Let M7 = {z ∈ Z | ∃k ∈ Z z = 7k} and let f : M7 → {0, 1, 2, 3, 4} be defined bythe rule f (z) = z mod 5. Prove or disprove each of the following:(i) f is an injective function.(ii) f is a surjective function.(iii) f is an invertible function
Question
Let M7 = {z ∈ Z | ∃k ∈ Z z = 7k} and let f : M7 → {0, 1, 2, 3, 4} be defined bythe rule f (z) = z mod 5. Prove or disprove each of the following:(i) f is an injective function.(ii) f is a surjective function.(iii) f is an invertible function
Solution
(i) f is not an injective function. An injective function is one where every element of the domain maps to a unique element of the codomain. In this case, the function f(z) = z mod 5 is not injective because multiple elements of M7 can map to the same element in {0, 1, 2, 3, 4}. For example, both 7 and 12 in M7 map to 2 in {0, 1, 2, 3, 4}.
(ii) f is a surjective function. A surjective function is one where every element of the codomain is mapped to by at least one element of the domain. In this case, the function f(z) = z mod 5 is surjective because every element in {0, 1, 2, 3, 4} is the remainder when a multiple of 7 is divided by 5. For example, 0 is the remainder when 10 (a multiple of 7) is divided by 5, 1 is the remainder when 7 (a multiple of 7) is divided by 5, and so on.
(iii) f is not an invertible function. An invertible function is one that is both injective and surjective, meaning it has a two-way mapping between the domain and the codomain. Since f is not injective, it cannot be invertible.
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