[size=150]A function f : Z → Z is defined as f (n) = 2n +1. [br][br]Verify whether this function is injective and whether it is surjective. [br][br]This function is injective. To see this, suppose m, n ∈ Z and f (m) = f (n).[br] [br]This means 2m+1 = 2n+1, from which we get 2m = 2n, and then m = n.[br] [br]Thus, f is injective. This function is not surjective. [br][br]To see this notice that f (n) is odd for all n ∈ Z. [br][br]So, given the (even) number 2 in the codomain Z, there is no n with f (n) = 2.[br][/size][br][br]

[size=150]a) Function Definition: The function [i]f: [/i]Z→Z is defined as f(n) = 2n + 1, where n is an integer.[br][br]b) Injective (One-to-One) Property:[br]A function is injective if each element in the domain maps to a unique element in the codomain. To verify if [i]f [/i]is injective, we need to show that whenever f(m) = f(n) for m, n[img width=13,height=20]file:///C:/Users/burgesss/AppData/Local/Temp/msohtmlclip1/01/clip_image002.png[/img] Z, then m = n.[br][br]Let m, n[img width=13,height=20]file:///C:/Users/burgesss/AppData/Local/Temp/msohtmlclip1/01/clip_image002.png[/img] Z such that f(m) = f(n). This means 2m + 1 = 2n + 1.[br]-By subtracting 1 from both sides of the equation, we get 2m = 2n.[br]-Further dividing both sides by 2, we obtain m = n.[br]Since m and n are equal, the function [i]f[/i] is injective.[br][br]c) Surjective (Onto) Property: A function is surjective if every element in the codomain has at least one[br]pre-image in the domain. To check if [i]f [/i]is surjective, we need to ensure that for every integer m in the codomain Z, there exists an integer n in the domain Z such that f(n) = m.[br][br]Notice that f(n) is odd for all integers n because it is in the form 2n + 1, where 2n is even, and adding 1[br]makes it odd. Now, consider the (even) number 2 in the codomain Z. There is no integer n that satisfies [br]f(n) =2, because f(n)is always odd and cannot be equal to an even number.[br][br]Since there is no integer n that maps to the (even) number 2 in the codomain, the function [i]f[/i] is not surjective.[br][br]In summary, the function [i]f: [/i]Z→Z defined as f(n) = 2n + 1 is injective but not surjective. It is injective because distinct integers in the domain have distinct images in the codomain. However, it is not surjective[br]because not every element in the codomain has a pre-image in the domain. Specifically, there is no integer n such that f(n) = 2.[/size]