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 $n\geq N$, $|a_n-a|\geq \epsilon$.
2012/11/26
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.