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Use KaTeX for client-side math rendering instead of MathJax

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Dimitri Lozeve 2019-08-18 11:31:44 +02:00
parent fe6d8d5839
commit 633507e193
26 changed files with 241 additions and 177 deletions
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_site/posts/peano.html
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@ -7,7 +7,16 @@
<title>Dimitri Lozeve - Peano Axioms</title>
<link rel="stylesheet" href="../css/default.css" />
<link rel="stylesheet" href="../css/syntax.css" />
<script src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.5/MathJax.js?config=TeX-MML-AM_CHTML" async></script>
<!-- KaTeX CSS styles -->
<link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/katex@0.11.0/dist/katex.min.css" integrity="sha384-BdGj8xC2eZkQaxoQ8nSLefg4AV4/AwB3Fj+8SUSo7pnKP6Eoy18liIKTPn9oBYNG" crossorigin="anonymous">
<!-- The loading of KaTeX is deferred to speed up page rendering -->
<script defer src="https://cdn.jsdelivr.net/npm/katex@0.11.0/dist/katex.min.js" integrity="sha384-JiKN5O8x9Hhs/UE5cT5AAJqieYlOZbGT3CHws/y97o3ty4R7/O5poG9F3JoiOYw1" crossorigin="anonymous"></script>
<!-- To automatically render math in text elements, include the auto-render extension: -->
<script defer src="https://cdn.jsdelivr.net/npm/katex@0.11.0/dist/contrib/auto-render.min.js" integrity="sha384-kWPLUVMOks5AQFrykwIup5lo0m3iMkkHrD0uJ4H5cjeGihAutqP0yW0J6dpFiVkI" crossorigin="anonymous" onload="renderMathInElement(document.body);"></script>
</head>
<body>
<header>
@ -81,26 +90,13 @@ then <span class="math inline">\(\varphi(n)\)</span> is true for every natural n
<p>First, we prove that every natural number commutes with <span class="math inline">\(0\)</span>.</p>
<ul>
<li><span class="math inline">\(0+0 = 0+0\)</span>.</li>
<li><p>For every natural number <span class="math inline">\(a\)</span> such that <span class="math inline">\(0+a = a+0\)</span>, we have:</p>
<span class="math display">\[\begin{align}
0 + s(a) &amp;= s(0+a)\\
&amp;= s(a+0)\\
&amp;= s(a)\\
&amp;= s(a) + 0.
\end{align}
\]</span></li>
<li><p>For every natural number <span class="math inline">\(a\)</span> such that <span class="math inline">\(0+a = a+0\)</span>, we have:</p></li>
</ul>
<p>By Axiom 5, every natural number commutes with <span class="math inline">\(0\)</span>.</p>
<p>We can now prove the main proposition:</p>
<ul>
<li><span class="math inline">\(\forall a,\quad a+0=0+a\)</span>.</li>
<li><p>For all <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> such that <span class="math inline">\(a+b=b+a\)</span>,</p>
<span class="math display">\[\begin{align}
a + s(b) &amp;= s(a+b)\\
&amp;= s(b+a)\\
&amp;= s(b) + a.
\end{align}
\]</span></li>
<li><p>For all <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> such that <span class="math inline">\(a+b=b+a\)</span>,</p></li>
</ul>
<p>We used the opposite of the second rule for <span class="math inline">\(+\)</span>, namely <span class="math inline">\(\forall a,
\forall b,\quad s(a) + b = s(a+b)\)</span>. This can easily be proved by another induction.</p>