Let $\Bbb R^+=\{x\in\Bbb R \mid x>0\}$ and $x,y\in\Bbb R^+$. We define multiplication operation $(\cdot)$ on $\Bbb R^+$ and addition operation $(+)$ on $\Bbb R$ by $$x\cdot y:=\inf\{r\cdot s\mid r,s\in\Bbb Q \text{ and } x<r \text{ and } y<s\}$$ $$x+ y:=\inf\{r+ s\mid r,s\in\Bbb Q \text{ and } x<r \text{ and } y<s\}$$
Prove that $$\forall x \in \Bbb R^+:x+0=x\cdot 1=x$$
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
- $x+0=x$.
By definition, we have:
$x=\inf \{r \mid r\in\Bbb Q,x<r\}=\inf A$
$x+0=\inf \{r+s \mid r,s\in\Bbb Q,x<r,0<s\}=\inf B$
$b\in B \implies b=r+s$ for some $r,s\in\Bbb Q$ such that $x<r,0<s$ $\implies b>r$ for some $r\in\Bbb Q$ such that $x<r$ $\implies b>r$ for some $r\in A$ $\implies \inf A \le \inf B \implies x \le x+0$.
Assume the contrary that $x<x+0$. By the fact that $\Bbb Q$ is dense in $\Bbb R$, there exist $p\in \Bbb Q$ such that $x<p<x+0$ and $r\in\Bbb Q$ such that $x <r<p$. Let $p=r+s>r$. Then $s\in\Bbb Q$ and $s>0$. It follows that $x+0>p$ where $p=r+s$ such that $r,s\in\Bbb Q,x<r,0<s$. This is a contradiction.
- $x\cdot 1=x$
By definition, we have:
$x=\inf \{r \mid r\in\Bbb Q,x<r\}=\inf A$
$x\cdot 1=\inf \{r\cdot s \mid r,s\in\Bbb Q,x<r,1<s\}=\inf B$
$b\in B \implies b=r\cdot s$ for some $r,s\in\Bbb Q$ such that $x<r,1<s$ $\implies b>r$ for some $r\in\Bbb Q$ such that $x<r$ $\implies b>r$ for some $r\in A$ $\implies \inf A \le \inf B \implies x \le x\cdot 1$.
Assume the contrary that $x<x\cdot 1$. By the fact that $\Bbb Q$ is dense in $\Bbb R$, there exist $p\in \Bbb Q$ such that $x<p<x\cdot 1$ and $r\in\Bbb Q$ such that $x <r<p$. Let $p=r\cdot s>r$. Then $s\in\Bbb Q$ and $s>1$. It follows that $x\cdot 1>p$ where $p=r\cdot s$ such that $r,s\in\Bbb Q,x<r,1<s$. This is a contradiction.