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.
2012/11/26
epsilon-delta proofs 4
In [Def-1] we defined a real number sequence $a_n$ approaching $a$. Although [Def-1] is much complicated, the reason why we should use it is that we were able to discuss precisely.
Using mathematical symbols, [Def-1] is also expressed as follows.
$a_n$ approaches $a$, when
\[ (\forall \epsilon>0)(\exists N\in \mathbb{N})(\forall n\in \mathbb{N})(n\geq N\rightarrow |a_n-a|< \epsilon) \]
where $\forall $ means 'all ' or 'any', $\exists $ means 'exist ' or 'some', and $\in $ means 'in elements of '.
Make a contrapositive statement of [Def-1] forcibly. Next sentence is the answer. If
\[ (\exists \epsilon>0)(\forall N\in \mathbb{N})(\exists n\in \mathbb{N}) (n\geq N \land |a_n-a|\geq \epsilon), \]
a real number sequence $a_n$ never approaches $a$, where $\land$ means 'and '.
That is, $a_n$ does not approach $a$ when for some $\epsilon>0 $, any $N>0$, there is $n>0$ such that $n\geq N$ and $|a_n-a|\geq \epsilon $.
The sequence $2.3888\cdots $ does not approach $2.4$. But the sequence $2.3888\cdots $ approaches $43/18$. One of the sequence which approaches $2.4$ is $2.3999\cdots $. These shows that the converging depends on the relation between a sequence $a_n$ and a value $a$.
We are interested in the case in which $a_n$ does not converge with any values. It is called diverging. In particular, if the greater $n$ becomes, the greater $a_n$ becomes, we say that $a_n$ approaches infinity, or the limit of $a_n$ is $\infty$ and expresses that
\[ \lim_{n\rightarrow \infty} a_n =\infty, \quad or \quad a_n\to\infty \quad (n\to \infty )\]
Next definition is equivalent to above.
[Def-3] $a_n$ approaches infinity, when for any $\epsilon >0$, there is $N>0$ such that if $n\geq N$, $|a_n-a|\geq \epsilon$.
Using mathematical symbols, [Def-1] is also expressed as follows.
$a_n$ approaches $a$, when
\[ (\forall \epsilon>0)(\exists N\in \mathbb{N})(\forall n\in \mathbb{N})(n\geq N\rightarrow |a_n-a|< \epsilon) \]
where $\forall $ means 'all ' or 'any', $\exists $ means 'exist ' or 'some', and $\in $ means 'in elements of '.
Make a contrapositive statement of [Def-1] forcibly. Next sentence is the answer. If
\[ (\exists \epsilon>0)(\forall N\in \mathbb{N})(\exists n\in \mathbb{N}) (n\geq N \land |a_n-a|\geq \epsilon), \]
a real number sequence $a_n$ never approaches $a$, where $\land$ means 'and '.
That is, $a_n$ does not approach $a$ when for some $\epsilon>0 $, any $N>0$, there is $n>0$ such that $n\geq N$ and $|a_n-a|\geq \epsilon $.
The sequence $2.3888\cdots $ does not approach $2.4$. But the sequence $2.3888\cdots $ approaches $43/18$. One of the sequence which approaches $2.4$ is $2.3999\cdots $. These shows that the converging depends on the relation between a sequence $a_n$ and a value $a$.
We are interested in the case in which $a_n$ does not converge with any values. It is called diverging. In particular, if the greater $n$ becomes, the greater $a_n$ becomes, we say that $a_n$ approaches infinity, or the limit of $a_n$ is $\infty$ and expresses that
\[ \lim_{n\rightarrow \infty} a_n =\infty, \quad or \quad a_n\to\infty \quad (n\to \infty )\]
Next definition is equivalent to above.
[Def-3] $a_n$ approaches infinity, when for any $\epsilon >0$, there is $N>0$ such that if $n\geq N$, $|a_n-a|\geq \epsilon$.
2012/11/17
epsilon-delta proofs 3
In [Def-1], we defined that a sequence of real numbers $a_n$ approached $a$. The definition that will bring mathematical complications to us has the advantages which make us possible to discuss precisely. Let a sequence $b_n$ be as follows.
$b_0=2.3$
$b_1=2.38$
$b_2=2.388$
$b_3=2.3888$
・・・・・・
Obviously in this sequence, as $b_{n+1}$ is always greater than $b_n$, the greater $n$ becomes, the greater $b_n$ becomes. Hence, it is not wrong that $b_n$ comes closer to $a=2.4$. But no one will be convinced that $b_n$ approaches $a=2.4$. The differences of $|b_n-a|$ are
$|2.3-2.4|=0.1$
$|2.38-2.4|=0.02$
$|2.388-2.4|=0.012$
$|2.3888-2.4|=0.0112$
・・・・・・
Look at above and guess more. No matter how small the differences are, it is definitely no doubt for us to understand that those do not become less than $0.01$, and $b_n$ does not come closer to $2.4$ over $2.39$. We never want to include a kind of sequences $b_n$ in sequences approaching $a=2.4$.
On the preceding calculation, we put $b=2.3888\cdots $, and multiply by $10$. Then, as $10b=23.888\cdots $, after subtracting both sides of first equation from this, we get a rational number $b=43/18$. It means that if $n\rightarrow \infty $, $b_n$ approaches $b=43/18$. $b$ is not $432/180=2.4$.
If we accept a definition of converging sequences by saying more and more or lower and lower, we also cannot eliminate from sequences approaching $a=2.4$ as follows.
$c_0=0.3$
$c_1=0.33$
$c_2=0.333$
$c_3=0.3333$
・・・・・・
We enough know that $c_n$ approaches $c=1/3$. Even these examples, it is clear that we have to eliminate such a kind of sequences by adopting a precise definition of converging sequences. This is [Def-1].
If we conform to [Def-1], for any $\epsilon >0$ less than $0.009$, as we are not able to find out a $N>0$ such that if $n\geq N$, $|a_n-a|< \epsilon$. As it goes against [Def-1], we can eliminate $b_n$ from the class of $a_n$. We can do $c_n$, too.
This is a reason why epsilon-delta definitions are nice, and we have to adopt [Def-1], although it's much complicated.
$b_0=2.3$
$b_1=2.38$
$b_2=2.388$
$b_3=2.3888$
・・・・・・
Obviously in this sequence, as $b_{n+1}$ is always greater than $b_n$, the greater $n$ becomes, the greater $b_n$ becomes. Hence, it is not wrong that $b_n$ comes closer to $a=2.4$. But no one will be convinced that $b_n$ approaches $a=2.4$. The differences of $|b_n-a|$ are
$|2.3-2.4|=0.1$
$|2.38-2.4|=0.02$
$|2.388-2.4|=0.012$
$|2.3888-2.4|=0.0112$
・・・・・・
Look at above and guess more. No matter how small the differences are, it is definitely no doubt for us to understand that those do not become less than $0.01$, and $b_n$ does not come closer to $2.4$ over $2.39$. We never want to include a kind of sequences $b_n$ in sequences approaching $a=2.4$.
On the preceding calculation, we put $b=2.3888\cdots $, and multiply by $10$. Then, as $10b=23.888\cdots $, after subtracting both sides of first equation from this, we get a rational number $b=43/18$. It means that if $n\rightarrow \infty $, $b_n$ approaches $b=43/18$. $b$ is not $432/180=2.4$.
If we accept a definition of converging sequences by saying more and more or lower and lower, we also cannot eliminate from sequences approaching $a=2.4$ as follows.
$c_0=0.3$
$c_1=0.33$
$c_2=0.333$
$c_3=0.3333$
・・・・・・
We enough know that $c_n$ approaches $c=1/3$. Even these examples, it is clear that we have to eliminate such a kind of sequences by adopting a precise definition of converging sequences. This is [Def-1].
If we conform to [Def-1], for any $\epsilon >0$ less than $0.009$, as we are not able to find out a $N>0$ such that if $n\geq N$, $|a_n-a|< \epsilon$. As it goes against [Def-1], we can eliminate $b_n$ from the class of $a_n$. We can do $c_n$, too.
This is a reason why epsilon-delta definitions are nice, and we have to adopt [Def-1], although it's much complicated.
2012/11/10
irrational numbers
I will give some examples about irrational numbers. Irrational numbers are numbers that cannot be expressed by dividing any integer $p$ by any integer $q$. According to this definition [Def-2], $p$ and $q$ are relative prime (integer) numbers.
In order to prove that a target number is irrational, we use a proof by contradiction. Let the target number be expressed by $p/q$. Then, if any contradiction will occur, it follows that the number is not a rational number. Here are some examples.
If $\sqrt{2}$ is rational, we are able to assume $\sqrt{2}=p/q$. Since $p^2=2q^2$, $p^2$ is an even number. In this case, $p$ must be an even number, too. If $p$ is an odd number, $p^2=(2m+1)^2=2(2m^2+2m)+1$, $p^2$ is also an odd number. Therefore, $p$ can not be an odd number.
Thus, as we are able to put $p=2m$, we get $2q^2=4m^2$. Both sides of this equation can be divided by 2. However, if $p$ and $q$ are relative prime numbers, $p^2$ and $q^2$ must be relative prime numbers. it is inconsistent with [Def-2] in which $p$ and $q$ are relative prime numbers. Therefore, $\sqrt{2}$ is an irrational number.
According to this method, we can prove that $\sqrt{3}$ is irrational. It is important that when $p^2=3q^2$, $p$ must be a multiple of 3. Suppose $p=3m+1$. Then $p^2=(3m+1)^2=3(3m^2+2m)+1$. Therefore $p^2$ is not a multiple of 3. If $p=3m+2$, also then $p^2=(3m+2)^2=3(3m^2+4m+1)+1$. $p^2$ is not a multiple of 3, either. Hence, we get $p=3m$. Similarly, since it is a contradiction of [Def-2], Hence $\sqrt{3}$ is an irrational number.
It is even more difficult to show that $\pi$ is an irrational number. However, it has already been successfully done by contradiction. In mathematics, since rational numbers are at most countable and irrational numbers are uncountable, irrational numbers take a greater part in real numbers. We are not able to figure out irrational numbers. For example, we do not know whether $\pi+e$ is an irrational number, or not.
In order to prove that a target number is irrational, we use a proof by contradiction. Let the target number be expressed by $p/q$. Then, if any contradiction will occur, it follows that the number is not a rational number. Here are some examples.
If $\sqrt{2}$ is rational, we are able to assume $\sqrt{2}=p/q$. Since $p^2=2q^2$, $p^2$ is an even number. In this case, $p$ must be an even number, too. If $p$ is an odd number, $p^2=(2m+1)^2=2(2m^2+2m)+1$, $p^2$ is also an odd number. Therefore, $p$ can not be an odd number.
Thus, as we are able to put $p=2m$, we get $2q^2=4m^2$. Both sides of this equation can be divided by 2. However, if $p$ and $q$ are relative prime numbers, $p^2$ and $q^2$ must be relative prime numbers. it is inconsistent with [Def-2] in which $p$ and $q$ are relative prime numbers. Therefore, $\sqrt{2}$ is an irrational number.
According to this method, we can prove that $\sqrt{3}$ is irrational. It is important that when $p^2=3q^2$, $p$ must be a multiple of 3. Suppose $p=3m+1$. Then $p^2=(3m+1)^2=3(3m^2+2m)+1$. Therefore $p^2$ is not a multiple of 3. If $p=3m+2$, also then $p^2=(3m+2)^2=3(3m^2+4m+1)+1$. $p^2$ is not a multiple of 3, either. Hence, we get $p=3m$. Similarly, since it is a contradiction of [Def-2], Hence $\sqrt{3}$ is an irrational number.
It is even more difficult to show that $\pi$ is an irrational number. However, it has already been successfully done by contradiction. In mathematics, since rational numbers are at most countable and irrational numbers are uncountable, irrational numbers take a greater part in real numbers. We are not able to figure out irrational numbers. For example, we do not know whether $\pi+e$ is an irrational number, or not.
2012/11/02
class letters of numbers
It is useful to define the following elements using symbols listed below.
[Def-2]
$\mathbb{N}$ : $1,2,3,\cdots $ are called natural numbers.
$\mathbb{Z}$ : Integers consist of natural numbers and negative natural numbers, including 0 and $\infty $ (i.e. $0, \pm 1, \pm 2,\cdots \pm \infty $).
$\mathbb{Q}$ : Rational numbers are numbers expressed by the fraction $p/q$ in which $p$ and $q$ are relative prime numbers and integers ($p,q\neq\pm\infty , q\neq 0$).
$\mathbb{R}$ : Real numbers consist of rational and irrational numbers.
Although '0' does not fall in any class of numbers as it means empty, we decide to include 0 among integers, for convenience. $\pm \infty $ is not a number, too. However, I am going to use $n\rightarrow \infty $ in the explanation of epsilon-delta techniques. Therefore, we also include $\pm \infty $ among integers at this time. (Later, we will extend real number system including $\pm \infty $. )
Now, we need to make rules for calculations in which $p$ and $q$ are $\pm \infty $. I think you already know that. If necessary, then I will provide additional comments on that issue. The largest class is real numbers. The relation of inclusion is as follows:
\[ \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \]
Perhaps you think the definition of irrational numbers is unclear. But please understand in this step that irrational numbers are not rational numbers, or in other words, they are numbers that cannot be expressed by dividing any $p$ by any $q$. In this case, $p$ and $q$ should be integers. You are familiar with $\sqrt{2}$, $\pi$, $e$, and so on. Nobody denies the existence of such irrational numbers.
[Def-2]
$\mathbb{N}$ : $1,2,3,\cdots $ are called natural numbers.
$\mathbb{Z}$ : Integers consist of natural numbers and negative natural numbers, including 0 and $\infty $ (i.e. $0, \pm 1, \pm 2,\cdots \pm \infty $).
$\mathbb{Q}$ : Rational numbers are numbers expressed by the fraction $p/q$ in which $p$ and $q$ are relative prime numbers and integers ($p,q\neq\pm\infty , q\neq 0$).
$\mathbb{R}$ : Real numbers consist of rational and irrational numbers.
Although '0' does not fall in any class of numbers as it means empty, we decide to include 0 among integers, for convenience. $\pm \infty $ is not a number, too. However, I am going to use $n\rightarrow \infty $ in the explanation of epsilon-delta techniques. Therefore, we also include $\pm \infty $ among integers at this time. (Later, we will extend real number system including $\pm \infty $. )
Now, we need to make rules for calculations in which $p$ and $q$ are $\pm \infty $. I think you already know that. If necessary, then I will provide additional comments on that issue. The largest class is real numbers. The relation of inclusion is as follows:
\[ \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \]
Perhaps you think the definition of irrational numbers is unclear. But please understand in this step that irrational numbers are not rational numbers, or in other words, they are numbers that cannot be expressed by dividing any $p$ by any $q$. In this case, $p$ and $q$ should be integers. You are familiar with $\sqrt{2}$, $\pi$, $e$, and so on. Nobody denies the existence of such irrational numbers.
2012/10/27
epsilon-delta proofs 2
Do you remember that real numbers consist of rational numbers, irrational numbers, and 0? And all rational numbers have fractional expressions $p/q$, in which $p$ is an integer, and $q$ is also an integer not zero. In calculations using fractions, the value of $p/q$ is a positive or negative integer, 0, or a terminating or recurring decimal.
As explained in the previous post, integers or terminating decimals are equal to a recurring decimal. As irrational numbers originally are numbers recurring infinitely, it follows that all real numbers except 0 also recurre infinitely.
Since it is possible to construct a real number sequence by analyzing the number having infinite digits, all real numbers can be expressed a real number sequence. For instance, when $a=2.4$, its sequence $a_n$ is as follows:
$a_0=2.3$
$a_1=2.39$
$a_2=2.399$
$a_3=2.3999$
$\cdots \cdots$
If we make $n$ infinite, then $a_{\infty }=a=2.4$. But if $n$ is finite, the more you increase $n$, the smaller the difference $|a_n-a|$ becomes. I will give the definition of $a_n$ approches $a$.
[Def-1] $a_n$ approches $a$, when for any $\epsilon >0$, there is a $N>0$ such that if $n\geq N$,
\[ |a_n-a|< \epsilon \]
Then we can write that:
\[ \lim_{n\rightarrow \infty }a_n=a \]
Or:
\[ a_n\rightarrow a\quad (n\rightarrow \infty ) \]
This is the most elementary expression of epsilon-delta techniques. The above example of $\lim_{n\rightarrow \infty }a_n=a$ is $2.3999\cdots =2.4$
As explained in the previous post, integers or terminating decimals are equal to a recurring decimal. As irrational numbers originally are numbers recurring infinitely, it follows that all real numbers except 0 also recurre infinitely.
Since it is possible to construct a real number sequence by analyzing the number having infinite digits, all real numbers can be expressed a real number sequence. For instance, when $a=2.4$, its sequence $a_n$ is as follows:
$a_0=2.3$
$a_1=2.39$
$a_2=2.399$
$a_3=2.3999$
$\cdots \cdots$
If we make $n$ infinite, then $a_{\infty }=a=2.4$. But if $n$ is finite, the more you increase $n$, the smaller the difference $|a_n-a|$ becomes. I will give the definition of $a_n$ approches $a$.
[Def-1] $a_n$ approches $a$, when for any $\epsilon >0$, there is a $N>0$ such that if $n\geq N$,
\[ |a_n-a|< \epsilon \]
Then we can write that:
\[ \lim_{n\rightarrow \infty }a_n=a \]
Or:
\[ a_n\rightarrow a\quad (n\rightarrow \infty ) \]
This is the most elementary expression of epsilon-delta techniques. The above example of $\lim_{n\rightarrow \infty }a_n=a$ is $2.3999\cdots =2.4$
2012/10/20
epsilon-delta proofs
In the first post, I want to pick up epsilon-delta techniques.
In high school, it has been well said that epsilon-delta techniques have been hardly understood since students are not used to the mathematical expression. Whereas in university, as the techniques suddenly appear in limit proofs, many students are very much confused. I think that the difficulty of epsilon-delta techniques basically owes to the continuity of real number. I am going to give some hints to help you understand them.
Well, a real numerical sequence $a_n$ comes close to a value without any end. For example,
$a_0=0$
$a_1=0.9$
$a_2=0.99$
$a_3=0.999$
・・・・・
$n$ of $a_n$ means the number of 9 after the decimal point. As n increases infinitely, we can prove that this $a_n$ becomes 1. The $a_n$ when $n\rightarrow \infty $ is provided below.
\[(1) \qquad a_{\infty }=0.999\cdots \]
We multiply both sides by 10. Then,
\[(2) \qquad 10a_{\infty }=9.999\cdots \]
subtract both sides of (1) from (2), and divide by 9,
\[ a_{\infty }=1 \]
Therefore,
\[ 1=0.999\cdots \]
If it is possible to multiply and subtract a real number $a_{\infty }$ which has no end, we have to accept this result.
Extending this result, we come to the fact that a number is equal to the number obtained by subtracting 1 from the last digit of the finite real number and by appending $999\cdots $. For example, if $a=2.4$, it is same as $a=2.3999\cdots $.
In conclusion, any finite real number has infinite real numerical sequences.
In high school, it has been well said that epsilon-delta techniques have been hardly understood since students are not used to the mathematical expression. Whereas in university, as the techniques suddenly appear in limit proofs, many students are very much confused. I think that the difficulty of epsilon-delta techniques basically owes to the continuity of real number. I am going to give some hints to help you understand them.
Well, a real numerical sequence $a_n$ comes close to a value without any end. For example,
$a_0=0$
$a_1=0.9$
$a_2=0.99$
$a_3=0.999$
・・・・・
$n$ of $a_n$ means the number of 9 after the decimal point. As n increases infinitely, we can prove that this $a_n$ becomes 1. The $a_n$ when $n\rightarrow \infty $ is provided below.
\[(1) \qquad a_{\infty }=0.999\cdots \]
We multiply both sides by 10. Then,
\[(2) \qquad 10a_{\infty }=9.999\cdots \]
subtract both sides of (1) from (2), and divide by 9,
\[ a_{\infty }=1 \]
Therefore,
\[ 1=0.999\cdots \]
If it is possible to multiply and subtract a real number $a_{\infty }$ which has no end, we have to accept this result.
Extending this result, we come to the fact that a number is equal to the number obtained by subtracting 1 from the last digit of the finite real number and by appending $999\cdots $. For example, if $a=2.4$, it is same as $a=2.3999\cdots $.
In conclusion, any finite real number has infinite real numerical sequences.
2012/10/17
about
Welcome to my blog, "WILL THAT BE ALL？"
This blog will contain posts on various mathematical subjects.
Nearly for 2 years, I have been posting memoranda of the probability theory on my website ( Mathematical Finance by Japanese), and have just finished it recently. Since then, I have been taking a break.
Though I have shut down a bulletin board on my website, I would like to respond to some questions, and write some articles which I gave up addressing on my website, as they are not related to the main issue.
For a change, I will also post just something in my life, impressions of books that I have finished reading recently, authors whom I have been interested in, some photos and pictures which I have taken, and so on.
Any feedback and constructive criticism would be very much appreciated.
This blog will contain posts on various mathematical subjects.
Nearly for 2 years, I have been posting memoranda of the probability theory on my website ( Mathematical Finance by Japanese), and have just finished it recently. Since then, I have been taking a break.
Though I have shut down a bulletin board on my website, I would like to respond to some questions, and write some articles which I gave up addressing on my website, as they are not related to the main issue.
For a change, I will also post just something in my life, impressions of books that I have finished reading recently, authors whom I have been interested in, some photos and pictures which I have taken, and so on.
Any feedback and constructive criticism would be very much appreciated.