# Prove that we an find a positive integer so that $1/2^n < \epsilon$

$${\bf Homework \; Problem}$$: 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?

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$$\frac1{2^n}\leqslant\frac1n<\varepsilon.$$

• Thanks Jose for response: in the first sentence i meant there is epsilon_0 that holds for all n – James Apr 19 '20 at 8:12

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.

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 $$\epsilon>\frac1{2^n}$$.

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 $$\ \dfrac{1}{2^n} \lt \dfrac{1}{\varphi}=\epsilon$$

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 $$\epsilon$$.

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