Proof of Least upper bound of $\frac{x}{x+1}$ The questions states:
Prove that there is a unique positive real number $m$ that has the following two properties.

*

*For every positive real number $x$, $\frac{x}{x+1}<m$.

*If $y$ is any positive real number with the property that for every positive real number $x$ , $\frac{x}{x+1}<y$ then $m\leq y$.

My concern is my proof of the second part as I am not sure if its valid or there is a clearer way of phrasing it.
For the first part I used $m=1$. My doubt is with my second portion
My proof:
Suppose for the sake of contradiction, that there exists a $y<m$ such that $\frac{x}{x+1}<y$. Then $0<\frac{x}{x+1}<y<1$. Thus $y$ can be written as $\frac{p}{q}$ where $p$ and $q$ are positive integers and $q>p$. As $p$ and $q$ are positive integers, then $q-p\geq1$. Thus $q+1\geq p+2$. So we have $\frac{x}{x+1} < \frac{p}{q} < \frac{p+1}{q+1} \leq \frac{p+1}{p+2}$. However  if we choose $x=p+1$ then $\frac{x}{x+1} = \frac{p+1}{p+2} \geq y$ , we arrive at a contradiction. So it follows that if $y$ is any positive real number with the property that for every positive real number $x$ , $\frac{x}{x+1}<y$ then $m\leq y$.
My specific questions are :

*

*Is it valid to express $\frac{p+1}{p+2}$ as $\frac{x}{x+1}$, or do I need to choose another variable $y$ to show that its the same thing.

*Is it needed to show $\frac{p}{q} < \frac{p+1}{q+1}$ ? Or does it make the proof too long winded?

*Is there a more succinct way of proving this?


$$  \dfrac{x}{x+1} < \dfrac{x+1}{x+1} \implies \dfrac{x}{x+1} < 1 $$
So 1 is an upper bound. Let $\epsilon < 1$ We will show that there is an $x$ for which
$\dfrac{x}{x+1} = 1-\epsilon$
\begin{align}
   \dfrac{x}{x+1} &= 1-\epsilon \\
   x(1-\epsilon) + 1(1-\epsilon) &= x \\
   -\epsilon x &= -(1-\epsilon) \\
   x &= \dfrac{1-\epsilon}{\epsilon}
\end{align}
So, for all positive \epsilon close to $0$, $x = \dfrac{1-\epsilon}{\epsilon}$ will give us $\dfrac{x}{x+1} = 1-\epsilon$.
It follows that $1$ is the LUB of $\dfrac{x}{x+1}$.
