Injective = 1-to-1 = no two inputs have same output = no repeated second element in relation = every horizontal line in the codomain intersects <= 1. Surjective = onto = every codomain elements mapped-to = every codomain element appears as a second element in relation (ie, range=codomain) = every horizontal line in the codomain intersects >= 1 Bijective = invertible = injective and surjective

2) Consider the logarithm function ln : (0, ∞) → R. Decide whether this function is injective and whether it is surjective.

Injective because it's an increasing function. All increasing functions are injective because horizontal line will intersect at mist once. Surjective because ln(x) goes to +/- infinity as n goes to +/- infinity, so every horizontal line crosses ln(x).

4) A function f : Z→Z×Z is defined as f(n) = (2n,n+3). Verify whether this function is injective and whether it is surjective.

f = { ..., (-1, (-2,2)), (0, (0,3)), (1, (2,4)), ... } Injective, output elements are increasing with input (ie, the 2n part is increasing) Not Surjective, no input produces (-1,0).

6) A function f : Z×Z → Z is defined as f(m,n) = 3n−4m. Verify whether this function is injective and whether it is surjective.

f = { ..., ((0, 0), 0), ((0,1), 3), ((1, 0), -4), ...} Not injective: f(3,4) and f(30,40) both map to 0. Surjective, to get output x, set n=m=-x and 3n-m=x. This works for all x, so every x is mapped to.

8) A function f:Z×Z→Z×Z is defined as f(m,n) = (m+n,2m+n). Verify whether this function is injective and whether it is surjective.

f = { ((0,0),(0,0)), ((0,1),(1,1)), ((1,0),(1,2)), } Surjective: Want (13,1). 1-13=-12 f(-12,25) = (13,1). This works for all desired outputs, so all outputs mapped to. Injective: The above process produces a different value for each input.

12) Consider the function θ : 􏰛{0,1}􏰜×N→Z defined as θ(a,b) = a−2ab+b. Is θ injective? Is it surjective? Bijective? Explain.

θ(0,b) = 0-(2)(0)(b)+b = b θ(1,b) = 1-(2)(1)(b)+b = 1-2b+b = 1-b

Not injective: θ(0,1) = 1 and θ(1,0) = 1 Surjective: θ(0,b) = b can produce all of N and θ(1,b) = 1-b can produce all of Z-N. Bijective? Not injective, so not bijective.