This is an exercise from Velleman's "How To Prove It".
Prove that there is a unique positive real number $m$ that has the following two properties:
For every positive real number $x$, $x/(x+1) <m$
If y is any positive real number with the property that for every positive real number $x$, $x/(x+1)<y$, then $m≤y$
$∃!m∈R^+[∀x∈R^+(x/(x+1)<m)∧∀y∈R^+[(∀x∈R^+(x/(x+1)<y))→m≤y]]$
Proof:
Existence: Let $m=1∈R^+$ . Now let $x∈R^+$ be arbitrary. Suppose $x/(x+1)≥m$ . Then $x≥mx+m=x+1$ , and $0≥1$ , which is a contradiction. Thus, $x/(x+1)<m$ . Now let $y∈R+$ be arbitrary such that $∀x∈R+(x/(x+1)<y)$ . Suppose that$ m=1>y$ . Now consider $x=y/(1−y)∈R^+$ , which is defined since $y<1$ .$ x/(x+1)=y/(1−y)/(y/(1−y+1))==y≥y$ . Thus, we have found an $x∈R^+$ such that $x/(x+1)≥y$ , which is a contradiction. It follows that $m≤y$ . This is from another post(Prove that there is a unique positive real number $m$ with two desired properties).
I was wondering how the choice of x=y/(1-y) was produced. I can follow it once it's been chosen, but before looking at the solutions, I would have had no idea how to strategically find one.
Thanks
You can produce $x$ by working backwards. A possible thought process might look like this:
We want to find $x$ where $\frac{x}{x+1} \ge y$ to show a contradiction. Let's write an equation that satisfies the inequality, then we can simply solve the equation. $\frac{x}{x+1} = y$ seems like a good choice. Solving: $$ x = yx+y\\ (1-y)x = y\\ x = \frac{y}{1-y} $$ Note that to be more clear and concise, the author presents $x = \frac{y}{1-y}$ first, then arrive at the contradiction.