Ytukyg

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

Then there exists a unique $winBbb R$ 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 $x,y,zinBbb R$.

   
   
   
   
   

   • $x+y=y+x$

   
   

   • $(x+y)+z=x+(y+z)$

   
   

   • $x+0=x$

   
   
   
   



2. If $rinBbb Q$ and $r>0$, there exists $ninBbb Nsetminus{0}$ such that $dfrac{1}{n}<r$.




3. For every $xin Bbb R$ and $ninBbb Nsetminus{0}$, there exist $r,sinBbb Q$ such that $r<xle s$ and $s-rledfrac{1}{n}$.

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

Existence

Let $w=inf{-smid sinBbb Q,s<x}$.

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

• By definition, $x+w=inf{r+pmid r,pinBbb Q,x<r,w<p}$.




• Substituting $-p$ for $p$, we get $x+w=inf{r-pmid r,-pinBbb Q,x<r,w<-p}=$ $inf{r-pmid r,pinBbb Q,x<r,w<-p}$.




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




• Hence $x+w=inf{r-smid r,sinBbb Q,x<r,s<x}=inf{r-smid r,sinBbb Q,s<x<r}$.

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

• We have $s<x<r implies r-s>0 implies x+w=inf{r-smid r,sinBbb Q,s<x<r} ge 0$.




• Assume the contrary that $x+w>0$. Since $Bbb Q$ is dense in $Bbb R$, there exists $pinBbb Q$ such that $0<p<x+w$. By Lemma 2, there exists $ninBbb Nsetminus{0}$ such that $dfrac{1}{n}<p$. By Lemma 3, there exist $r_0,s_0inBbb Q$ such that $r_0<x< s_0$ and $s_0-r_0ledfrac{1}{n}<p<x+w$. 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.

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

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

Then there exists a unique $winBbb R^+$ such that $xcdot w=1$

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 $x,y,zinBbb R^+$.

   
   
   
   
   

   • $xcdot y=ycdot x$

   
   

   • $xcdot (y+z)=xcdot y+xcdot z$

   
   

   • $(xcdot y)cdot z=xcdot (ycdot z)$

   
   

   • $xcdot 1=x$

   
   
   
   



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

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

Existence

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

We prove that $xcdot w=inf{r/smid r,sinBbb Q,0<s<x<r}$.

• By definition, $xcdot w=inf{rcdot pmid r,pinBbb Q,x<r,w<p}$.




• We substitute $1/p$ for $p$ and get $xcdot w=inf{rcdot (1/p)mid r,1/pinBbb Q,x<r,w<1/p}$ $=inf{r/pmid r,pinBbb Q,x<r,w<1/p}$.




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




• Hence $xcdot w=inf{r/smid r,sinBbb Q,x<r,0<s<x}=inf{r/smid r,sinBbb Q,0<s<x<r}$.

Second, we prove $xcdot w=1$.

• We have $0<s<r implies 1<r/s implies xcdot w=inf{r/smid r,sinBbb Q,0<s<x<r} ge 1$.




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

Uniqueness

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