Prove that the function \(f:\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}\) defined by \(f(x,{\text{ }}y) = (2x + y,{\text{ }}x + y)\) is a bijection.

to be a bijection it must be injective and surjective     R1

Note: This R1 may be awarded at any stage

suppose \(f(x,{\text{ }}y) = f(u,{\text{ }}v)\)     M1

\(2x + y = 2u + v\;\;\;({\text{ - i}})\)

\(x + y = u + v\;\;\;({\text{ - ii}})\)

\({\text{i - ii}} \Rightarrow x = u\)

\({\text{i - 2(ii)}} \Rightarrow  - y =  - v\)

\( \Rightarrow x = u,{\text{ }}y = v\)     A1

thus \((x,{\text{ }}y) = (u,{\text{ }}v)\) hence injective     A1

let \(2x + y = s\;\;\;({\text{ - i}})\)

\(x + y = t\;\;\;({\text{ - ii}})\)     M1

\({\text{i - ii }}x = s - t\)

\( \Rightarrow y = 2t - s\)

both \(x\) and \(y\) are integer if \(s\) and \(t\) are integer     R1

hence it is surjective     A1

hence  \(f\) is a bijection     AG

Note: Accept a valid argument based on matrices

Most candidates were able to show that \(f\) was an injection although some candidates appear to believe that it is sufficient to show that \(f(x,{\text{ }}y)\) is unique. A significant minority failed to show that \(f\) is a surjection and most candidates failed to note that it had to be checked that all values were integers. Some candidates introduced a matrix to define the transformation which was often a successful alternative method.