For every $n$ there exists $k_n \in \mathbb{N}$ such that $a+k_n/2^n$ is an upper bound while $a+(k_n-1)/2^n$ is not 
Let $ \mathcal{P} \subset  \mathbb{R}$,\ $\mathcal{P}\neq  \emptyset $ and let $b$ be an upper bound of  $\mathcal{P}$.
  
  
*
  
*Let $a \in \mathcal{P}$ and let $n\in \mathbb{N}^*$ Show that : 
  $$\exists\ m\in\mathbb{N} \text{ such that: } \quad a+\dfrac{m}{2^n}\geq b$$
  
*Deduce that : $\exists\ k_n \in \mathbb{N}\quad a+k_n\times\dfrac{1}{2^n}\ $  is  upper bound of  $\mathcal{P}$  while : $\ a+(k_{n}-1)\dfrac{1}{2^n}\ $ is not .
  

Answer for 1 question is here but my personal answer is :
(Q1)
For any integer $m$, we have equivalence $a+\frac{m}{2^n}\geq b\ \Leftrightarrow\ m\geq2^n(b-a)$
Now as ${\mathbb R}$ is Archimedean, there is a natural number $m$ higher than the actual  $2^n(b-a)$ )
I tired:
Q2) Question $1$ shows that there is at least one integer $m$ such that $a + \frac{m}{2^n}$ is an upper bound for ${\mathcal P}$ (any number greater than an upper bound for
 ${\mathcal P}$ is itself an upper bound for ${\mathcal P}$).
We can therefore consider the smallest integer  natural $m$ such that  $a + \frac{m}{2^n}$ is a upper bound for ${\mathcal P}$. let's call it $k_n $.
By definition, $a + \frac{k_n}{2^n}$ and therefore an upper bound for $a + \frac{m}{2^n} $.
Two cases:


*

*Let $k_n \geq 1$ therefore $k_n-1$ is also a Natural number and as it is strictly smaller than $k_n$,  $a + \frac{k_n-1}{2^n}$ can not be an upper bound for ${\mathcal P} $ (this would contradict the minimal nature of $k_n$)

*Let $k_n=0$ and $a +\frac{k_n-1}{2^n}=a-\frac{1}{2^n}<a$ and as $a\in {\mathcal P}$, this again shows that $a +\frac{k_n-1}{2^n}$ n 'isn't an upper bound of ${\mathcal P}$.
Am I right?
 A: For the first part you have $$m\geq\left(b-a\right)2^{n}$$
 but you are looking for $m\in\mathbb{N}$ so you can take $$m=\left\lceil \left(b-a\right)2^{n}\right\rceil$$ where $\left\lceil \left(b-a\right)2^{n}\right\rceil$ represents the smallest integer greater than $\left(b-a\right)2^{n}.$ So for example if $\left(b-a\right)2^{n}=6.2$ you take $m=7$ and if $\left(b-a\right)2^{n}\in\mathbb{N}$ is simply $m=\left(b-a\right)2^{n}.$ Note that is the smallest integer that verifies you inequality, so if you take $m-1$ you have $$a+\left(m-1\right)/2^{n}<b$$ as you want.
A: (Q1)
For any integer $m$, we have equivalence $a+\frac{m}{2^n}\geq b\ \Leftrightarrow\ m\geq2^n(b-a)$
Now as ${\mathbb R}$ is Archimedean, there is a natural number $m$ higher than the actual  $2^n(b-a)$ )
Q2) Question $1$ shows that there is at least one integer $m$ such that $a + \frac{m}{2^n}$ is an upper bound for ${\mathcal P}$ (any number greater than an upper bound for
 ${\mathcal P}$ is itself an upper bound for ${\mathcal P}$).
We can therefore consider the smallest integer  natural $m$ such that  $a + \frac{m}{2^n}$ is a upper bound for ${\mathcal P}$. let's call it $k_n $.
By definition, $a + \frac{k_n}{2^n}$ and therefore an upper bound for $a + \frac{m}{2^n} $.
Two cases:


*

*Let $k_n \geq 1$ therefore $k_n-1$ is also a Natural number and as it is strictly smaller than $k_n$,  $a + \frac{k_n-1}{2^n}$ can not be an upper bound for ${\mathcal P} $ (this would contradict the minimal nature of $k_n$)

*Let $k_n=0$ and $a +\frac{k_n-1}{2^n}=a-\frac{1}{2^n}<a$ and as $a\in {\mathcal P}$, this again shows that $a +\frac{k_n-1}{2^n}$ n 'isn't an upper bound of ${\mathcal P}$.
