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${\bf Homework \; Problem}$: Given any real number $\epsilon > 0$, prove ${\bf very}$ carefully that there exists a positive integer $n$ such that $\dfrac{1}{2^n} < \epsilon $

Attempt:

If there is some $\epsilon_0$ so that $\dfrac{1}{2^n} \geq \epsilon_0 $ then $\dfrac{1}{\epsilon_0} \geq 2^n $

Now, notice that $\log_2 (1/\epsilon_0+1) \in \mathbb{R}$ so that by the archimidean property of reals one can find some $n_0$ so that $n_0 > \log_2 (1/\epsilon_0+1)$. Thus,

$$ \dfrac{1}{\epsilon_0 } \geq 2^n > 2^{\log_2 (1/\epsilon_0+1)} = \dfrac{1}{\epsilon_0} + 1 $$

And this is false! Is this a correct and sufficient argument? Do I need to explain more?

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It is correct, but when you wrote that $\dfrac1{\varepsilon_0}\geqslant2^n$, you should have written that $\dfrac1{\varepsilon_0}\geqslant2^{n_0}$. And the first sentence needs a quantifier: If there is some $\varepsilon_0$ so that $(\forall n\in\Bbb N):\dfrac1{2^n}\geqslant\varepsilon_0$

Note that there is a simpler proof: take $n\in\Bbb N$ such that $n>\frac1\varepsilon$. Then$$\frac1{2^n}\leqslant\frac1n<\varepsilon.$$

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  • $\begingroup$ Thanks Jose for response: in the first sentence i meant there is epsilon_0 that holds for all n $\endgroup$ – James Apr 19 '20 at 8:12
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You can avoid logarithms.

Suppose $\forall n \in \mathbb N$ we have $\frac{1}{2^n}>\epsilon$. By the Archimedian Property, we can find a natural number $c$ such that $\epsilon \cdot c >\frac{1}{2^n}$. As $c$ is finite, thus $\exists j \in \mathbb N$ such that $c<2^j \iff \frac{1}{c}>\frac{1}{2^j}$ and thus $\epsilon > \frac{1}{2^nc}>\frac{1}{2^{n+j}}$ a contradiction.

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As $\epsilon>0$, also $\frac1\epsilon>0$. By the Archimedean property, there exists $n\in\Bbb N$ with $n>\frac1\epsilon$. We know (or prove by induction) that $2^n>n$. Then from $2^n>\frac1\epsilon>0$ we arrive at $\epsilon>\frac1{2^n}$.

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By Archimedean property of $\ \mathbb{R}$, for given $\ x,y\in\mathbb{R}, \exists n\in \mathbb{N}$ such that $\ nx\gt y$.
Now putting $\ y= \varphi$ and $\ x=1$ , we get
$\ n\gt\varphi \implies 2^n\gt n\gt \varphi$
Thus $\ \dfrac{1}{2^n} \lt \dfrac{1}{\varphi}=\epsilon$

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Since $n$ is a positive integer, so $\frac{1}{2^n}$ must also be positive.

Also, $\epsilon$ is given to be a positive real number.

So we have $\frac{1}{2^n}>0$ and $\epsilon>0$.

But note that $\frac{1}{2^x} \rightarrow 0$ as $x \rightarrow \infty$. So it can be any positive real number, including those less than $\epsilon$.

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  • $\begingroup$ why $\ \dfrac{1}{2^n}\in\mathbb{Z} \forall n\in\mathbb{N}$? and also why $\ \dfrac{1}{2^n}\lt 0$? $\endgroup$ – Manjoy Das Apr 19 '20 at 17:15
  • $\begingroup$ @ManjoyDas That is mistyping. I corrected it now. THANKS. $\endgroup$ – Hussain-Alqatari Apr 19 '20 at 21:49

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