Make CSS compatible with dark reader. Fix grammer.

Author: plzin

posts/linear-systems-mod-n.md

@@ -670,10 +670,12 @@ $$
 
 We are lucky here, because we can eliminate the 3 in the second column directly.
 In general this need not happen.
-We would then have to do the same procedure but with column operations, which might make entries in the row reappear.
+We would then have to do the same procedure but with column operations,
+which might make entries in the row reappear.
 This is not a problem, however.
-Important is, that the diagonal entry, which is also the pivot, is getting smaller
-in each iteration, so eventually we will be able to use it to eliminate both the row and column.
+The only thing that is important, is that the diagonal entry, which is also the pivot,
+is getting smaller in each iteration.
+So eventually we will be able to use it to eliminate both the row and column.
 Nevertheless, in this instance we can now use a single column operations to eliminate the 3.
 
 $$

posts/perm-poly.md

@@ -376,7 +376,7 @@ and subtracting the result from $F$.
 We can reduce the other coefficients of $F$ in a similar way.
 Let $l_i$ be the leading coefficient of $P_i$, i.e.$$l_i=\frac{n}{\gcd(n, i!)}$$
 which is a power of two.
-We can reduce the coefficient $a_i$ into the range $0\leq a_i\leq l_i$ by multiplying $P_i$ by $\lfloor \frac{a_i}{l_i}\rfloor$
+We can reduce the coefficient $a_i$ into the range $0\leq a_i < l_i$ by multiplying $P_i$ by $\lfloor \frac{a_i}{l_i}\rfloor$
 and subtracting the result from $F$.
 
 <div class="theorem">
@@ -423,7 +423,7 @@ But we wanted $2^c$ factors of two, i.e. an additional one, which $(w+2)!$ obvio
 # Inverting Binary Permutation Polynomials
 Let $P\in\P$ be a permutation polynomial.
 It is not immediately clear that the inverse permutation $p^{-1}$ is even representable by a polynomial.
-To see that it is, consider the group of all permutations on $n$ Elements $S_n$.
+To see that it is, consider the group of all permutations on $n$ elements $S_n$.
 We have $p^{-1}=p^{m-1}$ where $m$ is the order of $p$,
 i.e. the order of the subgroup generated by $p$, $m=|\langle p\rangle|$.
 Note that $p^n$ in this context isn't the product $p\cdot\ldots\cdot p$,
@@ -595,7 +595,7 @@ results from [Part 2](/posts/linear-systems-mod-n).
 
 I assume the idea is well-known, so I will go over it quickly.
 We have to choose the degree of the polynomial beforehand.
-Usually (i.e. over a field) $\deg P = m-1$, as this results in a system with a unique solution,
+Usually, (i.e. over a field), $\deg P = m-1$, as this results in a system with a unique solution,
 but in our case, we know that for any polynomial there is a reduced polynomial with $\deg P < d$,
 so we can always choose $\deg P = d-1$.
 

styles/global.scss

@@ -37,8 +37,7 @@
     --selection: #303030;
 }
 
-html,
-body {
+html {
     padding: 0;
     margin: 0;
     font-family: -apple-system, BlinkMacSystemFont, Segoe UI, Roboto, Oxygen, Ubuntu,