Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 1 / b

Since $x^2+1\ge 2|x|$, the image of a function is contained in $[-1/2,1/2]$. Both endpoints occur, at $x=-1$ and $x=1$. By continuity and the intermediate value theorem applied on $[-1,1]$, every intermediate value occurs too. HenceIt is not surjective onto $\mathbb{R}$ because, for example, $1$ is not attained. It is not injective because $f(2)=f(1/2)=2/5$ although $2\ne 1/2$.

If $0\le x<y$, then $\sqrt{x}<\sqrt{y}$ and $x<y$, so $\sqrt{x}+x<\sqrt{y}+y$. Thus the function is strictly increasing on its domain and is injective. Its values are nonnegative, so it is not surjective onto $\mathbb{R}$:If ${\mathbb{R}}_{+}$ means $(0,\infty )$ its image of a function is $(0,\infty )$; if it includes zero its image of a function is $[0,\infty )$. Indeed, for any $u>0$, set $t=(\sqrt{1+4u}-1)/2>0$ and $x=t^2$; then $\sqrt{x}+x=t+t^2=u$. This convention does not change the classification.

With the convention $\mathbb{N}=\{0,1,2,\dots \}$, the even inputs $2j$ map to $j$ for $j\ge 0$, while the odd inputs $2j+1$ map to $-(j+1)$ for $j\ge 0$. These two images of a function are disjoint and together equal $\mathbb{Z}$. Within either class there are no repeated outputs. More explicitly the inverse function isConsequently $f$ is both injective and surjective, hence a bijection. If instead $\mathbb{N}=\{1,2,\dots \}$, the input zero is absent: $f$ remains injective but its image of a function is $\mathbb{Z}\setminus \{0\}$, so it is not onto. The printed paper does not specify this convention; the zero-based convention gives the apparent intended bijection.