Mark/handouts
1
0
Fork 0
Code Pull Requests 1 Actions Packages Releases Activity

Intermediate handouts

Browse Source 
missing files
This commit is contained in:
Mark
2025-01-22 12:29:01 -08:00
parent dd4abdbab0
commit 80eef6b7dc
40 changed files with 30039 additions and 0 deletions
Download Patch File Download Diff File Expand all files Collapse all files

30
src/Intermediate/Proof Techniques/main.tex Executable file
View File

@ -0,0 +1,30 @@
% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{units}
\uptitlel{Intermediate}
\uptitler{\smallurl{}}
\title{Proof Techniques}
\subtitle{Prepared by Mark on \today{}}
% Default \implies is ugly
\let\implies\Rightarrow
\let\rimplies\Leftarrow
\let\iff\Leftrightarrow
\let\notimplies\nRightarrow
\begin{document}
\maketitle
\input{parts/0 intro}
\input{parts/1 contradiction}
\input{parts/2 induction}
\end{document}

6
src/Intermediate/Proof Techniques/meta.toml Normal file
View File

@ -0,0 +1,6 @@
[metadata]
title = "Proof Techniques"
[publish]
handout = false
solutions = true

203
src/Intermediate/Proof Techniques/parts/0 intro.tex Normal file
View File

@ -0,0 +1,203 @@
\section{Introduction}
\definition{}
A \textit{proof} is a mathematical argument that irrefutably
demonstrates the truth of a given proposition.
\vspace{2mm}
Every proof involves some sort of \textit{implication}, denoted $\implies$. \par
The statement \say{$A$ implies $B$} (written $A \implies B$), means that $B$ is true whenever $A$ is true.
\problem{}<trueimplies>
Which of the following are true? \par
\note{You don't need to provide a proof.}
\begin{itemize}
\item $x$ is prime $\implies$ $x$ is odd.
\item $x$ is real $\implies$ $x$ is rational.
\item $x$ is odd $\implies$ $x$ is prime.
\end{itemize}
\vfill
\problem{}
As you saw above, $A \implies B$ does not guarantee that $B \implies A$. \par
Find two new statements $A$ and $B$ so that $A \implies B$ but $B \notimplies A$. \par
\hint{\say{new} as in \say{not from \ref{trueimplies}}}
\begin{solution}
A fairly trite example is below. \par
Note that \say{$X$ is a square} is a subset of the statement \say{$X$ is a rectangle.}
\vspace{2mm}
$X$ is a square $\implies$ $X$ is a rectangle, but $X$ is a rectangle $\notimplies$ $X$ is a square.
\end{solution}
\vfill
\pagebreak
\definition{}<iffdef>
As we just saw, implication is one-directional. \par
The statements $A \implies B$ and $B \implies A$ are independent of one another. \par
\vspace{1mm}
If both are true, we write $A \iff B$. This can be read as \say{$A$ if and only if $B$.} \par
In text, \say{if and only if} is often abbreviated as iff. \par
\vspace{1mm}
Bidirectional implication is the strongest relationship we can have between two statements: \par
If $A \iff B$, $A$ and $B$ are \textit{equivalent.} They are always either both true or both false.
\definition{}
The \textit{floor} of $x$ is the largest integer $a$ so that $a \leq x$. This is denoted $\lfloor x \rfloor$. \par
The \textit{ceiling} of $x$ is the largest integer $a$ so that $a \geq x$. This is denoted $\lceil x \rceil$.
\generic{Property:}
If $b_1 \leq a \leq b_2$ and $b_1 = b_2$, we must have that $b_1 = a = b_2$. \par
\vspace{1mm}
Also, if $a \leq b$ and $a \geq b$, we must have that $a = b$. \par
This is a trick we often use when showing that two quantities are equal.
\problem{}
Although $A \iff B$ looks like a single statement, we often need to prove each direction separately. \par
Show that $x \in \mathbb{Z}$ iff $\lfloor x \rfloor = \lceil x \rceil$
\begin{solution}
\textbf{Forwards:} $x \in \mathbb{Z} ~\implies~ \lfloor x \rfloor = \lceil x \rceil$ \par
If $x \in \mathbb{Z}$, by definition we have that $\lfloor x \rfloor = x$ and $\lceil x \rceil = x$ \par
So, $\lfloor x \rfloor = \lceil x \rceil$
\vspace{2mm}
\textbf{Backwards:} $x \in \mathbb{Z} ~\rimplies~ \lfloor x \rfloor = \lceil x \rceil$ \par
Assume that $\lfloor x \rfloor = \lceil x \rceil$, and show that $x$ is an integer. \par
Note that $\lfloor x \rfloor \leq x \leq \lceil x \rceil$ (by definition) \par
Since $\lfloor x \rfloor = \lceil x \rceil$, we must have that $\lfloor x \rfloor = x = \lceil x \rceil$. \par
$\lfloor x \rfloor$ is an integer, so $x$ must be an integer.
\end{solution}
\vfill
\pagebreak
\problem{}
We don't always need to prove each direction of an iff statement separately. \par
\begin{itemize}[itemsep = 1mm]
\item Convince yourself that we can \say{chain} iffs together: \par
If we show that $A \iff B \iff C \iff D$, do we know that $A \iff D$?
\item Does this still work if $A \iff B \implies C \iff D$?
\item Show that $x^2 - 6x - 6 = 3 \iff x = 3$ by building a chain of iffs. \par
\hint{You remember how to factor quadratics, right?}
\end{itemize}
\begin{solution}
Does this still work if $A \iff B \implies C \iff D$? \par
Of course not. $D \notimplies A$ since $C \notimplies B$.
We can only conclude that $A \implies D$.
\linehack{}
$x^2 - 6x - 6 = 3$ \par
$\iff x^2 - 6x - 9 = 0$ \par
$\iff (x-3)^2 = 0$ \par
$\iff x-3 = 0$ \par
$\iff x = 0$
Note that this is a well-defined argument. Every step is an iff statement we can rigorously justify.
We're not hand-wavily \say{rearranging} one equation into another,
we're building a chain of implications that eventually bring us to our result.
This is the logic behind most algebraic proofs.
\end{solution}
\vfill
\problem{}
Another trick you may find useful is the \say{implication cycle.} \par
Convince yourself that if $A \implies B \implies C \implies D \implies A$, \par
we can conclude that $A$, $B$, $C$, and $D$ are equivalent. \par
\note{$A \iff B$ means that $A$ and $B$ are equivalent. See \ref{iffdef}.}
\vfill
\problem{Bonus}
Use an implication cycle to show that the following definitions of a \textit{squarefree integer} are equivalent.
%\hint{Show that $A \implies B \implies C \implies D \implies A$}
\begin{enumerate}
\item $n^2$ does not divide $q$ for any $n \in \mathbb{Z}^+$, $n \neq 1$
\item $p^2$ does not divide $q$ for any prime $p$
\item $q$ is a product of distinct primes
\item $q ~|~ n^k \implies q ~|~ n$ for all $n, k \in \mathbb{Z}^+$
\end{enumerate}
\begin{solution}
Assume $q$ has a square factor, so that $q = an^2$ for some $a, n \in \mathbb{Z}^+$ \par
By D, we know that $q ~|~ (an)^2 \implies q ~|~ an$ \par
But $q ~|~ an \implies an^2 ~|~ an$ \par
$\implies n = 1$
\vspace{2mm}
So, $q$ cannot have a square factor that isn`t 1.
\end{solution}
\vfill
\pagebreak
Often enough, proving a statement is simply a matter of \say{definition chasing,}
where we expand the symbols used in the statement we're proving, and then do a bit of
rearranging to arrive at the result we want.
\definition{}
Let $n, x \in \mathbb{Z}$. \par
We say \say{$n$ divides $x$} if $x = kn$ for some $k \in \mathbb{Z}$
\definition{}
Let $a, b \in \mathbb{Z}$ and $n \in \mathbb{Z}^+$. \par
We say \say{$a$ is congruent to $b$ modulo $n$} (and write $a \equiv_{n} b$) if $n$ divides $a - b$. \par
\definition{}
Let $a, b \in \mathbb{Z}$. We define $a ~\%~ b$ as the remainder of $a \div b$.
\problem{}
Let $a, b, n$ be positive integers. \par
Show that $a + b ~\equiv_n (a ~\%~ n) + (b ~\%~ n)$
\begin{solution}
$a + b ~\equiv_n (a ~\%~ n) + (b ~\%~ n)$ \par
\vspace{2mm}
...can be rewritten as... \par
$n$ divides $a + b - (a ~\%~ n) - (b ~\%~ n)$ \par
\vspace{2mm}
...which can be rearranged to... \par
$n$ divides $(a - (a ~\%~ n)) + (b - (b ~\%~ n))$
\vspace{2mm}
...which is clearly true, if you think about the meaning of \say{$n$ divides $x$} and \say{$a ~\%~ b$}.
\end{solution}
\vfill
\pagebreak

63
src/Intermediate/Proof Techniques/parts/1 contradiction.tex Normal file
View File

@ -0,0 +1,63 @@
\section{Proofs by Contradiction}
\definition{}
A very common proof technique is \textit{proof by contradiction}.
It works as follows:
\vspace{2mm}
Say we want to prove a statement $P$. Assume that $P$ is false, and show that this implies a false statement.
In other words, we show that $P$ can't \textit{not} be true. \par
If it's false, we either get a known impossibility ($1 = 2$, pigs fly, et cetera), \par
or we find that (not $P$) $\implies$ (not (not $P$)), which is a contradiction in itself.
\problem{}
Show that the set of integers has no maximum using a proof by contradiction.
\begin{solution}
Assume there is a maximal integer $x$. \par
$x + 1$ is also an integer. \par
$x + 1$ is larger than $x$, which contradicts our original assumption!
\vspace{2mm}
This is a \textit{proof by infinite descent}, a special type of proof by contradiction.\par
Such proofs have the following structure:
\begin{itemize}
\item Assume there is a smallest (or largest) object with property $X$.
\item Show that we have an even smaller object that has the same property $X$.
\end{itemize}
\end{solution}
\vfill
\definition{}
We say a number $x \in \mathbb{R}$ is \textit{rational} if we can write $x$ as $\nicefrac{p}{q}$, \par
where $p, q$ are integers with no common factors.
\problem{}
Show that $\sqrt{2}$ is irrational. \par
\hint{Start by chasing definitions. \say{Not irrational} $=$ \say{rational.}}
\begin{solution}
Suppose $\sqrt{2}$ is rational. Then, there exist $p, q$ so that $\sqrt{2} = \frac{p}{q}$.
\vspace{2mm}
Multiply by $q$ and square to find that $2q^2 = p^2$, which implies that $p^2$ is even. \par
This then implies that $p$ is even, \par
which implies that $p^2$ is divisible by 4, \par
which implies that $q^2$ is divisible by 2, \par
and thus we see that $q$ is also even.
\vspace{2mm}
$p$ and $q$ are both even, so they cannot be coprime. \par
Therefore, we cannot write $\sqrt{2}$ as $\nicefrac{p}{q}$ for comprime $p, q$, \par
and $\sqrt{2}$ is therefore irrational.
\end{solution}
\vfill
\pagebreak

240
src/Intermediate/Proof Techniques/parts/2 induction.tex Normal file
View File

@ -0,0 +1,240 @@
\section{Proofs by Induction}
\definition{}
The last proof technique we'll discuss in this handout is \textit{induction.} \par
This is particularly useful when we have a \say{countable} variable, usually an integer. \par
Let's say we're proving a statement $A$ for all positive integers $n$. \par
We'll write the special case \say{$A$ holds for $n$} as $A_n$.
\vspace{2mm}
A proof by induction consists of two parts: a \textit{base case} and a \textit{inductive step}. \par
The base case is usually fairly simple: we show that our statement holds for $n = 0$. \par
In other words, the base case shows that $A_0$ is true.
The inductive step is a bit more confusing: we show that if our statement holds for
$n$, it must hold for $n = 1$. In other words, we show that $A_n \implies A_{n + 1}$.
\vspace{2mm}
In this way, we build an infinite implication chain: \par
The base case proves that $A_0$. By the inductive step,
$A_0 \implies A_1$, $A_1 \implies A_2$, and so on. \par
We can thus conclude that $A_n$ is true for all $n \in \{0, 1, 2, 3, ...\}$
\problem{}
Proof by induction will make a bit more sense with an example. \par
Read and understand the following proof.
\begin{examplesolution}
Show that $1 + 2 + ... + n = \frac{n(n+1)}{2}$
\linehack{}
\textbf{Base case:} $n = 1$ \par
Substitute $n = 1$ into the hypothesis:
$1 \qe \frac{1(1 + 1)}{2}$ \par
but we have that
$\frac{1(1 + 1)}{2} = \frac{2}{2} = 1$,
so this is of course true.
\vspace{2mm}
\textbf{Inductive step:} \par
Now, we assume our hypothesis is true for $n$, \par
and show it is true for $n + 1$.
\vspace{2mm}
Write the hypothesis for $n + 1$:
$$
1 + 2 + ... + n + (n + 1) \qe \frac{(n+1)(n+2)}{2}
$$
We know that $1 + 2 + ... + n = \frac{n(n+1)}{2}$, so:
$$
1 + 2 + ... + n + (n + 1) = \frac{n(n+1)}{2} + (n+1)
$$
now we can complete the proof with some algebra:
$$
\frac{n(n+1)}{2} + (n+1) = \frac{n(n+1) + 2(n+1)}{2} = \frac{(n + 2)(n+1)}{2}
$$
So, we've shown that the $n^\text{th}$ case implies the $(n+1)^\text{th}$ case.\par
We're therefore done, the hypothesis is true for $n \in \{1, 2, 3, ...\}$
\end{examplesolution}
\vfill
\pagebreak
\problem{}
Why do we need a base case when constructing a proof by induction? \par
\hint{Try to prove that $n = n + 1$ for all $n \in \mathbb{Z}^+$}
\begin{solution}
Consider the following example: \par
Say we want to prove that $n = n + 1$ for all $n \in \mathbb{Z}^+$.
\vspace{2mm}
\textbf{Inductive step:}
Assume our hypothesis is true for $n$ (that is $n = n + 1$). Is this true for $n + 1$? \par
Adding 1 to both sides, we get $n+1=n+1+1$, which means that $(n+1)=(n+1)+1$,
which is the statement we wanted to prove. We've thus completed the inductive step!
\vspace{2mm}
Our problem is as follows: we've shown that $A_1 \implies A_2 \implies ...$,
but we have no reason to believe that $A_1$ is true. If it was, our hypothesis
would be correct---but since it isn't, this is not a complete proof.
\end{solution}
\vfill
\problem{}
Show that $1^2 + 2^2 + 3^3 + ... + n^2 = \frac{1}{6}(n)(n+1)(2n+1)$.
\begin{solution}
\textbf{Base case:}\par
$1^2 = \frac{1}{6}(1)(1+1)(2 + 1) = 1$, which is true.
\vspace{2mm}
\textbf{Induction:}\par
Assume $1^2 + ... + n^2$ satisfies the equation above. \par
$$
1^2 + 2^2 + ... + n^2 + (n+1)^2 =
\frac{(n)(n+1)(2n+1)}{6} + (n + 1)^2
$$
which is equal to
$$
\frac{(n)(n+1)(2n+1) + 6(n+1)^2}{6}
$$
now expand and factor to get
$$
\frac{(n+1)(n+2)(2(n+1)+1)}{6}
$$
\end{solution}
\vfill
\problem{}
Show that $2^n - 1$ is divisible by 3 for all odd $n$ \par
\hint{If $n$ is odd, the next odd number is $n + 2$.}
\begin{solution}
\textbf{Base case:} \par
$2^2 - 1 = 3$, which is divisible by 3..
\vspace{2mm}
\textbf{Induction:}\par
Assume $2^n - 1$ is divisible by 3. \par
$2^{(n+2)} - 1 = 4(2^n) - 1 = 4(2^n - 1) + 3$ \par
By our induction hypothesis, $(2^n - 1)$ has a factor of 3. \par
Therefore, $4(2^n - 1) + 3$ must also have a factor of 3.
\end{solution}
\vfill
\pagebreak
\definition{}
As you may already know, \say{n choose k} is defined as follows: \par
$$
\binom{n}{k} = \frac{n!}{k!(n-k)!}
$$
This counts the number of ways to choose $k$ things from a set of $n$,
disregarding the order of the chosen items.
\theorem{Pascal's Identity}
The binomial coefficient defined above satisfies the following equality:
$$
\binom{n+1}{k} = \binom{n}{k-1} \times \binom{n}{k}
$$
\problem{}<binomsum>
Using induction, show that
$$
\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n} = 2^n.
$$
\vfill
Note that although induction is a powerful proof technique, it usually leads to uninteresting results. \par
If we prove a statement using induction, we conclude that it is true---but we get very little insight on
\textit{why} it is true.
\vspace{2mm}
Alternative proofs are take a bit more work than inductive proofs, but they are much more valuable. \par
For example, see the proof of the statement in \ref{binomsum} on the next page.
\pagebreak
\begin{ORMCbox}{Alternative Proof}{ogrape!10!white}{ogrape}
Consider the following problem: \par
How many ways are there to write a number $x$ as an ordered sum of positive integers? \par
\note{
An \say{ordered sum} means that the order of numbers in the sum matters. \\
For example, if $x = 5$, we will consider $4 + 1$ and $1 + 4$ as distinct sums.
}
\vspace{2mm}
First, we'll think of $x$ as an array of $1$s which we want to group
into positive integers. If $x = 3$, We have $x = 1~1~1$, which we can
group as $1+1+1$,~ $(1+1) + 1$,~ $1 + (1+1)$,~ and $(1+1+1)$.
\linehack{}
\textbf{Solution 1:}\par
One way to solve this is to use the usual \say{stars and bars} method, \par
where we count the number of ways we can place $n$ \say{bars} between
$x$ \say{stars}. Each bar corresponds to a \say{$+$} in the array of ones. \par
\note[Note]{Convince yourself that there are $\binom{x-1}{n}$ ways to place $n$ bars
between $x$ objects.}
If we add the number of ways to write $x$ as a sum of $n \in \{1, 2, ..., x\}$ integers, we get:
$$
\sum_{n = 1}^{x-1} \binom{x-1}{n} = \binom{x-1}{0} + \binom{n}{1} + ... + \binom{x-1}{x-1}
$$
\linehack{}
\textbf{Solution 2:}\par
We could also observe that there are $x - 1$ places to put a \say{bar} in
the array of ones. This corresponds to $x - 1$ binary positions, and thus
$2^{x-1}$ ways to separate our array of $1$s with bars.
\linehack{}
\textbf{Conclusion:}\par
We've found that the number of ways to split $x$ can be written as either
$\sum_{n = 1}^{x-1} \binom{x-1}{n}$ or $2^{x-1}$,
and therefore $\sum_{n = 1}^{x-1} \binom{x-1}{n} = 2^{x-1}$.
\end{ORMCbox}
\pagebreak