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Chapter 10: Problem 4

Es ist \(a^{2}=\left|a^{2}\right|=|a|^{2} ; \sqrt{a^{2}}=|a| \quad(\) nicht \(=a\) )

Short Answer

Expert verified
The expressions given demonstrate how the square and absolute function ensures non-negative real values.

Step by step solution

01

Understanding Absolute Value

The absolute value of a number, denoted by \(|a|\), represents the non-negative value of \(a\). This ensures that any expression involving \(|a|\) will always be zero or positive.
02

Properties of Squares and Absolute Values

For any real number \(a\), it holds that \(a^2 = |a|^2\) because both expressions yield the non-negative value of \(a^2\). This is because squaring a negative is also positive, just as the square of the absolute value is positive.
03

Equality Explanation

The equality \(a^2 = \left|a^2\right| = |a|^2\) holds due to the reason explained above. Squaring any real number whether it is positive, negative or zero results in a non-negative number.
04

Square Root Relation

The square root of \(a^2\) is given by \(\sqrt{a^2} = |a|\). This is because the principal square root function itself always yields a non-negative result, hence the necessity of using \(|a|\) instead of just \(a\), which could be negative.
05

Conclusion of Explanation

Given that \(\sqrt{a^2} = |a| eq a\) reinforces that while squaring and then taking the square root of any \(a\) returns its absolute value, it does not necessarily return \(a\) itself if \(a\) were negative.

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Key Concepts

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

Properties of Squares
When we square any real number, the result is always non-negative, regardless of whether the original number was positive or negative. This is because multiplying two negative numbers results in a positive product. So, for a number \( a \), its square \( a^2 \) inherently becomes non-negative.
For example, if \( a = -3 \), then \( a^2 = (-3)^2 = 9 \). This behavior helps simplify expressions because you can always expect a positive result. It aligns with the absolute value squared, \(|a|^2\), since \(|a|\) is always non-negative by definition.
Square Root
The square root operation is essentially the inverse of squaring a number. However, taking the square root is a little tricky because it aims to return a value whose square gives the original input.
The square root of a number \( a^2 \) is denoted as \( \sqrt{a^2} \). Here, it becomes clear why the result is typically the absolute value \( |a| \). This is because the square root operation is designed to provide a non-negative result, which corresponds directly to the absolute value rather than potentially negative \( a \).
Non-negative Values
Non-negative values are all values that are either zero or positive. The absolute value of any real number \( a \) is written as \(|a|\), and it is defined to be non-negative.
This means that regardless of whether \( a \) itself was negative, zero, or positive, \(|a|\) will always be zero or greater.
By ensuring that values remain non-negative, these principles simplify mathematical operations and their subsequent results. This plays a significant role, especially when combined with operations like square rooting, which are constrained to non-negative outputs.
Principal Square Root
The principal square root refers to the non-negative square root of a number. In mathematics, especially when dealing with real numbers, the principal square root is the default choice for solving square root expressions.
For \( a^2 \), the principal square root is \( \sqrt{a^2} = |a| \). This notation emphasizes that whenever we square a number and find its square root, the outcome is the absolute value \( |a| \), ensuring the result is inherently non-negative. This property consistently guides mathematical solutions that require square roots, reflecting the inherent non-negativity of squaring operations and their inverses.

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Most popular questions from this chapter

Sei \(\rho\) reell Zeige: Zu jedem \(\varepsilon>0\) gibt es ein rationales \(r\) mit \(|\rho-r|<\varepsilon\)

Ist \(\mathrm{d}(a, b)\) irgendeine Metrik auf \(\mathbf{R}\), so ist auch \(\mathrm{d}_{1}(a, b):=\frac{\mathrm{d}(a, b)}{1+\mathrm{d}(a, b)}\) cine solche Mit d ist auch \(\mathrm{d}_{1}\) translationsinvariant Hinwe is: Aus \(0 \leqslant s \leqslant t\) folgt \(\frac{s}{1+s} \leqslant \frac{t}{1+t}\).

Man sagt, ein K枚rper K besitze eine (reelle) Bewertung, wenn jedem \(a \in \mathbf{K}\) eine reelle Zahl \(|a|\) (der,, Betrag"'von \(a\) ) so zugeordnet ist, da脽 die Betragsaxiome (B 1), (B 2) und (B 3) - mit \(|a|\) an Stelle von \(|a|\)-gelten \(\mathbf{R}\) ist also mittels seines kanonischen Betrages \(|a|\) ein bewerteter K枚rper, ebenso \(\mathbf{Q}\) und \(\mathbf{Q}(\sqrt{2})\); eine Bewertung von \(\mathbf{C}\) werden wir in A \(12.15\) einf眉hren. Auf \(\mathbf{Q}\) lassen sich neben dem kanonischen Betrag noch weitere Betr盲ge einf眉hren: Sei \(p\) eine feste Primzahl. Da jede nat眉rliche \(\mathrm{Zahl}>1\) nach A \(6.3\) ein Produkt aus Primzahlen und diese ,Primfaktorzerlegung" eindeutig ist", kann man jedes \(a \neq 0\) aus \(\mathbf{Q}\) in der Form \(a=\frac{s}{t} p^{n}\) mit eindeutig bestimmtem \(n \in \mathbf{Z}\) und ganzen Zahlen \(s, t\) darstellen, die nicht mehr durch \(p\) teilbar sind (f眉r \(p=5\) ist z B. \(250 / 15=(2 / 3) 5^{2}\), \(17 / 15=(17 / 3) 5^{-1} ;\) im ersten Falle ist \(n=2\), im zweiten \(n=-1\) ). Zeige: a) \(\mathbf{Q}\) wird durch \(|a|_{p}:=p^{-n}\) f眉r \(a \neq 0,|0|_{p}:=0\) ein bewerteter K枚rper \((\ldots p\)-adische Bewertung "von \(\mathbf{Q})\) b) Es gilt die,,versch盲rfte Dreiecksungleichung \(^{\prime *}|a+b|_{p} \leqslant \max \left(|a|_{p}|b|_{p}\right)\). c) \(|n|_{p} \leqslant 1\) f眉r alle \(n \in \mathrm{N}\) (die \(p\)-adische Bewertung ist,, nichtarchimedisch'), d) Durch \(\delta(a, b):=|a-b|_{p}\) wird auf \(\mathbf{Q}\) eine translationsinvariante Metrik definiert

Es ist \(|a x+b| \leqslant|\alpha x+\beta| \Leftrightarrow(a x+b)^{2} \leqslant(\alpha x+\beta)^{2}\) Dieser Umstand erlaubt es, die L枚sungsmenge f眉r die linke Ungleichung manchmal leichter zu finden als durch Fallunterscheidungen.

\(\max (a, b)=\frac{a+b+|a-b|}{2}, \quad \min (a, b)=\frac{a+b-\mid a-b}{2}\)
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