Mathematical-Reasoning · Unit 5 · Video 2 · Interactive Practice

Negation Flips the Quantifier: Reading 'For Every' and 'There Exists' in English

IKey Formulas

Statement	Its negation	Why
$\neg(\forall x \in U)[P(x)]$	$(\exists x \in U)[\neg P(x)]$	the truth set of $P(x)$ is not all of $U$ , so some $x \in U$ makes $P(x)$ false
$\neg(\exists x \in U)[P(x)]$	$(\forall x \in U)[\neg P(x)]$	the truth set of $P(x)$ is $\emptyset$ , so every $x \in U$ makes $P(x)$ false
$a \ge b$ , $\ a > b$ , $\ a = b$	$a < b$ , $\ a \le b$ , $\ a \ne b$	any two real numbers satisfy exactly one of $a < b$ , $a = b$ , $a > b$

Key Insight: switch the quantifier, keep the universal set, negate the predicate, then check that exactly one of the statement and its negation is true. A quantifier can hide in English: “the square of a real number is greater than $0$ ” and “if $x \in \mathbb{R}$ , then $x^2 > 0$ ” both mean $(\forall x \in \mathbb{R})(x^2 > 0)$ .

IIBuilding a Negation

A negation switches the quantifier, keeps the universal set and negates the predicate; any other edit fails.

Statement

💡 Challenge: for $(\exists x \in \mathbb{Q})(x^2 - 3x - 7 = 0)$ , which is false, build a candidate that is true but is still not its negation.

IIINegating a Comparison

An open sentence $Q(x)$ is the negation of $P(x)$ exactly when every $x$ makes one of them true, never both.

Compare P(x) relation Q(x) relation

IVQuiz Questions

Problem 1 · Apply the Theorem

Given: the statement $(\forall x \in \mathbb{R})(x^2 + 4 \ge 4x)$ . Which is its negation, written with no negation symbol?

✅ Correct! The

\forall

becomes

\exists

\mathbb{R}

stays, and

\neg(x^2 + 4 \ge 4x)

x^2 + 4 < 4x

❌ Not quite. This keeps the quantifier. It claims every real number fails the inequality, but the negation only needs one that fails, so

\forall

must become

\exists

❌ Close, but… the negation of

\ge

<

, not

\le

. At

x = 2

both

2^2 + 4 \ge 4 \cdot 2

and

2^2 + 4 \le 4 \cdot 2

are true (

8 = 8

), and a predicate and its negation are never both true at the same value.

❌ Not quite. The universal set must stay

\mathbb{R}

. Changing it to

\mathbb{Z}

makes a claim about different numbers.

Show solution

By the theorem on negations of quantified statements,

\neg(\forall x \in \mathbb{R})(x^2 + 4 \ge 4x) \equiv (\exists x \in \mathbb{R})\,\neg(x^2 + 4 \ge 4x) \equiv (\exists x \in \mathbb{R})(x^2 + 4 < 4x).

Check the truth values. Since $x^2 + 4 - 4x = (x - 2)^2 \ge 0$ for every real $x$ , the original statement is true, so its negation must be false. It is: no real $x$ makes $(x - 2)^2 < 0$ .

Option b is true (with $x = 2$ a witness), so it has the same truth value as the original statement and cannot be its negation.

Problem 2 · A Hidden Quantifier and the Equality Case

Given: the sentence “The square of a natural number is greater than $1$ ,” where $\mathbb{N} = \{1, 2, 3, \ldots\}$ . Which is its negation?

✅ Correct! The sentence means

(\forall x \in \mathbb{N})(x^2 > 1)

, so its negation is

(\exists x \in \mathbb{N})(x^2 \le 1)

. It is true with witness

x = 1

, and the original is false.

❌ Close, but… the negation of

>

\le

, not

<

. The equality case matters here:

1^2 = 1

is exactly what makes the original false, yet no natural number has square less than

1

, so this candidate is false along with the original.

❌ Not quite. “The square of a natural number…” hides a universal quantifier, and this candidate keeps it. It claims every natural number has square at most

1

, which

2^2 = 4

refutes, so it is false along with the original.

❌ Not quite. The quantifier switched, but the predicate was never negated. By the theorem the negation is

(\exists x \in \mathbb{N})\,\neg(x^2 > 1)

, and

\neg(x^2 > 1)

x^2 \le 1

Show solution

The sentence states no quantifier word, but it claims something about every natural number:

(\forall x \in \mathbb{N})(x^2 > 1).

Switch the quantifier, keep $\mathbb{N}$ , and negate the predicate ( $\neg(x^2 > 1)$ is $x^2 \le 1$ ):

(\exists x \in \mathbb{N})(x^2 \le 1).

In English: there exists a natural number whose square is less than or equal to $1$ .

Check: $x = 1$ gives $1^2 = 1 \le 1$ , so the negation is true, and the original is false because $1^2 = 1$ is not greater than $1$ . Exactly one of the two is true.

Problem 3 · Negate, Then Decide

Given: the statement $(\exists x \in \mathbb{Z})(x^2 - 5x + 6 < 0)$ . Find its negation with no negation symbol, then decide which of the two is true.

What is the negation?

Which one is true?

✅ Correct! The negation

(\forall x \in \mathbb{Z})(x^2 - 5x + 6 \ge 0)

is true, because no integer lies strictly between

2

and

3

; the original statement is false.

❌ Not quite. Switch

\exists

\forall

, keep

\mathbb{Z}

, and replace

<

by its negation

\ge

❌ Close, but… the negation of

<

\ge

, not

>

. At

x = 2

the value is

4 - 10 + 6 = 0

, which makes both

x^2 - 5x + 6 < 0

and

x^2 - 5x + 6 > 0

false.

❌ Not quite. Keeping

\exists

is the common slip. The negation must say that no integer works, which is a claim about every integer, so it needs

\forall

❌ Not quite. The universal set must stay

\mathbb{Z}

. Over

\mathbb{R}

the claim changes, and

x = \tfrac{5}{2}

even makes it false.

❌ Not quite. Check whether any integer makes

x^2 - 5x + 6

negative.

❌ Close, but…

\left(\tfrac{5}{2}\right)^2 - 5 \cdot \tfrac{5}{2} + 6 = -\tfrac{1}{4} < 0

, but

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

. A witness must come from the universal set.

❌ Not quite. At

x = 2

the value is

4 - 10 + 6 = 0

, and

0 < 0

is false, so

2

is not a witness.

❌ Not quite. The open sentence

x^2 - 5x + 6 < 0

has no truth value on its own, but a quantifier turns it into a statement, which is either true or false.

Show solution

Negation. By the theorem,

\neg(\exists x \in \mathbb{Z})(x^2 - 5x + 6 < 0) \equiv (\forall x \in \mathbb{Z})\,\neg(x^2 - 5x + 6 < 0) \equiv (\forall x \in \mathbb{Z})(x^2 - 5x + 6 \ge 0).

Truth value. Completing the square,

x^2 - 5x + 6 = \left(x - \tfrac{5}{2}\right)^2 - \tfrac{1}{4},

so $x^2 - 5x + 6 < 0$ exactly when $\left(x - \tfrac{5}{2}\right)^2 < \tfrac{1}{4}$ , that is, when $x$ is within $\tfrac{1}{2}$ of $\tfrac{5}{2}$ : $2 < x < 3$ . No integer lies strictly between $2$ and $3$ , so the truth set over $\mathbb{Z}$ is $\emptyset$ . The original statement is false and its negation is true.

Over $\mathbb{R}$ the original would be true, with witness $x = \tfrac{5}{2}$ : the universal set decides the truth value.

Problem 4 · A Compound Predicate

Given: the statement “There is a negative real number whose square is $4$ ,” written $(\exists x \in \mathbb{R})(x^2 = 4 \wedge x < 0)$ . Which is its negation, with no negation symbol?

✅ Correct! Switch

\exists

\forall

, keep

\mathbb{R}

, and negate the conjunction with De Morgan's law:

\neg(x^2 = 4 \wedge x < 0) \equiv x^2 \ne 4 \vee x \ge 0

❌ Close, but… De Morgan's law turns the negation of a conjunction into a disjunction:

\neg(P \wedge Q) \equiv \neg P \vee \neg Q

. This candidate is false at

x = -1

, which the true negation allows, since

(-1)^2 \ne 4

❌ Not quite. Keeping

\exists

is the common slip. The negation says no real number is both negative and a square root of

4

, which is a claim about every real number.

❌ Not quite. The negation of

x^2 = 4

x^2 \ne 4

, not

x^2 > 4

. At

x = -1

the true negation's predicate holds but this one's fails, since

1 > 4

and

-1 \ge 0

are both false.

Show solution

By the theorem, $\neg(\exists x \in \mathbb{R})[P(x)] \equiv (\forall x \in \mathbb{R})[\neg P(x)]$ with $P(x)$ the conjunction $x^2 = 4 \wedge x < 0$ . De Morgan's law negates the conjunction:

\neg(x^2 = 4 \wedge x < 0) \equiv \neg(x^2 = 4) \vee \neg(x < 0) \equiv x^2 \ne 4 \vee x \ge 0.

So the negation is

(\forall x \in \mathbb{R})(x^2 \ne 4 \vee x \ge 0).

In English: for each real number $x$ , $x^2 \ne 4$ or $x \ge 0$ .

Check: the original is true with witness $x = -2$ , since $(-2)^2 = 4$ and $-2 < 0$ . The negation is false because $x = -2$ breaks it: $(-2)^2 \ne 4$ is false and $-2 \ge 0$ is false. Exactly one of the two is true.

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