Let $$P(x)$$ be a predicate, and consider the following English sentence:
>"There is exactly one number for which $$P$$ is true."
1. Construct the negation of the statement in English.
2. Suppose $$P(x)$$ is the predicate $$(x^2 = x)$$. Find a domain for which the statement is true, and one for which it is false.
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**Part 1**
Construct the negation of the statement in English.
_Scratchwork_:
1. Convert to logical notation:
"There is a number for which P is true and for all other numbers, P is false."
$$\exists x(P(x) \land \forall y((y \ne x) \Rightarrow \neg P(y)))$$
2. Negate:
$$\begin{align*} \neg &(\exists x(P(x) \land \forall y((y \ne x) \Rightarrow \neg P(y)))) \\
& \equiv \forall x \neg(P(x) \land \forall y((y \ne x) \Rightarrow \neg P(y))) \\
& \equiv \forall x (\neg P(x) \lor \neg(\forall y((y \ne x) \Rightarrow \neg P(y)))) \\
& \equiv \forall x (\neg P(x) \lor \exists y \neg((y \ne x) \Rightarrow \neg P(y))) \\
& \equiv \forall x (\neg P(x) \lor \exists y ((y \ne x) \land P(y))) \\
\end{align*}$$
3. Translate back into English:
>"Either P is false for all numbers or there are multiple numbers for which P is true."
_Answer_: "Either P is false for all numbers or there are multiple numbers for which P is true."
_Alternative_:
$$\forall x (\neg P(x) \lor \exists y ((y \ne x) \land P(y))) \equiv \forall x (P(x) \Rightarrow \exists y ((y \ne x) \land P(y)))$$
i.e., "For every number x such that P is true, there is a different number y for which P is also true."
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**Part 2**
Suppose $$P(x)$$ is the predicate $$(x^2 = x)$$. Find a domain for which the statement is true, and one for which it is false.
>"There is exactly one number $$x$$ for which $$x^2 = x$$."
$$\exists x((x^2 = x) \land \forall y((y \ne x) \Rightarrow (y^2 \ne y)))$$
_Scratchwork:_
$$x^2 = x$$ is true only for $$x = 0$$ and $$x = 1$$.
_Answer:_
For the domain consisting of only $$0$$, i.e., $$\{0\}$$, the statement is true: there is only one number, and that number satisfies $$x^2 = x$$. Alternatively: $$\{0,2\}$$. Here, there is only one number satisfying $$x^2 = x$$.
For the domain $$\{0,1\}$$, the statement is false: both $$0$$ and $$1$$ satisfy $$x^2 = x$$. Alternatively, $$\{0,1,2\}$$, multiple numbers satisfying $$x^2 = x$$. Another alternative: $$\{2\}$$. No numbers satisfying $$x^2 = x$$.