Mathematical-Reasoning · Unit 11 · Video 3 · Interactive Practice

Writing a Proof for a Reader: Openings and Signposts

IKey Formulas

Guideline	Draft	Written for a reader
State the theorem	Prove that if `x` is an odd integer, then `x^2` is an odd integer.	Theorem. If $x$ is an odd integer, then $x^2$ is an odd integer.
Open with the assumption and the goal, using “we”	(no opening)	Proof. We assume that $x$ is an odd integer and will prove that $x^2$ is an odd integer.
Typeset, in complete sentences	`x=2n+1 so x^2=4n^2+4n+1=2*(2n^2+2n)+1 so odd`	Then $x^2 = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1$ .
Give each claim its status	We have $x = 2n + 1$ .	Since $x$ is an odd integer, there exists an integer $n$ such that $x = 2n + 1$ .

Key Insight: A reader should never have to ask what is assumed, what the goal is, which method is in use, or why a line holds. A proof by contrapositive or by contradiction says so in its first few lines.

IIFrom Draft to Proof

Every equation in the draft is correct; five guidelines turn it into a proof a reader can follow.

💡 Knowing your audience decides how many of these reasons to write out: a solutions manual for instructors leaves details to the reader that a manual for students shows.

IIIAnnounce the Method

Proofs by contrapositive and by contradiction both open by assuming the conclusion is false, so they must say so.

Method

IVQuiz Questions

Problem 1 · State the Theorem

Given: the exercise “Show that if $a$ and $b$ are even integers, then $a + b$ is even.” Which line states it as a theorem?

✅ Correct! The instruction is dropped and what is left is a declarative statement with the same meaning.

❌ Not quite. The label is right, but “Show that” is still an instruction to the student; a theorem is a declarative statement.

❌ Not quite. This is the converse, and it is false:

1 + 1 = 2

is even, yet

1

is odd. Rewording an exercise must not change what it says.

❌ Not quite. This describes what the writer plans to do, in the first person; a theorem states a fact.

Show solution

Drop the instruction “Show that” and label what is left. The hypothesis stays the hypothesis and the conclusion stays the conclusion:

Theorem. If $a$ and $b$ are even integers, then $a + b$ is an even integer.

Option (a) keeps the instruction, option (c) swaps hypothesis and conclusion (the converse, which fails for $a = b = 1$ ), and option (d) is a plan in the first person rather than a statement.

Problem 2 · The First Sentence

Given: Theorem. If $x$ and $y$ are odd integers, then $xy$ is an odd integer. Which first sentence of the proof follows every guideline for opening a proof?

✅ Correct! The first half tells the reader what may be used, the second half names the goal, and the pronoun is “we”.

❌ Not quite. This assumes the conclusion and aims at the hypothesis, so it opens a proof of the converse.

❌ Not quite. The content is right, but the convention is “we”: the reader is doing the mathematics with you.

❌ Not quite. This names what may be used but not the goal, so the reader cannot read the later lines as steps toward it.

Show solution

The proof opens with a statement of the assumptions followed by a statement of what will be proved, in the “we” voice:

Proof. We assume that $x$ and $y$ are odd integers and will prove that $xy$ is an odd integer.

The assumption is the hypothesis of the theorem and the goal is its conclusion. Reversing them (a) proves a different statement, “I” (b) breaks the pronoun convention, and stopping after the assumption (c) leaves the goal unstated.

Problem 3 · The Status of Each Claim

Given: Theorem. If $x$ and $y$ are odd integers, then $xy$ is an odd integer. Proof. (1) We assume that $x$ and $y$ are odd integers and will prove that $xy$ is an odd integer. (2) Since $x$ and $y$ are odd, there exist integers $m$ and $n$ such that $x = 2m + 1$ and $y = 2n + 1$ . (3) Then $xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1$ . (4) Since the integers are closed under addition and multiplication, $2mn + m + n$ is an integer. (5) Hence $xy$ is an odd integer.

(A) Where do the integers $m$ and $n$ come from?

(B) Why is $2mn + m + n$ an integer?

✅ Correct! Sentence (2) cites the assumption and the definition of odd; sentence (4) cites closure, a basic property of the integers from the reader's background.

❌ Not quite. Sentence (3) uses

m

and

n

; they were introduced one sentence earlier.

❌ Not quite. Sentence (1) says only that

x

and

y

are odd; it takes the definition of an odd integer to produce

m

and

n

❌ Not quite. Closure in sentence (4) shows that

2mn + m + n

is an integer;

m

and

n

already exist by then.

❌ Not quite. The assumption gives the integers

m

and

n

(through the definition of odd); combining them into

2mn + m + n

needs closure.

❌ Not quite. The definition of odd produced

m

and

n

; it says nothing about the expression

2mn + m + n

❌ Not quite. That is the goal of the proof, so using it as a reason here would be circular.

Show solution

Each sentence carries its status:

(1) the assumption and the goal;
(2) the assumption, unpacked by the definition of an odd integer (background), which introduces $m$ and $n$ ;
(3) algebra (background): $(2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1$ ;
(4) closure of the integers under addition and multiplication (background), so $2mn + m + n$ is an integer;
(5) the definition of an odd integer once more, which gives the conclusion.

Problem 4 · Name the Method

Given: Theorem. For each integer $n$ , if $3n + 1$ is even, then $n$ is odd. A proof contains the sentence “We assume that $n$ is an even integer and will prove that $3n + 1$ is an odd integer.” Which sentence should come before it, at the start of the proof?

✅ Correct! The contrapositive is “if

n

is even, then

3n + 1

is odd”, which is exactly what the given sentence sets out to prove.

❌ Not quite. A proof by contradiction would assume both that

3n + 1

is even and that

n

is even, then aim at a contradiction; this proof assumes only that

n

is even and aims at “

3n + 1

is odd”.

❌ Not quite. The converse is a different statement (true here, but a converse is not logically equivalent to the statement it comes from), so proving it does not prove the theorem; it would also assume that

n

is odd, not even.

❌ Not quite. The theorem concludes that

n

is odd, so a proof that opens by assuming

n

is even looks wrong until the method is named; say so in the first few lines.

Show solution

The theorem has the form “if $P$ , then $Q$ ” with $P$ : $3n + 1$ is even, and $Q$ : $n$ is odd. For an integer, “not odd” means even and “not even” means odd, so the contrapositive “if not $Q$ , then not $P$ ” reads: if $n$ is even, then $3n + 1$ is odd. The given sentence assumes not $Q$ and aims at not $P$ , so the proof is a proof of the contrapositive, and it should open by saying so.

Proof. We will prove this result by proving the contrapositive of the statement. We assume that $n$ is an even integer and will prove that $3n + 1$ is an odd integer. Since $n$ is even, there exists an integer $k$ such that $n = 2k$ . Then $3n + 1 = 6k + 1 = 2(3k) + 1$ . Since the integers are closed under multiplication, $3k$ is an integer, and hence $3n + 1$ is an odd integer.

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