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

The Know-Show Table: Proving Odd Times Odd Is Odd

IKey Formulas

Tool	Statement	Role in the proof
Odd integer (definition)	$a$ is odd when $a = 2k + 1$ for some integer $k$	Forward: unpacks $P$ into $x = 2m + 1$ , $y = 2n + 1$ . Backward: turns $Q$ into $xy = 2q + 1$
Distributive property	$x(y + z) = xy + xz$ and $(y + z)x = yx + zx$	Expands $(2m + 1)(2n + 1)$
Closure of the integers	Sums and products of integers are integers	Certifies that $2mn + m + n$ is an integer
The bridge	$(2m + 1)(2n + 1) = 2(2mn + m + n) + 1$	Links $P_1$ to $Q_1$ with $q = 2mn + m + n$

Key Insight: The goal $xy = 2q + 1$ asks only for some integer $q$ . Algebra writes $xy$ as $2(\text{something}) + 1$ , and closure proves the something is an integer.

IIWorking From Both Ends

Backward questions reshape the goal $Q$ , forward questions unpack the hypothesis $P$ , until the two ends meet.

Step	Know	Reason
$P$	$x$ and $y$ are odd integers.	Hypothesis
$P_1$	There exist integers $m$ and $n$ such that $x = 2m + 1$ and $y = 2n + 1$ .	Definition of an odd integer
$P_2$	$xy = (2m + 1)(2n + 1)$	Substitution
$P_3$	$xy = 4mn + 2m + 2n + 1$	Algebra (distributive, identity, commutative, associative)
$P_4$	$xy = 2(2mn + m + n) + 1$	Algebra (distributive)
$P_5$	$2mn + m + n$ is an integer.	Closure properties of the integers
⋮
$Q_1$	There exists an integer $q$ such that $xy = 2q + 1$ .
$Q$	$x \cdot y$ is an odd integer.	?
Step	Show	Reason

IIIOdd Times Odd: Pairs Plus One

Expanding $(2m + 1)(2n + 1)$ splits $xy$ into blocks that pair off, with exactly one left over.

x 7 y 5

💡 The picture needs positive $x$ and $y$ , but the algebra in $P_1$ through $P_5$ never uses positivity, so the proof covers every pair of odd integers, negatives included.

IVA Fresh Letter for Each New Object

Writing $x = 2q + 1$ with the goal's $q$ would claim $xy = x$ , which fails unless $y = 1$ .

x 7 y 5

VQuiz Questions

Problem 1 · The Backward Question

Given: the proposition “If $x$ is an even integer and $y$ is an odd integer, then $x + y$ is an odd integer.” Which statement is $Q_1$ , the answer to the backward question “how do we prove that an integer is odd?” applied to the conclusion?

✅ Correct! The definition of an odd integer, applied to

x + y

, gives the target form

x + y = 2q + 1

for some integer

q

; it goes in the next-to-last row.

❌ That is a forward step. It answers “what can we conclude from

P

?”, so it is

P_1

, at the top of the table, not

Q_1

❌ Not quite.

2q

is the form of an even integer; the conclusion says

x + y

is odd, so the target is

2q + 1

❌ Not quite. This uses

m

and

n

from the hypothesis, so it comes from working forward (substitution), not from asking when

Q

holds.

❌ Not quite. Ask how any integer is proved odd, then apply that answer to

x + y

Show solution

$P$ : $x$ is even and $y$ is odd. $Q$ : $x + y$ is odd.

The backward question in the abstract is “how do we prove that an integer is odd?”. The answer is the definition: write it as $2q + 1$ for some integer $q$ . Applied to $x + y$ :

Q_1:\ \text{there exists an integer } q \text{ such that } x + y = 2q + 1.

For completeness, working forward gives $x = 2m$ and $y = 2n + 1$ , so $x + y = 2m + 2n + 1 = 2(m + n) + 1$ , and $m + n$ is an integer by closure, so $q = m + n$ works.

Problem 2 · One Letter Too Few

Given: a proof of “if $x$ and $y$ are odd integers, then $x + y$ is even” whose step $P_1$ reads: “There exists an integer $k$ such that $x = 2k + 1$ and $y = 2k + 1$ .” What is wrong with this step?

✅ Correct! If

x = 2k + 1

and

y = 2k + 1

, then

x = y

. The hypothesis never said that, so

x

and

y

need different letters:

x = 2m + 1

y = 2n + 1

❌ Not quite. Any letter is fine for one integer, but using the same letter twice claims both integers come from the same

k

. Try

x = 3

y = 5

: no single

k

gives both.

❌ Not quite. The hypothesis gives

P

(“

x

and

y

are odd”); turning that into equations is exactly what the definition of an odd integer does.

❌ Not quite. The definition uses

2k + 1

, and the sign is not the problem here; the shared letter is.

Show solution

From $x = 2k + 1$ and $y = 2k + 1$ it follows that $x = y$ , so the step assumes something the hypothesis does not give. For $x = 3$ we would need $k = 1$ , and for $y = 5$ we would need $k = 2$ : one $k$ cannot do both.

The rule of thumb: a different symbol for each new object. The corrected proof:

x = 2m + 1,\quad y = 2n + 1 \ \Rightarrow\ x + y = 2m + 2n + 2 = 2(m + n + 1).

Since $m$ , $n$ and $1$ are integers, $m + n + 1$ is an integer by closure under addition, so $x + y$ is even.

Problem 3 · Filling the Know Rows

Given: the proposition “If $x$ is an odd integer, then $x^2 + 3x$ is an even integer,” where an integer is even when it equals $2q$ for some integer $q$ . Step $P_1$ is $x = 2m + 1$ for some integer $m$ . Complete the next steps.

After substitution and algebra, $x^2 + 3x$ equals:

So $x^2 + 3x = 2q$ with $q$ equal to:

The reason for “ $2m^2 + 5m + 2$ is an integer” is:

✅ Correct!

x^2 + 3x = 4m^2 + 10m + 4 = 2(2m^2 + 5m + 2)

, and closure makes

2m^2 + 5m + 2

an integer, so

x^2 + 3x

is even.

❌ Check the square.

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

; the distributive property produces the middle term

4m

❌ Check $3x$ .

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

: the

3

multiplies both terms.

❌ That is half the answer. This is the integer

q

that comes next;

x^2 + 3x

itself is twice it.

❌ Not quite. Expand

(2m + 1)^2 + 3(2m + 1)

term by term.

❌ Not quite. That is

x^2 + 3x

itself;

q

is what multiplies the

2

after factoring.

❌ Check the factoring. Every one of the three terms must be divided by

2

2(2m^2 + 5m + 2) = 4m^2 + 10m + 4

❌ Not quite. Factor

2

out of all of

4m^2 + 10m + 4

❌ Not quite. The hypothesis only says that

x

is odd; it says nothing about the new expression

2m^2 + 5m + 2

❌ Not quite. The definition of an odd integer produced

x = 2m + 1

P_1

; it does not tell us that

2m^2 + 5m + 2

is an integer.

❌ Not quite. The distributive property was used to factor out the

2

; whether the result is an integer is a separate question about integers.

❌ Not quite. Which fact says that products and sums of integers are integers?

Show solution

Backward: $Q_1$ is “there exists an integer $q$ such that $x^2 + 3x = 2q$ ”.

Forward: $P_1$ : $x = 2m + 1$ . Substituting,

x^2 + 3x = (2m + 1)^2 + 3(2m + 1) = (4m^2 + 4m + 1) + (6m + 3) = 4m^2 + 10m + 4.

Every term has a factor of $2$ :

x^2 + 3x = 2(2m^2 + 5m + 2).

Since $m$ is an integer, $m^2$ , $2m^2$ and $5m$ are integers (closure under multiplication) and so is $2m^2 + 5m + 2$ (closure under addition). With $q = 2m^2 + 5m + 2$ , $Q_1$ holds, so $x^2 + 3x$ is even. Check with $m = 1$ : $x = 3$ , $x^2 + 3x = 18 = 2 \cdot 9$ , and $2 + 5 + 2 = 9$ .

Problem 4 · The Same Method, a New Definition

Given: say an integer $a$ has form A when $a = 3k + 1$ for some integer $k$ . Prove: “If $x$ and $y$ have form A, then $xy$ has form A.”

The backward question gives $Q_1$ :

With $x = 3m + 1$ and $y = 3n + 1$ , the integer $q$ is:

✅ Correct!

xy = 9mn + 3m + 3n + 1 = 3(3mn + m + n) + 1

, and

3mn + m + n

is an integer by closure.

❌ That is a forward step. It comes from the hypothesis (

P_1

P_2

);

Q_1

comes from asking when an integer has form A.

❌ Not quite. That is the target for an odd integer; form A asks for

3q + 1

❌ Not quite. Apply the definition of form A to

xy

❌ Not quite.

3(9mn + 3m + 3n) + 1 = 27mn + 9m + 9n + 1

, which is not

xy

: the

3

still has to be factored out.

❌ Not quite. The

+1

stays outside:

3(3mn + m + n + 1) + 1

3

more than

xy

❌ Check the first term. Factoring

3

out of

9mn

leaves

3mn

, not

mn

3(mn + m + n) + 1 = 3mn + 3m + 3n + 1

, which is not

xy

❌ Not quite. Expand

(3m + 1)(3n + 1)

and factor

3

out of every term except the final

1

Show solution

Backward: “how do we prove an integer has form A?” By the definition, so $Q_1$ : there exists an integer $q$ with $xy = 3q + 1$ .

Forward: by the definition, $x = 3m + 1$ and $y = 3n + 1$ for integers $m$ and $n$ (two letters, two objects). Then

xy = (3m + 1)(3n + 1) = 9mn + 3m + 3n + 1 = 3(3mn + m + n) + 1.

By closure, $3mn + m + n$ is an integer, so $q = 3mn + m + n$ gives $Q_1$ , and $xy$ has form A. Check: $x = 4$ , $y = 7$ ( $m = 1$ , $n = 2$ ) give $xy = 28 = 3 \cdot 9 + 1$ , and $3 \cdot 2 + 1 + 2 = 9$ .

Solved: 0 / 4