Mathematical-Reasoning · Unit 2 · Video 1 · Interactive Practice

Odd Means 2n + 1: Why a Proof Starts From the Definition

IKey Formulas

Term	Definition	Example
Even integer	$a = 2n$ for some integer $n$	$-42 = 2(-21)$ , $\ 0 = 2 \cdot 0$
Odd integer	$a = 2n + 1$ for some integer $n$	$-11 = 2(-6) + 1$
Direct proof of $P \to Q$	Assume $P$ ; reach $Q$ using definitions and proven results	Assume $x, y$ odd; show $x \cdot y$ odd
Backward question	Under what conditions is $x \cdot y$ odd?	When $x \cdot y = 2q + 1$ for an integer $q$

Key Insight: “There exists an integer $n$ ” means one integer that works is enough, and each odd number gets its own letter: $x = 2m + 1$ and $y = 2n + 1$ .

IIFinding the Witness $n$

An integer $a$ is shown even or odd the moment one integer $n$ gives $a = 2n$ or $a = 2n + 1$ .

Integer a Witness n −5

IIIExamples Versus Counterexamples

A counterexample needs a true hypothesis and a false conclusion; agreeing examples, however many, prove nothing.

Claim

IVSetting Up the Direct Proof

Each odd number gets its own integer from the definition, and the definition of odd also names the goal.

x 5 y 3 Letters

VQuiz Questions

Problem 1 · Produce the Witness

Given: an integer $a$ is odd provided there exists an integer $n$ with $a = 2n + 1$ . Which equation shows, directly from this definition, that $-15$ is odd?

✅ Correct!

2(-8) + 1 = -16 + 1 = -15

, and

-8

is an integer, so

-15

meets the definition of odd.

❌ Not quite.

2(-7) + 1 = -14 + 1 = -13

, not

-15

. Solve

2n + 1 = -15

: subtract

1

first, then halve, which gives

n = -8

❌ Not quite. The equation is true, but

-7.5

is not an integer, and the definition of odd needs the form

2n + 1

with

n

an integer.

❌ Not quite. The equation is true, but its right side is not of the form

2n + 1

. It shows that

15 = 2 \cdot 7 + 1

is odd; the definition asks for an integer

n

with

-15 = 2n + 1

itself.

Show solution

We need one integer $n$ with $2n + 1 = -15$ . Then $2n = -16$ , so $n = -8$ , which is an integer. Check:

2(-8) + 1 = -16 + 1 = -15.

Option (a) uses $n = -7$ , the same trap as $n = -5$ for $-11$ in the video: $2(-7) + 1 = -13$ . Option (b) has $n = -7.5$ , not an integer. Option (d) is true arithmetic but shows only that $15$ is odd, written with a minus sign in front.

Problem 2 · Is It a Counterexample?

Given: the proposition “if $x$ and $y$ are odd integers, then $x \cdot y$ is an odd integer.” A classmate says: “ $x = 2$ , $y = 3$ gives $x \cdot y = 6$ , which is even, so the proposition is false.” Which response is correct?

✅ Correct! A conditional with a false hypothesis is true, so this pair cannot be a counterexample. A counterexample needs

x

and

y

both odd with

x \cdot y

even.

❌ Not quite. The first half is right, but

x = 5

y = 3

checks only one pair. The proposition is about every pair of odd integers, and no list of examples covers them all.

❌ Not quite. The hypothesis says

x

and

y

are odd. With

x = 2

the hypothesis is false, whatever

y

is.

❌ Not quite. One counterexample does disprove a conditional, but a counterexample must make the hypothesis true. Here

x = 2

is not odd, so the hypothesis is false.

Show solution

The proposition is a conditional $P \to Q$ with $P$ : “ $x$ and $y$ are odd integers” and $Q$ : “ $x \cdot y$ is an odd integer.” It is false for a pair only when $P$ is true and $Q$ is false.

For $x = 2$ , $y = 3$ : $2 = 2 \cdot 1$ is even, not odd, so $P$ is false. A conditional with a false hypothesis is true, so this pair is consistent with the proposition and tells us nothing about it.

A counterexample would need both $x$ and $y$ odd and $x \cdot y$ even. Checking pairs like $x = 5$ , $y = 3$ only confirms single cases; proving the proposition takes a direct proof that starts from the definition of odd.

Problem 3 · From the Witnesses to the Goal

Given: for one particular pair, $x = 2m + 1$ and $y = 2n + 1$ with $m = -4$ and $n = 3$ . Find $x \cdot y$ , then the integer $q$ with $x \cdot y = 2q + 1$ .

What is x · y?

What is q?

✅ Correct!

x = -7

y = 7

x \cdot y = -49

, and

2(-25) + 1 = -49

with

-25

an integer, so this product is odd.

❌ Check $x$ .

x = 2m + 1 = 2(-4) + 1 = -8 + 1 = -7

, not

-9

❌ Check the sign.

x = 2(-4) + 1 = -7

is negative and

y = 7

is positive, so the product is negative.

❌ Check $x$ . The odd form has a

+1

x = 2(-4) + 1 = -7

, not

2(-4) = -8

❌ Check the product. Find

x

and

y

first, then multiply.

❌ Check $q$ .

2(-24) + 1 = -47

. From

2q + 1 = -49

, subtract

1

first:

2q = -50

❌ Check $q$ .

-\tfrac{49}{2}

solves

2q = -49

, which drops the

+1

; and

q

must be an integer.

❌ Check $q$ .

2(24) + 1 = 49

, the product with the wrong sign. The product here is

-49

❌ Check $q$ . Solve

2q + 1 = x \cdot y

for

q

Show solution

Step 1: the numbers. $x = 2(-4) + 1 = -7$ and $y = 2(3) + 1 = 7$ .

Step 2: the product. $x \cdot y = (-7)(7) = -49$ .

Step 3: the backward goal. We need an integer $q$ with $2q + 1 = -49$ :

2q = -50, \qquad q = -25.

Check: $2(-25) + 1 = -50 + 1 = -49$ . Since $-25$ is an integer, $-49$ is odd. This settles one pair; the proof in the next video finds $q$ for every pair at once.

Problem 4 · Both Ends of a New Proof

Given: the proposition “if $a$ is an even integer and $b$ is an odd integer, then $a + b$ is odd.” Which is the correct setup of a direct proof: what we know after assuming the hypothesis, and what we need to show?

✅ Correct! The definitions of even and odd give the starting equations, each with its own letter, and the definition of odd answers the backward question.

❌ Not quite.

a + b = 2q

is the definition of even. The conclusion says

a + b

is odd, so the goal is

a + b = 2q + 1

❌ Not quite. Reusing

n

forces

b = a + 1

, an assumption the proposition never made: with

a = 4

and

b = 9

no single

n

works for both.

❌ Not quite. This assumes the conclusion and aims at the hypothesis. A direct proof starts from the hypothesis and arrives at the conclusion.

Show solution

Forward (hypothesis). Assume $a$ is even and $b$ is odd. By the definitions there is an integer $m$ with $a = 2m$ and an integer $n$ with $b = 2n + 1$ . Different letters, because nothing says the two integers are equal: for $a = 4$ , $b = 9$ they are $m = 2$ and $n = 4$ .

Backward (conclusion). Under what conditions can we conclude $a + b$ is odd? By the definition of odd, when there is an integer $q$ with $a + b = 2q + 1$ .

So the proof starts from $a = 2m$ , $b = 2n + 1$ and must reach $a + b = 2q + 1$ for some integer $q$ . For $a = 4$ , $b = 9$ : $13 = 2 \cdot 6 + 1$ , so $q = 6$ there.

Solved: 0 / 4

Odd Means 2n + 1: Why a Proof Starts From the Definition

IKey Formulas

IIFinding the Witness nnn

IIIExamples Versus Counterexamples

IVSetting Up the Direct Proof

VQuiz Questions

IIFinding the Witness $n$