10.17 Axioms of real numbers

Using the axiom of order \((\leq)\), prove the following properties for \(<\):

The main idea: We have Trichotomy and Compatibility.

The only difference is that there is no equal, thus we can consider two possibility and use contradiction!

(a) If \(a < b\) and \(b < c\), then \(a < c\).

By transitivity: if \(a \leq b\) and \(b \leq c\), then \(a \leq c\), which means if \(a <b~or~a=b\) and \(b \leq c~or~b=c\), then \(a < c~or~a=c\).

We know \(a=c\) if and only if \(a=b\) and \(b=c\)

But \(a<b\) and \(b<c\), then \(a\neq c\)

Thus \(a<c\)

(b) If \(a < b\), then \(a + c < b + c\).

Since \(a\leq b\) (means \(a=b\) or \(a<b\)), then \(a+c\leq b+c\)

Now \(a<b\), We know \(a+c=b+c\) if and only if \(a=b\)

We need to prove \(a+c\neq b+c\)

Suppose \(a+c=b+c\), then \(a=b\) which is contradiction to \(a<b\)

(c) If \(a < b\) and \(c > 0\), then \(ac < bc\).

Since \(a < b~or~a=b\) and \(c > 0~or~c=0\), then \(ac < bc ~or~ac=bc\)

We know \(ac=bc\) if and only if \(a=b\)

Thus we need to prove \(ac\neq bc\), since \(a\neq b\), then \(ac\neq bc\)
Show that \(c > 0\) if and only if \(-c < 0\) for all \(c \in \mathbb{R}\).

\(\Rightarrow\)) Since \(c>0\), then \(0<c\Rightarrow0+\left(-c\right)<c+\left(-c\right)\Rightarrow-c<0\)

\(\Leftarrow\)) Since \(-c<0\), then \(0>-c\Rightarrow0+c>-c+c\Rightarrow c>0\)
Sign rules: Show the following sentences. (a) If \(a > 0\) and \(b > 0\), then \(ab > 0\).

\(a>0 \Leftrightarrow 0<a\)

\(b>0\Leftrightarrow0<b\)

By axioms of order, \(0<ab\Leftrightarrow ab>0\)

(b) If \(a > 0\) and \(b < 0\), then \(ab < 0\).

\(a>0\)

\(b<0\)

By axioms of order, \(ab<a\cdot0\Leftrightarrow ab<0\) (c) If \(a < 0\) and \(b > 0\), then \(ab < 0\).

\(a<0\)

\(b>0\)

By axioms of order, \(ab<b\cdot0\Leftrightarrow ab<0\) (d) If \(a < 0\) and \(b < 0\), then \(ab > 0\)

\(a<0\Rightarrow -a>0\)

\(b<0\)

\((-a)b<0\cdot (-a)\)

\(-ab<0\Rightarrow -(-ab)>0\)

\(ab>0\)
Prove that if \(a < b\) and \(c < 0\), then \(ac > bc\)

\(c<0\Rightarrow -c>0\)

\(a<b\)

\(a\cdot (-c)<b\cdot (-c)\)

\(-ac<-bc\)

\(-ac+ac<-bc+ac\)

\(0<ac-bc\)

\(0+bc<ac-bc+bc\)

\(bc<ac\)

\(ac>bc\)
Show that: (a) \(a > 0\) if and only if \(a^{-1} > 0\).

\(\Rightarrow\)) Since \(a\cdot a^{-1}=1>0\) and \(a>0\), then we consider 3 cases.

If \(a^{-1}>0\), then \(a\cdot a^{-1}=1>0\) ok

If \(a^{-1}=0\), then \(a\cdot a^{-1}=0\neq1\) not

If \(a^{-1}<0\), then \(a\cdot a^{-1}=1<0\) not

Thus \(a^{-1}>0\)

\(\Leftarrow\)) Since \(a\cdot a^{-1}=1>0\) and \(a^{-1}>0\), then we consider 3 cases.

If \(a>0\), then \(a\cdot a^{-1}=1>0\) ok

If \(a=0\), then \(a\cdot a^{-1}=0\neq1\) not

If \(a<0\), then \(a\cdot a^{-1}=1<0\) not

Thus \(a>0\)

(b) \(a < 0\) if and only if \(a^{-1} < 0\).

\(\Rightarrow\)) Since \(a\cdot a^{-1}=1>0\) and \(a<0\), then we consider 3 cases.

If \(a^{-1}>0\), then \(a\cdot a^{-1}=1<0\) not

If \(a^{-1}=0\), then \(a\cdot a^{-1}=0\neq1\) not

If \(a^{-1}<0\), then \(a\cdot a^{-1}=1>0\) ok

Thus \(a^{-1}<0\)

\(\Leftarrow\)) Since \(a\cdot a^{-1}=1>0\) and \(a^{-1}<0\), then we consider 3 cases.

If \(a>0\), then \(a\cdot a^{-1}=1<0\) not

If \(a=0\), then \(a\cdot a^{-1}=0\neq1\) not

If \(a<0\), then \(a\cdot a^{-1}=1>0\) ok

Thus \(a<0\) (c) If \(0 < a < b\), then \(a^{-1} > b^{-1}\).

Since \(0<a<b\) and \(a^{-1}>0,b^{-1}>0\), then \(a^{-1}\cdot b^{-1}>0\)

Then \(0\cdot a^{-1}\cdot b^{-1}<a\cdot a^{-1}\cdot b^{-1}<b\cdot a^{-1}\cdot b^{-1}\Rightarrow0<1\cdot b^{-1}<1\cdot a^{-1}\Rightarrow0<b^{-1}<a^{-1}\)

Which means \(a^{-1}>b^{-1}\)
Prove that: (a) \(ab > 0\) if and only if \(a > 0\) and \(b > 0\) or \(a < 0\) and \(b < 0\).

\(\Rightarrow\)) If \(a>0\), \(a^{-1}>0\)

\(0<ab\Rightarrow 0\cdot a^{-1}<aba^{-1}\Rightarrow 0<b\)

If \(a<0\Rightarrow a^{-1}<0\), then \(0<ab\Rightarrow 0\cdot a^{-1}>aba^{-1}\Rightarrow 0>b\)

\(\Leftarrow\)) \(a>0\) and \(b>0\), by Sign rules, \(ab>0\)

\(a<0\) and \(b<0\), by Sign rules, \(ab>0\)

(b) \(ab < 0\) if and only if \(a > 0\) and \(b < 0\) or \(a < 0\) and \(b > 0\).
Find the solution set of the following inequalities: (a) \(4 - x < 3 - 3x\) (b) \(x^2 > 9\) (c) \(\frac{x - 1}{x + 1} > 0\) (d) \(\frac{1}{x} + \frac{1}{1 - x} > 0\)

Show that: (a) \(|a \cdot b| = |a| \cdot |b|\)

\(a\)	\(b\)	\(ab\)	\(\\|ab\\|\)	\(\\|a\\|\)	\(\\|b\\|\)	\(\\|a\\|\\|b\\|\)
\(\geq 0\)	\(\geq 0\)	\(\geq 0\)	\(ab\)	\(a\)	\(b\)	\(ab\)
\(\geq 0\)	\(<0\)	\(\leq 0\)	\(-ab\)	\(a\)	\(-b\)	\(-ab\)
\(<0\)	\(\geq 0\)	\(\leq 0\)	\(-ab\)	\(-a\)	\(b\)	\(-ab\)
\(<0\)	\(<0\)	\(>0\)	\(ab\)	\(-a\)	\(-b\)	\(ab\)

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

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

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

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

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

(c) \(||a| - |b|| \leq |a - b|\)

If \(|a|\leq |b|\), \(||a|-|b||=-|a|+|b|=|b|-|a|\leq|b-a|=|(-1)(a-b)=|-1||a-b|=|a-b|\)(using 8.b)

If \(|a|>|b|\), \(||a|-|b||=|a|-|b|\leq |a-b|\)

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

\(\Rightarrow\)) We prove \(-b\leq a\leq b\Rightarrow |a|\leq b\).

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

Then \(b\geq -a\Leftrightarrow -a\leq b\)

Thus \(a\leq b\) and \(-a\leq b\), we have \(|a|\leq b\)

\(\Leftarrow\)) We prove \(|a|\leq b\Rightarrow-b\leq a\leq b\).

If \(a\geq0\Rightarrow|a|=a\leq b\Rightarrow a\leq b\)

Since \(b\geq 0\Rightarrow -b\leq 0\Rightarrow-b\leq a\)

Then \(-b\leq a\leq b\)

If \(a<0\Rightarrow|a|=-a\leq b\Rightarrow-a\leq b\Rightarrow -b\leq a\)

Since \(a<0\) and \(b\geq 0\), then \(a\leq b\)

Therefore we prove it.
Show that the supremum and infimum of a set are unique.

Let \(s_1\) and \(s_2\) be two different supremum of \(A\)

Since \(s_1\) is the supremum of \(A\) and \(s_2\) is an upper bound, then \(s_1\leq s_2\)

Moreover, \(s_2\) is also the supremum of \(A\) and \(s_1\) is an upper bound of \(A\), then \(s_2\leq s_1\)

Then by Axiom of order, we have that \(s_1=s_2\)
(a) Prove that if \(M \in A\) is an upper bound for \(A\), then \(M = \sup A\).

Since \(M\) is an upper bound, \(M\) satisfies the first condition of supremum

Suppose \(b\) is another upper bound for \(A\), then \(\forall a\in A, a\leq b\)

In particular, \(M\in A\), then \(M\leq b\)

Therefore \(M=sup(A)\)

(b) (Workshop) Prove that if \(m \in A\) is a lower bound for \(A\), then \(m = \inf A\).

Since \(m\) is an lower bound, \(m\) satisfies the first condition of infimum

Suppose \(b\) is another lower bound for \(A\), then \(\forall a\in A, a\geq b\)

In particular, \(m\in A\), then \(m\geq b\)

Therefore \(m=inf(A)\)
Assume \(A \subseteq \mathbb{R}\) and \(A \neq \emptyset\). We denote \(-A := \{-a : a \in A\}\). (a) If \(A\) has a lower bound, then \(-A\) has an upper bound and \(\inf(A) = -\sup(-A)\).

Suppose \(d\) is a lower bound for \(A\), that means \(\forall a\in A,d\leq a\Rightarrow -a\leq -d,\forall -a\in -A\Rightarrow -d\) is an upper bound for \(-A\)

Since \(A\neq \emptyset\) and has a lower bound, by Axiom of completeness, then \(\exists inf\left(A\right)\)

We know that \(A\neq\emptyset\Rightarrow-A\neq\emptyset\) and \(-A\) has an upper bound, then \(\exists sup(-A)\)

Then we need to prove they are equal

Let's denote \(u=sup(-A)\), N.T.P \(inf(A)=-u\)

Since \(u\geq -a,\forall -a\in -A\) and \(\exists-a,u-\varepsilon<-a<u\)

Then \(-u\leq a,\forall a\in A\) and \(\exists a,\varepsilon-u<a<-u\) which is the definition of inf

(b) (Workshop) If \(A\) has an upper bound, then \(-A\) has a lower bound and \(\inf(-A) = -\sup(A)\).

Totally same...
Show that if \(\sup A < \sup B\), then there exists \(b \in B\) that is an upper bound of \(A\).

Let \(x\) be the \(sup(A)\) and \(y\) be the \(sup(B)\), \(\forall a\in A,\exists x\in A,x\geq a\) and \(\forall b\in B,\exists y\in B,y\geq b\)

Since \(sup(A)<sup(B)\), then \(x<y\)

Then \(\exists b \in B,s.t.\) \(b\geq a,\forall a\in A\)

Which means there exists \(b \in B\) that is an upper bound of \(A\).

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