For each function below, determine whether or not the function is injective and whether or not the function is surjective. (you do not need to justify your answer). (a) f: r + r given by f(x) = x2 (b) f: n + n given by f(n) = n2 (c) f: zx z → z given by f(n, k) = n +k

First, injective means that for every y, there is only one x such f(x) = y, and surjective means that if f is a function that goes from the set {x} to the set {y}, for every element y in the codomain there is at least one element x in the domain such f(x) = y

(a) f:R----> + R given by f(x) = x2 (i guess you written )

The function is not injective, because f(-2) = f(2), and is surjective, because the codomain is +R, and is always a positive real number.

(b) f:N----> + N given by f(n) = n2  (i guess you written )

As the naturals have no negative numbers, in this case the function is injective, but isn't surjective, because there is no number that when squared is equal to 5, for example.

(c) f: Zx Z → Z given by f(n, k) = n +k:

here the domain is of the form (n,k) and the function is n +k, so the numbers (z,w) and (w,z) return the same value when evaluated in f, then f is not injective. And is easy to see that f is surjective, because in the sum you can reach al the integers on the codomain.

no

step-by-step explanation:

∡ eia and ∡ kij are congruent

step-by-step explanation: