To answer this question, we should substitute the values of x in the predicate. Once we do that, the predicate is transformed into a proposition, and we can state the truth value.
So, let’s start answering the questions.
a) P(0)
P(x): “x=x2”
P(0): “0=02=0”
Answer: true.
b) P(1)
P(x): “x=x2”
P(1): “1=12=1”
Answer: true.
c) P(2)
P(x): “x=x2”
P(2): “2=22=4”
Answer: false.
d) P(−1)
P(x): “x=x2”
P(-1): “-1=(-1)2=1”
Answer: false.
e) ∃xP(x)
To answer this type of question, we can use the answer to the previous ones.
We already know that exists at least one value x, such that P(x) is true.
Such a value can be x=0.
Answer: true.
f) ∀xP(x)
In this case, we can use a similar approach to e).
From c), we know that exists at least on value x=2, such that P(x) is false.
Therefore, we can state that not for all values of x, P(x) is true.
Answer: false.
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