Question

Check that \(f: \mathbb{Z} \rightarrow \mathbb{Z}\) defined by \(f(n)=6-n\) is bijective. Then compute \(f^{-1}\).

Short answer

The function \(f(n) = 6 - n\) is bijective due to being both injective and surjective. The inverse function is \(f^{-1}(n) = 6 - n\).

Step-by-step solution

Understand Bijective Function

A function \(f: A \rightarrow B\) is said to be bijective (or one-to-one correspondence) if it is both one-to-one (injective) and onto (surjective). This means each 'b' in B is associated with exactly one 'a' in A.

Confirm Injectivity (One-to-One)

To confirm the given function \(f(n)=6-n\) is one-to-one, assume the function equals to two different integers \(m\) and \(p\). i.e. \(f(m) = f(p)\). If \(m = p\), then the function is injective. If the conditions are met, \(f(m) = f(p)\) gives us \(6 - m = 6 - p\). From this, it is clear that \(m = p\). Hence, the function is injective.

Confirm Surjectivity (Onto)

A function is surjective if for every element 'b' in the codomain B, there is at least one element 'a' in the domain A such that \(f(a) = b\). In this case, the codomain is integer \(\mathbb{Z}\), and the function is \(f(n) = 6 - n\). For all \(b \in \mathbb{Z}\), we can find \(n = 6 - b\) in domain \(\mathbb{Z}\). So, any integer 'b' can be written as \(6 - n\) and the function \(f(n) = 6 -n\) is surjective.

Compute the Inverse of the Function

If a function \(f\) is bijective, the inverse function \(f^{-1}\) exists. In this case, \(f(n) = 6 - n\) and \(n = 6 - f^{-1}(n)\) we isolate \(f^{-1}(n)\) to calculate the inverse function. Rearranging gives us the inverse function \(f^{-1}(n) = 6 - n\).