Quantifiers Quiz: Master Universal and Existential Reasoning

“Every dog is friendly” is ∀x(Dog(x) → Friendly(x)). Negating a universal gives an existential with a negated predicate: ∃x(Dog(x) ∧ ¬Friendly(x)), which in English is “Some dog is not friendly.”

For any real x, we can choose y = x − 1, which is strictly smaller than x. Since such a y exists for every x, the statement is true over ℝ.

Saying “no student failed” means that whenever x is a student, x is not among those who failed. This is captured by the universal conditional ∀x (Student(x) → ¬Failed(x)).

A “smallest” real number would be some x such that x ≤ y for all real y. Saying there is no such smallest number is written as ¬∃x ∀y (x ≤ y).

The quantifier ∀x marks this as a universal statement about all real numbers, asserting that every real x satisfies x² ≥ 0.

The symbol ∃ expresses that there is at least one natural number x whose square is 9, which makes it an existential statement (and in fact a true one, since x = 3 works).

A universal quantifier in front of a disjunction requires that for each element x, P(x) ∨ Q(x) is true. This means that for every x, at least one of the two predicates is satisfied.

The existential quantifier ∃x requires that there is at least one element in the domain for which the disjunction P(x) ∨ Q(x) is true, i.e., P(x) or Q(x) holds for that x.

Negating a universal turns it into an existential with a negated predicate: ¬∀x P(x) is equivalent to ∃x ¬P(x), expressing “not all” as “there exists at least one counterexample.”

Quantifier order usually changes meaning: “for every x there exists some y” does not generally mean the same as “there exists one y that works for all x.”

Given any real x, choosing y = −x makes x + y = 0. Since such a y exists for every x, the universally quantified statement is true over ℝ.

The equation x + y = y for all real y holds exactly when x = 0, since 0 + y = y for every y. Because such an x exists (namely 0), the existential statement is true.

If x + y = x were to hold for all real y, then y would have to be 0 for every y, which is impossible. No real x satisfies this for all y, so the existential statement is false.

The statement says that for each nonzero x, there exists some y such that x·y = 1. This is exactly what the conditional with an existential in option A expresses.

The existential quantifier ∃y is inside the scope of ∀x, so for each particular x we just require at least one suitable y. That y may depend on x and need not be the same for all x.

Negating a universal switches it to an existential and negates the predicate: not “P holds for all x” means “there is at least one x for which P fails.”

Negating an existential switches it to a universal with a negated predicate: not “there exists an x with P(x)” means “for every x, P(x) is false.”

This would require a single integer x that is greater than every integer y. The integers are unbounded above, so no such greatest integer exists and the statement is false.

For any integer x, we can take y = x + 1, which is always an integer larger than x. Thus every x has some y with x

The intended meaning is that for each integer x there exists an integer y one greater than x. This is expressed by the universally quantified ∀x ∃y (y = x + 1).