Using predicate logic to see if P(X) holds

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Suppose We have a set called ${A}$ and we let it equal to $A = \{a \in \mathbb{Z} \mid a^2 = 3\}$ and we let ${P(x)}$ be the predicate ${x \in \mathbb{Z} }$

Problems:

1.) ${\forall x \in A }, P(x)$

2.) ${\exists x \in A }, P(x)$

Reasoning:

For both of these problems, The first problem that I see is that equation ${a^2 = 3}$ will never be true for number in the integer set and and since ${a}$ is an integer, it has a blank set.

Thus I reason both can't hold.

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You have good progress. Certainly $A = \emptyset$, so that's good. So statements of the form $\forall x \in \emptyset\, P(x)$ are always true vacuously. On the other hand, statements of the form $\exists x \in \emptyset\, P(x)$ are always false, as they require finding an element of the empty set.