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.