It depends on what z and x are. Since you did not share that information, it is not possible to give a sensible answer.
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It is a particular 2-dimensional subset of 3-dimensional space.
It depends on what x and f are.
If a complex number z = (x, y) = x +iy where x and y are real numbers (and i is the imaginary root of -1), then RxR is is isomorphic to C. This means that the two sets are equivalent.
It is a subset of the Group G which has all the properties of a Group, namely that it is a set of elements (numbers) with a binary operation (addition) that combines any two elements in the set to form a third element which is also in the set. The Group satisfies four axioms: closure, associativity, identity and invertibility. The set of integers, Z, is a Group, with addition as the binary operation. [It is also a Ring, but that is not important here]. The set of all multiples of 7 is a subgroup of Z. Denote the subgroup by Z7. It is a Group because: Closure: If x and y are in Z7, then x = 7*p for some p in Z and y = 7*q for some q in Z. Then x + y = 7*p + 7*q = 7*(p+q) where p+q is in Z because Z is a Group. Therefore 7*(p+q) is in Z7. Associativity: If x (= 7p), y (= 7q) and z (= 7r) are in 7Z, then (x + y) + z = (7p + 7q) + 7r since these are in Z an Z is associative, = 7p + (7q + 7r) = x + (y + z). Identity: The additive identity is 0, since 0 + x = 0 + 7p = 7p since 0 is the additive identity in Z. Invertibility: If x = 7a is in Z7 then 7*(-a) is also in 7Z. If 7*(-a) is denoted by -x, then x + (-x) = 7a + 7*(-a) = 0 and so -x is the additive inverse of x. But there are elements of Z, for example, 2 which are not in Z7 so Z7 it is a proper subset of Z.
The set X is a proper subset of Y if Xcontains none or more elements from Y and there is at least one element of Y that is not in X.