deployment update

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Zheyuan Wu
2026-01-13 10:40:19 -06:00
parent 33c02dc4a8
commit e639e241c8
23 changed files with 2403 additions and 1775 deletions

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@@ -15,12 +15,20 @@ A binary operation (usually denoted by $*$) on a set $X$ is a function from $X\t
$+$ is a binary operation on $\mathbb{Z}$ or $\mathbb{R}$.
---
$\cdot$ is a binary operation on $\mathbb{Z}$ or $\mathbb{R}$.
division is not a binary operation on $\mathbb{Z}$ or $\mathbb{R}$.
---
division is not a binary operation on $\mathbb{Z}$ or $\mathbb{R}$. (Consider 0)
---
Generally, we can define a binary operation over sets whatever we want.
---
Let $X=\{a,b,c\}$ and we can define the table for binary operation as follows:
|*| a | b | c |
@@ -29,6 +37,8 @@ Let $X=\{a,b,c\}$ and we can define the table for binary operation as follows:
|b| b | c | c |
|c| a | b | c |
---
If we let $X$ be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$.
then $(f+g)(x)=f(x)+g(x)$,
@@ -80,12 +90,42 @@ Suppose $X=\{a,b,c\}$
is not associative, take $a,b,c$ as examples.
$$
a*(b*c)=a*c=b\neq (a*b)*c=b*c=c
$$
</details>
#### Theorem forAssociativity of Composition
#### Theorem for Associativity of Composition
(Associativity of Composition) Let S be a set and let $f,g$ and $h$ be functions from S to S. Then $(f\circ g)\circ h=f\circ(g\circ h)$.
> [!NOTE]
>
> There exists binary operation that is associative but not commutative.
>
> Consider $(f\circ g)$ where $f,g$ are functions over some set $X$.
>
> $(f\circ g)(x)=f(g(x))$ is generally not commutative but always associative.
>
> There exists binary operation that is commutative but not associative.
>
> Consider operation defined belows:
>
> $S=\{a,b,c\}$
>
> |*| a | b | c |
> |---|---|---|---|
> |a| a | b | b |
> |b| b | b | c |
> |c| b | c | c |
>
> Note that this operation is commutative since the table is symmetric on diagonal.
>
> This operation is not associative, take $a,b,c$ as examples.
>
> $a*(b*c)=a*c=b\neq (a*b)*c=b*c=c$
#### Definition of Identity element
An element $e\in X$ is called identity element if $a*e=e*a=a$ for all $a\in X$.
@@ -124,7 +164,11 @@ identity function $f(x)=x$ is the identity element of $(f\circ g)(x)=f(g(x))$.
>
> Suppose $X=\{a,b,c\}$
> |*| a | b | c |
|---|---|---|---|
|a| a | b | b |
|b| b | c | c |
|c| a | b | c |
> |---|---|---|---|
> |a| a | b | b |
> |b| b | c | c |
> |c| a | b | c |
>
> No identity element exists for this binary operation.

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@@ -1,4 +1,4 @@
# MatH4302 Modern Algebra
# Math4302 Modern Algebra
Prerequisites: Math 429 or permission of instructor