Question

Descrete structures

Determine whether each of these functions from z to z is one to
one.
a.) f(n) = n-1
b.) f(n) = n^2+1
c.) f(n) = n^3
d.) f(n) = [n/2]

Answer 1

We determined that the functions f(n) = n - 1 and f(n) = n^3 are one-to-one, while f(n) = n^2 + 1 and f(n) = [n/2] are not one-to-one by demonstrating that they fail to assign distinct outputs to distinct inputs.

Step-by-step explanation:

The student's question is asking to determine whether each of the given functions from Z (the set of all integers) to Z is one-to-one. A one-to-one function, also known as an injective function, assigns distinct outputs to distinct inputs. Let's analyze each function.

1. f(n) = n - 1: This function is one-to-one because for each integer n, there is a distinct integer n - 1. If f(a) = f(b), then a - 1 = b - 1, which implies a = b.
2. f(n) = n^2 + 1: This function is not one-to-one because it fails to assign distinct outputs for distinct inputs. For example, f(-1) = (-1)^2 + 1 = 2 and f(1) = 1^2 + 1 = 2, so f(-1) = f(1) but -1 ≠ 1.
3. f(n) = n^3: This function is one-to-one. Similar to the first case, if f(a) = f(b), then a^3 = b^3, which implies a = b, since the cube function is strictly increasing for all integers n.
4. f(n) = [n/2]: Assuming that the brackets denote flooring (taking the largest integer not greater than n/2), this function is not one-to-one. For example, f(2) = [2/2] = 1 and f(3) = [3/2] = 1, which means f(2) = f(3) but 2 ≠ 3.