(a) Using hint: given $a\in\mathbb{Z}_{+}$ , it easy to see that $1\in X$ , as $a+1\in\mathbb{Z}_{+}$ , and $X$ is inductive, as if $a+x\in\mathbb{Z}_{+}$ , then $a+(x+1)=(a+x)+1\in\mathbb{Z}_{+}$ . Hence, $\mathbb{Z}_{+}\subset X$ . (Note, that, in general, $X\neq\mathbb{Z}_{+}$ but showing that $\mathbb{Z}_{+}\subset X$ is enough for our purposes.)

(b) Similarly here, but using (a): define $X$ similarly to (a) but for multiplication, argue that $1\in X$ , which is kind of obvious, and then argue, using (a), that if $x\in X$ , i.e. if $a\cdot x\in\mathbb{Z}_{+}$ , then $x+1\in X$ , i.e. $a\cdot(x+1)=a\cdot x+a\in\mathbb{Z}_{+}$ .

(c) Using hint: $1\in X$ as $1-1=0\in\mathbb{Z}_{+}\cup\{0\}$ , and if $x\in X$ , then $x-1\in\mathbb{Z}_{+}\cup\{0\}$ , and $(x+1)-1=x=(x-1)+1\in\mathbb{Z}_{+}\subset\mathbb{Z}_{+}\cup\{0\}$ , implying $x+1\in X$ . Therefore, $X$ is inductive, and $\mathbb{Z}_{+}\subset X$ .

For (d) and (e) we first prove the following statement.

We can now prove (d) and (e) similar to (a) and (b). But first, using the hint, we prove (d) for $d=1$ . Let $A=\{c\in\mathbb{Z}|c-1,c+1\in\mathbb{Z}\}$ . Then, $0\in A$ as $0-1=-1$ , $0+1=1$ $\in\mathbb{Z}$ , and if $x\in A\subset\mathbb{Z}$ , then $x-1,x+1\in\mathbb{Z}$ , and $(x-1)+1=(x+1)-1=x\in\mathbb{Z}$ . Further, if $(x-1)>0$ then $(x-1)-1\in\mathbb{Z}_{+}\cup\{0\}\subset\mathbb{Z}$ (using (c)), if $(x-1)=0$ then $(x-1)-1=-1\in\mathbb{Z}$ , and if $(x-1)<0$ then $(x-1)=-n$ for some $n\in\mathbb{Z}_{+}$ , so that $(x-1)-1=-n-1=-(n+1)\in\mathbb{Z}$ . Similar, if $(x+1)>0$ then $(x+1)+1\in\mathbb{Z}_{+}\subset\mathbb{Z}$ , if $(x+1)=0$ then $(x+1)+1=1\in\mathbb{Z}$ , and if $(x+1)<0$ then $(x+1)=-n$ for some $n\in\mathbb{Z}_{+}$ , so that $(x+1)+1=-n+1=-(n-1)$ where, by (c), either $n-1=0$ or $n-1\in\mathbb{Z}_{+}$ , but in either case $(x+1)+1=-(n-1)\in\mathbb{Z}$ . Overall, $(x\pm1)\pm1\in\mathbb{Z}$ , so that $x-1,x+1\in A$ , and, by the theorem, $A=\mathbb{Z}$ .

Now, having (d) for the case $d=1$ , we prove that if $c\in\mathbb{Z}$ , then for every $d\in\mathbb{Z}$ , $c\dot{\pm}d\in\mathbb{Z}$ . We define $A=\{a\in\mathbb{Z}|c\dot{\pm}a\in\mathbb{Z}\}$ , argue that $0\in A$ as $c\dot{\pm}0=\begin{cases} 0\\ c\\ c \end{cases}\in\mathbb{Z}$ , and if $a\in A$ , then $a\pm1\in A$ as $c\dot{\pm}(a\pm1)=\begin{cases} ca & \pm c\\ c+a & \pm1\\ c-a & \mp1 \end{cases}\in\mathbb{Z}$ (note that, in (d), we use the proved case $d=1$ , and for multiplication in (e), we use (d)). Hence, we show that $A$ satisfies the conditions of the theorem above, and $A=\mathbb{Z}$ .