Let $x\in\Bbb R$. Then there exists a unique $w\in\Bbb R$ such that $x+w=0$

Question

Let $x,y\in\Bbb R$. We define addition operation $(+)$ on $\Bbb R$ by $$x+ y:=\inf\{r+ s\mid r,s\in\Bbb Q \text{ and } x<r \text{ and } y<s\}$$

Then there exists a unique $w\in\Bbb R$ such that $x+w=0$.

---

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

---

My attempt:

Lemma:

1. Let $x,y,z\in\Bbb R$.

   • $x+y=y+x$

   • $(x+y)+z=x+(y+z)$

   • $x+0=x$

2. If $r\in\Bbb Q$ and $r>0$, there exists $n\in\Bbb N\setminus\{0\}$ such that $\dfrac{1}{n}<r$.

3. For every $x\in \Bbb R$ and $n\in\Bbb N\setminus\{0\}$, there exist $r,s\in\Bbb Q$ such that $r<x\le s$ and $s-r\le\dfrac{1}{n}$.

It is obvious that $x=\inf \{r \mid r\in\Bbb Q,x<r\}$.

Existence

Let $w=\inf\{-s\mid s\in\Bbb Q,s<x\}$.

First, we prove that $x+w=\inf\{r-s \mid r,s\in\Bbb Q,s<x<r\}$.

• By definition, $x+w=\inf\{r+p\mid r,p\in\Bbb Q,x<r,w<p\}$.

• Substituting $-p$ for $p$, we get $x+w=\inf\{r-p\mid r,-p\in\Bbb Q,x<r,w<-p\}=$ $\inf\{r-p\mid r,p\in\Bbb Q,x<r,w<-p\}$.

• We have $p\in\Bbb Q$ and $w<-p \iff p\in\Bbb Q$ and $-p>-s$ for some $s\in\Bbb Q$ such that $s<x$ $\iff p\in\Bbb Q$ and $p<s$ for some $s\in\Bbb Q$ such that $s<x$ $\iff p=s$ for some $s\in\Bbb Q$ such that $s<x$.

• Hence $x+w=\inf\{r-s\mid r,s\in\Bbb Q,x<r,s<x\}=\inf\{r-s\mid r,s\in\Bbb Q,s<x<r\}$.

Second, we prove that $x+w=0$.

• We have $s<x<r \implies r-s>0 \implies x+w=\inf\{r-s\mid r,s\in\Bbb Q,s<x<r\} \ge 0$.

• Assume the contrary that $x+w>0$. Since $\Bbb Q$ is dense in $\Bbb R$, there exists $p\in\Bbb Q$ such that $0<p<x+w$. By Lemma 2, there exists $n\in\Bbb N\setminus\{0\}$ such that $\dfrac{1}{n}<p$. By Lemma 3, there exist $r_0,s_0\in\Bbb Q$ such that $r_0<x< s_0$ and $s_0-r_0\le\dfrac{1}{n}<p<x+w$. This is clearly a contradiction. Hence $x+w=0$.

Uniqueness

Assume that $x+w=0$ and $x+w'=0$. By Lemma 1, $w=w+0=w+(x+w')=w+(w'+x)=(w+w')+x=(w'+w)+x=w'+(w+x)=w'+(x+w)=w'+0=w'.$

Hence such $w$ is unique.

Answer 1

By a similar reasoning, I would like to present a proof of the below theorem.

---

Let $\Bbb R^+=\{x\in\Bbb R \mid x>0\}$ and $x,y\in\Bbb R^+$. We define multiplication operation $(\cdot)$ on $\Bbb R^+$ by $$x\cdot y:=\inf\{r\cdot s\mid r,s\in\Bbb Q \text{ and } x<r \text{ and } y<s\}$$

Then there exists a unique $w\in\Bbb R^+$ such that $x\cdot w=1$

---

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

---

My attempt:

Lemma:

1. Let $x,y,z\in\Bbb R^+$.

   • $x\cdot y=y\cdot x$

   • $x\cdot (y+z)=x\cdot y+x\cdot z$

   • $(x\cdot y)\cdot z=x\cdot (y\cdot z)$

   • $x\cdot 1=x$

2. For every $x,k\in\Bbb R$ such that $0<x,1<k$. There exists $s,r\in\Bbb Q$ such that $0<s<x<r$ and $r/s<k$.

Since $\Bbb Q$ is dense in $\Bbb R$, $x=\inf \{r \mid r\in\Bbb Q,x<r\}$.

Existence

Since $\Bbb Q$ is dense in $\Bbb R$, $x=\inf \{r \mid r\in\Bbb Q,x<r\}$. Let $w=\inf\{1/s\mid s\in\Bbb Q,0<s<x\}$.

We prove that $x\cdot w=\inf\{r/s\mid r,s\in\Bbb Q,0<s<x<r\}$.

• By definition, $x\cdot w=\inf\{r\cdot p\mid r,p\in\Bbb Q,x<r,w<p\}$.

• We substitute $1/p$ for $p$ and get $x\cdot w=\inf\{r\cdot (1/p)\mid r,1/p\in\Bbb Q,x<r,w<1/p\}$ $=\inf\{r/p\mid r,p\in\Bbb Q,x<r,w<1/p\}$.

• We have $p\in\Bbb Q$ and $w<1/p$ $\iff p\in\Bbb Q$ and $1/p>1/s$ for some $s\in\Bbb Q$ such that $0<s<x$ $\iff p\in\Bbb Q$ and $0<p<s$ for some $s\in\Bbb Q$ such that $0<s<x$ $\iff p=s$ for some $s\in\Bbb Q$ such that $0<s<x$.

• Hence $x\cdot w=\inf\{r/s\mid r,s\in\Bbb Q,x<r,0<s<x\}=\inf\{r/s\mid r,s\in\Bbb Q,0<s<x<r\}$.

Second, we prove $x\cdot w=1$.

• We have $0<s<r \implies 1<r/s \implies x\cdot w=\inf\{r/s\mid r,s\in\Bbb Q,0<s<x<r\} \ge 1$.

• Assume the contrary that $x\cdot w>1$. Since $\Bbb Q$ is dense in $\Bbb R$, there exists $p\in\Bbb Q$ such that $1<p<x\cdot w$. By Lemma 2, there exist $s,r\in\Bbb Q$ such that $0<s<x<r$ and $r/s<p<x\cdot w$. This is clearly a contradiction. Hence $x\cdot w=1$.

Uniqueness

Assume that $x\cdot w=x\cdot w'=1$. By Lemma 1, $w=w\cdot 1=w\cdot (x\cdot w')=w\cdot (w'\cdot x)=$ $(w\cdot w')\cdot x=(w'\cdot w)\cdot x=w'\cdot (w\cdot x)=w'\cdot (x\cdot w)=w'\cdot 1=w'.$ Hence such $w$ is unique.