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Chapter 2: Problem 13

Given a real number \(x \geq 0\), show that there exists a real number \(s \geq 0\) such that \(s^{2}=x\).

Short Answer

Expert verified
For any \(x \geq 0\), \(s = \sqrt{x} \geq 0\) exists such that \(s^2 = x\).

Step by step solution

01

Understanding the Problem

We need to show that for any non-negative real number \(x\), there exists a non-negative real number \(s\) such that \(s^2 = x\). This means finding the square root of \(x\).
02

Using the Definition of Square Root

The definition of a square root states that for any non-negative real number \(x\), the square root \(\sqrt{x}\) is a non-negative number \(s\) such that \(s^2 = x\). This \(s\) is unique and satisfies the condition we are seeking to prove exists.
03

Existence of Square Root for Non-negative Real Numbers

The existence of square roots for non-negative real numbers \(x\) is guaranteed by the completeness property of real numbers, which ensures that every non-negative real number has a non-negative square root.
04

Statement of Existence

Thus, based on mathematical definitions and properties, for every non-negative real number \(x\), there exists a real number \(s = \sqrt{x} \geq 0\) such that \(s^2 = x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Root
The concept of a square root is essential in understanding how we can determine a number that, when multiplied by itself, gives the original number. For non-negative real numbers, the square root is particularly important because it highlights a specific quality of real numbers. When we talk about the square root of a number, we refer to the unique non-negative number, denoted as \( \sqrt{x} \), which satisfies the equation \( s^2 = x \).
  • If you have a non-negative number like 9, its square root is 3, since \( 3^2 = 9 \).
  • The square root of 0 is also 0, because \( 0^2 = 0 \).
  • Notably, square roots always yield a non-negative result in the context of real numbers, aligning with the nature of being a non-negative real number itself.
In essence, the square root operation reverses the process of squaring a number, helping us discover an original number from its square.
Completeness Property
The completeness property of real numbers is a fundamental concept in mathematics that ensures the robustness and faultlessness of the real number line. This property is crucial in proving the existence of square roots for non-negative real numbers.
  • The completeness property states that every non-negative real number has a least upper bound or supremum, which ensures no "gaps" exist in the real number line.
  • Thanks to this property, we can guarantee that a non-negative real number has a square root, which is also a real number.
  • This means that if you take any real number that has been squared, you can "reverse" the operation to find its square root which lies within the real numbers.
Without completeness, the square root might not always be assured for every non-negative real number, making this property vital to understanding and utilizing square roots effectively in mathematical analysis.
Non-negative Real Numbers
Non-negative real numbers include all the numbers on the number line greater than or equal to zero. These are the numbers we interact with frequently in real-life contexts and many mathematical operations.
  • Non-negative real numbers are symbolized as \( x \geq 0 \), meaning they start from zero and extend positively towards infinity.
  • They include whole numbers like 0, 1, 2, rational numbers like \( \frac{1}{2} \), and irrational numbers like \( \sqrt{2} \).
  • This set of numbers is critical when addressing functions such as square roots, as square roots are only defined for these non-negative values.
Through operations like finding square roots, non-negative real numbers illustrate their versatility and reliability, offering a substantial foundation for various mathematical computations and theories.

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Most popular questions from this chapter

Show that for any positive integer \(k\) $$ \lim _{n \rightarrow \infty} \frac{1}{n^{k}}=0 $$

Evaluate $$ \lim _{n \rightarrow \infty} \frac{3 n^{5}+8 n^{3}-6 n}{8 n^{5}+2 n^{4}-31} $$ carefully showing each step.

Show that a nonincreasing sequence of real numbers either converges or diverges to negative infinity.

Suppose \(\sum_{i=m}^{\infty} a_{i}\) and \(\sum_{i=m}^{\infty} b_{i}\) are infinite series for which \(a_{i} \geq 0\) and \(b_{i}>0\) for all \(i \geq m .\) Show that if \(\sum_{i=m}^{\infty} a_{i}\) diverges and $$ \lim _{i \rightarrow \infty} \frac{a_{i}}{b_{i}}<+\infty $$ then \(\sum_{i=m}^{\infty} b_{i}\) diverges.

Suppose both \(\sum_{i=m}^{\infty} a_{i}\) and \(\sum_{i=m}^{\infty} b_{i}\) converge. Show that \(\sum_{i=m}^{\infty}\left(a_{i}+b_{i}\right)\) converges and $$ \sum_{i=m}^{\infty}\left(a_{i}+b_{i}\right)=\sum_{i=m}^{\infty} a_{i}+\sum_{i=m}^{\infty} b_{i} $$
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