Question

For each of the following functions, say if it is one-to-one and / or onto? Prove or disprove each statement.

(a) f:Z→Z with f(n) = n²+1 for n ∈ Z.

Answer 1

The function f(n) = n²+1 for n \u2208 Z is neither one-to-one nor onto. Not one-to-one is shown by equal outputs for negative and positive values of the same absolute value. Not onto is proven since outputs never achieve 0, for example.

Step-by-step explanation:

A student has asked if the function f:Z\u2192Z with f(n) = n²+1 for n \u2208 Z (where Z refers to the set of all integers) is one-to-one and/or onto. To determine if a function is one-to-one, we need to show that different inputs map to different outputs. However, for negative and positive values of n that are the same in absolute value, the output of the function will be the same. For example, f(-1) = (-1)²+1 = 2 and f(1) = (1)²+1 = 2. This proves that f(n) is not one-to-one. To determine if a function is onto, every element in the codomain Z must be an output of the function. The value f(n) = 0 is never achieved because n² is always non-negative, and thus n²+1 \u2265 1 for all n in Z. Therefore, f(n) is also not onto.