Answer:
a) The statement is also true, because a square is a geometric figure in the plane and a square is also a rectangle.
b) The statement is also true, because for example a rectangle with length 4 and width 6 is a possible rectangle that is not a square (as the length and the width are unequal).
c) The statement is true, because squares are a specific type of rectangle and thus all squares are rectangles.
Step-by-step explanation:
Negation ~P: not P
Conjunction p ∧ q: p and q
Conditional statement p --> q: if p, then q
Existential statement ∃xP(x) is true if and only one element x in the domain for which P(x) is true.
Universal statement ∀xP(x) is true if and only if P(x) is true for all values of x in the domain.
Domain=set of all geometric figures in the plane
Square(x)="x is a square"
Rect(x)="x is a rectangle"
(a)
∃x such that Rect(x) ∧ Square(x)
∃ mean "there exists"
In words, the given statement then means that: There exists a geometric figure in the plane that is a rectangle and that is a square.
The statement is also true, because a square is a geometric figure in the plane and a square is also a rectangle.
(b)
∃x such that Rect(x) ∧ ~ Square(x)
∃ mean "there exists"
In words, the given statement then means that: There exists a geometric figure in the plane that is a rectangle and that is not a square.
The statement is also true, because for example a rectangle with length 4 and width 6 is a possible rectangle that is not a square (as the length and the width are unequal).
(c)
∀x, Square(x) —> Rect(x)
∀ means "for every".
In words, the given statement means that: All squares are also rectangles.
The statement is true, because squares are a specific type of rectangle and thus all squares are rectangles.
