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Trance-0
2025-10-27 11:56:32 -05:00
parent 0d93eb43d3
commit fb1ffcd040
17 changed files with 219 additions and 134 deletions

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@@ -27,7 +27,8 @@
(Archimedean property) If $x,y\in \mathbb{R}$ and $x>0$, then $\exists n\in \mathbb{N}$ such that $nx>y$.
Proof
<details>
<summary>Proof</summary>
Suppose the property is false, then $\exist x,y\in \mathbb{R}$ with $x>0$ such that $\forall v\in \mathbb{N}$, nx\leq y$
@@ -39,7 +40,7 @@ This implies $(m+1)x>\alpha$
Since $(m+1)x\in \alpha$, this contradicts the fact that $\alpha$ is an upper bound of $A$.
QED
</details>
### $\mathbb{Q}$ is dense in $\mathbb{R}$
@@ -51,7 +52,8 @@ $$
x<\frac{m}{n}<y\iff nx<m<ny
$$
Proof:
<details>
<summary>Proof</summary>
Let $x,y\in\mathbb{R}$, with $x<y$. We'll find $n\in \mathbb{N},\mathbb{m}\in \mathbb{Z}$ such that $nx<m<ny$.
@@ -59,7 +61,7 @@ By Archimedean property, $\exist n\in \mathbb{N}$ such that $n(y-x)>1$, and $\ex
So $-m_2<nx<m_1$. Thus $\exist m\in \mathbb{Z}$ such that $m-1\leq nx<m$ (Here we use a property of $\mathbb{Z}$) We have $ny>1+nx\geq 1+(m-1)=m$
QED
</details>
### $\sqrt{2}\in \mathbb{R}$, $(\sqrt[n]{x}\in\mathbb{R})$