Denial of Existence/Examples/x less than or equal to 3
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Examples of Denial of Existence
Let $S \subseteq \R$ be a subset of the real numbers.
Let $P$ be the statement:
- $\exists x \in S: x \le 3$
The negation of $P$ is the statement written in its simplest form as:
- $\forall x \in S: x > 3$
Proof
\(\ds \) | \(\) | \(\ds \lnot \exists x \in S: x \le 3\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \forall x \in S: \lnot x \le 3\) | Denial of Universality | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \forall x \in S: x \not \le 3\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \forall x \in S: x > 3\) |
$\blacksquare$
Examples
Example where $S = \set {2, 3, 4}$
Let $P$ be the statement:
- $\exists x \in S: x \le 3$
and $\lnot P$ its negation:
- $\forall x \in S: x > 3$
Let $S = \set {2, 3, 4}$.
Then we have that:
- $P$ is true
and consequently:
- $\lnot P$ is false
Example where $S = \closedint 0 3$
Let $P$ be the statement:
- $\exists x \in S: x \le 3$
and $\lnot P$ its negation:
- $\forall x \in S: x > 3$
Let $S = \closedint 0 3$ where $\closedint \cdot \cdot$ denotes a closed real interval.
Then we have that:
- $P$ is true
and consequently:
- $\lnot P$ is false
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $1 \ \text{(ii)}$