The statement ∀x ∈ ℝ, x² ≥ 0 is claimed to be false. What is the minimum needed to refute it?
AShow that x² < 0 for most real numbers
BProvide a single real number x where x² < 0
CShow that no real number satisfies x² ≥ 0
DDemonstrate that the claim fails for infinitely many values
Disproving a universal claim ∀x P(x) requires exactly one counterexample — one value of x that makes P(x) false. 'Most' or 'infinitely many' is far more than necessary. Conversely, to *prove* a universal claim, you need an argument that works for an arbitrary x — checking any finite number of cases is never enough, no matter how many.
Question 2 Multiple Choice
Compare ∀ε>0 ∃δ>0 [P(ε,δ)] with ∃δ>0 ∀ε>0 [P(ε,δ)]. Why does the order of quantifiers matter?
AThe order doesn't matter — ∀ and ∃ always commute when they involve different variables
BIn ∀ε ∃δ, δ may depend on ε; in ∃δ ∀ε, one fixed δ must work for all ε — these express genuinely different claims
CThe first form is just notational convention; both mean the same thing logically
DThe second form is always stronger because the universal quantifier appears last
Order is critical because inner quantifiers can depend on outer ones. In ∀ε ∃δ, you choose δ after seeing ε, so δ is allowed to depend on ε. In ∃δ ∀ε, one fixed δ must serve every ε simultaneously — a strictly stronger claim that is usually false. This is exactly the distinction in the epsilon-delta limit definition: the δ must respond to each ε, not be chosen in advance for all of them.
Question 3 True / False
A predicate P(x) = 'x is prime' has a definite truth value even before x is specified.
TTrue
FFalse
Answer: False
A predicate with a free variable is not a proposition — it has no fixed truth value until a specific value is substituted for x. P(7) is true, P(4) is false, but P(x) by itself is neither true nor false. Quantifying the variable (∀x P(x) or ∃x P(x)) converts the predicate into a proposition with a definite truth value.
Question 4 True / False
To prove an existential statement ∃x P(x), it suffices to exhibit one concrete value of x that makes P(x) true.
TTrue
FFalse
Answer: True
Existential claims are proved by witness. Since ∃x P(x) only asserts that at least one value works, showing any single concrete example is a complete proof. This contrasts with universal claims, which require an argument valid for an arbitrary x. The asymmetry — ∀ needs general argument, ∃ needs one example — is essential to proof strategy.
Question 5 Short Answer
Why does swapping the order of ∀ and ∃ in a mathematical statement change its meaning? Give an example that illustrates the difference.
Think about your answer, then reveal below.
Model answer: In ∀x ∃y ..., the value of y can depend on x (you choose y after seeing x). In ∃y ∀x ..., one fixed y must work for every x simultaneously. Example: ∀x ∈ ℝ ∃y ∈ ℝ (y > x) is true — for any x, pick y = x + 1. But ∃y ∈ ℝ ∀x ∈ ℝ (y > x) is false — no single real number exceeds every real number.
The key is whether inner variables can depend on outer ones. When ∀ comes first, its variable is 'given,' and subsequent ∃ witnesses can be tailored to it. When ∃ comes first, its witness must be fixed before the ∀ variable is known. This distinction underlies virtually every definition in analysis and topology that uses nested quantifiers.