202004241649 Problem 2, Ch. 1 Sec. 1

Prove the following equalities:

(a) $|ab|=|a|\cdot |b|$ ;

(b) $|a|^2=a^2$ ;

(c) $\displaystyle{\bigg|\frac{a}{b}\bigg|=\frac{|a|}{|b|}}$ (where $b\neq 0$ );

(d) $\sqrt{a^2}=|a|$ .

Proof.

(a) (proof by cases)

i. When $a,b\geqslant 0$ :

$|ab|=ab=|a||b|$ .

ii. When $a,b< 0$ :

$|ab|=ab=(-a)(-b)=|a||b|$ .

iii. When $a\geqslant 0$ and $b<0$ :

$|ab|=-ab=a(-b)=|a||b|$ .

iv. When $a<0$ and $b\geqslant 0$ :

$|ab|=-ab=(-a)b=|a||b|$ .

QED

(b) (proof by induction)

By making a stronger claim

$P(n)$ : For any positive integer $n$ , $|a^n|=|a|^n$ .

Proof.

The trivial cases $n=0$ and $n=1$ are evident.

Consider the case $n=2$ ,

$\begin{aligned} |a^2| & = |a||a| \qquad \textrm{(by equality (a))} \\ & = |a|^2 \end{aligned}$

$P(2)$ is true.

Assume now that $P(n)$ is true,

$\begin{aligned} P(n+1): \qquad |a^{n+1}|& = |a^n\cdot a| \\ & = |a^n||a| \qquad \textrm{(by equality (a))}\\ & = |a|^n|a| \qquad \textrm{(by the assumption }P(n)\textrm{ is true)} \\ & = |a|^{n+1} \end{aligned}$

it can be seen that $P(n+1)$ is also true.

From the fact that $P(2)$ is true and by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$ .

It follows that $|a|^2=|a^2|=a^2$ holds.

QED

(c) (direct proof)

$\because |a|=\displaystyle{\bigg| \frac{a}{b}\cdot b\bigg|} =\bigg|\displaystyle{\frac{a}{b}}\bigg| |b|$ ,

where the second equality sign is due to equality (a),

$\therefore \displaystyle{\bigg| \frac{a}{b} \bigg| = \frac{|a|}{|b|}}$ .

QED

(d) (proof by definition)

The absolute value of a real number $a$ , denoted by $|a|$ , is defined by

$|a|= \begin{cases} x & \textrm{if } x\geqslant 0, \\ -x & \textrm{if }x<0 \end{cases}$

For any non-negative real number $a$ , the symbol $\sqrt{a}$ denotes the non-negative square root of $a$ .

QED

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