1009 lines
31 KiB
Text
1009 lines
31 KiB
Text
#import "@youwen/zen:0.1.0": *
|
|
|
|
#show: zen.with(
|
|
title: "Math 8 Course Notes",
|
|
author: "Youwen Wu",
|
|
date: "Winter 2025",
|
|
subtitle: [Taught by Matt Porter],
|
|
abstract: [
|
|
In the broad light of day mathematicians check their equations and their
|
|
proofs, leaving no stone unturned in their search for rigour. But, at night,
|
|
under the full moon, they dream, they float among the stars and wonder at the
|
|
miracle of the heavens. They are inspired. Without dreams there is no art, no
|
|
mathematics, no life.
|
|
#align(end, [-- Michael Atiyah])
|
|
],
|
|
)
|
|
|
|
#outline()
|
|
|
|
= Introduction
|
|
|
|
Math 8 is an introductory course on mathematical logic and the methods of
|
|
proof. In general it will be quite trivial for anyone somewhat familiar with
|
|
competition mathematics or proofs. If you are at UCSB, I highly recommend you
|
|
take this course as soon as possible to unlock the rest of the higher
|
|
mathematics offerings here (which are much more interesting).
|
|
|
|
= Course logistics
|
|
|
|
Everything in this section is information only valid for the Winter 2025 quarter with Professor Porter.
|
|
|
|
The textbook for the course is _Smith, Eggen, Andre. A Transition to Advanced
|
|
Mathematics. 8th ed_. #smallcaps[isbn:] `978-1-285-46326-1`. Chapters 1-5 will
|
|
be covered.
|
|
|
|
Lecture meets every M-W-F from 12:00 -- 12:50 in Phelps 1444. Recitation meets
|
|
M-W from 7:00 -- 7:50 in HSSB 1236.
|
|
|
|
== Homework
|
|
|
|
Homework is from textbook and is worth 30% of the grade, due on Gradescope.
|
|
Homework is due every W at 11:59 PM. LaTeX is recommended for typesetting but
|
|
of course we will be using Typst, the superior typesetting software for
|
|
mathematics.
|
|
|
|
Section and problem numbers should be clearly labeled and problems should be
|
|
done on a single column.
|
|
|
|
The lowest homework score will be dropped.
|
|
|
|
== Exams
|
|
|
|
Each exam is 20% of the grade. The final exam will replace the lowest of the
|
|
first two exam scores if it is higher.
|
|
|
|
= Lecture #datetime(year: 2025, month: 1, day: 6).display()
|
|
|
|
|
|
== Trivial preliminaries
|
|
|
|
Definitions barely worth considering. Included purely for posterity.
|
|
|
|
#definition("Proposition")[
|
|
A proposition is a sentence which is either true or false.
|
|
]
|
|
|
|
#example("Primes")[
|
|
The numbers 5 and 7 are prime.
|
|
]
|
|
|
|
#example("Not a proposition")[
|
|
$x^2 + 6x + 8 = 0$
|
|
]
|
|
|
|
Propositions may be stated in the formalism of mathematics using connectives,
|
|
as *propositional forms*.
|
|
|
|
#definition("Propositional forms")[
|
|
Let $P$ and $Q$ be propositions. Then:
|
|
|
|
+ The conjunction of $P$ and $Q$ is written $P and Q$ ($P$ and $Q$).
|
|
+ The disjunction of $P$ and $Q$ is written $P or Q$ ($P$ or $Q$) (here "or" is the inclusive or).
|
|
+ The negation of $P$ is written $not P$.
|
|
]
|
|
|
|
#definition("Tautology")[
|
|
A propositional form for which all of its values are true. In other words, a statement which is always true.
|
|
]
|
|
|
|
#definition("Contradiction")[
|
|
A propositional form for which all of its values are false. In other words, a statement which is always false.
|
|
]
|
|
|
|
#problem[Prove that $(P or Q) or (not P and not Q)$ is a tautology][
|
|
Trivial, omitted.
|
|
]
|
|
|
|
#example[Several denials of the statement "integer $n$ is even"][
|
|
- It is not the case that integer $n$ is even.
|
|
- Integer $n$ is not even.
|
|
- $n != 2m, forall m in ZZ$
|
|
- $n = 2m + 1, exists m in ZZ$
|
|
]
|
|
|
|
DeMorgan's Laws tell us how to distribute logical connectives across
|
|
parentheses.
|
|
|
|
#fact[DeMorgan's Laws][
|
|
+ $not (P or Q) = not P and not Q$
|
|
+ $not (P and Q) = not P or not Q$
|
|
]
|
|
|
|
#proof[
|
|
Trivially, by completing a truth table.
|
|
]
|
|
|
|
Also, propositional forms obey commutative, associative, distributive laws,
|
|
which can be trivially obtained from symbolic manipulations and will not be
|
|
restated. Together with the double negation law and the _law of the excluded
|
|
middle_, these comprise the axioms of a system of propositional logic.
|
|
|
|
#fact[
|
|
We abbreviate propositional forms by eliding parentheses, according to the rules:
|
|
|
|
+ $not$ is applied to the smallest proposition following it.
|
|
+ $and$ connects the smallest propositions surrounding it.
|
|
+ $or$ connects the smallest propositions surrounding it.
|
|
]
|
|
|
|
= Notes on Logic and Proofs, 1.2
|
|
|
|
_Prototypical example for this section:_ If $sin pi = 1$, then $6$ is prime.
|
|
|
|
#definition[
|
|
For a *antedecent* $P$ and *consequent* $Q$, the *conditional sentence* $P =>
|
|
Q$ is the proposition "If $P$, then $Q$."
|
|
]
|
|
|
|
#remark[
|
|
The statement $P => Q$ states $P$ _implies_ $Q$ and is only false if $P$ is
|
|
true and $Q$ is false, since this is the only case where $P$ did not imply $Q$.
|
|
]
|
|
|
|
A conditional may be true even when the antedecent and consequent are unrelated.
|
|
|
|
= Lecture #datetime(day: 8, month: 1, year: 2025).display()
|
|
|
|
== More propositional forms
|
|
|
|
#definition[
|
|
Let $P$ and $Q$ be propositions. The *biconditional sentence*
|
|
$ P <=> Q $
|
|
is true exactly when $P$ and $Q$ are both true or both false.
|
|
]
|
|
|
|
#example[Ways of stating $P <=> Q$][
|
|
- $P$ if and only if $Q$
|
|
- $P$ iff. $Q$
|
|
- $P$ is equivalent to $Q$
|
|
]
|
|
|
|
#exercise[
|
|
Translate each statement into symbols, where $a$ is a fixed real number.
|
|
|
|
+ $a > 5$ is sufficient for $a > 3$
|
|
+ $a > 3$ is necessary for $a > 5$
|
|
+ $a > 5$ only if $a > 3$
|
|
+ $|a| = -a$ whenever $a < 0$
|
|
+ $|a| = 2$ is necessary and sufficient for $a^2 = 4$
|
|
]
|
|
|
|
#definition[
|
|
#set enum(numbering: "a.")
|
|
Let $P$ and $Q$ be propositions.
|
|
|
|
+ The converse of $P => Q$ is $Q => P$
|
|
+ The contrapositive of $P => Q$ is $not Q => not P$
|
|
]
|
|
|
|
#theorem[
|
|
Let $P$ and $Q$ be propositions. Then:
|
|
]
|
|
|
|
#example[
|
|
If $f(x)$ is differentiable at $x = a$, then $f(x)$ is continuous at $x = a$.
|
|
|
|
$ P => Q $
|
|
|
|
+ $not Q => not P$: if $f$ is not continuous at $x=a$, then $f$ is not differentiable at $x = a$.
|
|
|
|
+ $Q => P$: if $f$ is continuous at $x = a$, then $f$ is diffferentiable at $x = a$.
|
|
]
|
|
|
|
#fact[
|
|
We apply our new logical connectives in the following order: $not, and, or, => , <=>$
|
|
]
|
|
|
|
#example[
|
|
Include parentheses to clarify the expression.
|
|
|
|
$ P or Q => not R <=> S and T $
|
|
]
|
|
|
|
#theorem[
|
|
#set enum(numbering: "a.")
|
|
For propositions $P$, $Q$, and $R$, the following are equivalent:
|
|
|
|
+ $P => Q "and" not P or Q$
|
|
+ $P <=> Q "and" (P => Q) and (Q => P)$
|
|
+ $not (P => Q) "and" P and not Q$
|
|
+ $not (P and Q) "and" P => not Q$
|
|
+ $not (P and Q) "and" Q => not P$
|
|
+ $P => (Q => R) "and" (P and Q) => R$
|
|
+ $P => (Q and R) "and" (P => Q) and (P => R)$
|
|
+ $(P or Q) => R "and" (P => R) and (Q => R)$
|
|
]
|
|
|
|
== Quantified statements
|
|
|
|
#definition[
|
|
A *predicate* or *open sentence* is a sentence involving one or more variables.
|
|
]
|
|
|
|
#example[
|
|
Consider the open sentence $P(x,y): x^2 + y^2 = 25$. Write a true and a false proposition.
|
|
|
|
$
|
|
P(3,-4) &: 3^2 + (-4)^2 = 25 &"(true)" \
|
|
P(2,0) &: 2^2 + 0^2 = 25 &"(false)" \
|
|
$
|
|
]
|
|
|
|
|
|
#definition[
|
|
The *universe* is the set of all objects available for substitution into an open sentence. Denoted $U$.
|
|
]
|
|
|
|
#definition[
|
|
A *truth set* is all objects in $U$ that make an open sentence true.
|
|
]
|
|
|
|
#example[
|
|
Let the universe be the set of all real numbers for the open sentence $P(x) : x^2 + x = 6$. Find the truth set.
|
|
|
|
$
|
|
U = RR \
|
|
"truth set:" {2,-3}
|
|
$
|
|
]
|
|
|
|
#definition[
|
|
Let $P(x)$ be an open sentence with variable $x$.
|
|
|
|
The *universal quantifier* is the sentence
|
|
|
|
$ forall x in U, P(x) $
|
|
|
|
The *existential quantifier* is the sentence:
|
|
$ exists x in U, P(x) $
|
|
|
|
The *unique existence quantifier* is the sentence
|
|
$ exists! x in U, P(x) $
|
|
]
|
|
|
|
#example[
|
|
Let the universe be the set of all real numbers and consider the open sentence
|
|
$ P(x) : x^2 + 1 >= 0 $
|
|
Consider the quantified sentence $forall x in U,P(x)$. Then
|
|
$ forall x in RR, x^2 + 1 >= 0 $
|
|
is a true statement.
|
|
|
|
However, if instead $U = CC$, then the sentence is false.
|
|
]
|
|
|
|
#example[
|
|
Let the universe be the set of all real numbers and consider the open sentence
|
|
|
|
$
|
|
Q(x) "where" x in ZZ \
|
|
R(x) "is a perfect square" \
|
|
$
|
|
Consider the quantified sentence
|
|
$ exists Q(x), R(x) $
|
|
]
|
|
|
|
#example[
|
|
Let the universe be the set of all real numbers and consider the open sentence
|
|
|
|
$ P(x,y) : y = x^3 + 4 $
|
|
|
|
Consider the quantified sentence
|
|
|
|
$ forall y in U, exists! x in U, P(x,y) $
|
|
|
|
It is true because $P(x,y)$ is injective (one-to-one).
|
|
]
|
|
|
|
= Missed a bunch of lecture :(
|
|
|
|
Probably not any important content, though.
|
|
|
|
= Lecture #datetime(day: 17, month: 1, year: 2025).display()
|
|
|
|
== Proof of a biconditional statement
|
|
|
|
To prove a biconditional statement of the form $P <=> Q$, we need to show
|
|
$ P => Q and Q => P $
|
|
|
|
#theorem("Fundamental Theorem of Arithmetic")[
|
|
$forall x in ZZ, x > 1$, $x$ can be written as a product of prime factors.
|
|
Additionally, these prime factors are unique, i.e. there is only one set of
|
|
prime factors that uniquely factorizes $x$. Hence it is sometimes called the
|
|
_unique factorization theorem_ or the _prime factorization theorem_.
|
|
]
|
|
|
|
#example[
|
|
Assume $p$ is prime. Then $p | b$ iff $p | b^2$.
|
|
|
|
#proof[
|
|
First let us show $p | b => p | b^2$. We know $b$ can be written as the
|
|
product of the prime $p$ and some other integer $n$.
|
|
$ b = p n $
|
|
Then
|
|
$ b^2 = p^2 n^2 $
|
|
which implies $p$ is a factor of and divides $b^2$. So,
|
|
$ p | b => p | b^2 $
|
|
|
|
Now let us show the converse, i.e. $p | b^2 => p | b$.
|
|
|
|
By the unique factorization theorem, $b^2$ can be written as a unique
|
|
product of primes, one of which is $p$. But we also know $b^2 = b dot b$
|
|
and so at least one $b$ must have a prime factor $p$. But $b$ has unique
|
|
prime factors (again by the same theorem) so $b$ always has a prime factor
|
|
$p$. Hence,
|
|
$ p | b^2 => p | b $
|
|
]
|
|
]
|
|
|
|
== Proof by contradiction
|
|
|
|
#definition[
|
|
A proof by contradiction of the statement $P$ proceeds by assuming $not P$,
|
|
then showing that this fact leads to a contradiction.
|
|
|
|
A proof by contradiction of $P => Q$ proceeds by assuming $P and not Q$, then
|
|
showing a contradiction, forcing that $P$ indeed implies $Q$.
|
|
]
|
|
|
|
#definition[
|
|
A real number $x in RR$ is called rational iff
|
|
$ exists p,q in ZZ, x = p / q $
|
|
]
|
|
|
|
#definition[
|
|
$x$ is irrational if it is not rational.
|
|
]
|
|
|
|
#example[
|
|
Prove that if $x$ is rational and $y$ is irrational, then $2x - y$ is irrational.
|
|
|
|
#proof[
|
|
Suppose that $x$ is rational and $y$ is irrational but $2x - y$ is
|
|
rational. Then $x = a / b$ and $2x - y = c/d$ where $a,b,c,d in ZZ$ and $b
|
|
!= 0, d != 0$.
|
|
|
|
Then
|
|
$
|
|
2a / b - y &= c / d \
|
|
y &= (2a) / b - c / d \
|
|
y = (2a d - b c) / (b d) &= m / n
|
|
$
|
|
which implies $y = m/n$ and therefore we have a contradiction. So $2x - y$
|
|
is irrational.
|
|
]
|
|
]
|
|
|
|
#fact[$sqrt(2)$ is irrational.]
|
|
|
|
The proof of this fact generalizes nicely to show that the square root of any
|
|
non-perfect square is irrational.
|
|
|
|
#example[
|
|
Prove that $sqrt(6)$ is irrational.
|
|
|
|
#proof[
|
|
Seeking a contradiction, suppose $sqrt(6)$ is rational. Then $sqrt(6) =
|
|
a/b, exists a,b in ZZ$ with $b != 0$.
|
|
|
|
|
|
Then
|
|
$ a^2 = 6b^2 = 3 (2b^2) $
|
|
Since $a^2 = a dot a$, $a$ has $j = 0,1,2, ...$ factors of $3$ in its
|
|
unique prime factorization, then $a^2$ has $2j$ factors of $3$, which is to
|
|
say that $a^2$ has an even $hash$ factors of 3.
|
|
|
|
Similarly, $b^2$ has an even $hash$ of 3, say $2k$, where $k = 0, 1, 2,
|
|
...$. Then $3(2b^2)$ has $2k + 1$ $hash$ factors of 3, an odd amount.
|
|
|
|
But $a^2$ = $3(b^2)$ so they must have the same factors, a contradiction.
|
|
Therefore $sqrt(6)$ must be irrational.
|
|
]
|
|
]
|
|
|
|
#exercise[
|
|
Prove that $sqrt(2)$ is irrational using the method above.
|
|
]
|
|
|
|
#exercise[
|
|
Show that $sqrt(15)$ is irrational.
|
|
]
|
|
|
|
#problem("Euclid's Theorem")[
|
|
Show the infinitude of primes (there are an infinite amount of prime numbers).
|
|
]<euclid>
|
|
|
|
#problem[
|
|
Show that in general, given any integer $n$ that is not a perfect square,
|
|
i.e. $ exists.not a in ZZ, n = a^2 $ $sqrt(n)$ is irrational.
|
|
]<perfectsquare>
|
|
|
|
== Proofs involving quantifiers
|
|
|
|
Many of our proofs up to this point have been of the form
|
|
$ forall x in U, P(x) => Q(x) $
|
|
|
|
To prove a statement of this form,
|
|
|
|
+ Let $x in U$.
|
|
+ Assume $P(x)$.
|
|
+ Show $Q(x)$.
|
|
+ Conclude $forall x in U, P(x) => Q(x)$.
|
|
|
|
#example[
|
|
Prove that $forall x,y in ZZ$, $2x + 14y != 3$.
|
|
|
|
#proof[
|
|
Seeking a contradiction, suppose instead $2x + 14y = 3$. Then
|
|
$2x $ is even, and $14y = 2(7y)$ is even, and therefore $2x + 14y$ is even.
|
|
But $3$ is odd, so they cannot be equal.
|
|
|
|
Alternatively simply write
|
|
$ x + 7y = 1.5 $
|
|
but the ring of $ZZ$ is closed under addition so we have a contradiction.
|
|
]
|
|
]
|
|
|
|
To prove a statement of the form
|
|
|
|
$ exists x in U, P(x) $
|
|
|
|
+ Find at least one $x in U$ that makes $P(x)$ true.
|
|
+ Conclude $exists x in U, P(x)$.
|
|
|
|
#example[
|
|
Prove that there is a natural number $N$ such that for all natural numbers $n > N$,
|
|
$ 1 / n < 0.02 $
|
|
We want $1/n < 0.02$, so the idea is to play around with this statement. Taking reciprocals,
|
|
$ n > 50 $
|
|
#proof[
|
|
Let $N = 50$. Then $forall n in NN$ and $n > N$,
|
|
$ 1 / n < 1 / 50 = 0.02 $
|
|
]
|
|
]
|
|
|
|
#exercise[
|
|
Prove that between any two rational numbers $x$ and $y$ there is another
|
|
rational number $z$.
|
|
]<rational-between>
|
|
|
|
= Lecture #datetime(day: 27, month: 1, year: 2025).display()
|
|
|
|
== Basic properties of sets
|
|
|
|
#definition[
|
|
A set is a collection of elements.
|
|
]
|
|
|
|
#definition[
|
|
The cardinality of a set is the amount of elements in the set.
|
|
]
|
|
|
|
#example[
|
|
Prove $A subset.eq B$ iff. $A sect B = A$.
|
|
|
|
#proof[
|
|
Assume $A subset.eq B$. Let $x in A sect B$. Then $x in A$ and $x in B$. So
|
|
$x in A$ and $A sect B subset.eq A$. Now let $x in A$. Since $A subset.eq
|
|
B$, then $x in B$. Thus $x in A$ and $x in B$, so $x in A sect B$. Then $A
|
|
subset.eq A sect B$.
|
|
|
|
Now we show the other direction. Assume $A sect B = A$. Then $x in A sect
|
|
B$ implies $x in A$ and $x in B$. In particular $forall x in A, x in B$.
|
|
Thus $a subset.eq B$.
|
|
]
|
|
]
|
|
|
|
#theorem[
|
|
Let $U$ be the universe, and let $A$ and $B$ be subsets of $U$. Then
|
|
+ $(A^c)^c = A$
|
|
+ $A union A^c = U$
|
|
+ $A sect A^c = emptyset$
|
|
+ $A - B = A sect B^c$
|
|
+ $A subset.eq B <=> B^c subset.eq A^c$
|
|
+ $A sect B = emptyset <=> A subset.eq B^c$
|
|
+ $(A union B)^c = A^c sect B^c$
|
|
+ $(A sect B)^c = A^c union B^c$
|
|
]<basic-sets>
|
|
|
|
#example[part of @basic-sets][
|
|
Show that $(A sect B)^c = A^c union B^c$.
|
|
|
|
$
|
|
&x in (A sect B)^c \
|
|
&<=> x in.not A sect B \
|
|
&<=> x in A^c, x in B^c \
|
|
&<=> x in A^c union B^c
|
|
$
|
|
]
|
|
|
|
#definition[
|
|
The Cartesian product of sets $A$ and $B$:
|
|
|
|
$ A times B = {(a,b) : a in A "and" b in B} $
|
|
]
|
|
|
|
#fact[
|
|
If $A$ has cardinality $m$, $B$ has cardinality $n$, then $A times B$ has
|
|
cardinality $n dot m$.
|
|
]
|
|
|
|
#example[
|
|
Prove that $A times emptyset = emptyset$.
|
|
|
|
#proof[
|
|
Suppose that $A times emptyset != emptyset$. Then $exists (a,b), a in A, b in
|
|
emptyset$. But $b in emptyset$ is a contradiction by its definition.
|
|
]
|
|
]
|
|
|
|
== Index families of sets
|
|
|
|
#definition[A set of sets is called a family or collection of sets.]
|
|
|
|
#definition[
|
|
Let $Delta$ be a nonempty index set such that $forall alpha in Delta, exists
|
|
A_alpha$. The family of sets $cal(A) = {A_alpha : alpha in Delta}$ is an
|
|
index family of sets.
|
|
]
|
|
|
|
#example[
|
|
Let $A in {1,3}$ and consider $cal(P)(A)$.
|
|
]
|
|
|
|
#example[
|
|
Define $A_n = (n,n+2)$, the open interval from $n$ to $n+2$, for each $n in
|
|
NN$.
|
|
|
|
$cal(A) = {(n, N + 2) : n in NN} = {A_n : n in NN}$
|
|
|
|
Let $Delta = NN$ and then $alpha in Delta <=> n in NN$.
|
|
]
|
|
|
|
#definition[
|
|
The union over $cal(A)$ is the set
|
|
$ union.big_(A in cal(A)) A = {x : x in A, exists A in cal(A)} $
|
|
]
|
|
|
|
#definition[
|
|
The intersection over $cal(A)$ is the set
|
|
$ sect.big_(A in cal(A)) A = {x : x in A, forall A in cal(A)} $
|
|
]
|
|
|
|
#example[
|
|
Let $cal(A) = {{1}, {1,2},{2,3}}$. Then
|
|
$
|
|
&union.big_(A in cal(A)) A = {1,2,3} \
|
|
§.big_(A in cal(A)) A = emptyset
|
|
$
|
|
]
|
|
|
|
#example[
|
|
Let $A_n = [-1/n,1/n]$, the closed interval from $-1/n$ to $1/n$ for each $N in NN$. Consider the family of sets $cal(A) = {A_n : n in NN}$. Then
|
|
|
|
$
|
|
&union.big_(A in cal(A)) A = union.big_(n=1)^infinity A_n = A_1 union A_2 union A_3 union dots.c = {x : x in [-1,1]} \
|
|
§.big_(A in cal(A)) A = sect.big_(n=1)^infinity A_n = 0
|
|
$
|
|
]
|
|
|
|
#example[
|
|
Let $A_n = [0,n)$ for each $n in NN$ and let $cal(A) = {A_n : n in NN}$. Then
|
|
|
|
$
|
|
&union.big_(A in cal(A)) A = [0, infinity] \
|
|
§.big_(A in cal(A)) = [0,1)
|
|
$
|
|
]
|
|
|
|
= 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
|
|
|
|
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)
|
|
$
|
|
]
|