auto-update(nvim): 2025-02-19 13:40:36

documents/by-course/math-8/course-notes/main.typ

@@ -596,101 +596,10 @@ emptyset$. But $b in emptyset$ is a contradiction by its definition.
   $
 ]
 
-= Solutions to selected exercises and problems
-
-Solutions to selected problems and exercises.
-
-#linebreak()
-
-*@euclid.* We begin by considering primes $p_1, p_2, ..., p_n$. Let $P = p_1 dot p_2 dot ... dot p_n$. Then let $q = P + 1$.
-
-Then if $q$ is prime, we have an additional prime not in the original list.
-
-Otherwise, $q$ is not prime and we have a unique prime factorization of $q$.
-Without loss of generality, take one such prime to be $p_k$. $p_k$ cannot be in
-the original list $p_1, p_2, ..., p_n$.
-
-If $p_k$ were in the original list, then since $P$ is divisible by $p_k$, and $P
-+ 1$ is also divisible by $p_k$, 1 must be divisible by $p_k$ which is
-impossible. So $p_k$ is a new prime.
-
-For completeness, let's finish the proof explicitly. Start with primes $p_1$,
-$p_2$. The method above implies the existence of another prime, which we denote
-$p_3$. Repeat this to find additional primes $p_(k+1)$.
-
-*@perfectsquare.* This is a generalization of the proof $sqrt(6)$ is
-irrational. Seeking a contradiction, suppose $sqrt(a)$ is irrational.
-
-$
-  exists p,q in ZZ, sqrt(a) = p / q \
-  p^2 = a q^2
-$
-
-Then by the fundamental theorem of arithmetic,
-
-$ a = b_1 dot b_2 dot ... dot b_n $
-
-where $b_i$ is prime.
-
-$ p^2 = (Pi^n_(i=1) b_1) dot q^2 $
-
-Notice all $b_i$ are unique (again by the same theorem) and without loss of
-generality, choose a $b_k$, $1 <= k <= n$.
-
-Then $p$ has $j = 1,2,...$ $hash$ of $b_k$ in its factors. Then $p^2$ has $2j$
-$hash$ of $b_k$. Similarly, $q$ has $L = 1,2,...$ $hash$ of $b_k$, and $q^2$
-has $2L$. Then $(Pi ^n _(i=1)) q^2$ has $2L + 1$ $hash$ of $b_k$. But
-
-$ p^2 = (Pi^n_(i=1)) q^2 $
-
-and by unique factorization they must have the same $hash$ of the prime factor $b_k$, so $sqrt(a)$ is irrational.
-
-Note that if $a$ was a perfect square, TODO
-
-*@rational-between.* Effectively we are asked to show that given $x,y in QQ$,
-where $x < y$, $exists z in QQ$ such that $x < z < y$.
-
-First, let us take the difference between $x$ and $y$.
-
-$
-  exists a,b,c,d in ZZ \
-  x = a / b \
-  y = c / d \
-  y - x = (b c - a d) / (b d)
-$
-
-So we know that
-
-$
-  y = x + (b c - a d) / (b d)
-$
-
-We see that if we can find any nonzero integer less than $(b c - a d)/(b d)$,
-adding it to $x$ gives us the number between $x$ and $y$ we desire. One option is
-
-$
-  (b c - a d) / (b d + 1) < (b c - a d) / (b d) \
-  x < x + (b c - a d) / (b d + 1) < x + (b c - a d) / (b d) = y \
-$
-
-So one such $z$ is
-
-$ z = x + (b c - a d) / (b d + 1) $
-
-#remark[
-  There is a minor hole in this proof, if $b d + 1 = 0$, $z$ is undefined. We
-  can easily avoid this by replacing all instances of $b d + 1$ with, say, $b d
-  + 2$, when $b d + 1 = 0$.
-]
-
-#remark[
-  A much easier and less roundabout proof is to take
-  $ z = (x + y) / 2 $
-  Then $(x + y)/2$ is obviously rational and strictly between $x$ and $y$.
-]
-
 = Induction
 
+== Weak and strong
+
 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
@@ -839,7 +748,9 @@ We demonstrate the WOP in an alternative proof of
   ]
 ]
 
-== Relations, partitions
+= Relations, partitions
+
+== Basic notions
 
 #definition[
   A relation on a set $A$ is a set of ordered pairs $(a,b)$ where $a,b in A$. A
@@ -1007,3 +918,199 @@ nonempty subsets of $A$ whose union is $A$.
     A = union.big_(x in A) overline(x)
   $
 ]
+
+= Lecture #datetime(day: 19, year: 2025, month: 2).display()
+
+== Functions
+
+Let $A$ and $B$ be sets. A relation $R$ from $A$ to $B$ is a subset $R subset.eq A times B$.
+
+#definition[
+  A *function* $f$ from $A$ to $B$ relates each element of $"Dom"(R)$ to to exactly
+  one element of $"Rng"(R)$.
+]
+
+#fact[
+  A function from $A$ to $B$ is written
+  $
+    f : A -> B
+  $
+]
+
+#fact[
+  $
+    (x,y) in f <=> f(x) = y <=> x f y
+  $
+]
+
+We say $f : A -> A$ is a function on $A$.
+
+#example[
+  Let $A = {1,2,3}$, $B = {x,y}$. A relation from $A$ to $B$ is a subset of $A
+times B$, which has 6 elements. $A$ and $B$ admit $2^6 = 64$ distinct
+  relations.
+
+  When asking about functions instead, we note that $f$ admits either $(1, x)$
+  or $(1, y)$, but not both and so on. So there are $2^3 = 8$ distinct
+  functions.
+]
+
+#definition[
+  A function $f : A -> B$ is called a *map/mapping*. $A$ is the *domain*, $B$ is
+  the *codomain*. The *range* of $f$ $"Rng"(f) = {y in B : (exists x in A)(f(x) =
+  y)}$.
+]
+
+More terminology. Let $f : A -> B$, $(a,b) in f$, $f(x) = y$. Then
+
+- $y$ is the *value* of $f$ at $x$
+- $y$ is the *image* of $x$ under $f$
+- $x$ is the *preimage* of $y$ under $f$
+
+#fact[
+  $f : A -> B = g : C -> D$ if and only if $f subset.eq g$ and $g subset.eq f$.
+]
+
+#theorem[
+  $f : A -> B = g : C -> D$ if and only if
+
+  1. $"Dom"(f) = "Dom"(g), ($A = C$)$
+  2. $f(x) = g(x)$, $forall x in A = C$
+]
+
+#definition[
+  The *identity function* on $A$, $I_A : A -> A$ is defined by $I_A (x) = x, forall x in A$.
+]
+
+#definition[
+  If $A subset.eq B$, the *inclusion function* $i : A -> B$ is defined by
+  $
+    i(x) = x, forall x in A
+  $
+  Sometimes it is written $iota : A arrow.r.hook B$.
+]
+
+Why consider the inclusion function rather than simply the identity on $A$? Say
+we have $iota : A -> B$ and $g : B -> C$. Then we can compose these $g compose
+iota$ which is an inclusion function $iota' : A -> C$.
+
+#definition[
+  Let $R$ be an equivalence relation on $A$. Recall $A\/R$ is the set of all
+  equivalence classes. The *canonical map* is
+
+  $
+    f : A -> A\/R, f(x) = overline(x), forall x in A
+  $
+]
+
+#example[
+  Consider the canonical map $f : ZZ - >ZZ_3 = {overline(0), overline(1),
+  overline(2)}$, $f(x) = overline(x)$. Then $f(2) = overline(2)$, $f(8) =
+  overline(2)$, and so on. So $f$ is not _one-to-one_.
+]
+
+== Logistics: exam 2
+
+- One non-proof free response on set operations and families (2.2-2.3, $union$,
+  $sect$, etc)
+- 4 proof based questions
+  - Set operations/families (2.2-2.3)
+  - Principle of mathematical induction, but no complete induction or WOP (2.4)
+  - Relations. $"Dom"(R)$, $"Rng"(R)$, etc. (3.1-3.2)
+  - Equivalence relations (3.3)
+- T/F questions
+- Extra credit question
+
+= Solutions to selected exercises and problems
+
+Solutions to selected problems and exercises.
+
+#linebreak()
+
+*@euclid.* We begin by considering primes $p_1, p_2, ..., p_n$. Let $P = p_1 dot p_2 dot ... dot p_n$. Then let $q = P + 1$.
+
+Then if $q$ is prime, we have an additional prime not in the original list.
+
+Otherwise, $q$ is not prime and we have a unique prime factorization of $q$.
+Without loss of generality, take one such prime to be $p_k$. $p_k$ cannot be in
+the original list $p_1, p_2, ..., p_n$.
+
+If $p_k$ were in the original list, then since $P$ is divisible by $p_k$, and $P
++ 1$ is also divisible by $p_k$, 1 must be divisible by $p_k$ which is
+impossible. So $p_k$ is a new prime.
+
+For completeness, let's finish the proof explicitly. Start with primes $p_1$,
+$p_2$. The method above implies the existence of another prime, which we denote
+$p_3$. Repeat this to find additional primes $p_(k+1)$.
+
+*@perfectsquare.* This is a generalization of the proof $sqrt(6)$ is
+irrational. Seeking a contradiction, suppose $sqrt(a)$ is irrational.
+
+$
+  exists p,q in ZZ, sqrt(a) = p / q \
+  p^2 = a q^2
+$
+
+Then by the fundamental theorem of arithmetic,
+
+$ a = b_1 dot b_2 dot ... dot b_n $
+
+where $b_i$ is prime.
+
+$ p^2 = (Pi^n_(i=1) b_1) dot q^2 $
+
+Notice all $b_i$ are unique (again by the same theorem) and without loss of
+generality, choose a $b_k$, $1 <= k <= n$.
+
+Then $p$ has $j = 1,2,...$ $hash$ of $b_k$ in its factors. Then $p^2$ has $2j$
+$hash$ of $b_k$. Similarly, $q$ has $L = 1,2,...$ $hash$ of $b_k$, and $q^2$
+has $2L$. Then $(Pi ^n _(i=1)) q^2$ has $2L + 1$ $hash$ of $b_k$. But
+
+$ p^2 = (Pi^n_(i=1)) q^2 $
+
+and by unique factorization they must have the same $hash$ of the prime factor $b_k$, so $sqrt(a)$ is irrational.
+
+Note that if $a$ was a perfect square, TODO
+
+*@rational-between.* Effectively we are asked to show that given $x,y in QQ$,
+where $x < y$, $exists z in QQ$ such that $x < z < y$.
+
+First, let us take the difference between $x$ and $y$.
+
+$
+  exists a,b,c,d in ZZ \
+  x = a / b \
+  y = c / d \
+  y - x = (b c - a d) / (b d)
+$
+
+So we know that
+
+$
+  y = x + (b c - a d) / (b d)
+$
+
+We see that if we can find any nonzero integer less than $(b c - a d)/(b d)$,
+adding it to $x$ gives us the number between $x$ and $y$ we desire. One option is
+
+$
+  (b c - a d) / (b d + 1) < (b c - a d) / (b d) \
+  x < x + (b c - a d) / (b d + 1) < x + (b c - a d) / (b d) = y \
+$
+
+So one such $z$ is
+
+$ z = x + (b c - a d) / (b d + 1) $
+
+#remark[
+  There is a minor hole in this proof, if $b d + 1 = 0$, $z$ is undefined. We
+  can easily avoid this by replacing all instances of $b d + 1$ with, say, $b d
+  + 2$, when $b d + 1 = 0$.
+]
+
+#remark[
+  A much easier and less roundabout proof is to take
+  $ z = (x + y) / 2 $
+  Then $(x + y)/2$ is obviously rational and strictly between $x$ and $y$.
+]
+