Suppose T is a set of Natural numbers.
C1: $2$ is the only prime number that divides elements of $T$
C2: $T$ is the set of all natural numbers that satisfy the quadratic equation $x^2+x+1=0$.
I'm trying to find out if C1 $\Rightarrow$ C2, and the other way around.
I know that C2 is the empty set, so I'm guessing C2 $\Rightarrow$ C1 because it's vacuously true?
Please help. Cheers.
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$$\forall (s, p) ~:~ (s \in T \and p \in \text{Primes} \and p\mid s) \rightarrow 2 = p \tag{C1}$$
$$T = \setcomp{x}{x \in \mathbb N \text{ and } x^2 + x + 1 = 0} \tag{C2}$$
From (C2), we know that $T$ is the empty set. So (C1) is:
$$\forall (s, p) ~:~ (s \in \emptyset \and p \in \text{Primes} \and p\mid s) \rightarrow 2 = p$$
$$\forall (s, p) ~:~ (\text{false} \and p \in \text{Primes} \and p\mid s) \rightarrow 2 = p$$
$$\forall (s, p) ~:~ \text{false} \rightarrow 2 = p$$
$$\forall (s, p) ~:~ \text{true}$$
$$\text{true}$$
So (C2) implies (C1).
On the other hand, if we assume (C1) (assuming T is a subset of the natural numbers so that divisibility is defined):
$$\forall (s, p) ~:~ (s \in T \and p \in \text{Primes} \and p\mid s) \rightarrow 2 = p$$ $$T \subseteq \{1,~ 2,~ 4,~ 8,~ 16,~ 32,~ \dots\}$$
Does $T \subseteq \{1,~ 2,~ 4,~ 8,~ 16,~ 32,~ \dots\}$ imply that $T = \emptyset$? Sometimes yes, sometimes no. For example:
$$\begin{array} {c|c|c} \hline \text{Value of T} & T \subseteq \{1,~ 2,~ 4,~ 8,~ 16,~ 32,~ \dots\} \rightarrow T = \emptyset & \text{..which is:} \\ \hline T = \{3\} & \text{false} \rightarrow \text{false} & \text{true} \\ T = \{4, 8\} & \text{true} \rightarrow \text{false} & \text{false} \\ T = \{\} & \text{true} \rightarrow \text{true} & \text{true} \\ \end{array}$$
So $(C1) \rightarrow (C2)$ is not fully defined.