Question

Consider the function \(\theta:\\{0,1\\} \times \mathbb{N} \rightarrow \mathbb{Z}\) defined as \(\theta(a, b)=a-2 a b+b .\) Is \(\theta\) injective? Is it surjective? Bijective? Explain.

Short answer

The function \( \theta(a, b)=a-2ab+b \) is surjective but is not injective. Therefore it is not bijective.

Step-by-step solution

Test for Injectiveness

For the function to be injective, it is important that every element in the range is the image of at most one element in the domain. For the given function, if \(a-2ab+b = a'-2a'b'+b'\) with \(a, a' \in \{0,1\}\) and \(b, b' \in \mathbb{N}\) then it can be simplified as \(a-b = a'-b'\) since \(2ab = 2a'b'\). There could be more than one pair of \(a, b\) or \(a', b'\) that satisfy this equation. Hence, the function is not injective.

Test for Surjectiveness

For the function to be surjective, every element of the co-domain \( \mathbb{Z} \) must be mapped by an element of the domain \( \{0,1\} \times \mathbb{N} \). It can be observed directly from the function definition that for any integer \( z \in \mathbb{Z} \), we can select \( a = 0 \) and \( b = z \) to obtain \( \theta(a,b)=z \), hence the function is surjective.

Summary

Summarizing the obtained results, the function \( \theta(a, b)=a-2ab+b \) is surjective but not injective.