In preceding post (real number system), we accepted the axiom for completeness [41] on real number system. For discussing this issue, some definitions or propositions will be required.
A set $M$ of real numbers is bounded above if there is an upper bound $a$ such that if $m\in M$, $m\leq a$.
As real number system is an ordered field, it is clear that if $M$ is bounded above, there is a lot of upper bound $a$. That is, there is a set of upper bound numbers, or a set by which $M$ is bounded above.
Similarly, a set $N$ of real numbers is bounded below if there is a lower bound $b$ such that if $n\in N$, $n\geq b$.
Of course, there is a set of lower bound numbers.
If $\alpha $ is an upper bound of $M$, but there is not any upper bound number less than $\alpha$, then $\alpha $ is the least upper bound number of $M$ and is called the supremum of $M$, and we write
\[ \alpha = \sup M \]
If $\beta$ is the greatest lower bound number of $N$ and the infimum of $N$,
\[ \beta = \inf N\]
You may recall $\max M$ or $\min N$. However, $\max M$ must be an element of $M$ and $\min N$ must be an element of $N$, too. Let us note that $\sup M$ may be an element of $M$, or may not be, and $\inf N$ may be an element of $N$, or may not be.
2012/12/27
2012/12/20
real number system
Real numbers are a set on which the operations of addition and multiplication are defined.
[axiom for addition]
[1] If $a,b\in \mathbb{R}$, $a+b\in \mathbb{R}$
[2] $a+b=b+a$
[3] $(a+b)+c=a+(b+c)$
[4] There is a distinct real number $0$ such that $a+0=a$
[5] For each $a$, there is a real number $-a$ such that $a+(-a)=0$
[axiom for multiplication]
[11] If $a,b\in \mathbb{R}$, $ab\in \mathbb{R}$
[12] $ab=ba$
[13] $(ab)c=a(bc)$
[14] There is a distinct real number $1$ such that $a1=a$
[15] For each $a$, there is a real number $1/a$ such that $a(1/a)=a$, where $a\neq 0$
As you have ever seen, operations of [2] and [12] are called commutative laws, and [3] and [13] are called associative laws. The following operation which shall be satisfied in real numbers is called the distributive law.
[distributive law]
[21] $a(b+c)=ab+ac$
A set which is satisfied the operations of addition [1]-[5], multiplication [11]-[15], and the distributive law [21] is called a field. Therefore, real number system has field properties.
And, real numbers are ordered by the relation "<" ( less than ) between every pair of elements.
[ordered relation]
[31] For each pair of real numbers $a$ and $b$, exactly one of the following is true:
$a=b$, $a<b$, or $b<a$
[32] If $a<b$ and $b<c$, then $a<c$
[33] If $a<b$, then $a+c<b+c$
[34] If $a<b$, then $ac<bc$, whenever $0<c$
$a=b$ in [31] means that $a$ is not less than $b$, and $b$ is not less than $a$. By the property [32], the relation $<$ is called transitive. Hence, real numbers are said to be a ordered field. As rational numbers satisfy above properties, that is also a ordered field. However, the following property is not satisfied in rational numbers. It is called completeness axiom.
[axiom for completeness]
[41] If a set of real numbers is bounded above, then it has a supremum.
Real number system has above all properties. Hence, real numbers are a ordered field satisfied completeness.
[axiom for addition]
[1] If $a,b\in \mathbb{R}$, $a+b\in \mathbb{R}$
[2] $a+b=b+a$
[3] $(a+b)+c=a+(b+c)$
[4] There is a distinct real number $0$ such that $a+0=a$
[5] For each $a$, there is a real number $-a$ such that $a+(-a)=0$
[axiom for multiplication]
[11] If $a,b\in \mathbb{R}$, $ab\in \mathbb{R}$
[12] $ab=ba$
[13] $(ab)c=a(bc)$
[14] There is a distinct real number $1$ such that $a1=a$
[15] For each $a$, there is a real number $1/a$ such that $a(1/a)=a$, where $a\neq 0$
As you have ever seen, operations of [2] and [12] are called commutative laws, and [3] and [13] are called associative laws. The following operation which shall be satisfied in real numbers is called the distributive law.
[distributive law]
[21] $a(b+c)=ab+ac$
A set which is satisfied the operations of addition [1]-[5], multiplication [11]-[15], and the distributive law [21] is called a field. Therefore, real number system has field properties.
And, real numbers are ordered by the relation "<" ( less than ) between every pair of elements.
[ordered relation]
[31] For each pair of real numbers $a$ and $b$, exactly one of the following is true:
$a=b$, $a<b$, or $b<a$
[32] If $a<b$ and $b<c$, then $a<c$
[33] If $a<b$, then $a+c<b+c$
[34] If $a<b$, then $ac<bc$, whenever $0<c$
$a=b$ in [31] means that $a$ is not less than $b$, and $b$ is not less than $a$. By the property [32], the relation $<$ is called transitive. Hence, real numbers are said to be a ordered field. As rational numbers satisfy above properties, that is also a ordered field. However, the following property is not satisfied in rational numbers. It is called completeness axiom.
[axiom for completeness]
[41] If a set of real numbers is bounded above, then it has a supremum.
Real number system has above all properties. Hence, real numbers are a ordered field satisfied completeness.
2012/12/14
epsilon-delta proofs 6
In preceding post, we worked on exercises in which a numerical sequence approached a value. Next case is famous as $f_n$ approaching $\infty$, that is, diverging.
Prove that $f_n=1+\frac{1}{2}+\cdots+\frac{1}{n}\rightarrow \infty$, when $n\rightarrow \infty $.
$f_1=1$
$f_2=1.5$
$f_3=1.8333\cdots $
$f_4=2.08333\cdots $
$f_5=2.28333\cdots $
$\cdots \cdots$
(proof)
\begin{eqnarray*}
f_n &=& 1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots \\
&>& 1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\cdots \\
&=& 1+\frac{1}{2}+\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)+\cdots \\
&=& 1+\frac{m}{2}\to \infty\quad (m\to \infty)
\end{eqnarray*}
At the first glance, may you have thought that $f_n$ would converge with any value?
Prove that $f_n=1+\frac{1}{2}+\cdots+\frac{1}{n}\rightarrow \infty$, when $n\rightarrow \infty $.
$f_1=1$
$f_2=1.5$
$f_3=1.8333\cdots $
$f_4=2.08333\cdots $
$f_5=2.28333\cdots $
$\cdots \cdots$
(proof)
\begin{eqnarray*}
f_n &=& 1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots \\
&>& 1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\cdots \\
&=& 1+\frac{1}{2}+\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)+\cdots \\
&=& 1+\frac{m}{2}\to \infty\quad (m\to \infty)
\end{eqnarray*}
At the first glance, may you have thought that $f_n$ would converge with any value?
2012/12/10
epsilon-delta proofs 5
<exercises>
1. When $n\rightarrow \infty $, what does $f_n=\frac{1}{n}$ ($n\in\mathbb{N}$) approach?
For any $\epsilon$, we are always able to set $N>\frac{1}{\epsilon}$. That is, $\frac{1}{N}<\epsilon $. Therefore, if $n\geq N$,
\[ 0<\frac{1}{n}\leq\frac{1}{N}<\epsilon \]
Hence, we proved
\[ \frac{1}{n}\rightarrow 0\quad (n\rightarrow \infty ) \]
2. Find $f_n=\frac{a_1+a_2+\cdots +a_n}{n}$, when $n\rightarrow \infty $ and $a_n\rightarrow a$.
(using the result of exercise-1)
\begin{eqnarray*}
f_n &=& \frac{(a_1-a)+(a_2-a)+\cdots +(a_n-a)+na}{n}\\
&=& (a_1-a)\frac{1}{n}+(a_2-a)\frac{1}{n}+\cdots +(a_n-a)\frac{1}{n}+a\rightarrow a\quad \left(\frac{1}{n}\rightarrow 0\right)
\end{eqnarray*}
(epsilon-delta proof)
\begin{eqnarray*}
|f_n-a| &=& \left| \frac{(a_1-a)+(a_2-a)+\cdots +(a_n-a)}{n}\right| \\
&=& \left| \frac{(a_1-a)+\cdots +(a_m-a)+(a_{m+1}-a)+\cdots +(a_n-a)}{n}\right|
\end{eqnarray*}
Based on [Def-1], for a $\epsilon>0$, we are able to choose a $N$ such that if $n\geq N$, $|a_n-a|<\frac{\epsilon}{2}$. Similarly, we can also choose $N_1>N$ such that if $n\geq N_1$,
\[ \left| \frac{\sum_{i=1}^m (a_i-a)}{n} \right|<\frac{\epsilon}{2} \]
Of course, for this $n$, $|a_n-a|<\frac{\epsilon}{2}$ is clear. Then
\[ \left| \frac{\sum_{i=m+1}^n (a_i-a)}{n} \right|<\frac{(n-m)}{n}\frac{\epsilon}{2} < \frac{\epsilon}{2} \]
Hence
\begin{eqnarray*}
|f_n-a| &<&\frac{\epsilon}{2}+ \frac{\epsilon}{2}=\epsilon \\
\end{eqnarray*}
A proof is completed.
1. When $n\rightarrow \infty $, what does $f_n=\frac{1}{n}$ ($n\in\mathbb{N}$) approach?
For any $\epsilon$, we are always able to set $N>\frac{1}{\epsilon}$. That is, $\frac{1}{N}<\epsilon $. Therefore, if $n\geq N$,
\[ 0<\frac{1}{n}\leq\frac{1}{N}<\epsilon \]
Hence, we proved
\[ \frac{1}{n}\rightarrow 0\quad (n\rightarrow \infty ) \]
2. Find $f_n=\frac{a_1+a_2+\cdots +a_n}{n}$, when $n\rightarrow \infty $ and $a_n\rightarrow a$.
(using the result of exercise-1)
\begin{eqnarray*}
f_n &=& \frac{(a_1-a)+(a_2-a)+\cdots +(a_n-a)+na}{n}\\
&=& (a_1-a)\frac{1}{n}+(a_2-a)\frac{1}{n}+\cdots +(a_n-a)\frac{1}{n}+a\rightarrow a\quad \left(\frac{1}{n}\rightarrow 0\right)
\end{eqnarray*}
(epsilon-delta proof)
\begin{eqnarray*}
|f_n-a| &=& \left| \frac{(a_1-a)+(a_2-a)+\cdots +(a_n-a)}{n}\right| \\
&=& \left| \frac{(a_1-a)+\cdots +(a_m-a)+(a_{m+1}-a)+\cdots +(a_n-a)}{n}\right|
\end{eqnarray*}
Based on [Def-1], for a $\epsilon>0$, we are able to choose a $N$ such that if $n\geq N$, $|a_n-a|<\frac{\epsilon}{2}$. Similarly, we can also choose $N_1>N$ such that if $n\geq N_1$,
\[ \left| \frac{\sum_{i=1}^m (a_i-a)}{n} \right|<\frac{\epsilon}{2} \]
Of course, for this $n$, $|a_n-a|<\frac{\epsilon}{2}$ is clear. Then
\[ \left| \frac{\sum_{i=m+1}^n (a_i-a)}{n} \right|<\frac{(n-m)}{n}\frac{\epsilon}{2} < \frac{\epsilon}{2} \]
Hence
\begin{eqnarray*}
|f_n-a| &<&\frac{\epsilon}{2}+ \frac{\epsilon}{2}=\epsilon \\
\end{eqnarray*}
A proof is completed.
2012/12/01
irrational numbers 2
<exercises>
1. $\sqrt{n^2+1}$ is an irrational number, in which $n\in N$.
[proof] If we put $\sqrt{n^2+1}=p/q$, $n^2=p^2/q^2-1$. We also know $n<\sqrt{n^2+1}<n+1$. Although $n^2$ is an integer number, obviously, $p^2/q^2-1$ is not an integer. These are a contradiction.
2. $\log_{10}2$ is an irrational number.
[proof] we set $\log_{10}2=p/q$, where $p,q\in \mathbb{Z}, q\ne 0$. Therefore $2^q=10^p$. However, $10^p=(2^p)(5^p)$. Two equations can never exist at the same time, because uniqueness of the factorization. As these show that there cannot be the initial definition of $\log_{10}2=p/q$, $\log_{10}2$ is an irrational number.
3. $e$ is an irrational number.
[proof] First of all, we accept the fact that
\[ e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{n!}+\cdots \]
Next, as always, let $e$ be $p/q>0$. If we make $n$ in above equation $q$,
the equation is decomposed to
\[ e=\left( \frac{1}{0!}+\cdots +\frac{1}{q!}\right) +\left( \frac{1}{(q+1)!}+ \frac{1}{(q+2)!}+\cdots \right) \]
Therefore we can get the equation
\[ q!\left( \frac{1}{(q+1)!}+\frac{1}{(q+2)!}+\cdots \right) =q!\left( \frac{p}{q}-\left( \frac{1}{0!}+\cdots +\frac{1}{q!}\right) \right) \]
Obviously the right side of this equation is an integer number.
The other hand, the left side is, as we are able to assume $q>1$,
\begin{eqnarray*}
\mbox{left side } &=& \frac{1}{(q+1)}+\frac{1}{(q+1)(q+2)}+\cdots \\
&<& \frac{1}{2}+\frac{1}{2^2}+\cdots =1
\end{eqnarray*}
Since the left side of the equation is not an integer number, a contradiction occurred.
1. $\sqrt{n^2+1}$ is an irrational number, in which $n\in N$.
[proof] If we put $\sqrt{n^2+1}=p/q$, $n^2=p^2/q^2-1$. We also know $n<\sqrt{n^2+1}<n+1$. Although $n^2$ is an integer number, obviously, $p^2/q^2-1$ is not an integer. These are a contradiction.
2. $\log_{10}2$ is an irrational number.
[proof] we set $\log_{10}2=p/q$, where $p,q\in \mathbb{Z}, q\ne 0$. Therefore $2^q=10^p$. However, $10^p=(2^p)(5^p)$. Two equations can never exist at the same time, because uniqueness of the factorization. As these show that there cannot be the initial definition of $\log_{10}2=p/q$, $\log_{10}2$ is an irrational number.
3. $e$ is an irrational number.
[proof] First of all, we accept the fact that
\[ e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{n!}+\cdots \]
Next, as always, let $e$ be $p/q>0$. If we make $n$ in above equation $q$,
the equation is decomposed to
\[ e=\left( \frac{1}{0!}+\cdots +\frac{1}{q!}\right) +\left( \frac{1}{(q+1)!}+ \frac{1}{(q+2)!}+\cdots \right) \]
Therefore we can get the equation
\[ q!\left( \frac{1}{(q+1)!}+\frac{1}{(q+2)!}+\cdots \right) =q!\left( \frac{p}{q}-\left( \frac{1}{0!}+\cdots +\frac{1}{q!}\right) \right) \]
Obviously the right side of this equation is an integer number.
The other hand, the left side is, as we are able to assume $q>1$,
\begin{eqnarray*}
\mbox{left side } &=& \frac{1}{(q+1)}+\frac{1}{(q+1)(q+2)}+\cdots \\
&<& \frac{1}{2}+\frac{1}{2^2}+\cdots =1
\end{eqnarray*}
Since the left side of the equation is not an integer number, a contradiction occurred.