Show that there exists $m\in\mathbb{N}$ such that $a+\frac{m}{2^n}\geq b$

Question

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

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

I tired

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 $2^n(b-a)$

am i right if there any other way ?

any help would be appreciated!

Answer 1

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 $2^n(b-a)$

Answer 2

What? Just solve for $m$. $m \geq (b-a)\cdot2^n$. In particular $m = (b-a)\cdot2^n$, or $m=(b-a)/2^n + 35$ will do.