diff --git a/documents/by-course/math-8/course-notes/main.typ b/documents/by-course/math-8/course-notes/main.typ index f67816e..3767471 100644 --- a/documents/by-course/math-8/course-notes/main.typ +++ b/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 $y$. +] +