Exercise 4.5

Main Problem.

(a)

Proof. Consider any $a\in {\mathbb{Z}}_{\text{+}}$ and define $X_a=\left\{x\in ℝ\mid a+x\in {\mathbb{Z}}_{\text{+}}\right\}$. We show that $X_a$ is inductive. First, since $a\in {\mathbb{Z}}_{\text{+}}$ we have that $a+1\in {\mathbb{Z}}_{\text{+}}$ since ${\mathbb{Z}}_{\text{+}}$ is inductive. Hence $1\in X_a$ by definition. Now suppose that $x\in X_a$ so that $a+x\in {\mathbb{Z}}_{\text{+}}$. Then we have $a+(x+1)=(a+x)+1\in {\mathbb{Z}}_{\text{+}}$ since $a+x\in {\mathbb{Z}}_{\text{+}}$ and ${\mathbb{Z}}_{\text{+}}$ is inductive. This shows by definition that $x+1\in X_a$ and therefore that $X_a$ is inductive. It follows that ${\mathbb{Z}}_{\text{+}}\subset X_a$ since ${\mathbb{Z}}_{\text{+}}$ is defined as the intersection of all inductive subsets of reals, of which $X_a$ is one.

Therefore, for any $a,b\in {\mathbb{Z}}_{\text{+}}$, we have that $b\in X_a$ since ${\mathbb{Z}}_{\text{+}}\subset X_a$. Thus by definition $a+b\in {\mathbb{Z}}_{\text{+}}$ as desired □

(b)

Proof. Consider any $a\in {\mathbb{Z}}_{\text{+}}$ and define $X_a=\left\{x\in ℝ\mid a\cdot x\in {\mathbb{Z}}_{\text{+}}\right\}$. We show that $X_a$ is inductive. To this end, we first have that $a\cdot 1=a\in {\mathbb{Z}}_{\text{+}}$ so that $1\in X_a$ by definition. Now suppose that $x\in X_a$ so that $\mathit{ax}\in {\mathbb{Z}}_{\text{+}}$. Then we have $a\cdot (x+1)=a\cdot x+a\cdot 1=\mathit{ax}+a\in {\mathbb{Z}}_{\text{+}}$ by part (a) since we know both $\mathit{ax}$ and $a$ are in ${\mathbb{Z}}_{\text{+}}$. Hence $x+1\in X_a$ by definition. This shows that $X_a$ is inductive so that again ${\mathbb{Z}}_{\text{+}}\subset X_a$.

Hence for any $a,b\in {\mathbb{Z}}_{\text{+}}$ we have that $b\in X_a$ since ${\mathbb{Z}}_{\text{+}}\subset X_a$. It then follows by definition that $a\cdot b\in {\mathbb{Z}}_{\text{+}}$ as desired. □

(c)

Proof. Let $X=\left\{x\in ℝ\mid x-1\in {\mathbb{Z}}_{\text{+}}z\right\}$, which we show is inductive. First, we have $1-1=1+(-1)=0$ so that clearly $1\in {\mathbb{Z}}_{\text{+}}z$ and hence $1\in X$. Now suppose that $x\in X$ so that $x-1\in {\mathbb{Z}}_{\text{+}}z$.

Case: $x-1\in \left\{0\right\}$. Then it must be that $x-1=0$, which clearly implies $x=1\in {\mathbb{Z}}_{\text{+}}$ since ${\mathbb{Z}}_{\text{+}}$ is inductive. Then $(x+1)-1=x+(1-1)=x+0=x\in {\mathbb{Z}}_{\text{+}}$ so that $(x+1)-1\in {\mathbb{Z}}_{\text{+}}z$ and therefore $x+1\in X$.

Case: $x-1\in {\mathbb{Z}}_{\text{+}}$. Then $(x+1)-1=x+(1-1)=x+((-1)+1)=(x-1)+1\in {\mathbb{Z}}_{\text{+}}$ since $x-1\in {\mathbb{Z}}_{\text{+}}$ and ${\mathbb{Z}}_{\text{+}}$ is inductive. Thus clearly $(x+1)-1\in {\mathbb{Z}}_{\text{+}}z$ so that $x+1\in X$ by definition.

Hence in both cases $x+1\in X$, which shows that $X$ is inductive, and so ${\mathbb{Z}}_{\text{+}}\subset X$. Therefore, for any $a\in {\mathbb{Z}}_{\text{+}}$, we have that also $x\in X$ since ${\mathbb{Z}}_{\text{+}}\subset X$. Then, by the definition of $X$, it follows that $a-1\in {\mathbb{Z}}_{\text{+}}z$ as desired. □

(d)

Proof. First we show that the set $X_c=\left\{x\in ℝ\mid c+x\in \mathbb{Z}\text{ and }c-x\in \mathbb{Z}\right\}$ is inductive for any $c\in \mathbb{Z}$. So consider any $c$ and $b$ in $\mathbb{Z}$ so that $c,b\in {\mathbb{Z}}_{\text{+}}z\cup {\mathbb{Z}}_{\text{−}}$.

Case: $b\in {\mathbb{Z}}_{\text{+}}$. Then $b+1\in {\mathbb{Z}}_{\text{+}}$ since ${\mathbb{Z}}_{\text{+}}$ is inductive and $b-1\in {\mathbb{Z}}_{\text{+}}z$ by part (c).

Case: $b=0$. Then $b+1=0+1=1\in {\mathbb{Z}}_{\text{+}}$ since it is inductive, and $b-1=0-1=-1\in {\mathbb{Z}}_{\text{−}}$ since $1\in {\mathbb{Z}}_{\text{+}}$.

Case: $b\in {\mathbb{Z}}_{\text{−}}$. Then $b=-a$ for $a\in {\mathbb{Z}}_{\text{+}}$, and we then have that $a+1\in {\mathbb{Z}}_{\text{+}}$ since ${\mathbb{Z}}_{\text{+}}$ is inductive. Hence $b-1=-a-1=-(a+1)\in {\mathbb{Z}}_{\text{−}}$. We also have that $a-1\in {\mathbb{Z}}_{\text{+}}z$ by part (c), from which it is trivial to show that $-(a-1)\in {\mathbb{Z}}_{\text{−}}\cup \left\{0\right\}$. Therefore $b+1=-a+1=-(a-1)\in {\mathbb{Z}}_{\text{−}}\cup \left\{0\right\}$.

Thus in all cases we have that $b+1$ and $b-1$ are in ${\mathbb{Z}}_{\text{+}}$ or $\left\{0\right\}$ or ${\mathbb{Z}}_{\text{−}}$ so that they are both in $\mathbb{Z}$, and so $1\in X_b$. Note that this is the case for any $b\in \mathbb{Z}$ so that it is clearly true for $c$, i.e. $1\in X_c$. Now suppose that $x\in X_c$ so that $c+x$ and $c-x$ are both in $\mathbb{Z}$. It then follows that $1\in X_{c+x}$ and $1\in X_{c-x}$ so that $c+(x+1)=(c+x)+1\in \mathbb{Z}$ and $c-(x+1)=(c-x)-1\in \mathbb{Z}$. This then shows that $x+1\in X_c$. Hence $X_c$ is inductive for any $c\in \mathbb{Z}$ so that ${\mathbb{Z}}_{\text{+}}\subset X_c$.

Now consider $c,d\in \mathbb{Z}$.

Case: $d\in {\mathbb{Z}}_{\text{+}}$. Then clearly $d\in X_c$ since ${\mathbb{Z}}_{\text{+}}\subset X_c$. Hence by definition $c+d$ and $c-d$ are both in $\mathbb{Z}$.

Case: $d=0$. Then $c+d=c+0=c\in \mathbb{Z}$ and $c-d=c-0=c\in \mathbb{Z}$.

Case: $d\in {\mathbb{Z}}_{\text{−}}$. Then by definition $d=-a$ for $a\in {\mathbb{Z}}_{\text{+}}$ so that $a\in X_c$ since ${\mathbb{Z}}_{\text{+}}\subset X_c$. Then $c+a$ and $c-a$ are both in $\mathbb{Z}$ by the definition of $X_c$. Hence $c+d=c+(-a)=c-a\in \mathbb{Z}$ and $c-d=c-(-a)=c+a\in \mathbb{Z}$.

Therefore we have shown that $c+d$ and $c-d$ are both integers in all cases, which is the desired result. □

(e)

Proof. For any $c\in \mathbb{Z}$, define $X_c=\left\{x\in ℝ\mid c\cdot x\in \mathbb{Z}\right\}$. We first show that $X_c$ is inductive for any such $c\in \mathbb{Z}$. We have $c\cdot 1=c\in \mathbb{Z}$ so that $1\in X_c$. Now suppose that $x\in X_c$ so that $c\cdot x\in \mathbb{Z}$. Then $c\cdot (x+1)=c\cdot x+c\cdot 1=c\cdot x+c\in \mathbb{Z}$ by part (d) since both $c\cdot x$ and $c$ are integers. This shows that $X_c$ is inductive so that ${\mathbb{Z}}_{\text{+}}\subset X_c$.

Now consider any $c,d\in \mathbb{Z}$.

Case: $d\in {\mathbb{Z}}_{\text{+}}$. Then $d\in X_c$ since ${\mathbb{Z}}_{\text{+}}\subset X_c$. Thus $c\cdot d\in \mathbb{Z}$.

Case: $d=0$. The $c\cdot d=c\cdot 0=0\in \mathbb{Z}$.

Case: $d\in {\mathbb{Z}}_{\text{−}}$. Then there is an $a\in {\mathbb{Z}}_{\text{+}}$ such that $d=-a$. Hence $a\in X_c$ since ${\mathbb{Z}}_{\text{+}}\subset X_c$, from which it follows that $c\cdot a\in \mathbb{Z}$. We then have $c\cdot d=c\cdot (-a)=-(c\cdot a)\in \mathbb{Z}$ as well by Lemma 1 .

Thus in all cases $c\cdot d\in \mathbb{Z}$ as desired. □