06. The modulus of a real number

Modulus

Let $a\in \mathbb{R}$ then

$$|a|=\{\begin{array}{ll}a& \text{ if }a>0\\ 0& \text{ if }a=0\\ -a& \text{ if }a<0\end{array}$$

Also called the absolute value of $a$
Well-defined due to the trichotomy property

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Properties of Modulus

Take $a,b,c\in \mathbb{R}$ then

1. $|-a|=|a|$
2. $|a|\ge 0$
3. $|a{|}^2=a^2$
4. $|ab|=|a||b|$
5. $-|a|\le a\le |a|$
6. If $c\ge 0$ then $|a|\le c⟺-c\le a\le c$
   Works similarly if we use $(<,>)$ instead of $(\le .\ge )$

Proof (not detailed)

1. By definition of modulus (do cases)
2. By definition of modulus (do cases)
3. Use definition and trichotomy (cases) and use $(-a)\cdot (-a)=a^2$
4. Use definition of modulus, trichotomy and use cases
5. If $a\ge 0$ then $-|a|\le 0\le a=|a|$
   If $a<0$ then $-|a|=a<0\le |a|$
   Hence$-|a|\le a\le |a|$
6. Proof $⟹$
   Suppose that $-c\le a\le c$
   Then by $(5)$, $-a\le c$ and $-c\le -|a|\le a\le |a|\le c$ and so done by transitivity ()

Proof $⟸$
Suppose that $-c\le a\le c$ then $-a\le c$ and $a\le c$
But $|a|=a$ or $|a|=-a$ so

$$|a|\le c$$

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02 - Triangle Inequality

Triangle Inequality

Take $a,b,c\in \mathbb{R}$

1. $|a+b|\le |a|+|b|$

2. $|a+b|\ge ||a|-|b||$
   Also called the reverse triangle inequality

Proof

1. As $-|a|\le a\le |a|$ and $-|b|\le b\le |b|$, by Properties of Modulus (5)
   Adding them together then

$$-(|a|+|b|)\le a+b\le |a|+|b|$$

Hence by Properties of Modulus (6) then

$$|a+b|\le |a|+|b|$$

2. By $(1)$ then

$$\begin{array}{rl}|a|& =|(a+b)+(-b)|\le |a+b|+|-b|=|a+b|+|b|\\ |b|& =|(b+a)+(-a)|\le |a+b|+|-a|=|a+b|+|a|\end{array}$$

Hence $|a|-|b|\le |a+b|$ and $|b|-|a|\le |a+b|$
Hence by Properties of Modulus (5)

$$|a+b|\ge ||a|-|b||$$

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