Mathematical-Reasoning · Unit 11 · Video 4 · Interactive Practice

Writing the Math in a Proof: Displays and a Clear Ending

IKey Formulas

Guideline	Written well
Display the important algebra inside a sentence, and write an unchanged side once	Using algebra, we obtain $\begin{aligned} x \cdot y &= (2m + 1)(2n + 1) \\ &= 4mn + 2m + 2n + 1 \\ &= 2(2mn + m + n) + 1. \end{aligned}$
Number only an equation that a later sentence refers to	$x = 2n + 1 \qquad \text{(1)}$ … then, using equation (1), we obtain …
Write logic symbols as English words	$\forall$ “for all”, $\exists$ “there exists”, $\ni$ “such that”, $\therefore$ “therefore”
Begin every sentence with a word, and say when the proof is complete	“An integer $n$ is odd provided …” “This completes the proof.” $\blacksquare$

Key Insight: Equations are part of the explanation, so they live inside English sentences: displayed when they carry the argument, numbered only when cited again, and followed by a clear signal that the proof is over.

IIVisualization 1 — Revising the Draft, One Guideline at a Time

Each guideline repairs one defect in the draft proof that the product of two odd integers is odd.

IIIVisualization 2 — Which Displays Get a Number?

A display earns a number only when a later sentence refers back to it.

💡 When a proof cites a number, the usual convention is a lowercase “equation (1)”; a capital “Equation (1)” is also acceptable, but less common.

IVVisualization 3 — From Logic Symbols to English

Each logic symbol has an English reading, while the equations themselves stay in symbols.

Statement

VQuiz Questions

Problem 1 · From Symbols to a Sentence

Given: the statement $(\forall n \in \mathbb{Z})(\exists m \in \mathbb{Z})(m > n)$ — find the version that follows the writing guidelines and keeps the meaning.

✅ Correct! The quantifiers keep their order, each logic symbol has become its English reading, and only the inequality

m > n

stays in symbols.

❌ The order of the quantifiers changed. This version claims a single integer

m

is larger than every integer

n

, including

n = m

, which is false. In the original,

m

may depend on

n

❌ Logic symbols are still in the prose. The symbol

\exists

should read “there exists” and

\ni

should read “such that”.

❌ Not quite. This sentence opens with the symbol

n

, and the words “for each” are gone, so it no longer says that the claim holds for every integer

n

Show solution

Read the symbols left to right: $(\forall n \in \mathbb{Z})$ is “for each integer $n$ ”, $(\exists m \in \mathbb{Z})$ is “there exists an integer $m$ ”, and the last bracket is the condition $m$ must meet, introduced by “such that”. This gives

“For each integer $n$ , there exists an integer $m$ such that $m > n$ .”

The statement is true: for each integer $n$ , the integer $m = n + 1$ satisfies $m > n$ . Swapping the quantifiers (option a) produces a different, false statement, since no integer $m$ satisfies $m > m$ . Option c keeps the symbols $\exists$ and $\ni$ in the prose, and option d begins with the symbol $n$ and drops the universal quantifier.

Problem 2 · Where a Sentence Begins

Given: four ways to write that the square of an odd integer is odd — find the one that follows the guidelines for symbols in prose.

✅ Correct! The sentence begins with words, contains no logic symbols, and the expression

x^2

sits after the words “the integer”.

❌ Not quite. This sentence opens with the expression

x^2

. Words in front of it, such as “For every odd integer

x

, the integer

x^2

…”, fix that.

❌ Not quite. The symbol

\forall

is a logic symbol in prose (write “for every”), and here it is also the first thing in the sentence.

❌ Close, but… the first sentence is fine, while the second one opens with the symbol

x^2

. Writing “The integer

x^2

is then odd.” would fix it.

Show solution

Two guidelines apply: a sentence never begins with a mathematical symbol, and logic symbols such as $\forall$ are written as English words.

Option a begins with $x^2$ . Option b begins with $\forall$ and uses it in place of the words “for every”. Option d is the pattern from the video: a good first sentence followed by one that opens with a symbol. Only option c begins with a word and keeps every logic idea in English:

“For every odd integer $x$ , the integer $x^2$ is odd.”

Problem 3 · Numbering the Right Equation

Given: a proof displays $x = 2k + 1$ , and a few lines later displays $x^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$ . Two paragraphs later it substitutes $x = 2k + 1$ again; the second display is never mentioned after it appears — find the numbering that follows the guidelines.

✅ Correct! The equation the proof points back to gets a number; the computation of

x^2

is still displayed, but it needs no number.

❌ Not quite. The display for

x^2

is never referred to again. An important equation that is never mentioned later is still displayed, but it gets no number.

❌ Not quite. Two paragraphs later, “the equation above” is ambiguous: the nearest display above is the computation of

x^2

, not

x = 2k + 1

. An equation a later step points back to gets a number.

❌ The numbers are on the wrong display. The

x^2

display is never referred to, so it needs no number, while

x = 2k + 1

, which is referred to, would be left without one.

Show solution

The rule is to number exactly the equations that a later step refers to. Here only $x = 2k + 1$ is cited again, so it is displayed as

x = 2k + 1 \qquad \text{(1)}

and the later step reads “Then, using equation (1), we obtain …”. The computation

\begin{aligned} x^2 &= (2k + 1)^2 \\ &= 4k^2 + 4k + 1 \\ &= 2(2k^2 + 2k) + 1 \end{aligned}

is important enough to display, but since nothing points back to it, it carries no number.

Problem 4 · Auditing a Draft

Given: this draft of a proof that the sum of two even integers is even:
Proof. $x, y$ even $\therefore$ $\exists\, m, n \in \mathbb{Z}$ $\ni$ $x = 2m$ , $y = 2n$ .
$x + y = 2m + 2n$
$x + y = 2(m + n)$
$m + n \in \mathbb{Z}$ $\therefore$ $x + y$ even
— find the guideline this draft does not break.

✅ Correct! The draft numbers nothing and never points back to an earlier equation, so this guideline is satisfied; each of the other four is broken.

❌ That one is broken. The symbols

\therefore

\exists

and

\ni

stand in the prose where “therefore”, “there exist” and “such that” belong.

❌ That one is broken. The left side

x + y

never changes, yet it is copied onto the second row of the algebra.

❌ That one is broken. The first sentence after “Proof.” opens with

x, y

, and the last line opens with the expression

m + n

❌ That one is broken. The draft ends on a fragment of symbols, with no closing sentence and no end mark such as

\blacksquare

Show solution

The draft breaks four guidelines: it uses $\therefore$ , $\exists$ and $\ni$ in prose, repeats the unchanged side $x + y$ , opens lines with $x, y$ and with $m + n$ , and stops without a closing sentence or end mark. It numbers nothing and refers back to nothing, so the numbering guideline is not broken. A revision that follows every guideline:

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

Proof. We assume that $x$ and $y$ are even integers and will prove that $x + y$ is an even integer. Since $x$ and $y$ are even, there exist integers $m$ and $n$ such that $x = 2m$ and $y = 2n$ . Using algebra, we obtain

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

Since $m$ and $n$ are integers, $m + n$ is an integer. Therefore, $x + y$ is an even integer. $\blacksquare$

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