Exercise 1161Logic, sets and proofs › Quantifiers and negationL1Added 3 Nov 2026

Negating the uniqueness quantifier

#negation#uniqueness#exists unique

Statement

Negate $\exists! x\in\mathbb{R},\ x^2=1$ .

Hint

$\neg(\exists!x,\ P(x))$ means: either no $x$ satisfies $P$ , or at least two do.

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Negation. Either $\nexists x,\ x^2=1$ or $\exists x\neq y$ with $x^2=y^2=1$ .

Concretely. $x=1$ and $x=-1$ both satisfy $x^2=1$ , so the original statement is false. The negation holds.

Conclusion. The negation is: "no real $x$ satisfies $x^2=1$ or at least two distinct reals do".