keeping things up to date

index.html

@@ -25,11 +25,16 @@
 
   <div> <!-- decription -->
     <p>posts about math, projects, niche interests, etc.</p>
-    <p>check out my <a href="https://github.com/apzzd">Github</a> and <a href="https://artofproblemsolving.com/community/user/600906">Art of Problem Solving</a> accounts, too.</p>
+    <p>check out my <a href="https://github.com/apzzd">Github</a>, <a href="https://artofproblemsolving.com/community/user/600906">Art of Problem Solving</a>, and <a href="https://tatoeba.org/en/user/profile/apzzd">Tatoeba</a> accounts, too.</p>
   </div>
 
   <div class="post-list"> <!-- list of posts -->
 
+   <div class="post-link-div"> <!-- a post link, with description -->
+    	<a href="./posts/billardtische-parkettieren-deutschpost-9-6-2025.html" class="post-link-head">billardtische und unendliche parkettieren (deutsch)</a>
+        <p class="post-link-descrip">ein sehr kurzes deutschpost über billardtische</p>
+    </div>
+
    <div class="post-link-div"> <!-- a post link, with description -->
     	<a href="./posts/reciprocals-and-zn-II-8-5-2025.html" class="post-link-head">reciprocals and $\mathbb{Z}_n$ (part II)</a>
         <p class="post-link-descrip">some dubious and potentially inconsistent reciprocal-finding strategies :/</p>
@@ -44,11 +49,6 @@
     	<a href="./posts/domino-problem-7-6-2025.html" class="post-link-head">arranging dominos</a>
         <p class="post-link-descrip">how many ways can you arrange some dominos on a rectangle, and what does that have to do with fibonacci numbers?</p>
     </div>
-
-   <div class="post-link-div"> <!-- a post link, with description -->
-    	<a href="./posts/billardtische-parkettieren-deutschpost-9-6-2025.html" class="post-link-head">billardtische und unendliche parkettieren (deutsch)</a>
-        <p class="post-link-descrip">ein sehr kurzes deutschpost über billardtische</p>
-    </div>
 
   </div>
 

posts/billardtische-parkettieren-deutschpost-9-6-2025.html

@@ -29,7 +29,7 @@
 
   <div class="post-body"> <!-- body of text -->
 
-   Stellen Sie vor, dass Sie einen Kugel auf einem völlig reibungsfreien Billardtisch haben, und Sie möchten vorhersagen, wann er in einem bestimmten Position befindet sich (in einer Ecktasche, zum Beispiel) nachdem er in einer bestimmten Richtung gerollt wurde und gegen die Rahmen des Billardtischs abgeprallt wurde. Eine nette Strategie für dieses Problem besteht darin, Reflexionen von der Tisch zu fliesen, oder einen unendlich großen Tisch zu machen, so dass die Flugbahn des Kugel vereinfacht wird.
+   Stellen Sie vor, dass Sie einen Kugel auf einem völlig reibungsfreien Billardtisch haben, und Sie möchten vorhersagen, wann er in einem bestimmten Position befindet sich (in einer Ecktasche, zum Beispiel) nachdem er in einer bestimmten Richtung gerollt wurde und gegen die Rahmen des Billardtischs abgeprallt wurde. Eine nette Strategie für dieses Problem besteht darin, Reflexionen von der Tisch zu parkettieren, oder einen unendlich großen Tisch zu machen, so dass die Flugbahn des Kugel vereinfacht wird.
 
    <br><br>
 

posts/reciprocals-and-zn-II-8-5-2025.html

@@ -35,28 +35,27 @@
     Anyway, here we go...
     <br><br>
     <h2>the “vector conversion” strategy</h2>
-    <br><br>
     Say we have the set $\mathbb{Z}_n$ where $n$ isn’t prime. We can split up $n$ into $2$ or more factors: say we have just $2$, $a$ and $b$. Then we can represent a number $x$ in $\mathbb{Z}_n$ as a vector:
     <br><br>
-    $$\langle x \mod a, x \mod b \rangle$$
+    $$\langle x \bmod a, x \bmod b \rangle$$
     <br><br>
     This is the same as:
     <br><br>
-    $$\langle x \mod a, 0 \rangle + \langle 0, x \mod b \rangle$$
+    $$\langle x \bmod a, 0 \rangle + \langle 0, x \mod b \rangle$$
     <br><br>
     The first component of the vector is in $\mathbb{Z}_a$, and the second is in $\mathbb{Z}_b$.
     <br><br>
-    Now we have a slightly simpler problem - we need to find a vector to multiply each term by in order to end up with $\langle 1 , 0 \rangle$ and $\langle 0 , 1 \rangle$, respectively. To do this, we just need to find the reciprocals of $x \mod a$ and $x \mod b$. Once we’ve done that, we add those vectors to get a new vector,
+    Now we have a slightly simpler problem - we need to find a vector to multiply each term by in order to end up with $\langle 1 , 0 \rangle$ and $\langle 0 , 1 \rangle$, respectively. To do this, we just need to find the reciprocals of $x \bmod a$ and $x \bmod b$. Once we’ve done that, we add those vectors to get a new vector,
     <br><br>
     $$\langle c , d \rangle,$$
     <br><br>
-    where $c$ is the reciprocal of $x \mod a$ in $\mathbb{Z}_a$, and $d$ is the reciprocal of $x \mod b$ in $\mathbb{Z}_b$.
+    where $c$ is the reciprocal of $x \bmod a$ in $\mathbb{Z}_a$, and $d$ is the reciprocal of $x \bmod b$ in $\mathbb{Z}_b$.
     <br><br>
     Note that when we’re multiplying vectors with this strategy, we’re multiplying them component-wise.
     <br><br>
-    Now, we just need to convert $\langle c , d \rangle$ back into a number in $\mathbb{Z}_n$. We’ve got to reverse the operation we did in order to convert a number into a vector - so, we need to find some number $x_2$ such that $x_2 \equiv c \mod a$ and $x_2 \equiv d \mod b$.
+    Now, we just need to convert $\langle c , d \rangle$ back into a number in $\mathbb{Z}_n$. We’ve got to reverse the operation we did in order to convert a number into a vector - so, we need to find some number $x_2$ such that $x_2 \equiv c \bmod a$ and $x_2 \equiv d \bmod b$.
     <br><br>
-    If we’re trying to find lots of reciprocals in $\mathbb{Z}_n$ using this process, it’d be worth our time to simplify the problem further. Instead of trying to convert back into $\mathbb{Z}_n$ for every single possible vector $\langle c , d \rangle$, we can convert the unit vectors $\langle 1 , 0 \rangle$ and $$\langle 0 , 1 \rangle$$ instead, and add multiples of them. In this case, we’d find this:
+    If we’re trying to find lots of reciprocals in $\mathbb{Z}_n$ using this process, it’d be worth our time to simplify the problem further. Instead of trying to convert back into $\mathbb{Z}_n$ for every single possible vector $\langle c , d \rangle$, we can convert the unit vectors $\langle 1 , 0 \rangle$ and $\langle 0 , 1 \rangle$ instead, and add multiples of them. In this case, we’d find this:
     <br><br>
     $$c \langle 1 , 0 \rangle + d \langle 0 , 1 \rangle$$
     <br><br>
@@ -68,7 +67,7 @@ <h2>the “vector conversion” strategy</h2>
     <br><br>
     We start by converting this number to a vector, where the first component is in $\mathbb{Z}_{13}$ and the second is in $\mathbb{Z}_4$ (because $13 \times 4 = 52$):
     <br><br>
-    $$5 \rightarrow \langle 5 \mod 13 , 5 \mod 4 \rangle = \langle 5 , 1 \rangle$$
+    $$5 \rightarrow \langle 5 \bmod 13 , 5 \bmod 4 \rangle = \langle 5 , 1 \rangle$$
     <br><br>
     Next, we find the reciprocal of this vector:
     <br><br>
@@ -77,20 +76,19 @@ <h2>the “vector conversion” strategy</h2>
     Now we can convert $\langle 1 , 0 \rangle$ and $\langle 0 , 1 \rangle$ to numbers in $\mathbb{Z}_{52}$... with a bit of calculation, we find that $\langle 1 , 0 \rangle \rightarrow 14$ and $\langle 0 , 1 \rangle \rightarrow 13$.
     <br><br>
     Finally, we can convert the reciprocal in vector form into a number in $\mathbb{Z}_{52}$:
-    $\langle 8 , 1 \rangle = 8 \langle 1 , 0 \rangle + 1 \langle 0 , 1 \rangle \rightarrow (8 \times 14 + 1 \times 13) \mod 52 = 21$
+    $\langle 8 , 1 \rangle = 8 \langle 1 , 0 \rangle + 1 \langle 0 , 1 \rangle \rightarrow (8 \times 14 + 1 \times 13) \bmod 52 = 21$
     <br><br>
     We can check our answer very quickly:
     $$21 \times 5 = 105 = (2 \times 52) + 1$$
     <br><br>
     <h2>the “powers of x” strategy</h2>
-    <br><br>
     Another strategy I found works for prime values of $n$ and is much simpler, but also much less reliable. 
     <br><br>
-    Given a system $\mathbb{Z}_n$, find the smallest possible power $p$ such that $x^p \equiv 1 \mod n$. If you write out a list of powers of $x \mod n$, you should see every possible remainder exactly once before you reach $x^p$.
+    Given a system $\mathbb{Z}_n$, find the smallest possible power $p$ such that $x^p \equiv 1 \mod n$. If you write out a list of powers of $x \bmod n$, you should see every possible remainder exactly once before you reach $x^p$.
     <br><br>
-    Now, you just have to “group” the powers in your list into pairs, such that the product of the powers of $x$ is $x^p$. The numbers that they’re congruent to $\mod n$ are reciprocals.
+    Now, you just have to “group” the powers in your list into pairs, such that the product of the powers of $x$ is $x^p$. The numbers that they’re congruent to $\bmod n$ are reciprocals.
     <br><br>
-    Now, this strategy is unreliable because one value of $x$ doesn’t work for all prime values of $n$ - if a list of powers of $x \mod n$ doesn’t include all possible remainders exactly once, it won’t work. 
+    Now, this strategy is unreliable because one value of $x$ doesn’t work for all prime values of $n$ - if a list of powers of $x \bmod n$ doesn’t include all possible remainders exactly once, it won’t work. 
     <br><br>
     <hr>
     <br><br>

style.css

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+	padding: 20px;
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-	margin-top: 1%;
+    margin-top: 2%;
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