Typesetting Math with $\KaTeX$

2019-05-27T00:00:00+00:00

Proof supports the typesetting mathematics using $\KaTeX$. You can use $ .. $ to write formulas following the $\LaTeX$ math syntax:

The function $\nu$ is written inline, but can also be in a block:
\\[ \nu: \mathbb{P}^{1} \to \mathbb{P}^{3} \\]

The function $\nu$ is written inline, but can also be in a block: \[ \nu: \R\PP^{1} \to \PP^{3}. \]

Multiple math formats work:

\$5\times 5\$ gives $5\times 5.$
$5\times 5$ gives $5\times 5.$
\\[5\times 5.\\] gives: \[5\times 5.\]
$$5\times 5$$ (on a separate line) gives

$$ 5\times 5. $$

Note: while $$ 5\times 5 $$ works, it will enclose the math in a separate block, which is undesirable.

$\KaTeX$ supports a long list of fonts and functions, but you can also add $\LaTeX$ macros in <script> tags. For instance:

<script>
macros["\\f"] = "\\mathscr{F}"
</script>

$$ 0\to\Gamma_Y(X,\f')\to\Gamma_Y(X,\f)\to\Gamma_Y(X,\f'')\to 0. $$

$\KaTeX$ even has support for commutative diagrams, though for now only using the amscd syntax, which does not allow diagonal arrows for instance.

\begin{CD} A @>{a}>> B \\ @V{b}VV @VV{c}V \\ C @>{d}>> D \end{CD}

So do numbered equations, in equation, align, and gather.

\begin{equation} \begin{split} a &=b+c\\ &=e+f \end{split} \end{equation}

To use either of these, your input needs to be excluded from Markdown processing like this:

<div markdown=0>
\begin{CD}
...
\end{CD}
</div>

If you are not familiar with writing mathematics in $\LaTeX$, I highly recommend reading and bookmarking this Wikibooks reference and using it to typeset problem sets or exercises to practice. Another great reference is this “short” guide by the AMS. To learn more about $\LaTeX$ itself, D.R.Wilkins’ Primer is a decent place to start.

def hello(name="World"):
    print(f"Hello, {name}!")

hello("Codex")

AMS-$\LaTeX$ Theorem Styles

2019-05-26T00:00:00+00:00

Proof supports $\LaTeX$ with AMS extensions, which provide various features to facilitate writing math formulas and to give the general structure and appearance of a mathematics article or book.

In particular, theorem styles including definition, theorem, lemma, proof, and a few more¹ can be declared as an environment using this template:

**Theorem**{:.label #Lagrange}
  _(Lagrange's Theorem)_
  Let $$G$$ be a finite group, and let $$H$$ be a subgroup
  of $$G$$.  Then the order of $$H$$ divides the order of $$G$$.
{:.theorem}

Various components of this template can be skipped or customized, so experiment with them. For instance, to use your own enumeration, use {:.label*} instead.

An Example

The following exposition is taken from here. Note the minimal changes required

Lagrange’s Theorem

Definition Let $H$ be a subgroup of a group $G$. A left coset of $H$ in $G$ is a subset of $G$ that is of the form $xH$, where $x \in G$ and $xH = \{ xh : h \in H \}$. Similarly a right coset of $H$ in $G$ is a subset of $G$ that is of the form $Hx$, where $Hx = \{ hx : h \in H \}$ {::comment} note: the start at end disables numbering

Note that a subgroup $H$ of a group $G$ is itself a left coset of $H$ in $G$.

Lemma Let $H$ be a subgroup of a group $G$, and let $x$ and $y$ be elements of $G$. Suppose that $xH \cap yH$ is non-empty. Then $xH = yH$.

Proof Let $z$ be some element of $xH \cap yH$. Then $z = xa$ for some $a \in H$, and $z = yb$ for some $b \in H$. If $h$ is any element of $H$ then $ah \in H$ and $a^{-1}h \in H$, since $H$ is a subgroup of $G$. But $zh = x(ah)$ and $xh = z(a^{-1}h)$ for all $h \in H$. Therefore $zH \subset xH$ and $xH \subset zH$, and thus $xH = zH$. Similarly $yH = zH$, and thus $xH = yH$, as required.

Lemma Let $H$ be a finite subgroup of a group $G$. Then each left coset of $H$ in $G$ has the same number of elements as $H$.

Proof Let $H = \{ h_1, h_2,\ldots, h_m\}$, where $h_1, h_2,\ldots, h_m$ are distinct, and let $x$ be an element of $G$. Then the left coset $xH$ consists of the elements $x h_j$ for $j = 1,2,\ldots,m$. Suppose that $j$ and $k$ are integers between $1$ and $m$ for which $x h_j = x h_k$. Then $h_j = x^{-1} (x h_j) = x^{-1} (x h_k) = h_k$, and thus $j = k$, since $h_1, h_2,\ldots, h_m$ are distinct. It follows that the elements $x h_1, x h_2,\ldots, x h_m$ are distinct. We conclude that the subgroup $H$ and the left coset $xH$ both have $m$ elements, as required.

Theorem (Lagrange’s Theorem) Let $G$ be a finite group, and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$.

Proof Each element $x$ of $G$ belongs to at least one left coset of $H$ in $G$ (namely the coset $xH$), and no element can belong to two distinct left cosets of $H$ in $G$ (see Lemma). Therefore every element of $G$ belongs to exactly one left coset of $H$. Moreover each left coset of $H$ contains $|H|$ elements (see Lemma). Therefore $|G| = n |H|$, where $n$ is the number of left cosets of $H$ in $G$. The result follows.

The full list is: theorem lemma problem proposition conjecture corollary definition example exercise remark note. ↩︎