Category: B. Demidovich’s (Ed.) Problems in Mathematical Analysis

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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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202004241533 Problem 1, Ch. 1 Sec. 1

Prove that if a and b are real numbers then

||a|-|b||\leqslant |a-b|\leqslant |a|+|b|.

Proof.

As |a| is non-negative and -|a| non-positive, one has

Eq. (1):

-|a|\leqslant a\leqslant |a|

Eq. (2):

-|b|\leqslant b\leqslant |b|

Combining Eq. (1) and Eq. (2),

-(|a|+|b|)\leqslant a+b\leqslant |a|+|b|,

or, Eq. (3): (the triangle inequality)

|a+b|\leqslant |a|+|b|.

Applying the triangle inequality to |a-b|, one gets

Eq. (4):

|a-b|=|a+(-b)|\leqslant |a|+|-b|=|a|+|b|,

or, Eq. (4)’:

|a-b|\leqslant |a|+|b|.

Applying the triangle inequality to |a|=|(a-b)+b|, one gets

Eq. (5):

|a|=|(a-b)+b|\leqslant |a-b|+|b|,

or, Eq. (5)’:

|a|-|b|\leqslant |a-b|.

Applying the triangle inequality to |b|=|(b-a)+a|, one gets

Eq. (6):

|b|=|(b-a)+a|\leqslant |b-a|+|a|,

Roughwork.

\begin{aligned} |b|-|a| & \leqslant |b-a| \\|b|-|a| & \leqslant |a-b| \\-|a-b| & \leqslant |a|-|b|\end{aligned}

or, Eq. (6)’:

-|a-b| \leqslant |a|-|b|

Combining Eq. (5)’ and Eq. (6)’:

-|a-b| \leqslant |a|-|b| \leqslant |a-b|,

or, Eq. (7):

||a|-|b||\leqslant |a-b|.

Combining Eq. (4)’ and Eq. (7), one obtains readily

||a|-|b||\leqslant |a-b|\leqslant |a|+|b|.

QED