# What is the negation of this sentence

What is the negation of the following case:

Case A: The function $$f$$ has a unique rational fixed point $$a$$.

I would say the function has:

• multiple rational fixed points or
• no fixed point at all or
• at least one irrational fixed point

Not sure if we can make this more "compact".

• @China: I don't agree with this answer. My point of view is explained in my answer. – Taroccoesbrocco Aug 3 at 10:59
• Well, my answer matches better to your 2. than your 1. So in fact we agree on the negation... – zwim Aug 3 at 13:30
• @Taroccoesbrocco: Thank you. "Unique" refers to "rational fixed point" – Germany Aug 3 at 14:46
• @zwim - I'm not sure to understand your comment: If $f$ has an irrational fixed point and a rational fixed point, then your version of the negation is true (because of point 3) but my version of the negation (see point 2 in my answer) is false. So, your negation and my negation are not equivalent, and only one of them is correct. – Taroccoesbrocco Aug 3 at 15:35
• @China - Then the correct negation is in point $(2)$ of my answer, and not in point $(1)$ (or in zwim's answer). – Taroccoesbrocco Aug 3 at 15:45

The sentence can be interpreted in two ways, depending on whether "unique" refers to "fixed point" or "rational fixed point":

1. "Unique" refers to "fixed point": The function $$f$$ has a unique fixed point $$a$$, and moreover $$a$$ is rational. The logical form of this sentence is \begin{align} \exists a : a \in \mathbb{Q} \land a = f(a) \land \forall z (z = f(z) \to z = a) \end{align} In this case, the negation is: the function $$f$$ either has a unique irrational fixed point, or several fixed points, or no fixed points at all (this is equivalent to what @zwim said in his/her answer).

2. "Unique" refers to "rational fixed point": The function $$f$$ has exactly one rational fixed point (and possibly many irrational fixed points). The logical form of this sentence is: \begin{align} \exists a : a \in \mathbb{Q} \land a = f(a) \land \forall z ((z \in \mathbb{Q} \land z = f(z)) \to z = a) \end{align} In this case the negation is: the function $$f$$ has either several rational fixed points or no rational fixed points at all (but it might have irrational fixed points).

Note that the negation of the interpretation $$(2)$$ is more restrictive than the negation of the interpretation $$(1)$$: for instance, if $$f$$ has a irrational fixed point and a rational fixed point, then the negation of $$(1)$$ is true but the negation of $$(2)$$ is false.

Unlike @zwim, in my opinion the intended meaning of the sentence should be $$(2)$$, and not $$(1)$$, unless contextual information (which is missing in the question of the OP) says differently.