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Chapter 2: Problem 15

Modify the argument in Theorem \(2.4 .7\) to show that if \(a>0\), then there exists a positive real number \(z\) such that \(z^{2}=a\)

Short Answer

Expert verified
The modification of the theorem 2.4.7 shows that for any positive real number a, there exists a positive real number z such that \(z^{2} = a\). This conclusion was reached by reasoning based on properties of real numbers and the nature of square functions.

Step by step solution

01

Use of exhaustive property of real numbers

Since \(a > 0\), there exists a set of positive real numbers \(b\) such that \(b^{2} < a\). This set must have an upper bound since for every real number \(a\) there is a square of a number \(b\) s.t. \(b^{2} > a\). This is due to the completeness or exhaustive property of real numbers.
02

Apply the Completeness Theorem

By the completeness property of real numbers, the supremum or least upper bound of the set exists. Let this least upper bound be \(z\). Now there are two scenarios that can be considered.
03

Argument for the first scenario

Let's say \(z^{2} < a\). This would contradict the fact that \(z\) is the least upper bound, because \(z + \epsilon\) (for some small \(\epsilon > 0\)) would also be an upper bound of the set. That means there would be a number which squares to a value less than \(a\) but greater than \(z^{2}\), contradicting the definition of \(z\) as our least upper bound.
04

Argument for the second scenario

The other possible scenario is that \(z^{2} > a\). By the same reasoning, we reach a contradiction. If \(z^{2} > a\), then \(z - \epsilon\) (for some small \(\epsilon > 0\)) will square to a value less than \(a\), contradicting again the claim that \(z\) is the supremum.
05

Conclude the proof

Since we have reached contradictions in both possible scenarios, we can conclude that the only possibility left is that \(z^{2} = a\). Thus, for any positive real number \(a\), there exists a positive real number \(z\) such that \(z^{2} = a\). This concludes our modified proof.

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Most popular questions from this chapter

If \(x, y, z \in \mathbb{R}\) and \(x \leq z\), show that \(x \leq y \leq z\) if and only if \(|x-y|+|y-z|=|x-z|\). Interpret this geometrically.

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

Find the decimal representation of \(-\frac{2}{7}\).

Modify the argument in Theorem \(2.4 .7\) to show that there exists a positive real number \(y\) such that \(y^{2}=3\)

Let \(S_{2}=\\{x \in \mathbb{R}: x>0\\} .\) Does \(S_{2}\) have lower bounds? Does \(S_{2}\) have upper bounds? Does inf \(S_{2}\) exist? Does sup \(S_{2}\) exist? Prove your statements.
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