Case analysis

This chapter critically examines various methodologies for case analysis, a fundamental skill in constructing rigorous mathematical proofs and solving complex problems. Mastery of these techniques, including identifying hidden assumptions, executing exhaustive case splits, and applying constraint-based elimination, is essential for demonstrating advanced logical reasoning and problem-solving capabilities required for the BS Hons course.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Hidden assumptions | | 2 | Exhaustive case split | | 3 | Constraint-based elimination |

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We begin with Hidden assumptions.

Part 1: Hidden assumptions

Hidden Assumptions

Overview

Many incorrect arguments fail not because of bad algebra, but because they silently use something that was never given. These silent steps are called hidden assumptions. In olympiad and CMI-style reasoning, finding the hidden assumption is often the difference between a valid proof and a convincing-looking mistake. ---

Learning Objectives

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Core Idea

Examples:

• dividing by an expression without checking whether it is zero

• taking square roots and assuming signs

• assuming a diagram is to scale

• assuming variables are integers when only real numbers were given

• assuming a process always terminates

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Very Common Hidden Assumptions

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Domain Assumptions

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Hidden Assumptions in Equations

This is one of the standard examples of a hidden assumption. ---

Hidden Assumptions in Geometry

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Hidden Assumptions in Counting and Number Theory

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Minimal Worked Examples

Example 1 Claim: If $\qquad x^2=9$, then $\qquad x=3$. This argument has a hidden assumption that $x$ is positive. The correct conclusion is $\qquad x=\pm 3$ --- Example 2 Claim: If two triangles look symmetric in a diagram, then the corresponding sides are equal. This uses a hidden assumption that the figure is drawn accurately enough to prove symmetry. A diagram may suggest symmetry without it being given. ---

How to Test for Hidden Assumptions

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following is a hidden assumption in the step “divide both sides by $x-2$”?" options=["$x$ is positive","$x-2\ne 0$","$x$ is an integer","$x>2$"] answer="B" hint="Division requires the divisor to be nonzero." solution="To divide by $x-2$, we must know that $\qquad x-2 \ne 0$ So the hidden assumption is that $x \ne 2$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="A student claims that from $x^2=16$ it follows that $x=4$. How many real solutions were missed?" answer="1" hint="Check all real square roots of $16$." solution="The equation $\qquad x^2=16$ has the real solutions $\qquad x=4 \text{ and } x=-4$ The student kept only $x=4$, so exactly one real solution was missed. Hence the answer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following are common hidden assumptions?" options=["Assuming a denominator is nonzero before dividing","Assuming a diagram is not to scale","Assuming variables are integers when the argument uses parity","Assuming an expression has a square root only when it is nonnegative"] answer="A,C,D" hint="One option is the opposite of the usual geometry warning." solution="1. True.

False. The common hidden assumption is the opposite: assuming the diagram is to scale.

True.

True.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Explain why the argument “from $a=b$, divide by $a-b$” is invalid." answer="Because $a-b=0$ when $a=b$, so division is not allowed." hint="Substitute the assumption $a=b$ directly into the divisor." solution="If $\qquad a=b$ then $\qquad a-b=0$ So any step that divides by $a-b$ is actually dividing by zero, which is undefined. Therefore the argument is invalid. Hence the flaw is exactly the hidden assumption that $\qquad a-b \ne 0$." ::: ---

Summary

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Part 2: Exhaustive case split

Exhaustive Case Split

Overview

An exhaustive case split solves a problem by dividing all possibilities into separate, non-overlapping cases and handling each one. This is a core proof and problem-solving method in olympiad mathematics. The power of the method comes from two requirements:

• every possible situation must belong to some case,

• no case should overlap confusingly with another.

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Learning Objectives

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Core Idea

Good case splits are:

• complete,

• simple,

• easy to analyze,

• logically clean.

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Standard Ways to Split Cases

These are the most natural splits in contest-style problems. ---

How to Check Exhaustiveness

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Minimal Worked Examples

Example 1 Show that for any integer $n$, the number $n(n+1)$ is even. Split into two cases:

• if $n$ is even, then $n(n+1)$ is even

• if $n$ is odd, then $n+1$ is even, so $n(n+1)$ is even

These two cases are exhaustive because every integer is either even or odd. --- Example 2 Solve $|x|=x$. Split into cases:

• if $x\ge 0$, then $|x|=x$, so every such $x$ works

• if $x<0$, then $|x|=-x\ne x$ unless $x=0$, but $0$ is not in this case

Hence the solution set is $\qquad x\ge 0$ ---

Case Splits in Inequalities

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Case Splits in Counting

The total is then the sum of the counts of all cases. ::: ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following is an exhaustive case split for an integer $n$?" options=["$n>0$ or $n<0$","$n$ even or $n$ odd","$n<5$ or $n>5$","$n$ prime or $n$ composite"] answer="B" hint="An exhaustive split must include every integer." solution="Every integer is either even or odd, and not both. So this is exhaustive and disjoint. Option A misses $n=0$. Option C misses $n=5$. Option D misses $1$ and also does not cover all integers cleanly. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="How many sign-based cases are needed to split a real variable $x$ exhaustively?" answer="3" hint="Think of positive, zero, and negative." solution="A real number can be:

• positive,

• zero,

• negative.

So the exhaustive sign-based split has $\boxed{3}$ cases." ::: :::question type="MSQ" question="Which of the following are good properties of a case split?" options=["Every valid input belongs to some case","Cases should ideally be disjoint","It is acceptable to ignore a rare case if the main argument works","A good split should simplify the reasoning"] answer="A,B,D" hint="A proof must cover all valid cases." solution="1. True.

True.

False. A proof must cover every valid case.

True.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Prove that for every integer $n$, the number $n(n+1)$ is even." answer="True by splitting into the cases $n$ even and $n$ odd." hint="Use parity as the case split." solution="We split into two cases. Case 1: $n$ is even. Then $n(n+1)$ is a product containing an even factor, so it is even. Case 2: $n$ is odd. Then $n+1$ is even, so again $n(n+1)$ is a product containing an even factor, hence it is even. These two cases are exhaustive because every integer is either even or odd. Therefore, for every integer $n$, the number $\boxed{n(n+1)}$ is even." ::: ---

Summary

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Part 3: Constraint-based elimination

Constraint-based Elimination

Overview

Constraint-based elimination is a structured problem-solving method in which impossible cases are removed step by step until the valid possibilities become small enough to analyze completely. It is especially useful in logic puzzles, integer problems, digit problems, and finite search problems. ---

Learning Objectives

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Core Idea

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Main Sources of Elimination

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Standard Workflow

This avoids both overcounting and missed cases. ---

Minimal Worked Example

Example 1 Find all positive integers $x,y$ such that $\qquad x+y=10,\quad x>y,\quad x \text{ is even}$ Since $x>y$ and $x+y=10$, we get $\qquad x>5$ Since $x$ is even and positive and at most $9$, the possibilities are $\qquad x=6,\ 8$ Then:

• if $x=6$, then $y=4$

• if $x=8$, then $y=2$

So the only solutions are $\qquad (6,4),\ (8,2)$ ---

Why Elimination Works

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Organizing Cases Correctly

For example, in digit problems:

• first split by last digit if divisibility is involved,

• not by random digit choice.

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="A positive integer $n$ is less than $20$, odd, divisible by $3$, and leaves remainder $1$ when divided by $4$. Which of the following is $n$?" options=["$9$","$15$","$13$","$17$"] answer="A" hint="Check the constraints one by one." solution="Among the options, only $9$ is odd, divisible by $3$, and satisfies $\qquad 9 \equiv 1 \pmod 4$. Therefore the correct option is $\boxed{A}$." ::: :::question type="NAT" question="Let $x,y$ be positive integers with $x+y=12$, $x>y$, and $x$ odd. How many possible values can $x$ take?" answer="3" hint="Use $x>6$ first." solution="Since $x>y$ and $x+y=12$, we must have $\qquad x>6$. Also $x$ is odd and positive, so the possible values are $\qquad 7,9,11$. Thus there are $\boxed{3}$ possible values of $x$." ::: :::question type="MSQ" question="Which of the following are standard valid elimination tools?" options=["Parity","Bounds","Divisibility","Ignoring one constraint temporarily and never returning to it"] answer="A,B,C" hint="Think of standard filters." solution="Parity, bounds, and divisibility are all standard elimination tools. Ignoring a constraint permanently is invalid. Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Find all positive integers $n<30$ such that $n$ is divisible by $4$ and $n$ leaves remainder $1$ when divided by $3$." answer="$4,16,28$" hint="List multiples of $4$ below $30$ and test mod $3$." solution="The multiples of $4$ less than $30$ are $\qquad 4,8,12,16,20,24,28$. Now reduce each mod $3$: $\qquad 4\equiv 1,\ 8\equiv 2,\ 12\equiv 0,\ 16\equiv 1,\ 20\equiv 2,\ 24\equiv 0,\ 28\equiv 1 \pmod 3$. Hence the required integers are $\boxed{4,16,28}$." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="When proving a statement for all integers $n$, which case split is guaranteed to be exhaustive and non-overlapping in the most fundamental sense?" options=["$n>0$, $n=0$, $n<0$", "$n$ is even, $n$ is odd", "$n$ is prime, $n$ is composite, $n=1$", "$n \equiv 0 \pmod 3$, $n \equiv 1 \pmod 3$, $n \equiv 2 \pmod 3$"] answer="$n$ is even, $n$ is odd" hint="Consider the definition of integers and the most basic way to partition them." solution="The parity split ($n$ is even or $n$ is odd) covers all integers without overlap. While other options like modulo 3 are also exhaustive, parity is often considered the most fundamental and universally applicable for general integer proofs, as it directly relates to the definition of integers."
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:::question type="NAT" question="Given three distinct positive integers $a, b, c$ such that $a+b+c=10$ and $a<b<c$. How many distinct sets $\{a,b,c\}$ are possible?" answer="4" hint="Systematically list possibilities for the smallest integer $a$, applying all constraints at each step. Determine the maximum possible value for $a$ first." solution="We require $a, b, c$ to be distinct positive integers with $a<b<c$ and $a+b+c=10$.
Since $a<b<c$, the smallest possible values are $a \ge 1$, $b \ge a+1$, $c \ge b+1$.
Thus, $a + (a+1) + (a+2) \le a+b+c=10$, which implies $3a+3 \le 10$, so $3a \le 7$, meaning $a \le 7/3$.
Since $a$ is a positive integer, $a$ can only be 1 or 2.

Case 1: $a=1$.
Then $b+c=9$ and $1<b<c$.
If $b=2$, then $c=7$. Set: $\{1,2,7\}$.
If $b=3$, then $c=6$. Set: $\{1,3,6\}$.
If $b=4$, then $c=5$. Set: $\{1,4,5\}$.
(If $b \ge 5$, then $c \le 4$, which violates $b<c$.)

Case 2: $a=2$.
Then $b+c=8$ and $2<b<c$.
If $b=3$, then $c=5$. Set: $\{2,3,5\}$.
(If $b=4$, then $c=4$, which violates $b<c$.)

Thus, there are 4 distinct sets: $\{1,2,7\}, \{1,3,6\}, \{1,4,5\}, \{2,3,5\}$."
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:::question type="MCQ" question="A proof states: 'If $x^2 = 9$, then $x=3$.' What hidden assumption, if any, makes this statement incomplete or potentially false in a general context?" options=["The domain of $x$ is real numbers.", "The problem implies $x$ must be positive.", "The value 9 is a perfect square.", "The equation $x^2=9$ has a solution." ] answer="The problem implies $x$ must be positive." hint="Consider all possible solutions to $x^2=9$ in the standard number system for $x$." solution="In the standard real number system, $x^2=9$ implies $x=3$ or $x=-3$. The statement 'If $x^2=9$, then $x=3$' is only true if there's an implicit (hidden) assumption that $x$ must be positive (e.g., $x$ represents a length or magnitude). Without this hidden assumption, the statement is incomplete as it misses the $x=-3$ case, making it potentially false as a general implication."
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What's Next?