Mathematical-Reasoning · Unit 1 · Video 2 · Interactive Practice

Quantifiers, Conjectures and Counterexamples: How Mathematicians Explore

IKey Formulas

Sentence form	True when	One example can prove
$P(x)$ with $x$ unknown	never decided: an open sentence has no truth value	nothing (not a statement)
There exists $x$ such that $P(x)$	at least one value makes $P(x)$ true	TRUE, by one witness
For all $x$ , $P(x)$	every value makes $P(x)$ true	FALSE, by one counterexample
If $x$ is …, then $Q(x)$	every such $x$ gives $Q(x)$ (a hidden “for all”)	FALSE, by one counterexample

Key Insight: The open sentence $(a+b)^2 = a^2 + b^2$ is true with “there exist” in front ( $a = 1$ , $b = 0$ ) and false with “for all” in front ( $a = 2$ , $b = 3$ gives $25 \neq 13$ ). Only the quantifier changed.

IIVisualization 1 — One Equation, Two Quantifiers

One pair $(a, b)$ settles “there exist” when it works, and settles “for all” when it fails.

Quantifier

IIIVisualization 2 — Testing $\sin(2x) = 2\sin x$

Matches at $x = 0$ and $x = \pi$ cannot prove $\sin(2x) = 2\sin x$ for every $x$ ; one mismatch disproves it.

x π/4

IVVisualization 3 — Evidence Is Not Proof

A “for all” claim over infinitely many cases survives any number of passing tests, until one fails.

Conjecture Tests 3

VQuiz Questions

Problem 1 · Statement or Not?

Which of these sentences is a statement?

✅ Correct! The “if … then” acts as a hidden “for every”: it claims something about every odd integer, so it is true or false (here, true).

❌ Not quite. This is an open sentence: its letters are unknown and no quantifier says which values it is about, so it has no truth value yet.

❌ Not quite. “Solve …” is a directive. It tells us to do something and asserts nothing, so it is neither true nor false.

Show solution

A statement must be either true or false.

(a) and (d) are open sentences: $x$ , $a$ , $b$ are unknown and nothing quantifies them. (Putting “for all real $a, b$ ” in front of (d) would make it a false statement; “there exist real $a, b$ ” would make it a true one.)
(b) is a directive, not a claim.
(c) contains a letter, but the “if … then” quantifies it: the claim is made about every odd integer $x$ . So it is a statement.

Answer: (c).

Problem 2 · Three Matches Are Not Enough

Given: “For all real numbers $x$ , $x^2 \ge x$ .” A student checks $x = 2$ , $x = 3$ and $x = -1$ : $4 \ge 2$ , $9 \ge 3$ , $1 \ge -1$ . Which conclusion is correct?

✅ Correct! One value that fails is enough to make a “for all” statement false, and

x = \tfrac{1}{2}

fails.

❌ Not quite. Three matches are evidence, never proof, for a claim about infinitely many real numbers. Look between

0

and

1

❌ Not quite. The words “for all real numbers

x

” quantify

x

, so the sentence is a statement.

❌ Not quite. Checking everything is needed to prove it true by examples. To prove it false, a single counterexample suffices.

Show solution

Rewrite the inequality: $x^2 \ge x \iff x^2 - x \ge 0 \iff x(x - 1) \ge 0$ .

The product $x(x-1)$ is negative exactly when $0 \lt x \lt 1$ , so any $x$ in that interval is a counterexample. Take $x = \tfrac{1}{2}$ :

\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4} < \tfrac{1}{2}.

The statement is false. The student’s three tests all happened to lie outside $(0, 1)$ .

Problem 3 · Find a Witness

Given: “There are integers $x$ and $y$ such that $3x + 5y = 1$ .”

Which pair is a witness?

What does that witness establish?

✅ Correct!

3(2) + 5(-1) = 6 - 5 = 1

, and that single pair proves the “there are” statement true.

❌ Not quite. Substitute the pair into

3x + 5y

and check whether you get exactly

1

❌ Close, but check the sign.

3(-2) + 5(1) = -6 + 5 = -1

, not

1

. Flip both signs.

❌ Not quite. A failing pair disproves a “for all” statement. A “there are” statement is false only if every pair fails.

❌ Not quite. It is the “for all” statements over infinitely many cases that need a proof. A “there are” statement is proved by exhibiting one witness.

❌ Not quite. Once

x

and

y

are quantified, the sentence is one statement with one truth value; it is not “true for” particular values.

Show solution

Check each pair in $3x + 5y$ :

$(1, -1)$ : $3 - 5 = -2$
$(-1, 2)$ : $-3 + 10 = 7$
$(-2, 1)$ : $-6 + 5 = -1$
$(2, -1)$ : $6 - 5 = 1$ ✓

So $(x, y) = (2, -1)$ is a witness. “There are integers $x$ and $y$ such that …” claims that at least one pair works, and we have exhibited one, so the statement is true. Other witnesses exist too, such as $(-3, 2)$ : $-9 + 10 = 1$ , but one is enough.

Problem 4 · A Double-Angle Conjecture

Conjecture: for all real numbers $x$ , $\cos(2x) = 2\cos x - 1$ .

Which value of $x$ is a counterexample?

Using $\cos(2x) = 2\cos^2 x - 1$ , exactly when do the two sides agree?

✅ Correct!

x = \pi

gives

1 \neq -3

, and the factored form

2\cos x(\cos x - 1) = 0

explains why

0

\tfrac{\pi}{2}

and

2\pi

all matched.

❌ Not quite. That value is a match: both sides are equal there. Evaluate

\cos(2x)

and

2\cos x - 1

separately at each option.

❌ Not quite. That is the answer for the sine conjecture in the video. Here, at

x = \pi

we have

\cos x = -1

, and the sides are

1

and

-3

❌ Not quite. Set

2\cos^2 x - 1 = 2\cos x - 1

, move everything to one side and factor.

Show solution

Step 1: Test the options.

$x = 0$ : $\cos 0 = 1$ and $2(1) - 1 = 1$ (match)
$x = \tfrac{\pi}{2}$ : $\cos \pi = -1$ and $2(0) - 1 = -1$ (match)
$x = \pi$ : $\cos 2\pi = 1$ and $2(-1) - 1 = -3$ , and $1 \neq -3$ (counterexample)
$x = 2\pi$ : $\cos 4\pi = 1$ and $2(1) - 1 = 1$ (match)

So $x = \pi$ is a counterexample and the conjecture is false.

Step 2: Explain the matches with prior knowledge. Replace $\cos(2x)$ by the identity:

2\cos^2 x - 1 = 2\cos x - 1 \iff 2\cos^2 x - 2\cos x = 0 \iff 2\cos x\,(\cos x - 1) = 0.

The sides agree exactly when $\cos x = 0$ or $\cos x = 1$ , that is, at $x = \tfrac{\pi}{2} + k\pi$ or $x = 2k\pi$ . That covers all three matches, and $x = \pi$ is in neither set.

Solved: 0 / 4

Quantifiers, Conjectures and Counterexamples: How Mathematicians Explore

IKey Formulas

IIVisualization 1 — One Equation, Two Quantifiers

IIIVisualization 2 — Testing sin⁡(2x)=2sin⁡x\sin(2x) = 2\sin xsin(2x)=2sinx

IVVisualization 3 — Evidence Is Not Proof

VQuiz Questions

IIIVisualization 2 — Testing $\sin(2x) = 2\sin x$