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# Completeness of the real numbers

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<span class="definition">Deddkind Cut</span>

Let $A,B$ be two subsets of an ordered set $P$
We say that $(A,B)$ is a Deddkind cut of $P$ if $A,B$ satisfy:

$A\not=\varnothing\not=B$
$A\cup B=P,A\cap B=\varnothing$
$\forall a\in A,b\in B\implies a<b$

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<span class="definition">Deddkind Cut on real number</span>
Let $(A,B)$ be a Deddkind Cut on $\mathbb{R}$

Then exactly one of the following statements hold :
1. $\max A$ exists and $\min B$ doesn't exist.
2. $\max A$ doesn't exist and $\min B$ exists.

where $\max A$ means
$x=\max A$ if and only if $x\in A$ and $a\ge x\implies a=x$
and $\min B$ means
$y=\min B$ if and only if $y\in B$ and $b\le y\implies b=y$

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<span class="definition">Least upper bound theorem</span>
(Equivalent proposition on $\mathbb{R}$)

Let $S$ be a non-empty subset of $\mathbb{R}$

If $S$ is bounded above, then $S$ has a least upper bound.
i.e. $\sup S$ exists.
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<span class="definition">Nested Intervals Theorem</span>

Let $\{I_n\}_{n=1}^{\infty}=\{[a_n,b_n]\}_{n=1}^{\infty}$ be a sequence satisfying
(1) $\forall n\in\mathbb{N},I_{n+1}\subset I_n$
(2) $\displaystyle\bigcap_{n=1}^{\infty}I_n:=[\displaystyle\lim_{n\to\infty}a_n,\displaystyle\lim_{n\to\infty}b_n]\not=\varnothing$

If $\displaystyle\lim_{n\to\infty}(b_n-a_n):=\displaystyle\lim_{n\to\infty}\text{length}(I_n)=0$, then
$\exists !r\in\mathbb{R},s.t.\{r\}=\displaystyle\bigcap_{n=1}^{\infty}I_n$

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<span class="definition">Monotonic Convergence Sequence</span>

A monotonic sequence is a sequence that is either increasing or decreasing.

A monotoinc sequence is bounded if and only if it converges.
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## Rational numbers are not complete

We can use the above 4 axioms to show that $\mathbb{Q}$ is not complete.

We form a Deddkind Cut on the number $p$ where $p^2=2$.

Let $A=\{q\in\mathbb{Q}\mid(q\le0)\lor(q^2\le2)\},B=\{q\in\mathbb{Q}\mid (q>0)\land(q^2>2)\}$
Check: $A\cap B=\varnothing$
$A\cap B=\{q\in\mathbb{Q}\mid[(q\le0)\lor(q^2\le2)]\land[(q>0)\land(q^2>2)]\}$
$[(q\le0)\lor(q^2\le2)]\land[(q>0)\land(q^2>2)]\equiv[(q\le0)\land(q>0)\land(q^2>2)]\lor[(q^2\le2)\land(q>0)\land(q^2>2)]$
Since $q\le0\land q>0$ and $(q^2\le2)\land(q^2>2)$ are always false, there is no $q\in\mathbb{Q}$ such that $q\in A\cap B$
So $A\cap B=\varnothing$

Check: $A\cup B=\mathbb{Q}$
$A\cup B=\{q\in\mathbb{Q}\mid[(q\le0)\lor(q^2\le2)]\lor[(q>0)\land(q^2>2)]\}$
$[(q\le0)\lor(q^2\le2)]\lor[(q>0)\land(q^2>2)]\equiv[(q\le0)\lor(q^2\le2)\lor(q>0)]\land[(q^2>2)\lor(q^2\le2)\lor(q>0)]$

Since $q\le0\lor q>0$ and $q^2>2\lor q^2\le2$ are both tautologies, $\forall q\in\mathbb{Q}\iff q\in A\cup B$
So $A\cup B=\mathbb{Q}$

Check: $\forall a\in A,b\in B,a<b$

Let $q_1\in A,q_2\in B$
Case 1:$q_1\le0$
Since $q_2>0$, $q_2>q_1$
Case 2:${q_1}^2\le2$
Since ${q_2}^2>2$ and $q_2>0$, ${q_2}^2>{q_1}^2\implies q_2>q_1$

So $(A,B)$ forms a Deddkind Cut on the number $p$ where $p^2=2$.

Now we prove that $\max{A}$ and $\min{B}$ don't exist.

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Suppose $\alpha\in A$
$\alpha\in A\iff(\alpha\le 0)\lor(\alpha^2\le 2)$

If $\alpha\le 1$ and then take $1\in A$, since $1^2\le 2$, so such $\alpha$ must not be the greatest element of $A$.
So $\alpha>1$ and $\alpha^2\le2$
Since $\not\exists q\in\mathbb{Q},s.t.\ q^2=2$, $\alpha^2<2$

Claim: Let $x=\dfrac{2}{\dfrac{\alpha}{2}+\dfrac{1}{\alpha}}$(How do we know?), then $x\in A$ and $x>\alpha$

(How do we know?)

$\dfrac{\dfrac{2}{\alpha}+\alpha}{2}\ge\sqrt{2}$ and $\alpha<\sqrt{2}$, so the equality won't hold.

We can take the reciprocal of $\dfrac{\dfrac{2}{\alpha}+\alpha}{2}$ then $\dfrac{2}{\dfrac{2}{\alpha}+\alpha}<\dfrac{\sqrt{2}}{2}$
So we multiply both sides by $2$, we get a number $x$ that satisfies $x^2<2$

Then we can check that $x>\alpha$, then the proof will be done.

$x\in\mathbb{Q}$ is clear
$x-\alpha=\dfrac{2}{\dfrac{\alpha}{2}+\dfrac{1}{\alpha}}-\alpha=\dfrac{4\alpha}{\alpha^2+2}-\alpha=\dfrac{\alpha(2-\alpha^2)}{\alpha^2+1}$
Since $1<\alpha$ and $\alpha^2<2$, $2-\alpha^2>0$
Therefore $x-\alpha=\dfrac{\alpha^2(3-\alpha)}{\alpha^2+1}>0\iff x>\alpha$

To show that $x^2<2$, we prove that $\dfrac{x^2}{2}<1$

$\dfrac{x^2}{2}=\dfrac{2}{(\dfrac{\alpha}{2}+\dfrac{1}{\alpha})^2}=\dfrac{2}{\dfrac{\alpha^2}{4}+1+\dfrac{1}{\alpha^2}}$

Since $\dfrac{\alpha^2}{4}+\dfrac{1}{\alpha^2}\ge2\sqrt{\dfrac{1}{4}}=1$
Suppose that $\dfrac{\alpha^2}{4}+\dfrac{1}{\alpha^2}=1$, then $\dfrac{\alpha^2}{4}=\dfrac{1}{\alpha^2}$ would lead to a contradiction saying that $\alpha^2=2$
So $\dfrac{\alpha^2}{4}+\dfrac{1}{\alpha^2}>1\implies\dfrac{\alpha^2}{4}+\dfrac{1}{\alpha^2}+1>2$
Therefore $\dfrac{x^2}{2}<1\iff x^2<2$

So $x\in A$ and $x>\alpha$.

Hence $\max{A}$ doesn't exist.
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Suppose $\min B$ exists, call it $\beta$

Let $y=\dfrac{1}{\beta}+\dfrac{\beta}{2}\in\mathbb{Q}$
Since $y\ge2\sqrt{\dfrac{1}{2}}=\sqrt{2}$ and $\dfrac{1}{\beta}\not=\dfrac{\beta}{2}$, $y^2>2$

And since $\beta^2>2\implies\beta>\dfrac{2}{\beta}\implies\dfrac{\beta}{2}>\dfrac{1}{\beta}\implies\beta>\dfrac{1}{\beta}+\dfrac{\beta}{2}=y$

So $y\in B$ and $y<\beta$, $\beta\not=\min{B}$ is a contradiction.

Therefore $\min{B}$ doesn't exist.

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Thus we know that if we form a Deddkind cut on $\mathbb{Q}$, the completeness doesn't necessarily hold.

# Why does any real number can be represented in decimal?

For this problem, we first define what exactly is the decimal representation of a number.

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<span class="definition">Def</span>

Let $x$ be a number, the decimal representation (if legal) of $x$ is
$x=\pm N.a_1a_2a_3\dots$
where $N\in\mathbb{N},a_k\in\{0,1,2,3,4,5,6,7,8,9\},\forall k\in\mathbb{N}$

This representation can be written as the infinite series

$x=\pm N+(\displaystyle\sum_{k=1}^{\infty}a_k(10)^{-k})$

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Notation $\mathbb{R}_{>0}=\{x\in\mathbb{R}\mid x>0\}$

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<span class="theorem">Thm1.1</span>

$\forall x\in\mathbb{R}_{>0},\exists n\in\mathbb{N},s.t.\ nx>1$

<span class="theorem">Cor1</span>Let $y\in\mathbb{R}$
$\forall x\in\mathbb{R}_{>0},\exists n\in\mathbb{N},s.t.\ nx>y$

<span class="theorem">Cor2</span>

$\forall x\in\mathbb{R}_{>0},\exists N\in\mathbb{N},s.t.\ \forall n\in\mathbb{N}\land n>N,n>x$

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$pf:$

Suppose $\exists x\in\mathbb{R}_{>0},s.t.\ \forall n\in\mathbb{N},nx\le 1$

Let $S=\{nx\mid n\in\mathbb{N}\}$
Since $\forall n\in\mathbb{N},nx\le 1$, $\sup S$ exists, call it $\alpha$.
$\alpha-x<\alpha$, so $\exists m\in\mathbb{N},s.t.\alpha-x<mx$.
But $(m+1)x\in S$ ($S$ is inductive) and $(m+1)x>\alpha$ is a contradiction.
Therefore $\forall x\in\mathbb{R}_{>0},s.t.\ \exists n\in\mathbb{N},nx>1$

For corollary 1, simply replace $1$ with $y$.

For corollary 2

$\forall x\in\mathbb{R}_{>0},\exists N\in\mathbb{N},s.t.\ N>x$
Since $n>N\implies n>N>x$.
Therefore $\forall x\in\mathbb{R}_{>0},\exists N\in\mathbb{N},s.t.\ \forall n\in\mathbb{N}\land n>N,n>x$
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<span class="theorem">Thm1.2</span>
$\forall x\in\mathbb{R}_{>0},\exists n\in\mathbb{N},s.t.\ x\in[n-1,n]$
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$pf:$

For $1>0,\exists m\in\mathbb{N},s.t. 1\times m=m>x$

Let $S=\{m\in\mathbb{N}\mid m>x\}$
By well-ordereing principle of the natrual numbers, suppose $n$ is the least element of $S$
So $n-1\not\in S\iff n-1\le x$

Therefore $\forall x\in\mathbb{R}_{>0},\exists n\in\mathbb{N},s.t.\ x\in[n-1,n]$.

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<span class="theorem">Thm1.3</span>
$\forall x\in\mathbb{R},\exists k\in\mathbb{Z},s.t.\ x\in[k-1,k]$
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This theorem is trivial since we can simply multiply the inequality by $-1$ to obtain the case where $x<0$

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<span class="theorem">Thm1.4</span>

If a number $x$ can be written as a unique decimal representation, then $x\in\mathbb{R}$
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$pf:$

It sufficies to prove that $0.a_1a_2\dots a_n\dots$ is a real number, since any real number can be written as the sum of an integer and $0.a_1a_2\dots a_n\dots$ by theorem 1.3.
Since $\mathbb{R}$ is a field and $\mathbb{Z}\subseteq\mathbb{R}$, so if $0.a_1a_2\dots a_n\dots$ is a real number, then $x$ is a real number.

Let $S_n:=\displaystyle\sum_{k=1}^{n}\dfrac{a_k}{10^k}$

Then $S_n$ is a monotonic convergence sequence.

Check: $S_n$ is non-decreasing
Since $S_n-S_{n-1}=\dfrac{a_n}{10^n}$ and $a_n\in\{0,1,2,3,4,5,6,7,8,9\}$
We have that $S_n-S_{n-1}\ge0$, hence $S_n$ is monotonic.

Check: $S_n$ is bounded
Since $0\le\dfrac{a_k}{10^k}\le\dfrac{9}{10^k}$
$0\le S_n\le\displaystyle\sum_{k=1}^{n}\dfrac{9}{10^k}=\dfrac{\dfrac{9}{10}(1-\dfrac{1}{10^n})}{1-\dfrac{1}{10}}$
Hence $\displaystyle\lim_{n\to\infty}0\le\lim_{n\to\infty}S_n\le\lim_{n\to\infty}\dfrac{\dfrac{9}{10}(1-\dfrac{1}{10^n})}{1-\dfrac{1}{10}}=1$, $S_n$ is bounded above.
Therefore by the completeness of the real number, we know that $\displaystyle\lim_{n\to\infty}S_n=\displaystyle\sum_{k=1}^{\infty}\dfrac{a_k}{10^k}$ converges to a real number.
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