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Chapter 2: Problem 1

If \(a, b \in \mathbb{R}\) and \(b \neq 0\), show that: (a) \(|a|=\sqrt{a^{2}}\), (b) \(|a / b|=|a| /|b|\).

Short Answer

Expert verified
The properties \(|a| = \sqrt{a^{2}}\) and \(|a / b| = |a| / |b|\) have been proven using the definitions and properties of absolute values in real numbers.

Step by step solution

01

Proof of \(|a| = \sqrt{a^{2}}\)

An absolute value of a number \(a\), denoted as \(|a|\), is defined as \(a\) if \(a \geq 0\), and \(-a\) if \(a < 0\). We will now separate this proof into two cases based on the sign of \(a\). \n Case 1: If \(a \geq 0\), then by definition, \(|a| = a\), and \(a^{2} = a*a = a^{2}\). So, \(|a| = \sqrt{a^{2}}\) because square root of \(a^{2}\) is \(a\). \n Case 2: If \(a < 0\), then by definition, \(|a| = -a\). So, \((-a)^{2} = -a*-a=a^{2}\). Thus, \(|a| = \sqrt{a^{2}}\) as square root of \(a^{2}\) is \(a\) or \(-a\). This concludes the proof for part (a).
02

Proof of \(|a / b| = |a| / |b|\)

The absolute value of the ratio \(a/b\) where \(b\) is not zero can be shown to be equal to the ratio of the absolute values of \(a\) and \(b\) by using definitions and properties of absolute values.\n To prove this property, we can write the left side of the equation as \(|a / b| = \sqrt{(a/b)^{2}}\), by using the result of part (a). Square of \(a/b\) is \(a^{2}/b^{2}\). Therefore, the left side of equation is \(\sqrt{a^{2}/b^{2}}\). On the right side, \(|a| = \sqrt{a^{2}}\) and \(|b| = \sqrt{b^{2}}\), so \(|a|/|b| = \sqrt{a^{2}} / \sqrt{b^{2}} = \sqrt{a^{2} / b^{2}}\). Therefore, \(|a / b| = |a| / |b|\). This concludes the proof for part (b).

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Most popular questions from this chapter

Find all \(x \in \mathbb{R}\) that satisfy the following inequalities: (a) \(|4 x-5| \leq 13\), (b) \(\left|x^{2}-1\right| \leq 3\).

Let \(X\) be a nonempty set, and let \(f\) and \(g\) be defined on \(X\) and have bounded ranges in \(\mathbb{R}\). Show that $$ \sup \\{f(x)+g(x): x \in X\\} \leq \sup (f(x): x \in X\\}+\sup \\{g(x): x \in X\\} $$ and that $$ \inf \\{f(x): x \in X\\}+\inf \\{g(x): x \in X\\} \leq \inf \\{f(x)+g(x): x \in X\\} $$ Give examples to show that each of these inequalities can be either equalities or strict inequalities.

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Prove that if \(a, b \in \mathbb{R}\), then th rom (a) \(-(a+b)=(-a)+(-b)\) (b) \((-a) \cdot(-b)=a \cdot b\) (c) \(1 /(-a)=-(1 / a)\) (d) \(-(a / b)=(-a) / b\) if \(b \neq 0\)

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 \(S_{1}=0\).
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