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I am currently trying to solve example 3.7.2 from Velleman's how to prove it. Here is the problem:

Prove that there is a unique positive real number m that has the following two properties:

(a) For every positive real number $x$, $x/(x+1)<m$

(b)​ If y is any positive real number with the property that for every positive real number $x$, $x/(x+1)<y$ then $m\leq y$

I usually don't have problems with existence. So I guessed here that $m=1$.

I then assumed that if there was a $y=m-\epsilon=1-\epsilon$ and $\epsilon>0$

Then I could show that there exist $x$ such that property $x/(x+1)< y$ doesn't hold for this choice of $y$. So I used the contrapositive here and $y$ must be greater than $1$.

Now I tried to prove uniqueness. I always have trouble doing these. So I tried the following. If we assume there is a $y=1+\epsilon$ then there is a smaller $y'=1+\epsilon/2$ which satisfies the property (a). But for property (b) this halving can then be done indefinitely such that we arrive at $y=1$. But somehow I feel this is not a valid proof for uniqueness. Could somebody point out what is wrong here or offer a better proof of uniqueness?

Velleman, Daniel J.. How to Prove It (A Structured Approach) (S.171). Cambridge University Press. Kindle-Version.

eeqesri
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    Suppose $\tilde m$ also satisfies properties (a) and (b). Apply property (b) to $\tilde m$ and $m$, implying $m\leq \tilde m$ and $\tilde m \leq m$. As for your argument, the intuition is correct, it just needs to be formalized. One way to write it would be as follows: Suppose to the contrary that there is a $y \neq 1$ that satisfies (a) and (b). Obviously $y >1$ (since $x/(x+1) \uparrow 1$ as $x\to\infty$), say $y= 1+\epsilon$. But then $1+\epsilon/2$ also satisfies (a) and $1+\epsilon/2 < y$, implying that any $y>1$ cannot satisfy (b), i.e. uniqueness. – Andrew Aug 14 '23 at 16:18
  • Oh wow that first part is a really elegant proof. It is much simpler than what I had thought. – eeqesri Aug 14 '23 at 16:29

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