diff --git a/documents/by-course/math-8/pset-3/main.typ b/documents/by-course/math-8/pset-3/main.typ
index c810b99..bb4dc51 100644
--- a/documents/by-course/math-8/pset-3/main.typ
+++ b/documents/by-course/math-8/pset-3/main.typ
@@ -276,3 +276,350 @@ The right to left direction is very easy in this case. By directly plugging in $
 ]
 
 *12a.*
+
+F. This is a proof that $not A => B$ is a contradiction, which is not
+sufficient to show $A => B$.
+
+= 1.6
+
+*1b.*
+
+#proof[
+  Consider $m = 1$, $n = -1$. Then $15(1) + 12(-1) = 3$.
+]
+
+*1c.*
+
+Rewrite $2m + 4n = 7$ as $2(m + 2n) = 7$. Notice that $2(m + 2n) = 2k, exists k
+in nonzero$ and therefore is an even number for any $m$, $n$. But 7 is not
+even. So no choices of $m$ and $n$ can create a number equal to 7. So no $m$ or
+$n$ exists.
+
+*4a.*
+
+#proof[
+  Consider $x = 2$. Then $x^2 + x + 41 = 46$, which is not prime. So it's false.
+]
+
+*4b.*
+
+#proof[
+  For any $x$, choose $y = -1 dot x$. Then $x + (-x) = x(1 - 1) = 0x = 0$. This
+  is also trivially true when considering our usual field of reals because the
+  existence of $y$ such that $x + y = 0$ is an axiom.
+]
+
+*4c.*
+
+Consider $x = 2$ and $y = 1$. Then $1 > 2$ is false.
+
+*4d.*
+
+Consider $a = 10$, $b = 2$, $c = 5$. Then $a | b c$ because $10 | 2 dot 5$ but
+10 does not divide either 2 or 5.
+
+*6a.*
+
+#proof[
+  Rewriting the inequality,
+  $
+    1 / n <= 1\
+    1 <= n
+  $
+  This is true for all $n in NN$ by the definition of $NN$.
+]
+
+*6b.*
+
+#proof[
+  Rewriting,
+  $
+    1 / n < 0.13 \
+    1 < 0.13n \
+    n > 1 / 0.13
+  $
+  So such an $M$ is $1/0.13$, because all naturals greater than $1/0.13$ are
+  greater than $1/0.13$.
+]
+
+*6e.*
+
+#proof[
+  $
+    forall n in NN, n + 1 > n
+  $
+]
+
+*6f.*
+
+$
+  forall k in ZZ, m = -k, k + m = 0 \
+  forall n in NN, 0 <= 0 < n
+$
+
+*6i.*
+
+First consider $K = 10$, and therefore
+
+$
+  forall r > 10, r^2 > 100
+$
+Then note that our inequality is equivalent to $r^2 > 100$, which is true for
+all $r$.
+$
+  1 / (r^2) < 0.01 \
+  r^2 > 100 \
+$
+So a $K$ exists, namely $K = 10$.
+
+*6k.*
+
+Consider $M = 51$. Then $forall r > 51, 2r > 102$ and $2r > 100$ is always
+true.
+
+$
+  1 / (2r) < 0.01 \
+  2r > 100
+$
+
+So such an $M$ exists, namely, $M = 51$.
+
+= 2.1
+
+*4a.*
+
+False.
+
+*4b.*
+
+True.
+
+*4c.*
+
+False.
+
+*4d.*
+
+True.
+
+*4e.*
+
+True.
+
+*4f.*
+
+False.
+
+*4g.*
+
+True.
+
+*4h.*
+
+False.
+
+*4i.*
+
+False.
+
+*4j*.
+
+True.
+
+*5a.*
+
+True.
+
+*5b.*
+
+True.
+
+*5c.*
+
+True.
+
+*5d.*
+
+True.
+
+*5e.*
+
+False.
+
+*5f.*
+
+True.
+
+*5g.*
+
+True.
+
+*5h.*
+
+True.
+
+*5i.*
+
+False.
+
+*5j.*
+
+True.
+
+*5k.*
+
+True.
+
+*5L.*
+
+True.
+
+*6a.*
+
+$
+  A = {1,2}, B = {1,2,3}, C = {1,2,4}
+$
+
+*6b.*
+
+$
+  A = B = C
+$
+
+A particular example might be $A = {1}, B = {1}, C = {1}$.
+
+*6c.*
+
+$
+  A = {1}, B = {1,2}, C = {3}
+$
+
+*6d.*
+
+$
+  A = {1,2}, B = {1,2,3}, C = {5,6,7}
+$
+
+*8.*
+
+Theorem 2.1.1: $forall A, B, C, A subset.eq B and B subset.eq C => A subset.eq C$.
+
+#proof[
+  $
+    forall a in A, a in B \
+    forall a in A, exists b in B, a = b \
+    forall b in B, b in C \
+    forall a in A, a in C => A subset.eq C
+  $
+]
+
+*11a.*
+
+#proof[
+  $
+    {x in RR : 3 / 4 x - 2 > 10} \
+    = {x in RR : x > 16 } \
+    = (16, infinity)
+  $
+]
+
+*11b.*
+
+#proof[
+  $
+    {x in RR : |x - 4| = 2|x| - 2} \
+    = {x in RR : |x - 4| = 2|x| - 2} \
+  $
+
+  Now we consider various cases. Consider $x >= 4$. Then
+  $
+    {x in RR : x - 4 = 2x - 2} \
+    = {x in RR : x - 4 = 2x - 2} \
+    = {x in RR : x = -2} \
+  $
+  But in this case $x >= 4$. Since there is no $x >= 4$ such that $x = -2$,
+  this is actually $emptyset$.
+
+  Now consider $0 <= x < 4$.
+  $
+    {x in RR : 4 - x = 2x - 2} \
+    = {x in RR : x = 2} \
+    = {2}
+  $
+
+
+  Now consider $x < 0$. Then
+  $
+    {x in RR : 4 - x = -2x - 2} \
+    {x in RR : x = -6}
+  $
+  So the set contains $6$ when $x < 0$.
+
+  And since these inequalities span all $x in RR$, the only members of
+  the set are ${2, -6}$.
+]
+
+*11c.*
+
+#proof[
+  $
+    {x in RR : 2|x+3| + x = 0}
+  $
+  For $x >= -3$, we have
+  $
+    {x in RR : 2(x + 3) + x = 0} \
+    = {x in RR : x = -2} \
+    = {-2}
+  $
+  For $x < -3$, we have
+  $
+    {x in RR : 2(3 - x) + x = 0} \
+    = {x in RR : x = 6 } \
+    = {6}
+  $
+  And since these inequalities partition $RR$ the original set is ${-2, 6}$.
+]
+
+*11d.*
+
+$
+  {x in RR : |x| = 6 - |2x|}
+$
+
+Consider $x >= 0$. Then
+$
+  {x in RR : x = 6 - 2x} \
+  = {x in RR : x = 2}
+$
+Consider $x < 0$. Then
+$
+  {x in RR : -x = 6 + 2x} \
+  = {x in RR : x = -2}
+$
+So
+$
+  {x in RR : |x| = 6 - |2x|} = {-2, 2}
+$
+
+*11e.*
+
+$
+  {x in RR : |x + 3| <= -4x - 2}
+$
+
+Consider $x >= -3$. Then
+$
+  {x in RR : x <= -1} \
+  = (-infinity, -1]
+$
+
+Now consider $x < -3$ Then
+$
+  {x in RR : 3 - x <= -4x - 2} \
+  = {x in RR : x <= -5 / 3} \
+  = (-infinity, -5 / 3]
+$
+But $(-infinity, -5/3) union (-infinity, 1] = (-infinity, 1]$ so that is our
+answer.
+
+*11f.*