Mathematical-Reasoning · Unit 9 · Video 2 · Interactive Practice

Choosing Exhaustive Cases and Writing a Case Proof

IKey Formulas

Formula	Name	Use in a case proof
$(P \vee Q) \to R \equiv (P \to R) \wedge (Q \to R)$	Proof by cases	Prove the conclusion separately in each case; the cases must cover every object the hypothesis allows
$n = 2m$ or $n = 2m + 1$ for some $m \in \mathbb{Z}$	Parity split	Every integer is even or odd; two integers $m$ , $n$ give $2 \times 2 = 4$ cases
$x > 0$ , $x = 0$ , $x < 0$ (or $x \geq 0$ , $x < 0$ )	Sign split	Every real number is positive, zero or negative; the boundary value $0$ must land in some case
$a = 0$ or $a \neq 0$	Guarding a step	Dividing by $a$ is valid only in the case $a \neq 0$

Key Insight: Cases may overlap, but they may never leave a gap: every object that satisfies the hypothesis has to fall into at least one case, and each case should open with its own label ("Case 1", "In the case where …").

IIVisualization 1 — Do the Cases Leave a Gap?

Cases may overlap, but every object allowed by the hypothesis must fall into at least one case.

Hypothesis

IIIVisualization 2 — Both Cases of $n^2 - 5n + 7$

Every integer is even or odd, and each case's algebra ends in the form $2(\text{integer}) + 1$ .

n 6

In the case where $n$ is even, there is an integer $m$ with $n = 2m$ .

\begin{aligned} n^2 - 5n + 7 &= (2m)^2 - 5(2m) + 7 \\ &= 4m^2 - 10m + 6 + 1 \\ &= 2(2m^2 - 5m + 3) + 1 \end{aligned}

In the case where $n$ is odd, there is an integer $m$ with $n = 2m + 1$ .

\begin{aligned} n^2 - 5n + 7 &= (2m + 1)^2 - 5(2m + 1) + 7 \\ &= 4m^2 - 6m + 3 \\ &= 2(2m^2 - 3m + 1) + 1 \end{aligned}

💡 Each bracket is an integer because sums, differences and products of integers are integers, so in both cases $n^2 - 5n + 7$ has the form $2k + 1$ with $k \in \mathbb{Z}$ .

IVVisualization 3 — The Step That Chooses the Cases

If $ab = 0$ , dividing by $a$ is valid only when $a \neq 0$ , so that condition and its failure become the cases.

Point on Other coordinate a = 2

VQuiz Questions

Problem 1 · Exhaustive Cases for a Real Number

Given: a proposition whose hypothesis is " $x$ is a real number." Which list of cases exhausts every possibility for $x$ ?

✅ Correct! Every real number satisfies

x \geq 0

x \leq 0

. The number

0

satisfies both, and an overlap is allowed.

❌ Not quite. The real number

0

is neither positive nor negative, so these two cases say nothing about

x = 0

❌ Not quite. The real number

1

satisfies neither

x > 1

nor

x < 1

, so

x = 1

falls into no case.

❌ Not quite. A number such as

x = -\tfrac{1}{2}

is not positive, not

0

, and not less than

-1

: every

x

with

-1 \leq x < 0

is left out.

Show solution

The cases exhaust the hypothesis exactly when every real number satisfies at least one of them. Test each list at its boundary values.

(a) $x = 0$ satisfies neither $x > 0$ nor $x < 0$ : a gap.
(b) $x = 1$ satisfies neither $x > 1$ nor $x < 1$ : a gap.
(c) If $x$ is not $\geq 0$ , then $x < 0$ , so $x \leq 0$ . Every real number is covered; $x = 0$ is covered twice, which is allowed.
(d) $x = -\tfrac{1}{2}$ satisfies none of $x > 0$ , $x = 0$ , $x < -1$ : a gap covering all of $-1 \leq x < 0$ .

Only list (c) is exhaustive.

Problem 2 · The Missing Parity Case

Given: a proof about all integers $m$ and $n$ uses three cases. Case 1: $m$ and $n$ are both even. Case 2: $m$ and $n$ are both odd. Case 3: $m$ is even and $n$ is odd. Which pair $(m, n)$ falls into none of the three cases?

✅ Correct! Here

m = 5

is odd and

n = 8

is even. That mixed case,

m

odd with

n

even, is missing from the list, so the proof says nothing about this pair.

❌ Not quite. Both

2

and

4

are even, so

(2, 4)

is in Case 1.

❌ Not quite. Both

3

and

5

are odd, so

(3, 5)

is in Case 2.

❌ Not quite. Here

m = 4

is even and

n = 7

is odd, which is exactly Case 3.

Show solution

Each of $m$ and $n$ is even or odd, so there are $2 \times 2 = 4$ parity pairs: (even, even), (even, odd), (odd, even), (odd, odd). The three cases cover (even, even), (odd, odd) and (even, odd). The pair (odd, even) is missing.

$(2, 4)$ is (even, even): Case 1.
$(3, 5)$ is (odd, odd): Case 2.
$(4, 7)$ is (even, odd): Case 3.
$(5, 8)$ is (odd, even): no case.

The two mixed cases are different cases. Unless the proof adds a fourth case, $m$ odd and $n$ even, it does not prove the statement for every pair of integers.

Problem 3 · Finishing the Odd Case

Given: the proposition "for each integer $n$ , $n^2 + 3n + 5$ is odd," proved by cases. In the case where $n$ is odd, there is an integer $m$ with $n = 2m + 1$ . Which line correctly finishes this case?

✅ Correct!

(2m + 1)^2 + 3(2m + 1) + 5 = 4m^2 + 10m + 9 = 2(2m^2 + 5m + 4) + 1

, and

2m^2 + 5m + 4

is an integer.

❌ Not quite. This comes from writing

3(2m + 1)

6m + 1

. Distribute fully:

3(2m + 1) = 6m + 3

❌ Not quite. This comes from squaring

2m + 1

4m^2 + 1

. The middle term is missing:

(2m + 1)^2 = 4m^2 + 4m + 1

❌ Not quite. This is the even case: substituting

n = 2m

gives

4m^2 + 6m + 5 = 2(2m^2 + 3m + 2) + 1

. In this case

n = 2m + 1

Show solution

Proof. Let $n$ be an integer. We will prove that $n^2 + 3n + 5$ is odd by examining the case where $n$ is even and the case where $n$ is odd.

In the case where $n$ is even, there exists an integer $m$ such that $n = 2m$ . Then

n^2 + 3n + 5 = 4m^2 + 6m + 5 = 2(2m^2 + 3m + 2) + 1.

In the case where $n$ is odd, there exists an integer $m$ such that $n = 2m + 1$ . Then

\begin{aligned} n^2 + 3n + 5 &= (4m^2 + 4m + 1) + (6m + 3) + 5 \\ &= 4m^2 + 10m + 9 \\ &= 2(2m^2 + 5m + 4) + 1. \end{aligned}

Sums and products of integers are integers, so $2m^2 + 3m + 2$ and $2m^2 + 5m + 4$ are integers. In both cases $n^2 + 3n + 5$ is odd. ∎

Check: $n = 3$ gives $m = 1$ and $2(2 + 5 + 4) + 1 = 23 = 9 + 9 + 5$ .

Problem 4 · Cases That Guard a Division

Given: the proposition "for all real numbers $x$ and $y$ , if $xy = x$ , then $x = 0$ or $y = 1$ ." A proof wants to divide both sides of $xy = x$ by $x$ .

Which two cases cover every real number $x$ and make the division by $x$ valid wherever it is used?

In the case where $x \neq 0$ , what does dividing both sides of $xy = x$ by $x$ give?

✅ Correct! The condition the step needs,

x \neq 0

, and its failure,

x = 0

, are the cases. In the case

x = 0

the conclusion already holds; in the case

x \neq 0

the division gives

y = 1

❌ Not quite. These cases leave out

x = 0

, the very value where the division fails, so the proof says nothing about it.

❌ Not quite. These cases cover every

y

, but they say nothing about whether

x \neq 0

, so they do not make the division by

x

valid.

❌ Not quite. These cases do not exhaust the possibilities:

x = 2

y = 1

satisfies

xy = x

but falls into neither case.

❌ Not quite. That is the pattern of

ab = 0

. Here the right side is

x

, not

0

\frac{xy}{x} = \frac{x}{x}

gives

y = 1

❌ Not quite. Dividing by

x

removes

x

from both sides; it says nothing about the value of

x

. The result is

y = 1

❌ Not quite. The right side

x

divided by

x

1

, not

x

, so the result is

y = 1

Show solution

Proof. Let $x$ and $y$ be real numbers, and assume that $xy = x$ . We will prove that $x = 0$ or $y = 1$ by considering two cases: (1) $x = 0$ , and (2) $x \neq 0$ . Every real number $x$ falls into one of these.

In the case where $x = 0$ , the conclusion " $x = 0$ or $y = 1$ " is true because its first part is true. There is nothing to prove.

In the case where $x \neq 0$ , the number $\frac{1}{x}$ exists. Multiplying both sides of $xy = x$ by $\frac{1}{x}$ gives

\frac{1}{x} \cdot xy = \frac{1}{x} \cdot x, \quad\text{so}\quad y = 1.

So in both cases $x = 0$ or $y = 1$ . ∎

The cases were chosen to match the condition the division step needs, exactly as in the proof that $ab = 0$ implies $a = 0$ or $b = 0$ .

Solved: 0 / 4

Choosing Exhaustive Cases and Writing a Case Proof

IKey Formulas

IIVisualization 1 — Do the Cases Leave a Gap?

IIIVisualization 2 — Both Cases of n2−5n+7n^2 - 5n + 7n2−5n+7

IVVisualization 3 — The Step That Chooses the Cases

VQuiz Questions

IIIVisualization 2 — Both Cases of $n^2 - 5n + 7$