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

What Is a Mathematical Statement? (And a First Look at If-Then)

IKey Formulas

Idea	Form	Example
Statement (proposition)	Declarative sentence, definitely true or false, not both	$3 \cdot 4 + 7 = 19$ (true), $3 \cdot 5 + 7 = 19$ (false)
Open sentence	Contains an unknown variable: no truth value, not a statement	$3x + 7 = 19$
“There exists” closes it	There exists an integer $x$ such that …	There exists an integer $x$ such that $3x + 7 = 19$ (true, $x = 4$ )
Conditional	If $P$ , then $Q$ ( $P$ = hypothesis, $Q$ = conclusion)	False only when $P$ is true and $Q$ is false

Key Insight: Opinions, questions and open sentences are not statements — but a false sentence is. False is a definite truth value.

IIVisualization 1 — Statement or Not?

A statement is a declarative sentence with a definite truth value: true or false, not both.

Sentence 1 / 11

IIIVisualization 2 — From Open Sentence to Statement

An open sentence has no truth value until $x$ is given a value or bound by “there exists.”

Sentence x ?

💡 Challenge: find the one value that makes $2x + 5 = 10$ true, then decide the truth value of “there exists an integer $x$ such that $2x + 5 = 10$ .”

IVVisualization 3 — When Is an If-Then False?

When is “If it is raining, then Laura is at the theater” false?

VQuiz Questions

Problem 1 · Spot the Statement

Given: four sentences — which one is a statement?

✅ Correct!

2 \cdot 6 + 1 = 13 \neq 12

, so the sentence is false — and false is a definite truth value, so it is a statement.

❌ Not quite.

x

is unknown, so

x^2 - 9 = 0

is an open sentence: neither true nor false as written.

❌ Not quite. A question asks; it asserts nothing, so it has no truth value.

❌ Not quite. “Beautiful” is a matter of opinion, so the sentence has no definite truth value.

Show solution

Test each sentence for a definite truth value:

$x^2 - 9 = 0$ : $x$ is unknown — an open sentence, not a statement.
“Is 17 a prime number?”: a question, not a statement.
$2 \cdot 6 + 1 = 12$ : $2 \cdot 6 + 1 = 12 + 1 = 13 \neq 12$ , so it is false — a statement.
“17 is a beautiful number.”: an opinion, not a statement.

Being false does not disqualify a sentence; only the lack of a definite truth value does.

Problem 2 · Adding “There Exists”

Given: the sentence “There exists an integer $x$ such that $4x - 6 = 9$ .” — decide whether it is a statement and, if so, its truth value.

Is it a statement?

What is its truth value?

✅ Correct! The words “there exists” turn the open sentence into a statement, and this one is false: the only solution

x = \tfrac{15}{4}

is not an integer.

❌ Not quite. The

x

is no longer unknown-and-free: the sentence claims that some integer works, which is either so or not.

❌ Not quite. A false sentence is still a statement — all a statement needs is a definite truth value.

❌ Not quite. Ask whether the sentence is definitely true or definitely false.

❌ Close, but…

x = \tfrac{15}{4}

does solve the equation, but the sentence asks for an integer, and

\tfrac{15}{4}

is not one.

❌ Not quite. Check it:

4 \cdot 4 - 6 = 16 - 6 = 10 \neq 9

❌ Close, but check the algebra. From

4x - 6 = 9

you add

6

to both sides:

4x = 15

, not

4x = 3

❌ Not quite. Solve

4x - 6 = 9

and ask whether the solution is an integer.

Show solution

Statement? Either an integer satisfying $4x - 6 = 9$ exists or none does, so the sentence has a definite truth value: it is a statement.

Truth value: solve the equation.

4x - 6 = 9 \;\Longrightarrow\; 4x = 15 \;\Longrightarrow\; x = \tfrac{15}{4} = 3.75

The only solution is not an integer. Checking the neighbours confirms it: $4 \cdot 3 - 6 = 6$ and $4 \cdot 4 - 6 = 10$ , and neither is $9$ . So the sentence is a false statement.

Problem 3 · Asserting an Answer

Given: the sentence “The equation $2x^2 + 3x + 5 = 0$ has two real number solutions.” — which description is right?

✅ Correct! The sentence asserts an answer about the equation, so it is a statement; a negative discriminant means no real solutions, so it is false.

❌ Not quite. The sentence is about the equation as a whole — how many real solutions it has — and that number is definite.

❌ Close, but check the sign.

4ac = 4 \cdot 2 \cdot 5 = 40

is subtracted:

b^2 - 4ac = 9 - 40

❌ Not quite. A statement needs a definite truth value, not one we already know; the discriminant settles it.

Show solution

The sentence is declarative and asserts a definite fact about the equation, so it is a statement. To find its truth value, use the discriminant with $a = 2$ , $b = 3$ , $c = 5$ :

b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 5 = 9 - 40 = -31

Since $-31 < 0$ , the equation has no real solutions — let alone two. The sentence is a false statement.

Compare the video's $3x^2 - 5x - 7 = 0$ : there $b^2 - 4ac = 25 + 84 = 109 > 0$ , so the same kind of sentence was true.

Problem 4 · Reading an If-Then

Given: the conditional “If an integer $n$ is not divisible by $2$ , then $n^2$ is odd.” — identify its hypothesis and the one situation that would make it false.

What is the hypothesis?

Which situation would make the conditional false?

✅ Correct! A conditional is false only when the hypothesis is true and the conclusion is false:

n

not divisible by

2

, yet

n^2

even.

❌ Not quite. “

n^2

is odd” follows the word then: it is the conclusion.

❌ Close, but… the hypothesis is everything after if, including the word not.

❌ Not quite. The hypothesis is the part after the word if.

❌ Not quite. Here the hypothesis is false, and the conditional makes no promise about integers divisible by

2

— so it is not false.

❌ Not quite. Hypothesis true and conclusion true: the promise is kept.

❌ Not quite. Look for the situation where the promise is broken.

Show solution

Hypothesis (after if): “ $n$ is not divisible by $2$ .” Conclusion (after then): “ $n^2$ is odd.”

Like the Laura sentence, the conditional is a promise about the hypothesis. It is broken — the conditional is false — only when the hypothesis is true and the conclusion is false: $n$ is not divisible by $2$ but $n^2$ is even.

In fact that situation never happens: if $n = 2k + 1$ , then $n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$ is odd. Showing that the one failing situation cannot occur is exactly what proving the conditional means.

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