Decide if each function described is injective, surjective, bijective, or none of these, and justify your decision. (

*For parts (e)-(g), note that $2\mathbb{Z}$ represents the set of all even integers.*)$f:\mathbb{N}\rightarrow\mathbb{N}$ where $f(n)=n+1$

$f$ is injective, since $f(a)=f(b) \rightarrow a+1=b+1 \rightarrow a=b$;

$f$ is not surjective, as $\not\exists a \in \mathbb{N} : f(a)=0$;

Since $f$ is not surjective, it can't be bijective.$f:\mathbb{Z}\rightarrow\mathbb{Z}$ where $f(n)=n+1$

$f$ is injective, since $f(a)=f(b) \rightarrow a+1=b+1 \rightarrow a=b$;

$f$ is surjective, as for any integer $n$, $f(n-1)=n$ and $(n-1)$ is an integer;

Since $f$ is both injective and surjective, it is also bijective.$f:\mathbb{N}\rightarrow\mathbb{N}$ where $f(n)=n^2$

$f$ is injective, since for $a,b \in \mathbb{N}$, we have $f(a)=f(b) \rightarrow a^2 = b^2 \rightarrow a = b$

(recall $a,b \in \mathbb{N} \rightarrow a,b \gt 0$);

$f$ is not surjective, as $\not\exists a \in \mathbb{N} : f(a)=2$;

Since $f$ is not surjective, it can't be bijective.$f:\mathbb{Z}\rightarrow\mathbb{Z}$ where $f(n)=n^2$

$f$ is not injective as $f(1)=f(-1)$ (among others);

$f$ is not surjective, as $\not\exists a \in \mathbb{N} : f(a)=2$;

Since $f$ is not surjective or injective, it can't be bijective.$f:\mathbb{Z} \rightarrow 2\mathbb{Z}$ where $f(n) = 2n+2$

$f$ is injective, as $f(a)=f(b) \rightarrow 2a+2=2b+2 \rightarrow 2a=2b \rightarrow a=b$;

$f$ is surjective, as for any $2k \in 2\mathbb{Z}$, note $f(k-1) = 2(k-1)+2 = 2k$ and $(k-1) \in \mathbb{Z}$;

Since $f$ is both injective and surjective, it is also bijective.$f:\mathbb{N} \rightarrow 2\mathbb{Z}$ where $f(n) = 2n+2$

$f$ is injective, as $f(a)=f(b) \rightarrow 2a+2=2b+2 \rightarrow 2a=2b \rightarrow a=b$;

$f$ is not surjective, as $\not\exists a \in \mathbb{N} : f(a)=0$ (among others);

Since $f$ is not surjective, it can't be bijective.$f:2\mathbb{Z} \rightarrow \mathbb{N}$ where $f(n) = \displaystyle{\frac{|n|}{2}}$

$f$ is not injective as $f(2)=f(-2)$ (among others);

$f$ is surjective as for any $n \in \mathbb{N}$, note $f(2n) = n$ and $2n \in 2\mathbb{Z}$;

Since $f$ is not injective, it can't be bijective.$f:\mathbb{Q} \rightarrow \mathbb{Q}$ where $f(n) = \displaystyle{\frac{1}{x^2+1}}$

$f$ is not injective as $f(1)=f(-1)$ (among others);

$f$ is not surjective as $\not\exists a \in \mathbb{N} : f(a)=0$;

Since $f$ is not injective or surjective, it can't be bijective.

Find a bijection between the sets $\{1,2,3,4,\ldots\}$ and $\{7,10,13,16,\ldots\}$.

Find a bijection between the real numbers in the interval $[0,\infty)$ and the real numbers in the interval $(0,\infty)$.

Find a bijection between the real numbers in the interval $[0,1]$ and the real numbers in the interval $(0,1)$.

Describe a bijection between the set of points on a unit square and the set of points on a semi-circle.

Describe a bijection between the set of points making up a line segment and the set of point making up a cube.