Question

A function \(f: \mathbb{Z} \rightarrow \mathbb{Z}\) is defined as \(f(n)=2 n+1\). Verify whether this function is injective and whether it is surjective.

Short answer

The function \(f(n) = 2n + 1\) is injective, but it is not surjective.

Step-by-step solution

Check for Injectiveness

Assume two different integers \(a\) and \(b\), such that \(a \neq b\). We need to show that \(f(a) \neq f(b)\). Let's compute \(f(a)\) and \(f(b)\):\(f(a) = 2a + 1\) and \(f(b) = 2b + 1\).We can clearly see that if \(a \neq b\), then \(f(a) \neq f(b)\). Thus, the function \(f(n)=2n+1\) is injective.

Check for Surjectiveness

To deem the function surjective, every element \(y\) in the codomain (in this case the set of integers \(\mathbb{Z}\)), there must exist an element \(x\) in the domain (also \(\mathbb{Z}\)) such that \(f(x) = y\). We can solve the equation \(f(x) = y\) for \(x\): \(2x + 1 = y \Rightarrow x = \frac{y-1}{2}\). For all odd integers \(y\), \(x\) is an integer, satisfying the condition for surjectiveness. However, for all even integers \(y\), \(x\) is not an integer, so the condition is not satisfied. Therefore, the function \(f(n) = 2n + 1\) is not surjective.