10.18 Absolute value

Prove:

(a) \(|a| = | - a|\)

\(a\)	\(-a\)	\(\\|a\\|\)	\(\\|-a\\|\)
\(>0\)	\(<0\)	\(a\)	\(-(-a)=a\)
\(<0\)	\(>0\)	\(-a\)	\(-a\)
\(=0\)	\(=0\)	\(0\)	\(0\)

Since we consider all the cases, from the table we can conclude that \(|a|=|-a|\) is true.

(b) \(\left|\frac{1}{a}\right|= \frac{1}{|a|}\), if \(a \neq 0\)

\(a\)	\(\frac1a\)	\(\\|a\\|\)	\(\frac{1}{\left\\|a\right\\|}\)	\(\\|\frac{1}{a}\\|\)
\(>0\)	\(>0\)	\(a\)	\(\frac1a\)	\(\frac1a\)
\(<0\)	\(<0\)	\(-a\)	\(-\frac{1}{a}\)	\(-\frac1a\)

Since we consider all the cases, from the table we can conclude that \(\left|\frac{1}{a}\right|= \frac{1}{|a|}\), if \(a \neq 0\)is true.

(c) \(\frac{|a|}{|b|} = \left|\frac{a}{b}\right|\), if \(b \neq 0\)

If \(a\neq 0\)

\(a\)	\(b\)	\(\frac{a}{b}\)	\(\\|a\\|\)	\(\\|b\\|\)	\(\frac{\left\\|a\right\\|}{\left\\|b\right\\|}\)	\(\\|\frac{a}{b}\\|\)
\(>0\)	\(>0\)	\(>0\)	\(a\)	\(b\)	\(\frac{a}{b}\)	\(\frac{a}{b}\)
\(>0\)	\(<0\)	\(<0\)	\(a\)	\(-b\)	\(-\frac{a}{b}\)	\(-\frac{a}{b}\)
\(<0\)	\(>0\)	\(<0\)	\(-a\)	\(b\)	\(-\frac{a}{b}\)	\(-\frac{a}{b}\)
\(<0\)	\(<0\)	\(>0\)	\(-a\)	\(-b\)	\(\frac{a}{b}\)	\(\frac{a}{b}\)

If \(a=0\), \(\frac{|a|}{|b|}=\frac{|0|}{|b|}=\frac{0}{|b|}=0\) and \(\left|\frac{a}{b}\right|=|\frac{0}{b}|=\left|0\right|=0\)

Thus \(\frac{|a|}{|b|} = \left|\frac{a}{b}\right|\)

Since we consider all the cases, from the table we can conclude that \(\frac{|a|}{|b|} = \left|\frac{a}{b}\right|\), if \(b \neq 0\) is true.

(d) \(|a + b + c| \leq |a| + |b| + |c|\)

We can use triangular inequality, then \(|a+b+c|=|(a+b)+c|\leq |a+b|+|c|\leq|a|+|b|+|c|\)

(e) \(|a - b| \leq |a| + |b|\)

Based on triangular inequality, we can let replace \(b\) with \(-b\) and get \(|a|+|-b|\geq|a+\left(-b\right)|\)

Then since \(|-b|=|b|\) (Homework 1.a), \(|a|+|b|\geq |a-b|\)

And using the definition of inequality, \(|a - b| \leq |a| + |b|\)

Let \(a, b \in \mathbb{R}\) and \(b \geq 0\), then \(|a| \geq b\) if and only if \(a \leq -b\) or \(a \geq b\).

\(\Rightarrow\)) N.T.P. \(|a|\geq b\Rightarrow a\leq -b\) or \(a\geq b\)

If \(a\geq0\), \(|a|=a\geq b\)

If \(a<0\), \(|a|=-a\geq b\Rightarrow-a\cdot\left(-1\right)\leq b\cdot\left(-1\right)\Rightarrow a\leq-b\)

Thus \(a \leq -b\) or \(a \geq b\)

\(\Leftarrow\)) N.T.P. \(a\leq-b\) or \(a\geq b\Rightarrow|a|\geq b\)

Since \(a\leq -b\), then \(-1\cdot a\geq -1\cdot(-b)\Rightarrow-a\geq b\Rightarrow b\leq -a\)

Since \(a\geq b\), then \(b\leq a\)

Thus \(b\leq |a|\)
Prove that if \(a \in A\) is a lower bound for \(A\), then \(a = \inf A\).

Since \(a\) and \(a'\) are lower bounds for \(A\)\, then \(a\leq x,\forall x\in A\) and \(a^{\prime}\leq x,\forall x\in A\)

Since \(a\in A\), then \(a'\leq a\).

Thus by definition of infimum, \(a=inf(A)\)
Show that for any bounded set \(A\), \(\inf A \leq \sup A\).

Since \(A\) is bounded, then it has upper bounds and lower bounds.

By the axiom of completeness, \(A\) has a infimum and supremum.

Let denote the infimum of \(A\) as \(m\), the supremum of \(A\) as \(M\)

Then \(\forall a\in A,\exists m\in A,m\leq a\) and \(\forall a\in A,\exists M\in A,M\geq a\)

Then \(m\leq a\leq M\), which is \(\inf A \leq \sup A\)
Compute the supremum and infimum. Justify:

(a) \(\{x \in \mathbb{Q} : x \geq 0 \text{ and } x^2 \leq 10\}\)

\(inf=0\) and \(sup=\sqrt{10}\)

(b) \(\left\{\frac{n^2 + 2}{n^2 + 1} : n \in \mathbb{N}\right\}\)

\(inf=1\) and \(sup=\frac32\)

(c) \(\left\{\frac{3n}{n + 2} : n \in \mathbb{N}\right\}\)

\(inf=1\) and \(sup=3\)