Solution

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.)

  1. $f:\mathbb{N}\rightarrow\mathbb{N}$ where $f(n)=n+1$

  2. $f:\mathbb{Z}\rightarrow\mathbb{Z}$ where $f(n)=n+1$

  3. $f:\mathbb{N}\rightarrow\mathbb{N}$ where $f(n)=n^2$

  4. $f:\mathbb{Z}\rightarrow\mathbb{Z}$ where $f(n)=n^2$

  5. $f:\mathbb{Z} \rightarrow 2\mathbb{Z}$ where $f(n) = 2n+2$

  6. $f:\mathbb{N} \rightarrow 2\mathbb{Z}$ where $f(n) = 2n+2$

  7. $f:2\mathbb{Z} \rightarrow \mathbb{N}$ where $f(n) = \displaystyle{\frac{|n|}{2}}$

  8. $f:\mathbb{Q} \rightarrow \mathbb{Q}$ where $f(n) = \displaystyle{\frac{1}{x^2+1}}$


  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.

  2. $f$ 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.

  3. $f$ 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.

  4. $f$ 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.

  5. $f$ 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.

  6. $f$ 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.

  7. $f$ 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.

  8. $f$ 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.