**Mathematical Analysis Problem****Problem Statement:**Prove that for every real number $ x $, there exists a sequence $ \{ a_n \}_{n=1}^\infty $ converging to $ x $ such that $ a_n $ is irrational for all $ n $.**Explanation:**This problem involves constructing a sequence of irrational numbers that converges to any given real number $ x $. This requires knowledge of sequences, limits, and properties of real and irrational numbers. The goal is to demonstrate that no matter what real value is chosen, it's possible to find a sequence made entirely of irrational numbers that approaches this value as closely as desired.

Name: **Mathematical Analysis Problem****Problem Statement:**Prove that for
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Description: **Mathematical Analysis Problem****Problem Statement:**Prove that for every real number $ x $, there exists a sequence $ \{ a_n \}_{n=1}^\infty $ converging to $ x $ such that $ a_n $ is irrational for all $ n $.**Explanation:**This proble

Let, x∈ℝ. To show that for every x∈ℝ there exists a sequence ann=1∞ that converges to x such that an…

Answered: Prove that for every real number z,…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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**Mathematical Analysis Problem**

**Problem Statement:**

Prove that for every real number $ x $, there exists a sequence $ \{ a_n \}_{n=1}^\infty $ converging to $ x $ such that $ a_n $ is irrational for all $ n $.

**Explanation:**

This problem involves constructing a sequence of irrational numbers that converges to any given real number $ x $. This requires knowledge of sequences, limits, and properties of real and irrational numbers. The goal is to demonstrate that no matter what real value is chosen, it's possible to find a sequence made entirely of irrational numbers that approaches this value as closely as desired.

Expert Solution

Step 1

Let, $x \in ℝ$ .

To show that for every $x \in ℝ$ there exists a sequence ${\{a_{n}\}}_{n = 1}^{\infty}$ that converges to $x$ such that $a_{n}$ is irrational for all $n$ .

For, $x \in ℝ$ , $x < x + \frac{1}{n}$ for all $n \in ℕ$ .

By density theorem, for any two real numbers, $x, y$ such that $x < y$ then there exists $z \in T$ , where $T$ is the set of all irrational numbers, such that $x < z < y$ .

So, let $a_{n}$ be the irrational number, such that, $x < a_{n} < x + \frac{1}{n}$ for all $n \in ℕ$ .

Then by Squeeze Theorem,

Let $\{a_{n}\}, \{b_{n}\} and \{c_{n}\}$ be sequence of real numbers such that, for some positive integer, $a_{n} \leq b_{n} \leq c_{n}$ , for all $n \in ℕ$ , and

$\lim (a_{n}) = \lim (c_{n}) = l$ , then, $\{b_{n}\}$ is convergent and $\lim_{} (b_{n}) = l$

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