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

Question

Let $ \mathcal{P} \subset \mathbb{R}$,\ $\mathcal{P}\neq \emptyset $ and let $b$ be an upper bound of $\mathcal{P}$.

1. Let $a \in \mathcal{P}$ and let $n\in \mathbb{N}^*$ Show that : $$\exists\ m\in\mathbb{N} \text{ such that: } \quad a+\dfrac{m}{2^n}\geq b$$
2. 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?

Answer 1

(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 ${\mathcal P}$.

Answer 2

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.