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Chapter 2: Problem 13

If \(y>0\), show that there exists \(n \in \mathbb{N}\) such that \(1 / 2^{n}

Short Answer

Expert verified
There exists \(n\) such that \(1 / 2^{n}<y\).

Step by step solution

01

Foundation of the Proof

From the Archimedean principle, since \(y > 0\) there exists \(n \in \mathbb{N}\) such that \(1/n < y\). That is, \(n > 1/y\).
02

Logarithmic Transformation

Taking the natural logarithm to both sides of the inequality \(n > 1/y\), we get \(\ln(n) > -\ln(y)\).
03

Comparison with log 2

Now, \(\ln(n)\) is always less than \(n\). We also know that \(\ln(2) < 1\). So, we can say that, \(n = \ln(2^{n}) < 2^{n}\). Therefore, if \(n > -\ln(y)\), then \(2^{n} > 1/y\).
04

Inverting and Final Proof

Inverting both sides of the aforementioned inequality obtains \(1 / 2^{n}<y\). Thus, there exists a natural number \(n\) such that \(1 / 2^{n}<y\).

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Most popular questions from this chapter

Let \(S_{1}:=\\{x \in \mathbb{R}: x \geq 0\\}\). Show in detail that the set \(S\), has lower bounds, but no upper bounds. Show that inf \(S_{1}=0\).

If \(0

Let \(a, b \in \mathbb{R}\), and suppose that for every \(\varepsilon>0\) we have \(a \leq b+\varepsilon .\) Show that \(a \leq b\).

Solve the following equations, justifying each step by referring to an appropriate property or theorem. (a) \(2 x+5=8\), (b) \(x^{2}=2 x\), (c) \(x^{2}-1=3\), (d) \((x-1)(x+2)=0\).

Let \(S \subseteq \mathbb{R}\) be nonempty. Prove that if a number \(u\) in \(\mathbb{R}\) has the properties: (i) for every \(n \in \mathbb{N}\) the number \(u-1 / n\) is not an upper bound of \(S\), and (ii) for every number \(n \in \mathbb{N}\) the number \(u+1 / n\) is an upper bound of \(S\), then \(u=\sup S .\) (This is the converse of Exercise 2.3.8.)
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