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feat: various content improvements
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cuppajoeman committed Oct 30, 2024
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7 changes: 5 additions & 2 deletions html/algebra/galois_theory/ideals.html
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Ideal
</div>
<div class="content">
An ideal in a crone \( \left( R, \oplus, \otimes \right) \) is a subset \( I \) containing \( 0 _ R \) such that
An ideal in a <a class="knowledge-link" href="/algebra/galois_theory/rings.html#definition-crone">crone</a> \( \left( R, \oplus, \otimes \right) \) is a subset \( I \) containing \( 0 _ R \) such that
<ul>
<li>\( a, b \in I \) implies that \( a - b \in I \)</li>
<li>\( a \in I \) and \( r \in R \) implies that \( r \otimes a \in I \)</li>
</ul>
</div>
</div>
<p>
Note that sometimes we say it is a left ideal withen \( r a \in I \) and a right ideal when \( a r \in I \)
</p>
<div class="corollary" id="corollary-trivial-ideal">
<div class="title">
Trivial Ideal
</div>
<div class="content">
Suppose \( R \) is a crone, then \( \left\{ 0 _ R \right\} \) is an ideal in \( R \)
Suppose \( R \) is a <a class="knowledge-link" href="/algebra/galois_theory/rings.html#definition-crone">crone</a>, then \( \left\{ 0 _ R \right\} \) is an ideal in \( R \)
</div>
<div class="proof"></div>
</div>
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16 changes: 8 additions & 8 deletions html/algebra/groups/binary_operations_and_groups.html
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Binary Operation
</div>
<div class="content">
Let \( S \) be a set, a binary operation on \( S \) is a <span class="knowledge-link" data-href="/fundamentals/functions.html#definition-function">function</span> \( \star : S \times S \to S \) where we denote \( \star \left ( a , b \right ) \) as \( a \star b \) using infix notation
Let \( S \) be a set, a binary operation on \( S \) is a <a class="knowledge-link" href="/fundamentals/functions.html#definition-function">function</a> \( \star : S \times S \to S \) where we denote \( \star \left ( a , b \right ) \) as \( a \star b \) using infix notation
</div>
</div>
<p>Note the distinction between binary operation and binary relation, a binary operation is a function, which is a binary relation with some extra conditions.</p>
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Commutative Binary Operation
</div>
<div class="content">
A <span class="knowledge-link" data-href="/algebra/groups/binary_operations_and_groups.html#definition-binary-operation">binary operation</span> on a set \( S \) such that for any \( a , b \in S \), then \( a \star b = b \star a \) is called commutative
A <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-binary-operation">binary operation</a> on a set \( S \) such that for any \( a , b \in S \), then \( a \star b = b \star a \) is called commutative
</div>
</div>
<div class="definition" id="definition-associative-binary-operation">
<div class="title">
Associative Binary Operation
</div>
<div class="content">
A <span class="knowledge-link" data-href="/algebra/groups/binary_operations_and_groups.html#definition-binary-operation">binary operation</span> on a set \( S \) such that for any \( a , b , c \in S \), then \( \left ( a \star b \right ) \star c = a \star \left ( b \star c \right ) \) is called associative
A <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-binary-operation">binary operation</a> on a set \( S \) such that for any \( a , b , c \in S \), then \( \left ( a \star b \right ) \star c = a \star \left ( b \star c \right ) \) is called associative
</div>
</div>
<div class="definition" id="definition-left-fold">
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Bracket Placement Doesn't Matter for Associative Binary Operations
</div>
<div class="content">
Suppose that \( \star \) is an <span class="knowledge-link" data-href="/algebra/groups/binary_operations_and_groups.html#definition-associative-binary-operation">associative binary operation</span>, then for any \( n \in \mathbb{Z}^{\ge 3} \), then no matter how one places the brackets in the product \( x_{1} \star x_{2} \star \ldots \star x_{n} \) it always evaluates to the same value
Suppose that \( \star \) is an <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-associative-binary-operation">associative binary operation</a>, then for any \( n \in \mathbb{Z}^{\ge 3} \), then no matter how one places the brackets in the product \( x_{1} \star x_{2} \star \ldots \star x_{n} \) it always evaluates to the same value
</div>
<div class="proof">
<p>In order to prove this statement we have to prove that no matter the bracketing we get the same value, but what exactly is this value? It must be the result of <i>some</i> bracketing so to make our life easier let's select a left-most bracketing of the form \[ \left( \left( x _ 1 \star x _ 2 \right) \star x _ 3 \star \cdots \right) \star x _ k \]</p>
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Composition Law
</div>
<div class="content">
A composition law is an <span class="knowledge-link" data-href="/algebra/groups/binary_operations_and_groups.html#definition-associative-binary-operation">associative binary operation</span>
A composition law is an <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-associative-binary-operation">associative binary operation</a>
</div>
</div>
<div class="definition" id="definition-identity-element">
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Group
</div>
<div class="content">
We say that \( \left( G, \star \right) \) where \( G \) is a set and \( \star \) is a <span class="knowledge-link" data-href="/algebra/groups/binary_operations_and_groups.html#definition-composition-law">composition law</span> on \( G \), when there exists an element \( e \in G \) satisfying the following properties:
We say that \( \left( G, \star \right) \) where \( G \) is a set and \( \star \) is a <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-composition-law">composition law</a> on \( G \), when there exists an element \( e \in G \) satisfying the following properties:
<ul>
<li>\( e \) is an <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-identity-element">identity element</a> on \( G \)
</li>
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Given a group \( G \) and \( a , b \in G \), then \( \left ( a \star b \right )^{- 1} = b^{- 1} \star a^{- 1} \)
</div>
<div class="proof">
For ease let \( c = \left ( a \star b \right )^{- 1} \), <span class="knowledge-link" data-href="/algebra/groups/binary_operations_and_groups.html#definition-inverse">so that</span> \( \left ( a \star b \right ) \star c = c \star \left ( a \star b \right ) = e \) <span class="knowledge-link" data-href="/algebra/groups/binary_operations_and_groups.html#definition-composition-law">since we know</span> \( \left ( a \star b \right ) \star c = a \star \left ( b \star c \right ) = e \) , and thus \( a^{- 1} \star \left ( a \star \left ( b \star c \right ) \right ) = a^{- 1} \star e \), thus \( \left ( a^{- 1} \star a \right ) \star \left ( b \star c \right ) = a^{- 1} e \), which simplifies to \( e \star \left ( b \star c \right ) = a^{- 1} \), so we can see that \( a^{- 1} = b \star c \), as needed
For ease let \( c = \left ( a \star b \right )^{- 1} \), <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-inverse">so that</a> \( \left ( a \star b \right ) \star c = c \star \left ( a \star b \right ) = e \) <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-composition-law">since we know</a> \( \left ( a \star b \right ) \star c = a \star \left ( b \star c \right ) = e \) , and thus \( a^{- 1} \star \left ( a \star \left ( b \star c \right ) \right ) = a^{- 1} \star e \), thus \( \left ( a^{- 1} \star a \right ) \star \left ( b \star c \right ) = a^{- 1} e \), which simplifies to \( e \star \left ( b \star c \right ) = a^{- 1} \), so we can see that \( a^{- 1} = b \star c \), as needed
</div>
</div>
<div class="proposition" id="proposition-cancellation">
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Abelian
</div>
<div class="content">
Given a <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-group">group</a>, if it's <span class="knowledge-link" data-href="/algebra/groups/binary_operations_and_groups.html#definition-composition-law">composition law</span> is <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-commutative-binary-relation">commutative</a>, then we say that the group is <b>abelian</b>
Given a <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-group">group</a>, if it's <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-composition-law">composition law</a> is <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-commutative-binary-relation">commutative</a>, then we say that the group is <b>abelian</b>
</div>
</div>
<div class="proposition" id="proposition-integers-with-addition-are-abelian">
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60 changes: 45 additions & 15 deletions html/algebra/linear/vector_spaces/basis.html
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<div class="definition">
<div class="definition" id="definition-basis">
<div class="title">
Basis
</div>
<div class="content">
Given a vector space \( V \) the set \( \mathcal{ B } \subseteq V \) is said to be a basis if \( B \) is <a class="knowledge-link" href="/algebra/linear/vector_spaces/span_and_linear_independence.html#definition-linearly-independent">linearly independent</a> and \( B \) <a class="knowledge-link" href="/algebra/linear/vector_spaces/span_and_linear_independence.html#definition-spanning-set">spans</a> \( V \)
</div>
</div>
<p>
The plural of basis is bases.
</p>
<div class="proposition" id="proposition-the-size-of-a-basis-is-unique">
<div class="title">The Size of a Basis is Unique</div>
<div class="content">
Suppose that \( \mathcal{ B }, \mathcal{ C } \) are bases for \( V \) then \( \left\lvert \mathcal{ B } \right\rvert = \left\lvert \mathcal{ C } \right\rvert \)
</div>
<div class="proof">
TODO: Add the proof here.
</div>
</div>
<div class="proposition" id="proposition-every-vector-space-has-a-basis">
<div class="title">
basis
Every Vector Space has a Basis
</div>
<div class="content">
For any vector space \( V \) there is a <a class="knowledge-link" href="/algebra/linear/vector_spaces/basis.html#definition-basis">basis</a> for it
</div>
<div class="proof">
TODO <!-- https://math.stackexchange.com/questions/189693/every-vector-space-has-a-basis-proof-using-zorns-lemma-in-linear-algebra-->
<!-- https://math.stackexchange.com/questions/466879/prove-that-every-vector-space-has-a-basis-->
<!-- https://math.stackexchange.com/questions/4085932/proof-that-every-vector-space-has-a-basis-->
<!-- https://math.stackexchange.com/questions/2785769/proof-of-every-finite-dimensional-vector-space-has-a-finite-basis-->
</div>
</div>
<div class="definition" id="definition-dimension-of-a-vector-space" >
<div class="title">Dimension of a Vector Space</div>
<div class="content">
if the vector space is denoted by \( X \), then for ease we will denote the basis by a calligraphic version of that character \( \mathcal{X} \)
Suppose that \( V \) is a vector space then since <a class="knowledge-link" href="/algebra/linear/vector_spaces/basis.html#proposition-every-vector-space-has-a-basis">it has a basis</a> \( \mathcal{ B } \) then we define:
\[
\operatorname{ dim } \left( V \right) = \left\lvert \mathcal{ B } \right\rvert
\]
Note that the above defintion makes sense as <a class="knowledge-link" href="/algebra/linear/vector_spaces/basis.html#proposition-the-size-of-a-basis-is-unique">the size of all bases are the same</a>
</div>
</div>
<div class="definition" id="definition-finite-dimensional-vector-space" >
<div class="title">Finite Dimensional Vector Space</div>
<div class="content">
A vector space \( V \) is said to be finite dimensional if <a class="knowledge-link" href="/algebra/linear/vector_spaces/basis.html#definition-dimension-of-a-vector-space">\( \operatorname{ dim } \left( V \right) \)</a> is finite
</div>
</div>
<div class="definition">
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By fixing any new basis of \( V \) (TODO: prove that for any vector space there is always a basis that spans it) \( b_1, \ldots b_n \), then by concatenating it with \(S\) we obtain \( ( s_1, ..., s_k, b_1, ..., b_n ) \). Then for each \( i \in [n] \) (in order) (TODO: this is an algorithm), remove \(v_i\) from the ordered set iff \( v_i \) is in the span of all the earlier elements in the set. In particular, once we have checked and kept \( n - k \) of the \(v - i \) (TODO: say why that's guarenteed), we can discard the remaining \( v_i \), leaving a basis: \[ (s_1, ..., s_k, v_{i_1}, , \ldots v_{i_{n - k}} ) \]
</div>
</div>
<div class="proposition" id="proposition-every-vector-space-has-a-basis">
<div class="title">
Every Vector Space has a Basis
</div>
<div class="content"></div>
<div class="proof">
TODO <!-- https://math.stackexchange.com/questions/189693/every-vector-space-has-a-basis-proof-using-zorns-lemma-in-linear-algebra-->
<!-- https://math.stackexchange.com/questions/466879/prove-that-every-vector-space-has-a-basis-->
<!-- https://math.stackexchange.com/questions/4085932/proof-that-every-vector-space-has-a-basis-->
<!-- https://math.stackexchange.com/questions/2785769/proof-of-every-finite-dimensional-vector-space-has-a-finite-basis-->
</div>
</div>
<div class="exercise" id="exercise-extending-a-basis">
<div class="title">
Extending a Basis
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20 changes: 17 additions & 3 deletions html/algebra/linear/vector_spaces/span_and_linear_independence.html
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Linear Combination
</div>
<div class="content">
A linear combination of the vectors \( \left\{ v _ 1, \ldots, v _ m \right\} \) of vectors in \( V \) \[ a _ 1 v _ 1 + \ldots + a_m v_m \] where \( a_1, \ldots, a_m \in \mathbb{F} \)
A linear combination of the vectors \( \left\{ v _ 1, \ldots, v _ m \right\} \) of vectors in a <a class="knowledge-link" href="/algebra/linear/vector_spaces/vector_spaces.html#definition-vector-space-over-a-field">vector space</a> \( V \) is:
\[
a _ 1 v _ 1 + \ldots + a_m v_m
\]
where \( a_1, \ldots, a_m \in \mathbb{F} \)
</div>
</div>
<div class="definition" id="definition-span">
<div class="title">
Span
</div>
<div class="content">
The set of all linear combinations of the vectors \( \left\{ v _ 1, \ldots, v _ m \right\} \subseteq V\) called the span of \( \left\{ v_1, \ldots, v_m \right\} \) and we define the notation: \[ \operatorname{ span } \left( \left\{ v_1, ..., v_m \right\} \right) := \left\{ a_1 v_1 + \ldots + a_m v_m: a_1, ..., a_m \in \mathbb { F } \right\} \] we define \( \operatorname{ span } \left( \emptyset \right) := \left\{ 0 \right\} \)
The set of all linear combinations of the vectors \( \left\{ v _ 1, \ldots, v _ m \right\} \subseteq V\) (where \( V \) is a <a class="knowledge-link" href="/algebra/linear/vector_spaces/vector_spaces.html#definition-vector-space-over-a-field">vector space</a>) called the span of \( \left\{ v_1, \ldots, v_m \right\} \) and we define the notation: \[ \operatorname{ span } \left( \left\{ v_1, ..., v_m \right\} \right) := \left\{ a_1 v_1 + \ldots + a_m v_m: a_1, ..., a_m \in \mathbb { F } \right\} \] we define \( \operatorname{ span } \left( \emptyset \right) := \left\{ 0 \right\} \)
</div>
</div>
<div class="definition" id="definition-linearly-independent">
<div class="title">
Linearly Independent
</div>
<div class="content">
A set of vectors \( \left\{ v _ 1, \ldots, v _ m \right\} \subseteq V\) is said to be linearly independent if the only solution to \( \sum _ { i = 1 } ^ m a_i v_i = 0 \) is \( \forall i \in \left\{ 1, ..., m \right\}, a_i = 0 \). We define \( \emptyset \) to be linearly independent.
A set of vectors \( \left\{ v _ 1, \ldots, v _ m \right\} \subseteq V\) is said to be linearly independent if the only solution to
\[
\sum _ { i = 1 } ^ m a_i v_i = 0
\]
is \( \forall i \in \left\{ 1, ..., m \right\}, a_i = 0 \). We define \( \emptyset \) to be linearly independent.
</div>
</div>
<div class="definition" id="definition-spanning-set" >
<div class="title">Spanning Set</div>
<div class="content">
Suppose that \( V \) is a vector space, then we say that \( S \) is a <b>spanning set of \( V \)</b> if <a class="knowledge-link" href="/algebra/linear/vector_spaces/span_and_linear_independence.html#definition-span">\( \operatorname{ span } \left( S \right) = V \)</a>
</div>
</div>
<div class="proposition" id="proposition-linearly-independent-iff-unique-representation">
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13 changes: 12 additions & 1 deletion html/algebra/linear/vector_spaces/vector_spaces.html
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Vector Space over a Field
</div>
<div class="content">
A <b>vector space over a field \( F \)</b> is a non-empty set \( V \) with a binary operation \( \oplus \) on \( V \) and a binary function \( \otimes : F \times V \to V \) such that the following hold for any \( a, b \in F \) and \( u , v \in V \)
A <b>vector space over a field \( F \)</b> is a non-empty set \( V \) with a <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-binary-operation">binary operation</a> \( \oplus \) on \( V \) and a binary function \( \otimes : F \times V \to V \) such that the following hold for any \( a, b \in F \) and \( u , v \in V \)
<ul>
<li>\( \oplus \) is associative and commutative</li>
<li>There is a <a class="knowledge-link" href="/algebra/groups/binary_operations_and_groups.html#definition-identity-element">identity element</a> \( 0 _ V \) with respect to \( \oplus \)
Expand All @@ -19,3 +19,14 @@
</ul>
</div>
</div>
<div class="definition" id="definition-subspace-of-a-vector-space" >
<div class="title">Subspace of a Vector Space</div>
<div class="content">
A subset \( W \) of a <a class="knowledge-link" href="/algebra/linear/vector_spaces/vector_spaces.html#definition-vector-space-over-a-field">vector space \( V \) over a field \( F \)</a> is called a <b>subspace</b> of \( V \) if \( W \) itself is a vector space over \( F \) with the same operations of vector addition and scalar multiplication as \( V \). Specifically, \( W \) is a subspace if:
<ol>
<li> \( 0_V \in W. \)</li>
<li> \( \forall u, v \in W, u + v \in W\)</li>
<li> \( \forall u \in W, c \in F, c \cdot u \in W\)</li>
</ol>
</div>
</div>
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