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typo fix and add extra contents

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Trance-0
2025-02-22 16:35:08 -06:00
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commit 7699fcaea1
6 changed files with 94 additions and 26 deletions
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pages/Math4111/Math4111_L4.md
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@@ -2,7 +2,7 @@
## Review
1. Let $F$ be a field. Let $a,b,c,...,z\in F$ . Using he field axioms, simplify
1. Let $F$ be a field. Let $a,b,c,...,z\in F$ . Using he field axioms, simplify
$$
(x-a)(x-b)(x-c)...(x-z)
@@ -10,7 +10,7 @@
$x\in F$, it must be at least one $0$ in the product...
2. Suppose $A,B\subset\mathbb{R}$. Suppose $A$ and $B$ are nonempty and bounded above,$A\subset B$. WHat can you say about $sup\ A$ and $sup\ B$? Please justify.
2. Suppose $A,B\subset\mathbb{R}$. Suppose $A$ and $B$ are nonempty and bounded above,$A\subset B$. WHat can you say about $\sup A$ and $\sup B$? Please justify.
$$
\forall x\in A, x\in B. sup\ A\leq sup\ B
@@ -31,7 +31,7 @@ Proof
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$
Let $A=\{nx:n\in\mathbb{N}\}$. Then $A\neq\phi$ (Since $x\in A$) and $A$ is bounded above by $y$. Since $\mathbb{R}$ has LUBP, $sup\ A$ exists. Let $\alpha=sup\ A$.
Let $A=\{nx:n\in\mathbb{N}\}$. Then $A\neq\phi$ (Since $x\in A$) and $A$ is bounded above by $y$. Since $\mathbb{R}$ has LUBP, $sup\ A$ exists. Let $\alpha=\sup A$.
$x>0\implies \alpha-x<\alpha$, $\alpha-x$ is not an upper bound of $A$. (Since $\alpha$ is the LUB of $A$) $\implies \exist m\in \mathbb{N}$ such that $mx>\alpha-x$ by definition of $A$.