Consider the following: h={(1,2),(2,4),(3,6),(-1,-2),(-2,-4)}.
a) Explain why h: ℤ --> ℤ does not describe h correctly.
This does not describe the function h correctly because the range of the function h seems to be finite or limited to the set of even integers while the codomain of function h is the set of integers. Since these two aren't equal to each other, then we can't say that h is a function from the set of integers to the set of integers. Is this the correct answer?
b) If the co-domain of h is the set {-4, -2, 0, 2, 4,... , 6}, is h onto? Why or why not?
If the co-domain of h is the set given, then this function would not be onto because the range of h is not equal to the codomain given. While each input goes to exactly one input, implying that it is one to one, this function is not onto because not every single element in the codomain was met. Is this a good explanation?
a) Yes, your explanation is correct. The function h={(1,2),(2,4),(3,6),(-1,-2),(-2,-4)} maps integers to integers. However, the range of h only includes even integers, while the codomain includes all integers. Since the range and the codomain are not equal, we cannot say that h: ℤ --> ℤ describes the function h correctly.
b) Yes, your explanation is also correct. If the co-domain of h is the set {-4, -2, 0, 2, 4, ..., 6}, this means that the possible outputs of h are limited to these values. In this case, the function h is not onto because not every element in the co-domain has a corresponding pre-image in the domain. Since there is no input that maps to the values -4, 0, or 6, for example, the function h cannot be considered onto.