Questions: Existential Quantifier and Existence Statements

Question 1

Which of the following correctly proves the statement 'There exists an integer x such that x² = x'?

• A. Show that x = 0.5 satisfies x² = x = 0.25, so the statement holds
• B. Exhibit x = 0 and verify 0² = 0, confirming the predicate is satisfied
• C. Show that for all integers x² ≥ x, which is consistent with existence
• D. Assume no such integer exists and derive a contradiction

Answer: (B)

A constructive existence proof requires exhibiting a specific witness in the correct domain and verifying the predicate. Here, x = 0 (an integer) satisfies 0² = 0 ✓. Also x = 1: 1² = 1 ✓. Option A fails because 0.5 is not an integer — domain matters. Option C is true but does not establish existence of a specific case. The most direct proof of ∃x P(x) is to name an x and verify P(x) holds.

Question 2

What is the correct negation of the statement '∃x P(x)' (there exists an x satisfying P)?

• A. ∃x ¬P(x) — there exists an x that does not satisfy P
• B. ¬∃x P(x) — written with the negation outside but not simplified
• C. ∀x ¬P(x) — for all x, P does not hold
• D. ∀x P(x) — for all x, P holds

Answer: (C)

The negation of an existential statement is a universal statement: ¬(∃x P(x)) ≡ ∀x ¬P(x). 'It is not the case that some x satisfies P' is exactly 'every x fails to satisfy P.' This equivalence — that 'there exists no x with P' means 'all x lack P' — is fundamental to logic and essential for proofs by contradiction. Option A (∃x ¬P(x)) would mean 'some x fails,' which is weaker than 'no x succeeds.'

Question 3

To prove an existential statement ∃x P(x), you should usually find and exhibit a specific concrete example.

• T. True
• F. False

Answer: False

There are two valid strategies: constructive proofs (exhibit a specific witness) and nonconstructive proofs (establish existence without identifying which element satisfies P). For example, the Intermediate Value Theorem proves a continuous function has a zero without pinning down exactly where. Nonconstructive proofs using contradiction, counting arguments, or topological theorems are fully rigorous. The misconception that existence always requires an explicit example is a common source of confusion.

Question 4

The statement ∃x (x² = 2) is true over the real numbers but false over the rational numbers.

• T. True
• F. False

Answer: True

The truth value of a quantified statement depends on the domain. √2 is irrational, so no rational number squares to 2 — ∃x (x² = 2) is false over ℚ. But √2 ∈ ℝ, so the statement is true over ℝ. This illustrates why specifying the domain is essential: the same formula can be true in one number system and false in another. Forgetting to anchor a quantified statement to a domain leaves it meaningless.

Question 5

What is the logical relationship between ∃x P(x) and ∀x P(x)? If the universal statement is true, what can you conclude about the existential one?

Model answer: ∀x P(x) implies ∃x P(x): if P holds for every element in the domain, it certainly holds for at least one. But ∃x P(x) does not imply ∀x P(x): existence of one satisfying element says nothing about the rest. The universal is strictly stronger. Their negations swap quantifiers: ¬(∀x P(x)) ≡ ∃x ¬P(x), and ¬(∃x P(x)) ≡ ∀x ¬P(x).

Understanding this asymmetry is critical for proofs. To disprove a universal claim ∀x P(x), you only need one counterexample (∃x ¬P(x)). To prove an existential claim ∃x P(x), you only need one witness. The difficulty of a claim depends on which direction you are working: one example proves existence but one example cannot establish universality.