If $x>0$ , show that there exists $n \in \mathbb{N}$ such that $\dfrac{1}{2^n}
If $x>0$ , show that there exists $n \in \mathbb{N}$ such that $\dfrac{1}{2^n}<x$
Here what I did,

*

*Proof by contrapostive


*If for all $n \in \mathbb{N}$, $\space$ $\dfrac{1}{2^n} \geq x$ then $x \leq 0$,


*Since for all $n \in \mathbb{N}$,  $\dfrac{1}{2^n} \geq x$


*taking $ n\to \infty$ we have  $x \leq 0$ by the comparison theorem


*therefor If $x>0$ ,  there exists $n \in \mathbb{N}$ such that $\dfrac{1}{2^n}<x$
Can anyone verify this? if this not correct please correct it?
 A: For $n = \left\lfloor \dfrac{1}{x} \right\rfloor + 1$ we have :
$$\dfrac{1}{x} < n = \underbrace{1 + \ldots + 1}_{n \text{ times}} \leq 1 + 2 + \cdots + 2^{n - 1} = 2^n$$
So :
$$\dfrac{1}{2^n} < x$$
A: Binomial expansion:
$2^n=(1+1)^n=$
$1+n+n(n-1)/2!+.....> n. $
Archimedean principle:
There is a $n \in \mathbb{N}$ s.t.
$n>1/x.$
The result follows.
A: Your proof is correct (assuming you can use the comparison theorem) but the presentation could be better.
For instance you write,

*

*Proof by contrapositive

*If for all $n\in \mathbb{N}$, $\frac{1}{2^{n}}\geq x$ then $x\leq0$,

the second point reads like a fact which you already know (which is not the case).
If you use words, you can express the above points without ambiguity, for example as follows:

We want to do a proof by contrapositive. Therefore we assume that
for all $n\in \mathbb{N}$, $\frac{1}{2^{n}}\geq x$ and
show that $x\leq 0$.

The rest of your proof should also be written in a single text with words that connect previous sentences and explain your reasoning. Don't use lists unnecessarily, they can ruin the flow of reading and often require further explanations to avoid ambiguity.
With all that in mind your proof can be written

We want to do a proof by contrapositive. Therefore we assume that
for all $n\in \mathbb{N}$, $\frac{1}{2^{n}}\geq x$ and
show that $x\leq 0$. Since $\frac{1}{2^{n}}$ approches zero as $n\to \infty$, our assumption that $\frac{1}{2^{n}}\geq x$ for all $n\in \mathbb{N}$ implies by the comparison theorem that $x\leq 0$, which is what we wanted to prove.

