The NEGATION of $\forall x\, P(x)$ is
A$\forall x\, \neg P(x)$
B$\exists x\, \neg P(x)$
C$\exists x\, P(x)$
D$\neg \exists x\, P(x)$
Answer & Solution
Correct answer: B. $\exists x\, \neg P(x)$
1. DeMorgan's laws for quantifiers:
- $\neg \forall x\, P(x) \equiv \exists x\, \neg P(x)$
- $\neg \exists x\, P(x) \equiv \forall x\, \neg P(x)$
2. To negate 'all $x$ have property $P$', it suffices to find ONE $x$ that fails.
3. Example: negate 'every prime is odd'. The negation is 'there exists a prime that is not odd' — which is true, since 2 is even.
4. Option A is too strong. Options C, D have different meanings.
_Source: Oscar Levin, "Discrete Mathematics: An Open Introduction", §0.3 (Quantifiers and DeMorgan)._
Related questions
A PREDICATE $P(x)$ isHow many ways to put 10 IDENTICAL balls into 4 DISTINCT boxes (each box can hold 0 or moreHow many DIFFERENT subsets of size 3 can be selected from a set of 7 elements?In how many ways can 7 people sit in a ROW?$\binom{n}{k} = \binom{n}{n-k}$. This identity reflectsHow many DIFFERENT 5-bit binary strings are there?The PIGEONHOLE PRINCIPLE states that if $n$ pigeons are placed into $m$ holes with $n > m$A function $f: A \to B$ is INJECTIVE (one-to-one) if