Write the statement using quantifiers

Question

I have a statement

S: "No positive number $x$ satisfies the equation $f(x) = 5$."

Can I write the statement S as

$\nexists x \mid f(x) = 5$.

Answer 1

Some equivalent ways of writing the statement include: $$ \forall x \in \mathbb{R}( x\geq 0 \implies \lnot (f(x)=5)) $$ $$ \forall x \in \mathbb{R}( x\geq 0 \implies f(x)\neq 5) $$ $$ \forall x \in \mathbb{R}_+(f(x)\neq 5) $$

Answer 2

I have a statement: $S$: "No positive number x satisfies the equation $f(x)=5$ ."

Can I write the statement S as: $\not{∃}x\mid f(x)=5$

That says "there does not exists anything that satisfies $f(x)=5$."   Which is fine as long as the domain of discourse has already been established as "positive numbers".   Otherwise you should include the restriction.

(PS: While still viable, usage of $\;\not \exists\;$ appears to be somewhat depreciated in favour of $\;\neg\exists\;$.)

Some ways to do so are: $$\neg \exists x: \big(x\in \Bbb R \wedge x>0 \wedge f(x)=5\big) \\[1ex] \neg \exists x : \big(x\in \Bbb R_+ \wedge f(x)=5\big) \\[2ex] \neg \exists x\in \Bbb R : \big(x>0 \wedge f(x)=5 \big) \\[1ex] \neg \exists x\in \Bbb R_+ : \big( f(x)=5 \big) $$

nullUser's statements using the universal quantifier are also equivalent, via the dual negation rule: $$\neg\exists x:(x\in \mathcal A\wedge P(x)) \;\iff\; \forall x:(x\in \mathcal A \to \neg P(x))$$

Answer 3

S: "No positive number $x$ satisfies the equation $f(x) = 5$."

i.e. It is not case that there exists $ x\in \mathbb{R}^{+} $ such that $f(x) = 5$.

$ \urcorner (\exists x\in \mathbb{R}^{+} \mid f(x) = 5)\equiv \forall x\in \mathbb{R}^{+} ( f(x) \neq 5)$

Therefore for each $ x\in \mathbb{R}^{+} $, $f(x) \neq 5$.