Mathematical-Reasoning · Unit 2 · Video 3 · Interactive Practice

How to Write a Proof: Six Writing Guidelines, Then Three Proofs About Sums

IKey Formulas

Result	Algebra	Conclusion
Definitions	even: $x = 2n$ ; odd: $x = 2n + 1$	for some integer $n$
even + even	$2m + 2n = 2(m + n)$	even, $q = m + n$
even + odd	$2m + (2n + 1) = 2(m + n) + 1$	odd, $q = m + n$
odd + odd	$(2m + 1) + (2n + 1) = 2(m + n + 1)$	even, $q = m + n + 1$

Key Insight: Each proof makes the same three moves: a definition writes $x$ and $y$ with integers $m$ and $n$ , algebra puts $x + y$ in the form $2q$ or $2q + 1$ , and closure of the integers under addition shows that $q$ is an integer.

IIRevising a Draft — The Six Writing Guidelines

A proof is not complete until it is well written, and each line of this draft breaks one guideline.

Guideline 1 · State the theorem

Prove that if $x$ is an odd integer, then $x^2$ is an odd integer.

Guideline 2 · Assumptions + goal

Proof. Squaring $2n + 1$ gives the answer.

Guideline 3 · Use 'we'

I know that there is an integer $n$ such that $x = 2n + 1$ .

Guideline 5 · Display equations

$x^2 = (2n + 1)^2$	squaring
$x^2 = 4n^2 + 4n + 1$	expanding
$x^2 = 2(2n^2 + 2n) + 1$	factoring

Guideline 4 · Typeset the math

Since n is an integer and the integers are closed under multiplication and addition, 2n^2 + 2n is an integer, so x^2 is odd by the definition of an odd integer.

Guideline 6 · Say when done

(no closing sentence)

IIIKnow-Show Tables for the Three Sums

Backward from the goal, forward from the hypothesis: the table is finished when step P3 matches step Q1.

Proposition Step 1

Step	Know / Show	Reason
$P$	$x$ and $y$ are even integers.	Hypothesis
$P1$	$x = 2m$ and $y = 2n$ for some integers $m$ , $n$	Definition of an even integer
$P2$	$x + y = 2m + 2n$	Substitution
$P3$	$x + y = 2(m + n)$	Distributive property
$P4$	$m + n$ is an integer.	Closure under addition
$Q1$	There is an integer $q$ with $x + y = 2q$ .	Use $q = m + n$
$Q$	$x + y$ is an even integer.	Definition of an even integer

Step	Know / Show	Reason
$P$	$x$ is an even integer and $y$ is an odd integer.	Hypothesis
$P1$	$x = 2m$ and $y = 2n + 1$ for some integers $m$ , $n$	Definitions of even and odd integers
$P2$	$x + y = 2m + (2n + 1)$	Substitution
$P3$	$x + y = 2(m + n) + 1$	Regrouping, distributive property
$P4$	$m + n$ is an integer.	Closure under addition
$Q1$	There is an integer $q$ with $x + y = 2q + 1$ .	Use $q = m + n$
$Q$	$x + y$ is an odd integer.	Definition of an odd integer

Step	Know / Show	Reason
$P$	$x$ and $y$ are odd integers.	Hypothesis
$P1$	$x = 2m + 1$ and $y = 2n + 1$ for some integers $m$ , $n$	Definition of an odd integer
$P2$	$x + y = (2m + 1) + (2n + 1)$	Substitution
$P3$	$x + y = 2(m + n + 1)$	Collecting terms, distributive property
$P4$	$m + n + 1$ is an integer.	Closure under addition ( $1$ is an integer)
$Q1$	There is an integer $q$ with $x + y = 2q$ .	Use $q = m + n + 1$
$Q$	$x + y$ is an even integer.	Definition of an even integer

💡 Leave out step P4 and the proof has a gap: the definition asks for two times an integer, so the proof must say why $q$ is one.

IVWhy the Parity Rules Hold — Pairs of Dots

An even number splits into pairs; an odd number leaves one dot over, and two leftover dots make a new pair.

x 3 y 7

💡 The dots check one pair of values at a time; the letters $m$ and $n$ in the proof cover every pair of integers at once, negative integers included.

VQuiz Questions

Problem 1 · The Opening Line

Given: an exercise reads “Prove that if $x$ is an even integer, then $x^2$ is an even integer.” Which is the best first line of the written proof?

✅ Correct! Guideline 1: the exercise is reworded as a simple declarative statement, hypothesis and conclusion intact.

❌ Not quite. That copies the exercise. “Prove that” is an instruction, not a statement; reword it as a theorem.

❌ Not quite. This starts computing before the reader knows what is being proved, and it never states its assumption or goal (guidelines 1 and 2).

❌ Close, but… the hypothesis has been dropped.

x^2

is not even for every

x

(take

x = 3

); the theorem is the whole conditional.

Show solution

Guideline 1 asks for a carefully worded statement of the result, as a simple declarative sentence rather than the exercise's instruction:

Theorem. If $x$ is an even integer, then $x^2$ is an even integer.

After a blank line comes Proof. with the assumptions and goal (guideline 2): “We assume that $x$ is an even integer and will prove that $x^2$ is an even integer.”

Problem 2 · Find the Gap

Given: a student's proof: “We assume that $x$ and $y$ are even integers. Then $x = 2m$ and $y = 2n$ for some integers $m$ and $n$ , so $x + y = 2m + 2n = 2(m + n)$ . Therefore $x + y$ is even.” What must be added to make it a complete proof?

✅ Correct! The definition of even needs

x + y = 2q

with

q

an integer; closure under addition is what shows

q = m + n

qualifies.

❌ Not quite. An example checks one case and proves nothing in general; the argument with

m

and

n

already covers every case, once its gap is closed.

❌ Not quite.

x

and

y

need separate letters: writing both as

2n

would force

x = y

, and the proof would cover only that case.

❌ Close, but…

x + y = 2(m + n)

shows even only if

m + n

is an integer, and the proof never says why it is.

Show solution

By definition, $x + y$ is even when $x + y = 2q$ for some integer $q$ . The algebra gives $x + y = 2(m + n)$ , so $q = m + n$ , but the reader still needs a reason that $q$ is an integer.

The missing sentence: “Since the integers are closed under addition, $m + n$ is an integer, and the last equation shows that $x + y$ is an even integer.” This is step P4 of the know-show table.

Problem 3 · Even Plus Odd

Given: $x$ is even and $y$ is odd, so $x = 2m$ and $y = 2n + 1$ for some integers $m$ and $n$ . Prove that $x + y$ is odd by answering both parts.

Which integer $q$ makes $x + y = 2q + 1$ ?

Which fact shows that $q$ is an integer?

✅ Correct!

x + y = 2(m + n) + 1

with

m + n

an integer by closure, so

x + y

is odd.

❌ Not quite.

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

, which is

2

more than

x + y = 2m + 2n + 1

❌ Not quite.

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

, not

2m + 2n + 1

: the

2

was not factored out.

❌ Not quite.

2 \cdot \tfrac{x + y}{2} + 1 = x + y + 1

, not

x + y

. Here

\tfrac{x + y}{2} = m + n + \tfrac{1}{2}

, which is not an integer, and

q

must be written in terms of

m

and

n

❌ Not quite. That definition is what turns

x + y = 2q + 1

into “odd” at the very end; it cannot also show that

q

is an integer.

❌ Not quite. The distributive property rewrites

2m + 2n

2(m + n)

, but it says nothing about whether

m + n

is an integer.

❌ Not quite. One example proves nothing about all integers

m

and

n

❌ Check $q$ . Expand

2q + 1

and compare it with

2m + 2n + 1

❌ Check the reason.

q

is a sum of two integers.

Show solution

Theorem. If $x$ is an even integer and $y$ is an odd integer, then $x + y$ is an odd integer.

Proof. We assume that $x$ is even and $y$ is odd, and will prove that $x + y$ is odd. By the definitions, there exist integers $m$ and $n$ such that $x = 2m$ and $y = 2n + 1$ . Using algebra, we obtain

\begin{aligned} x + y &= 2m + (2n + 1) \\ &= 2(m + n) + 1. \end{aligned}

Since the integers are closed under addition, $q = m + n$ is an integer, so $x + y = 2q + 1$ is an odd integer by the definition of an odd integer. This proves that if $x$ is even and $y$ is odd, then $x + y$ is odd. $\blacksquare$

Problem 4 · A New Proposition

Given: the theorem “if $x$ is an odd integer, then $3x + 5$ is an even integer.” The proof writes $x = 2m + 1$ for an integer $m$ . Which integer $q$ gives $3x + 5 = 2q$ ?

✅ Correct!

3(2m + 1) + 5 = 6m + 8 = 2(3m + 4)

, and

3m + 4

is an integer by closure.

❌ Close, but… the

3

must multiply both terms:

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

, not

6m + 1

❌ Not quite.

6m + 8

3x + 5

itself;

q

is what remains after factoring out the

2

❌ Not quite.

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

, but

3x + 5 = 6m + 8

Show solution

Proof. We assume that $x$ is an odd integer and will prove that $3x + 5$ is an even integer. Since $x$ is odd, there exists an integer $m$ such that $x = 2m + 1$ . Using algebra, we obtain

\begin{aligned} 3x + 5 &= 3(2m + 1) + 5 \\ &= 6m + 8 \\ &= 2(3m + 4). \end{aligned}

Since the integers are closed under multiplication and addition, $3m + 4$ is an integer, so $3x + 5$ is an even integer by the definition of an even integer. This proves that if $x$ is an odd integer, then $3x + 5$ is an even integer. $\blacksquare$

Check with $x = 3$ ( $m = 1$ ): $3 \cdot 3 + 5 = 14 = 2 \cdot 7$ , and $3m + 4 = 7$ .

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