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Absolute value

Further topics · Chapter A.1 · Lecture notes (PDF) · Exercises: Absolute value · Slides (PDF)

1. Definition

Definition 1: absolute value

The absolute value \(|a|\) of a real number \(a\) is the following non-negative real number:

\[\begin{equation} |a| = \begin{cases} a & {\rm if~~} a \ge 0\\ -a & {\rm if~~} a < 0 \end{cases} \label{ass_1} \end{equation}\]

From the definition of absolute value it immediately follows that:

\[\begin{equation} |a| < \varepsilon \Longleftrightarrow -\varepsilon < a < \varepsilon, \qquad \forall \varepsilon > 0, a \in \mathbb{R} \label{ass_2} \end{equation}\]

and, similarly, for non-strict inequalities:

\[\begin{equation} |a| \le \varepsilon \Longleftrightarrow -\varepsilon \le a \le \varepsilon, \qquad \forall \varepsilon \ge 0, a \in \mathbb{R} \label{ass_2b} \end{equation}\]

2. Triangle inequality

Observation 1: triangle inequality

\[\begin{equation} |b + c| \le |b| + |c|, \qquad \forall b,c \in \mathbb{R} \label{ass_3} \end{equation}\]
Proof

We write the two relations:

\[ -|b| \le b \le |b|, \quad \quad -|c| \le c \le |c| \]

and add member to member:

\[ -(|b| + |c|) \le b + c \le |b| + |c| \]

So for \(\eqref{ass_2b}\), with \(\varepsilon=|b|+|c|\ge 0\) and \(a=b+c\), it follows \(\eqref{ass_3}\). □

  • The triangle inequality is also used in the following form:

    \[\begin{equation} |d - e| \le |d-f| + |e-f| \qquad \forall d,e,f \in \mathbb{R} \label{ass_4} \end{equation}\]

    To obtain it, it suffices to set in \(\eqref{ass_3}\):

    \[ b = d - f, \quad c = f-e \]

    and we get:

    \[ |d - f +f-e|=|d -e| \le |d - f| + |f-e| = |d - f| + |e-f| \]

    since:

    \[ |f - e| = |e-f| \qquad \forall e,f \in \mathbb{R} \]

  • Finally, another form of the triangle inequality is:

    \[\begin{equation} |g| \le |g-h| + |h| {\rm ~~~~~that~is~~~~~} |g| - |h| \le |g-h| \qquad \forall g,h \in \mathbb{R} \label{ass_AA} \end{equation}\]

    To obtain it, it suffices to set in \(\eqref{ass_3}\):

    \[ b = g - h, \quad c = h \]

Observation 2: reverse triangle inequality

\[\begin{equation} \big|\, |g| - |h| \,\big| \le |g-h|, \quad \forall g,h \in \mathbb{R} \label{ass_5} \end{equation}\]
Proof

From \(\eqref{ass_AA}\), we have:

\[ |g| - |h| \le |g-h| \qquad \forall g,h \in \mathbb{R} \]

Similarly exchanging \(g\) for \(h\) in \(\eqref{ass_AA}\) we get:

\[ |h| - |g| \le |h-g| = |g-h| {\rm ~~~~that~is~~~~} |g| - |h| \ge -|g-h| \]

then we have

\[ -(|g-h|) \le |g| - |h| \le |g-h| \qquad \forall g,h \in \mathbb{R} \]

Accordingly from \(\eqref{ass_2b}\), with \(\varepsilon = |g-h| \ge 0\) and \(a= |g| - |h|\), we get the reverse triangle inequality. □

  • The triangle inequality \(\eqref{ass_3}\) can easily be generalized to the case of \(k\) addends:

    \[\begin{equation} \left| \sum_{i=1}^k b_i \right| \le \sum_{i=1}^k |b_i|. \label{ass_6} \end{equation}\]

  • The following immediate properties also hold:

    \[\begin{equation} |b\:c| = |b| \: |c|, \qquad \left| \frac{b}{c}\right|= \frac{|b|}{|c|}, \qquad |-b|=|b| \qquad \forall b,c \in \mathbb{R} ~ (c \neq 0 {\rm ~in~the~quotient}). \label{ass_7} \end{equation}\]

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