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)} does not describe h correctly as a function from the set of integers to the set of integers. The range of h is limited to the set of even integers, while the codomain is the set of all integers. Since the range and codomain do not match, the function h cannot accurately be described by h: ℤ --> ℤ.
b) Your explanation is partially correct. If the codomain of h is given as the set {-4, -2, 0, 2, 4,..., 6}, then h is not onto. A function is onto if every element in the codomain is hit by the function, meaning that for every output value in the codomain, there exists at least one input value that maps to it. In the given function h, not every element in the codomain set {-4, -2, 0, 2, 4,..., 6} is covered by the range of h. The range of h is only the set {2, 4, 6}, so it does not include {-4, -2, 0}. Therefore, h is not onto.