Replace MathJax with MathML

config/mkdocs-common.yml

@@ -159,7 +159,6 @@ theme:
 extra_css:
   - assets/stylesheets/extra.css?v=3.17.0
 extra_javascript:
-  - assets/javascripts/mathjax.js
   - assets/javascripts/randomize-element.js
   - assets/javascripts/resolution.js
 

docs/android.md

@@ -202,7 +202,18 @@ A few more tips for purchasing a Google Pixel:
 - If you're after a bargain on a Pixel device, we suggest buying an "**a**" model, just after the next flagship is released. Discounts are usually available because Google will be trying to clear their stock.
 - Consider price beating options and specials offered at physical stores.
 - Look at online community bargain sites in your country. These can alert you to good sales.
-- Google provides a list showing the [support cycle](https://support.google.com/nexus/answer/4457705) for each one of their devices. The price per day for a device can be calculated as: $\text{Cost} \over \text {EOL Date}-\text{Current Date}$, meaning that the longer use of the device the lower cost per day.
+- Google provides a list showing the [support cycle](https://support.google.com/nexus/answer/4457705) for each one of their devices. The price per day for a device can be calculated as:
+    <math xmlns="http://www.w3.org/1998/Math/MathML" display="inline" class="tml-display" style="display:inline math;">
+      <mfrac>
+        <mtext>Cost</mtext>
+        <mrow>
+          <mtext>End of Life Date</mtext>
+          <mo>−</mo>
+          <mtext>Current Date</mtext>
+        </mrow>
+      </mfrac>
+    </math>
+    , meaning that the longer use of the device the lower cost per day.
 - If the Pixel is unavailable in your region, the [NitroPhone](https://shop.nitrokey.com/shop) can be shipped globally.
 
 ## General Apps

docs/basics/passwords-overview.md

@@ -82,11 +82,62 @@ We recommend using [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_l
 
 To demonstrate how strong diceware passphrases are, we'll use the aforementioned seven word passphrase (`viewable fastness reluctant squishy seventeen shown pencil`) and [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_large_wordlist.txt) as an example.
 
-One metric to determine the strength of a diceware passphrase is how much entropy it has. The entropy per word in a diceware passphrase is calculated as $\text{log}_2(\text{WordsInList})$ and the overall entropy of the passphrase is calculated as $\text{log}_2(\text{WordsInList}^\text{WordsInPhrase})$.
-
-Therefore, each word in the aforementioned list results in ~12.9 bits of entropy ($\text{log}_2(7776)$), and a seven word passphrase derived from it has ~90.47 bits of entropy ($\text{log}_2(7776^7)$).
-
-The [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_large_wordlist.txt) contains 7776 unique words. To calculate the amount of possible passphrases, all we have to do is $\text{WordsInList}^\text{WordsInPhrase}$, or in our case, $7776^7$.
+One metric to determine the strength of a diceware passphrase is how much entropy it has. The entropy per word in a diceware passphrase is calculated as <math>
+  <mrow>
+    <msub>
+      <mtext>log</mtext>
+      <mn>2</mn>
+    </msub>
+    <mo form="prefix" stretchy="false">(</mo>
+    <mtext>WordsInList</mtext>
+    <mo form="postfix" stretchy="false">)</mo>
+  </mrow>
+</math> and the overall entropy of the passphrase is calculated as: <math>
+  <mrow>
+    <msub>
+      <mtext>log</mtext>
+      <mn>2</mn>
+    </msub>
+    <mo form="prefix" stretchy="false">(</mo>
+    <msup>
+      <mtext>WordsInList</mtext>
+      <mtext>WordsInPhrase</mtext>
+    </msup>
+    <mo form="postfix" stretchy="false">)</mo>
+  </mrow>
+</math>
+
+Therefore, each word in the aforementioned list results in ~12.9 bits of entropy (<math>
+  <mrow>
+    <msub>
+      <mtext>log</mtext>
+      <mn>2</mn>
+    </msub>
+    <mo form="prefix" stretchy="false">(</mo>
+    <mn>7776</mn>
+    <mo form="postfix" stretchy="false">)</mo>
+  </mrow>
+</math>), and a seven word passphrase derived from it has ~90.47 bits of entropy (<math>
+  <mrow>
+    <msub>
+      <mtext>log</mtext>
+      <mn>2</mn>
+    </msub>
+    <mo form="prefix" stretchy="false">(</mo>
+    <msup>
+      <mn>7776</mn>
+      <mn>7</mn>
+    </msup>
+    <mo form="postfix" stretchy="false">)</mo>
+  </mrow>
+</math>).
+
+The [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_large_wordlist.txt) contains 7776 unique words. To calculate the amount of possible passphrases, all we have to do is <math>
+  <msup>
+    <mtext>WordsInList</mtext>
+    <mtext>WordsInPhrase</mtext>
+  </msup>
+</math>, or in our case, <math><msup><mn>7776</mn><mn>7</mn></msup></math>.
 
 Let's put all of this in perspective: A seven word passphrase using [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_large_wordlist.txt) is one of ~1,719,070,799,748,422,500,000,000,000 possible passphrases.