Mathematical-Reasoning · Unit 5 · Video 1 · Interactive Practice

What Makes a Quantified Statement True: For Every, There Exists, and Their Negations

IKey Formulas

Form	Name	True exactly when
$(\forall x \in U)(P(x))$	Universal quantifier: “for every”, “for each”	$\{x \in U \mid P(x)\} = U$ : every element of $U$ makes $P(x)$ true
$(\exists x \in U)(P(x))$	Existential quantifier: “there exists”, “there is at least one”	$\{x \in U \mid P(x)\} \neq \emptyset$ : at least one element of $U$ makes $P(x)$ true
$(\forall x \in \mathbb{Z})(x \text{ is a multiple of } 2)$ negates to $(\exists x \in \mathbb{Z})(x \text{ is not a multiple of } 2)$	Negating “for every”	the statement is false, its negation true, since $3$ is not a multiple of $2$
$(\exists x \in \mathbb{Z})(x^3 > 0)$ negates to $(\forall x \in \mathbb{Z})(x^3 \le 0)$	Negating “there exists”	the statement is true ( $1^3 = 1 > 0$ ), its negation false

Key Insight: only a quantifier turns an open sentence into a statement. $\sin(2x) = 2\sin x\cos x$ has truth set $\mathbb{R}$ and is still an open sentence; $(\forall x \in \mathbb{R})(\sin(2x) = 2\sin x\cos x)$ is a true statement. “There exists” means at least one, never exactly one.

IIOne Truth Set, Two Quantifiers

A $\forall$ statement is true when the truth set is all of $U$ ; an $\exists$ statement, when the truth set is nonempty.

Predicate P(x) Quantifier Test value x 1/2

IIINegating a Quantified Statement

Negation swaps $\forall$ and $\exists$ , negates the predicate, and gives the opposite truth value.

Statement Step 1

💡 The next video states this pattern as a theorem and proves it with truth sets.

IVQuiz Questions

Problem 1 · An Existential Statement over the Integers

Given: the statement $(\exists x \in \mathbb{Z})(2x + 1 = 6)$ . Which verdict and reason are correct?

✅ Correct! The only solution of

2x + 1 = 6

\tfrac{5}{2}

, which is not an integer, so the truth set over

\mathbb{Z}

is empty and the existential statement is false.

❌ Not quite.

\tfrac{5}{2}

does solve the equation, but the universal set is

\mathbb{Z}

and

\tfrac{5}{2} \notin \mathbb{Z}

❌ Not quite. The phrase

(\exists x \in \mathbb{Z})

quantifies

x

, so this is a statement and has a truth value.

❌ Right verdict, wrong reason. One value that fails cannot make an existential statement false; you must show that no integer works.

❌ Not quite. Solve the equation, then check whether the solution lies in the universal set

\mathbb{Z}

Show solution

An existential statement is true exactly when its truth set is not empty. Solve the predicate:

2x + 1 = 6 \;\Longrightarrow\; 2x = 5 \;\Longrightarrow\; x = \tfrac{5}{2}.

That is the only solution, and $\tfrac{5}{2} \notin \mathbb{Z}$ . So

\{x \in \mathbb{Z} \mid 2x + 1 = 6\} = \emptyset,

and the statement is false. Checking a single value such as $x = 0$ is not enough: that only shows $0$ is outside the truth set.

Problem 2 · Statement or Open Sentence?

Given: the universal set is $\mathbb{R}$ for every variable. Which of the following is a statement, and is it true?

Which one is a statement?

What is its truth value?

✅ Correct! Only the quantified sentence is a statement, and it is true because at least one real number (in fact two) satisfies

x^2 = 9

❌ Not quite. Every real number makes this identity true, so its truth set is

\mathbb{R}

, but nothing quantifies

x

: it is still an open sentence.

❌ Not quite. Nothing quantifies

x

in that sentence, so it is an open sentence with a truth set, not a statement.

❌ Not quite. “There exists” means at least one, that is, one or more; two solutions are more than enough.

❌ Not quite. One failing value cannot make an existential statement false; the truth set only has to contain at least one element.

❌ Not quite. Find the truth set of

x^2 = 9

over

\mathbb{R}

and ask whether it is empty.

Show solution

The test is whether the variable is quantified. In (a), (b) and (d) nothing quantifies $x$ , so each is an open sentence with a truth set:

(a) is an identity: its truth set is $\mathbb{R}$ , yet it is an open sentence;
(b) has truth set $\{\tfrac{14}{3}\}$ ;
(d) has truth set $\{x \in \mathbb{R} \mid x \ge 0\}$ .

Only (c) is a statement. Its predicate gives $x^2 = 9 \iff x = 3$ or $x = -3$ , so

\{x \in \mathbb{R} \mid x^2 = 9\} = \{-3, 3\} \neq \emptyset,

and the existential statement is true. It claims at least one solution, so having two does not hurt.

Problem 3 · A Quantified Conditional

Given: the open sentence “If $x^2 < 9$ , then $x > 0$ ” with universal set $\mathbb{R}$ . Find its truth set, then decide the statement $(\forall x \in \mathbb{R})(\text{if } x^2 < 9 \text{, then } x > 0)$ .

What is the truth set of the open sentence?

Is the quantified statement true?

✅ Correct! The truth set leaves out

-3 < x \le 0

, so it is not all of

\mathbb{R}

, and

x = -1

is one of the values it leaves out.

❌ Not quite. That is where the conclusion is true; the conditional is also true wherever the hypothesis

x^2 < 9

is false.

❌ Not quite. That is where the conditional fails (hypothesis true, conclusion false), the opposite of the truth set.

❌ Not quite. That is where the hypothesis is false; the conditional is also true wherever the conclusion

x > 0

is true.

❌ Not quite. A conditional is true when its hypothesis is false or its conclusion is true; combine those two sets.

❌ Not quite. One value that works cannot make a universal statement true; every real number would have to be in the truth set.

❌ Right verdict, wrong reason. At

x = 4

the hypothesis is false, so the conditional is true there;

4

is in the truth set.

❌ Not quite. A nonempty truth set makes an existential statement true; a universal statement needs the truth set to be all of

\mathbb{R}

❌ Not quite. Compare the truth set from the first part with

\mathbb{R}

Show solution

A conditional is true when its hypothesis is false or its conclusion is true.

The hypothesis $x^2 < 9$ is false when $x^2 \ge 9$ , that is, when $x \le -3$ or $x \ge 3$ .
The conclusion $x > 0$ is true when $x > 0$ .

Together these give $x \le -3$ or $x > 0$ (the ray $x \ge 3$ is already inside $x > 0$ ):

\{x \in \mathbb{R} \mid x \le -3\} \cup \{x \in \mathbb{R} \mid x > 0\}.

The conditional fails exactly when $-3 < x \le 0$ . For instance, $x = -1$ gives $(-1)^2 = 1 < 9$ (hypothesis true) but $-1 > 0$ is false (conclusion false). The truth set is not all of $\mathbb{R}$ , so the universal statement is false.

Problem 4 · Negate, Then Decide

Given: the statement $(\forall x \in \mathbb{Z})(x^2 > x)$ . Find its negation, then decide which of the two is true.

What is the negation?

Which of the pair is true?

✅ Correct! The quantifier

\forall

became

\exists

, the predicate

x^2 > x

became

x^2 \le x

, and the integer

1

makes the negation true, so the original is false.

❌ Not quite. The quantifier must change too: the negation of a claim about every integer says that at least one integer breaks it.

❌ Not quite. The quantifier changed, but the predicate must be negated as well: not

x^2 > x

x^2 \le x

❌ Not quite. “Not greater than” means “less than or equal to”, so

x^2 = x

must be included.

❌ Not quite. Change

\forall

\exists

and negate the predicate.

❌ Not quite. One integer that works cannot make a universal statement true; test small integers such as

0

and

1

❌ Not quite. A statement and its negation always have opposite truth values.

❌ Not quite. A statement and its negation always have opposite truth values, so exactly one of them is true.

❌ Not quite. Look for an integer whose square is not larger than itself.

Show solution

The statement claims that every integer satisfies $x^2 > x$ . Its negation says that this claim fails, so at least one integer does not satisfy $x^2 > x$ . A number that is not greater than another is less than or equal to it, so the negation is

(\exists x \in \mathbb{Z})(x^2 \le x).

Now test small integers: $1^2 = 1 \le 1$ (and also $0^2 = 0 \le 0$ ). So the truth set of $x^2 \le x$ over $\mathbb{Z}$ is not empty, and the negation is true. The original statement is therefore false: its truth set leaves out $0$ and $1$ , so it is not all of $\mathbb{Z}$ .

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