁹/₄₄ ⇒ 44 = 2² · 11¹ ⇒ Repeats (0.2045)

B. High-Frequency Deceptive Numbers

C. Algebraic Parity Invariants

• Square Invariant: n² ≡ n (mod 2) (An integer and its square share the identical parity).
• Consecutive Pair Invariant: n(n + 1) is always Even for all n ∈ ℤ (Every pair of consecutive integers contains at least one factor of 2).
• Consecutive Triplet Invariant: n³ – n = (n – 1)n(n + 1) is always divisible by 6 (Contains at least one multiple of 2 and one multiple of 3).

4. Standard Algorithmic Protocols (Decision Logic)

Protocol 1: Systematic Rationality & Decimal Classification

When analyzing any given number N:

1. Algebraic Representation Check:
   • Can N be represented as ^p/_q with p, q ∈ ℤ and q ≠ 0?
   • Yes: N ∈ ℚ (Rational). Proceed to Step 2.
   • No: N ∈ ℚ^c (Irrational). Proceed to Step 3.
2. Rational Decimal Expansion Analysis:
   • Reduce ^p/_q completely such that gcd(p, q) = 1.
   • Factor the denominator: q = 2^a · 5^b · k.
   • If k = 1, the decimal expansion is Terminating.
   • If k > 1 (contains prime factors other than 2 or 5), the decimal expansion is Non-Terminating Recurring.
3. Irrational / Radical Form Analysis:
   • If N = √k for an integer k:
   • If k = m² for some m ∈ ℤ, then N = m ∈ ℚ (Rational).
   • If k is not a perfect square, N ∈ ℚ^c (Irrational).

Step-by-Step Numerical Example:
Classify the value N = ²¹/₂₈₀.

• Step 1: Reduce to lowest terms: gcd(21, 280) = 7 ⇒ ^(21/7)/_(280/7) = ³/₄₀.
• Step 2: Factor the simplified denominator: 40 = 2³ · 5¹.
• Step 3: Since all prime factors belong to {2, 5}, N produces a terminating decimal (0.075).

Protocol 2: Parity and Zero-Product Consecutive Integer Resolution

For problems constrained by P × Q × R = 0 over consecutive integers P < Q < R:

1. Zero-Product Condition:
   P · Q · R = 0 ⇒ 0 ∈ {P, Q, R}
2. Enumerate Feasible Consecutive Triplets:
   • Case 1 (R = 0): Triple A = {-2, -1, 0}
   • Case 2 (Q = 0): Triple B = {-1, 0, 1}
   • Case 3 (P = 0): Triple C = {0, 1, 2}
3. Evaluate Parity Constraints on the Sum:
   • Sum(A) = -2 + (-1) + 0 = -3 (Odd)
   • Sum(B) = -1 + 0 + 1 = 0 (Even)
   • Sum(C) = 0 + 1 + 2 = 3 (Odd)
4. Parity Elimination:
   Match the required sum parity (remembering that 0 is Even) to select the unique valid set.

Step-by-Step Numerical Example:
Find three consecutive integers whose product is 0 and whose sum is strictly odd and positive.

• Step 1: Identify the three candidate triplets: {-2, -1, 0}, {-1, 0, 1}, and {0, 1, 2}.
• Step 2: Calculate the sums: -3, 0, and +3.
• Step 3: Apply the constraint (Sum > 0 and Sum is Odd) ⇒ triplet {0, 1, 2} is the unique solution with P = 0, Q = 1, R = 2.

Protocol 3: Quantitative Terminology Decoder

To avoid boundary-value errors when translating word problems into algebraic domains:

1. Look for the Prefix “Non-“:
   • Non-Negative: Condition is x ≥ 0 ⇒ Includes 0 ({0, 1, 2, …}).
   • Non-Positive: Condition is x ≤ 0 ⇒ Includes 0 ({0, -1, -2, …}).
2. Check for Strict Positivity / Negativity:
   • Positive: Condition is x > 0 ⇒ Excludes 0 ({1, 2, 3, …}).
   • Negative: Condition is x < 0 ⇒ Excludes 0 ({-1, -2, -3, …}).
3. Check Parity Descriptors:
   • Even Integer: Condition is x ∈ 2ℤ ⇒ Includes 0 ({…, -2, 0, 2, …}).

5. Standard Question Typologies & Analytical Solutions

Type A: Parity & Multi-Variable Constraints

Core Principle: Zero is Even; Parity of Consecutive Integers

Question:
Let P, Q, R be three consecutive integers such that P < Q < R. If P × Q × R = 0 and P + Q + R is an even integer, what is the value of Q?
(a) -1
(b) 0
(c) 1
(d) Cannot be determined

Analytical Solution:

• Step 1 (Constraint Setup): Since P, Q, R are consecutive integers and their product is zero, the number 0 must be one of the three integers.
• Step 2 (Exhaustive Case Listing):
  Window 1: P = -2, Q = -1, R = 0
  Window 2: P = -1, Q = 0, R = 1
  Window 3: P = 0, Q = 1, R = 2
• Step 3 (Sum and Parity Evaluation):
  Window 1: P + Q + R = -2 + (-1) + 0 = -3 (Odd)
  Window 2: P + Q + R = -1 + 0 + 1 = 0 (Even, since 0 = 2 × 0)
  Window 3: P + Q + R = 0 + 1 + 2 = 3 (Odd)
• Step 4 (Conclusion): Only Window 2 satisfies the condition that the sum is an even integer. In this window, the middle integer is Q = 0.

Correct Answer: (b) 0
Exam Strategy: When a product of consecutive integers is 0, immediately list the three potential triplets containing 0. Evaluate the parity of the sums to eliminate invalid windows in seconds.

Type B: Statement Evaluation & Set Closure

Core Principle: Set Inclusions and Irrational Counterexamples

Question:
Consider the following statements:
1. Every integer is a rational number, but every rational number is not an integer.
2. The sum of two irrational numbers is always an irrational number.
3. π is an irrational number, whereas ²²/₇ is a rational number.

Which of the statements given above is/are correct?
(a) 1 and 2 only
(b) 1 and 3 only
(c) 3 only
(d) 1, 2, and 3

Analytical Solution:

• Evaluation of Statement 1: Any integer z ∈ ℤ can be written as ^z/₁, where z, 1 ∈ ℤ and 1 ≠ 0, so ℤ ⊂ ℚ. Conversely, fractions such as ¹/₂ ∈ ℚ do not belong to ℤ. Statement 1 is Correct.
• Evaluation of Statement 2: Consider the two irrational numbers x = 3 + √5 and y = 3 – √5.
  x + y = (3 + √5) + (3 – √5) = 6 = ⁶/₁ ∈ ℚ
  Because the sum can be rational, the sum of two irrationals is not always irrational. Statement 2 is Incorrect.
• Evaluation of Statement 3: π is non-terminating and non-repeating (transcendental irrational), whereas ²²/₇ is explicitly a ratio of two integers with a repeating decimal expansion (3.142857). Statement 3 is Correct.

Correct Answer: (b) 1 and 3 only
Exam Strategy: Statements containing universal quantifiers like “always” or “never” regarding irrational operations can frequently be disproven using conjugate pairs (a + √b) and (a – √b).

Type C: Boundary Counting & Non-Negative Constraints

Core Principle: Zero Inclusion in Non-Negative Sets

Question:
If x and y are non-negative integers such that x + y = 3, how many distinct ordered pairs (x, y) are possible?
(a) 2
(b) 3
(c) 4
(d) Infinite

Analytical Solution:

• Step 1 (Domain Interpretation): The term “non-negative integers” means x, y ∈ {0, 1, 2, 3, …}.
• Step 2 (Systematic Enumeration):
  (x, y) ∈ {(0, 3), (1, 2), (2, 1), (3, 0)}
  This gives exactly 4 distinct ordered pairs.
• Step 3 (Trap Analysis): If the stem had specified “positive integers”, x, y ≥ 1, which would leave only (1, 2) and (2, 1) (2 pairs). The inclusion of zero provides 2 additional boundary pairs.

Correct Answer: (c) 4
Exam Strategy: For equations of the form x + y = n:
• Number of solutions in non-negative integers = n + 1
• Number of solutions in positive integers = n – 1

Type D: Algebraic Expressions & Rational-Irrational Invariants

Core Principle: Preservation of Irrationality Under Non-Zero Operations

Question:
Let a be a non-zero rational number and b be an irrational number. Consider the following statements:
1. a + b is irrational.
2. a × b is irrational.
3. b² is always irrational.

Which of the statements given above is/are correct?
(a) 1 and 2 only
(b) 1 and 3 only
(c) 2 and 3 only
(d) 1, 2, and 3

Analytical Solution:

• Statement 1: If a + b = q ∈ ℚ, then b = q – a ∈ ℚ, which contradicts b ∈ ℚ^c. Thus, a + b ∈ ℚ^c. Statement 1 is Correct.
• Statement 2: If a · b = q ∈ ℚ, then since a ≠ 0, we have b = ^q/_a ∈ ℚ, which is a contradiction. Thus, a · b ∈ ℚ^c. Statement 2 is Correct.
• Statement 3: Let b = √3 ∈ ℚ^c. Then:
  b² = (√3)² = 3 ∈ ℚ
  Since b² can be rational, Statement 3 is Incorrect.

Correct Answer: (a) 1 and 2 only
Key Invariant: Operations between ℚ \ {0} and ℚ^c yield irrational outcomes (ℚ ± ℚ^c = ℚ^c and ℚ^* × ℚ^c = ℚ^c). However, operations within ℚ^c itself (ℚ^c ± ℚ^c or ℚ^c × ℚ^c) are indeterminate.

Type E: Parity Proofs & Definitions

Core Principle: Structural Form of Even Numbers and Parity of Zero

Question:
Consider the following statements:
• Statement-I: If n is an even integer, then n² + n is always even.
• Statement-II: Zero is not an even integer but is a whole number.

Which one of the following options is correct?
(a) Both Statement-I and Statement-II are correct
(b) Statement-I is correct but Statement-II is incorrect
(c) Statement-I is incorrect but Statement-II is correct
(d) Both Statement-I and Statement-II are incorrect

Analytical Solution:

• Analysis of Statement-I: Let n = 2k where k ∈ ℤ.
  n² + n = (2k)² + 2k = 4k² + 2k = 2(2k² + k)
  Since (2k² + k) ∈ ℤ, the expression n² + n is divisible by 2 and is therefore always even. (Alternatively, Even² + Even = Even + Even = Even). Statement-I is Correct.
• Analysis of Statement-II: 0 = 2(0), so 0 satisfies the definition of an even integer. It is also a whole number ({0, 1, 2, …}). The assertion that “Zero is not an even integer” is false. Statement-II is Incorrect.

Correct Answer: (b) Statement-I is correct but Statement-II is incorrect

Type F: Terminating Decimal Recognition

Core Principle: Denominator Prime Factorization

Question:
Which of the following fractions will produce a terminating decimal expansion?
1. ¹³/₈₀
2. ⁷/₂₄
3. ³/₁₆₀

Select the correct answer:
(a) 1 and 2 only
(b) 1 and 3 only
(c) 2 and 3 only
(d) 1, 2, and 3

Analytical Solution:
Ensure all fractions are reduced, then factor each denominator:

1. ¹³/₈₀: gcd(13, 80) = 1. The denominator factors as 80 = 2⁴ · 5¹. Prime factors are exclusively from {2, 5} ⇒ Terminating (0.1625).
2. ⁷/₂₄: gcd(7, 24) = 1. The denominator factors as 24 = 2³ · 3¹. Contains the prime factor 3 ⇒ Non-terminating recurring (0.2916).
3. ³/₁₆₀: gcd(3, 160) = 1. The denominator factors as 160 = 2⁵ · 5¹. Prime factors are exclusively from {2, 5} ⇒ Terminating (0.01875).

Correct Answer: (b) 1 and 3 only

6. The Examiner Trap Matrix

7. Golden Rules & Instant Exam Triggers