auto-update(nvim): 2025-02-19 18:44:31

documents/by-course/math-8/pset-6/main.typ

@@ -200,3 +200,66 @@ T z$ so $T$ is transitive. Therefore $T$ is an equivalence relation on $RR$.
 - $overline(0)$ is given by all $y in RR$ where $sin(y) = 0$, so ${pi n : n in ZZ}$
 - $overline(pi/2)$ is given by all $y in RR$ where $sin(y) = sin(pi/2) = 1$, so ${pi/2 + 2pi n : n in ZZ}$
 - $overline(pi/4)$ is given by all $y in RR$ where $sin(y) = sin(pi/4)$, so ${pi/4 + 2pi n, (3pi)/4 + 2pi n : n in ZZ}$
+
+== 7
+
+If $p/q R s/t$, then $p t = q s$ and hence $q s = p t$. Therefore $s/t R p/q$,
+so $R$ is symmetric. $p/q R p/q$ iff. $p t = p t$ so $R$ is reflexive. $p/q
+R s/t$ and $s/t R a/b$ iff. $p t = q s$ and $s b = a t$ iff. $s = (p t)/q$ iff.
+$(b p t)/q = a t$ iff. $b p = a q$ iff. $p/q R a/b$. So $R$ is transitive and
+is an equivalence relation. The elements of the equivalence class of $2/3$ are
+all of the rationals that reduce to $2/3$, e.g. $4/6$, $6/9$, etc.
+
+= 3.3
+
+== 2
+
+=== a
+
+No, $cal(P)$ is not pairwise disjoint as ${2,3}sect{3,4} = {3}$.
+
+=== c
+
+Yes, each element of $cal(P)$ is pairwise disjoint and their total union is
+$A$.
+
+=== e
+
+No, $cal(P)$ is not a subset of the power set of $A$ (its elements are not
+subsets of $A$). Instead $cal(P)$ is simply equal to $RR$.
+
+== 3
+
+=== a
+
+Proposition: ${{-x,x} : x in NN union {0}}$ is a partition of $ZZ$.
+
+#proof[
+  First note that the union of each set is $ZZ$, since it contains every
+  positive integer (definition of $NN$) as well as every negative integer, and
+  0.
+  $
+    union.big_(i in NN) {-i, i} = {0} union {-1,1} union {-2,2} union dots.c = ZZ
+  $
+  Then, note that each integer $n$ has exactly one additive inverse $-n$. So
+  each set ${-n,n}$ for all $n in NN$ is pairwise disjoint. Therefore it is a
+  partition.
+]
+
+== 7
+
+=== a
+
+These are the equivalence classes of $NN$ under relation "has the same number
+of digits."
+
+=== c
+
+These are the equivalence classes of $RR$ under the relation $a R b$ iff.
+
+== 9
+
+=== d
+
+$R = {(5,1), (1,5), (2,4),(4,2),(3,3), (5,5), (1,1),(2,2), (3,3),(4,4),(5,5)}$
+