diff --git a/documents/by-course/math-4b/course-notes/main.typ b/documents/by-course/math-4b/course-notes/main.typ
index e2c5949..992818f 100644
--- a/documents/by-course/math-4b/course-notes/main.typ
+++ b/documents/by-course/math-4b/course-notes/main.typ
@@ -835,3 +835,65 @@ To find gamma, we can simply use inverse trig functions.
 
   Which makes $cos(5t + pi - arctan(1/2))$ negative.
 ]
+
+= Lecture #datetime(day: 18, year: 2025, month: 2).display()
+
+== Linear systems of differential equations
+
+Consider a matrix $A$ and solution vector $x$.
+$
+  x'(t) = A(t) x(t)
+$
+
+#fact[Superposition principle for linear system][
+  If $x^((1)) (t)$ and $x^((2)) (t)$ are solutions, then
+  $
+    c_1 x^((1)) (t) + c_2 x^((2)) (t)
+  $
+  are also solutions.
+]
+
+=== Homogenous case
+
+This is when $g(t) = 0$. I want to solve for $arrow(x)' = A arrow(x)$, where
+$
+  arrow(x) = vec(mu_1 (t), mu_2 (t), dots.v, mu_n (t))
+$
+Similar to the scalar case, we look for solutions in terms of exponential
+functions. Guessing
+
+$
+  arrow(x) (t) = e^(r t) arrow(v) \
+  arrow(x)'(t) = r e^(r t) arrow(v) = A e^(r t) arrow(v) \
+  r arrow(v) = A arrow(v) \
+  (A - r I) arrow(v) = 0, arrow(v) != 0
+$
+
+If there are non-trivial solutions then $A$ must have determinant 0.
+
+$
+  det (A - r I) = 0
+$
+
+This is the eigenvalue equation for $A$!
+
+#example[
+  Find a fundamental set of solutions. FIrst find the eigenvalues and
+  corresponding eigenvectors of $A$.
+
+  $
+    x'(t) = A x(t), A = mat(2,1;1,2)
+  $
+
+  Find eigenvalues of $A$.
+
+  $
+    A - lambda I = mat(2 - lambda, 1; 1, 2- lambda) \
+    |a - lambda I| = (2 - lambda)^2 - 1 = 0 \
+    2 - lambda = plus.minus 1 \
+    lambda = cases(1,3)
+  $
+
+  Now we want the eigenvectors for our eigenvalues. Find an eigenvector
+  corresponding to $lambda_1 = 1$.
+]
diff --git a/documents/by-course/math-6a/course-notes/main.typ b/documents/by-course/math-6a/course-notes/main.typ
index b13702b..9d06b77 100644
--- a/documents/by-course/math-6a/course-notes/main.typ
+++ b/documents/by-course/math-6a/course-notes/main.typ
@@ -201,3 +201,75 @@ $
   (diff F) / (diff x) = diff / (diff z) (3x^2 + 5 y z + z^3) = diff / (diff z) z^3 = 0 + (5 (diff y) / (diff x) z + 5y) + 3z^2 \
   (diff y) / (diff z) = - (5y + 3z^2) / (5z)
 $
+
+= Lecture #datetime(day: 18, year: 2025, month: 2).display()
+
+== Critical points
+
+When optimizing in 2D, the strategy depends on whether we're
+
+- optimizing for all of $RR^2$ (or a region in $RR^2$)
+- optimizing on a constraint (like a curve through $RR^n$)
+
+We find critical points where the tangent plane is "flat": $m_x = 0$ and $m_y =
+0$.
+
+We classify critical points using the determinant of the gradient.
+
+$
+  D = f_(x x) f_(y y) = f_(x y)^2
+$
+
+- if $D >0$ and $f_(x x) (x_0, y_0) > 0$, then $f(x_0, y_0)$ is a relative minimum.
+- if $D > 0$ and $f_(x x) (x_0, y_0) < 0$, $f(x_0, y_0)$ is a relative maximum.
+- if $D < 0$ then $f(x_0, y_0)$ is neither and we call it a saddle point.
+- if $D = 0$ then we don't know
+
+== Lagrange multipliers
+
+Optimizing constrained curves. Idea: navigate along the curve and look for where
+the directional derivative is zero.
+
+#example[
+  Find the highest and lowest points on $f(x,y) = 81x^2 + y^2$ with the
+  constraint $x^2 + y^2 = 0$.
+
+  For notational purposes, we'll call $g(x,y) = 4x^2 + y^2$ and keep in mind
+  we're looking for $g(x,y) = 9$.
+
+  1. Find the gradients of $f$ and $g$.
+    $
+      vec(f_x,f_g) &= vec(162x,2y) \
+      vec(g_x,g_y) &= vec(8x,2y)
+    $
+  2. We want to find the points where the gradients "align", i.e. we want these
+  vectors to be parallel, that is:
+  $
+    arrow(F) = lambda arrow(G) \
+    162x = 8x dot lambda \
+    2y = 2y dot lambda
+  $
+  Remember to keep the constraint!
+  $
+    4x^2 + y^2 = 9
+  $
+  Breaking it down into cases,
+  $
+    162 = 8lambda => lambda = 81 / 4 "for" x != 0
+  $
+  which implies
+  $
+    2y (lambda - 1) 0 \
+    y = 0
+  $
+  since $lambda - 1$ is nonzero. So if $x$ is nonzero, $y$ must be zero.
+
+  $
+    4x^2 = 9 \
+    x = plus.minus 3 / 2
+  $
+  Now consider when $x = 0$. Then $y = plus.minus 3$. So our critical points
+  are $(plus.minus 3/2, plus.minus 3)$. Finally, just plug in these 4
+  critical points into $f$ and find the biggest/smallest.
+
+]
diff --git a/documents/by-course/math-8/course-notes/main.typ b/documents/by-course/math-8/course-notes/main.typ
index a6e2894..f67816e 100644
--- a/documents/by-course/math-8/course-notes/main.typ
+++ b/documents/by-course/math-8/course-notes/main.typ
@@ -688,3 +688,322 @@ $ z = x + (b c - a d) / (b d + 1) $
   $ z = (x + y) / 2 $
   Then $(x + y)/2$ is obviously rational and strictly between $x$ and $y$.
 ]
+
+= Induction
+
+Induction is a way to prove that a statement is true for all $NN$. Formally, it
+is introduced like this: say you have a set $S subset.eq NN$ such that $1 in
+S$. Then, if $n in S => n + 1 in S$, $S = NN$. Usually we plug $n$ into a
+proposition $P$ to make it more useful for proving statements. If $(forall n in
+S)(P(n))$, and we can show $S = NN$ by induction, then $P(n)$ is true for all
+$n in NN$.
+
+Induction consists of two main parts. First, we have to show that $1 in S$.
+This is the _base case_. Then, we assume the _inductive hypothesis_, $n in S$.
+If we can prove, using the inductive hypothesis, that $n in S$ implies that $n
++ 1$ in $S$, then we'll have sufficient justification for $S = N$. When proving
+a statement in practice, if $S$ is the set of all inputs to $P(n)$ that make it
+true, we show that $P(1)$ is true ($1 in S$) and then show that $P(n) =>
+P(n+1)$, which implies $(forall n in NN) P(n)$.
+
+There are two equivalent formulations of induction, the principle of
+mathematical induction (PMI, also called _weak induction_), and the principle
+of complete induction (PCI, also called _strong induction_). They are
+equivalent, but in some cases one is much more straightforward to show than the
+other.
+
+The previous strategy we discussed was _weak induction_, named such because we
+need only show $P(1) and P(n) => P(n+1)$. _Strong induction_ requires us to
+show a much stronger result, but also allows us to make a stronger assumption.
+
+#definition[Strong induction][
+  If ${1,2,3,...,n-1} subset.eq S => n in S$, then $S = NN$.
+]
+
+Essentially in strong induction we assume that every natural up to $n$ is in
+$S$, and if we can show that this implies $n$ is also in $S$, we have $S = NN$.
+Note that in this case we do not have to show $1 in S$, but this is implied.
+When proving a proposition $P$, we have to prove $P(1) and P(2) and ... and
+P(n-1) => P(n)$, but we can assume the strong inductive hypothesis $P(1) and
+P(2) and ... and P(n-1)$.
+
+#example[Game theory of Nim][
+  In the game of _Nim_, two players take turns taking coins from two piles of
+  $n$ coins each. If there are $m$ coins in a pile, a player may choose to take
+  $1,2,3,...,m$ coins, *except* $m - 1$ coins. That is, a player may take any
+  nonzero amount, but may not leave exactly one coin left in the pile. The
+  player to take the last coin in play wins.
+
+  Example game: if there are 6 coins on the left and 8 coins on the right,
+  player 1 goes first and takes 4 coins from the left. Player 2 takes the
+  remaining 2 coins on the left. Player 1 takes all 8 coins from the right and
+  wins.
+
+  Prove that Player 2 has a winning strategy in all games of _Nim_.
+
+  This is actually a somewhat unexpected way in which strong induction can be
+  very useful. Let $S$ be a set of the natural numbers where Player 2 has a
+  winning strategy when there are $n in S$ coins left in both piles.
+
+  Suppose $m$ is a natural number and ${1,2,3,...,m-1} subset.eq S$. This is
+  our inductive hypothesis, we assume that Player 2 has a winning strategy for
+  when the game starts with $1,2,3,...,m-1$ coins in each pile. Now we seek to
+  show that this implies that Player 2 has a winning strategy when there are
+  $m$ coins in the pile. For the special case when $m = 1$, Player 1 takes 1
+  coin from either pile and Player 2 takes both, winning the game. Otherwise,
+  for $m > 1$, if Player 1 takes $m$ coins from a pile, Player 2 takes all the
+  coins in the other pile, winning. If Player 1 takes $j$ coins from either
+  pile where $1 <= j < m - 1$, leaving $m - j > 1$ coins in the pile, Player 2
+  can take $j$ coins from the other pile, leaving both piles with $m - j$
+  coins. This is essentially a new game of Nim with $m - j$ coins in each pile.
+  Then $m - j in {1,2,3,...,m-1}$ since $j <= 1 < m-1$, so by our inductive
+  hypothesis Player 2 has a winning strategy from this point on. Therefore
+  Player 2 has a winning strategy when $m$ coins are in each pile.
+  Since ${1,2,3,...,m-1} subset.eq S => m in S$, by strong induction, $S = NN$
+  and Player 2 has a winning strategy for all possible games of _Nim_.
+]
+
+#example[A hint for the fundamental theorem of arithmetic][
+  Prove that all natural numbers can be expressed as the product of primes.
+
+  #proof[
+    Let $m$ be a natural number. Note that $m = 2$ is prime so its the product
+    of itself and 1. Now assume that all natural numbers $n$ such that $1 < n <
+    m$ can be written as the product of primes. Then $m$ is either prime, in
+    which case its prime factors are $1 dot m$, or $(exists s,t in
+    {1,2,...,m-1})(m = s t)$. Then by our inductive hypothesis both $s$ and $t$
+    can be written as the product of primes, so $m$ is also the product of
+    primes. By strong induction, we conclude all natural numbers can be written
+    as the product of primes.
+  ]
+]<fundamental-thm-arithmetic-hint>
+
+== Well ordering principle
+
+This is another property that characterizes $NN$.
+
+#fact[Well ordering principle][
+  Every nonempty subset of $NN$ has a smallest element.
+]
+
+We demonstrate the WOP in an alternative proof of
+@fundamental-thm-arithmetic-hint.
+
+#example[
+  All natural numbers can be expressed as the product of primes.
+
+  #proof[
+    Suppose there exists some non-empty set $S subset.eq NN$ that contains all
+    of the natural numbers which cannot be expressed as a product of primes.
+    By the WOP, there is a least number in this set, $T in S$. $T$ cannot be
+    prime because it would be the product of $1 dot T$, so it must be
+    composite. Then $T = s t$ for some natural numbers $s,t$, which are less
+    than $T$. But $T$ is the least element of $S$ so neither $s$ nor $t$ are in
+    $S$, and therefore they both have prime factorizations. But this implies
+    that $T$ also has a prime factorization, a contradiction. Therefore no such
+    set $S$ can exist, and all natural numbers have a prime factorization.
+  ]
+]
+
+#theorem[The division algorithm][
+  For all integers $a$ and $b$, with $a != 0$, there exist unique integers $q$
+  and $r$ such that $b = a q + r$ and $0 <= r < |a|$.
+
+  #proof[
+    Let us consider the case where $a > 0$. Then we have a set
+
+    $
+      S = {b - a k, k in ZZ, b - a k >= 0}
+    $
+
+    It follows that we should exclude 0 from $S$ because otherwise if $0 in S$,
+    $b - a k = 0$ for some integer $k$, and $a | b$, so $b = a dot b/a$.
+
+    In our restricted $S$, there is a least element, call it $r$. Then for some
+    integer $q$, $a q + r = b$. We know that $r > 0$ so let's show $r < |a|$.
+    Since we're considering only $a > 0$ for simplicity, we just show $r < a$.
+    If $r >= a$, then either $0 in S$ when $r = a$ because $a q + a = b$
+    implies that $a | b$, or, when $r > a$, there exists another element in
+    $S$, $b - a (q + 1)$. This is a contradiction since we assumed $b - a q$
+    was the smallest element in $S$ and clearly $b - a q - a$ is smaller.
+    Therefore $r < |a|$.
+
+    Now we show the uniqueness of $q, r$. Suppose there were other integers
+    satisfying the properties of $q$ and $r$, $q'$ and $r'$. Then we have
+
+    $
+      r' >= r
+    $
+    without loss of generality (otherwise just relabel). Then $a q + r = a q' +
+    r'$, so $a(q - q') = r' - r$. Then $a$ divides $r' - r$.
+  ]
+]
+
+== Relations, partitions
+
+#definition[
+  A relation on a set $A$ is a set of ordered pairs $(a,b)$ where $a,b in A$. A
+  relation from a set $A$ to a set $B$ is set of ordered pairs $(a,b)$ where $a
+  in A$ and $b in B$.
+]
+
+#abuse[
+  If $R$ is a relation from $A$ to $B$, for $a in A$ and $b in B$ where $(a,b)
+  in R$, we can abbreviate it $a R b$.
+]
+
+#definition[
+  For any set $A$, the identity relation on $A$ is the set
+  $
+    I_A = {(a,a) : a in A}
+  $
+]
+
+#definition[
+  The domain of the relation $R$ from $A$ to $B$ is the set
+  $
+    "Dom"(R) = {x in A : (exists y in B)(x R y)} \
+  $
+  The range is
+  $
+    "Rng"(R) = {y in B : (exists x in A)(x R y)}
+  $
+]
+
+So the domain of $R$ is the set of all first coordinates and the range is the
+set of all second coordinates.
+
+#theorem[
+  1. $"Dom"(R^(-1)) = "Rng"(R)$
+  2. $"Rng"(R^(-1)) = "Dom"(R)$
+]
+
+#definition[
+  Let $R$ be a relation from $A$ to $B$, let $S$ be a relation from $B$ to $C$.
+  The composite of $R$ and $S$ is
+
+  $
+    S compose R = {(a,c) : (exists b in B)((a,b) in R and (b,c) in S)}
+  $
+]
+
+== Equivalence relations
+
+Let $A$ be a set and $R$ a relation on $A$.
+
+We say $R$ is reflexive on $A$ if $forall x in A, x R x$. $R$ is symmetric if
+$forall x,y in A$, if $x R y$, then $y R x$. $R$ is transitive if $forall x,y,z
+in A$, if $x R y$ and $y R z$, then $x R z$.
+
+#theorem[
+  Let $A$ be a set. For the power set $cal(P)(A)$, the relation "is a subset of"
+  is reflexive on $cal(P)(A)$ and transitive but not symmetric.
+]
+
+#definition[
+  A relation $R$ on a set $A$ is an equivalence relation on $A$ if $R$ is
+  reflexive on $A$, symmetric, and transitive.
+]
+
+We make an equivalence relation when we think of objects as being related by
+having the same property.
+
+An equivalence relation on a set divides the set into subsets of related
+elements.
+
+#definition[
+  Let $R$ be an equivalence class on a set $A$. For $x in A$, the equivalence
+  class of modulo $R$ (or $x mod R$) is the set
+
+  $
+    overline(x) = {y in A : x R y}
+  $
+
+  Each element of $overline(x)$ is a *representative* of the class. The set
+  $
+    A \/ R = {overline(x) : x in A}
+  $
+  of all equivalence classes is called $A "modulo" R$.
+]
+
+#theorem[
+  Let $R$ be an equivalence relation on a nonempty set $A$. For all $x$ and $y$
+  in $A$,
+
+  1. $x in overline(x)$ and $overline(x) subset.eq A$
+  2. $x R y$ if and only if $overline(x) = overline(y)$
+  3. $x cancel(R) y$ if and only if $overline(x) sect overline(y) = emptyset$
+]
+
+== Congruence relations
+
+Now we define congruence relations on $ZZ$ and show that congruence mod $m$ is
+an equivalence relation. We describe its equivalence classes.
+
+#definition[
+  Let $m$ be a fixed positive integer. For $x,y in ZZ$, we say $x$ is congruent
+  to $y$ modulo $m$ and write $x = y (mod m)$ if $m$ divides $(x-y)$. The number
+  $m$ is called the modulus of the congruence.
+]
+
+#theorem[
+  For every fixed positive integer $m$, congruence modulo $m$ is an equivalence
+  relation on $ZZ$.
+]
+
+#proof[
+  1. Reflexivity. Let $x$ be an integer. We show $x = x (mod m)$. $m dot 0 = 0 = x - x$, $m | x - x$.
+  2. For symmetry, suppose $x = y (mod m)$. Then $m | x - y$. Thus $(exists k)(x - y = k m)$. But $-(x-y) = -(k m)$, or $y - x = (-k) m$. So $m | y - x$ so $y = x (mod m)$
+  3. Suppose $x = y (mod m)$ and $y = z (mod m)$. Thus $m | x - y$ and $m | y - z$. Thus $m | x - y + y - z <=> m | x - z$ so $x = z (mod m)$. So congruence modulo $m$ is transitive.
+]
+
+#definition[
+  The set of equivalence classes for the relation congruence modulo $m$ is
+  denoted $ZZ_m$.
+]
+
+#theorem[
+  Let $m$ be a fixed positive integer. Then
+
+  1. for integers $x$ and $y$, $x = y (mod m)$ if and only if the remainder when $x$ is divided by $m$ equals the remainder when $y$ is divided by $m$
+  2. $ZZ_m$ consists of $m$ distinct equivalence classes.
+]
+
+The equivalence classes of $ZZ_m$ $overline(0), overline(1), ...,
+overline(m-1)$ which are exactly all of the possible remainders when integers
+are divided by $m$. For this reason the elements of $ZZ_m$ are sometimes called
+the residue classes modulo $m$.
+
+== Partitions
+
+#definition[
+  Let $A$ be a nonempty set. $cal(P)$ is a partition of $A$ if $cal(P)$ is a set of subsets of $A$ such that
+  1. if $X in cal(P)$, then $X != emptyset$
+  2. if $X in cal(P)$ and $Y in cal(P)$, then $X = Y$ or $X sect Y = emptyset$
+  3. $union.big_(X in cal(P)) X = A$
+]
+
+In other words a partition of a set $A$ is a pairwise disjoint collection of
+nonempty subsets of $A$ whose union is $A$.
+
+#theorem[
+  If $R$ is an equivalence relation on a nonempty set $A$, then $A\/R$ is a
+  partition of $A$.
+]
+
+#proof[
+  Every equivalence class $overline(x)$ is a subset of $A$ and is nonempty
+  because it contains $x$. Any two equivalence classes are either equal or
+  disjoint. Also,
+  $
+    union.big_(x in A) overline(x) subset.eq A
+  $
+  because each $overline(x) subset.eq A$. To prove $A subset.eq union.big _(x in A) overline(x)$, suppose $y in A$. Because $y in overline(y)$, then
+  $
+    y = union.big_(x in A) overline(x)
+  $
+  Thus,
+  $
+    A = union.big_(x in A) overline(x)
+  $
+]