Moved Proof techniques
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203
Misc/Proof Techniques/parts/0 intro.tex
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Misc/Proof Techniques/parts/0 intro.tex
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\section{Introduction}
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\definition{}
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A \textit{proof} is a mathematical argument that irrefutably
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demonstrates the truth of a given proposition.
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\vspace{2mm}
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Every proof involves some sort of \textit{implication}, denoted $\implies$. \par
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The statement \say{$A$ implies $B$} (written $A \implies B$), means that $B$ is true whenever $A$ is true.
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\problem{}<trueimplies>
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Which of the following are true? \par
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\note{You don't need to provide a proof.}
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\begin{itemize}
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\item $x$ is prime $\implies$ $x$ is odd.
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\item $x$ is real $\implies$ $x$ is rational.
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\item $x$ is odd $\implies$ $x$ is prime.
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\end{itemize}
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\vfill
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\problem{}
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As you saw above, $A \implies B$ does not guarantee that $B \implies A$. \par
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Find two new statements $A$ and $B$ so that $A \implies B$ but $B \notimplies A$. \par
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\hint{\say{new} as in \say{not from \ref{trueimplies}}}
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\begin{solution}
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A fairly trite example is below. \par
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Note that \say{$X$ is a square} is a subset of the statement \say{$X$ is a rectangle.}
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\vspace{2mm}
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$X$ is a square $\implies$ $X$ is a rectangle, but $X$ is a rectangle $\notimplies$ $X$ is a square.
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\end{solution}
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\vfill
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\pagebreak
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\definition{}<iffdef>
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As we just saw, implication is one-directional. \par
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The statements $A \implies B$ and $B \implies A$ are independent of one another. \par
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\vspace{1mm}
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If both are true, we write $A \iff B$. This can be read as \say{$A$ if and only if $B$.} \par
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In text, \say{if and only if} is often abbreviated as iff. \par
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\vspace{1mm}
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Bidirectional implication is the strongest relationship we can have between two statements: \par
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If $A \iff B$, $A$ and $B$ are \textit{equivalent.} They are always either both true or both false.
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\definition{}
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The \textit{floor} of $x$ is the largest integer $a$ so that $a \leq x$. This is denoted $\lfloor x \rfloor$. \par
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The \textit{ceiling} of $x$ is the largest integer $a$ so that $a \geq x$. This is denoted $\lceil x \rceil$.
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\generic{Property:}
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If $b_1 \leq a \leq b_2$ and $b_1 = b_2$, we must have that $b_1 = a = b_2$. \par
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\vspace{1mm}
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Also, if $a \leq b$ and $a \geq b$, we must have that $a = b$. \par
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This is a trick we often use when showing that two quantities are equal.
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\problem{}
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Although $A \iff B$ looks like a single statement, we often need to prove each direction separately. \par
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Show that $x \in \mathbb{Z}$ iff $\lfloor x \rfloor = \lceil x \rceil$
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\begin{solution}
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\textbf{Forwards:} $x \in \mathbb{Z} ~\implies~ \lfloor x \rfloor = \lceil x \rceil$ \par
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If $x \in \mathbb{Z}$, by definition we have that $\lfloor x \rfloor = x$ and $\lceil x \rceil = x$ \par
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So, $\lfloor x \rfloor = \lceil x \rceil$
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\vspace{2mm}
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\textbf{Backwards:} $x \in \mathbb{Z} ~\rimplies~ \lfloor x \rfloor = \lceil x \rceil$ \par
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Assume that $\lfloor x \rfloor = \lceil x \rceil$, and show that $x$ is an integer. \par
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Note that $\lfloor x \rfloor \leq x \leq \lceil x \rceil$ (by definition) \par
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Since $\lfloor x \rfloor = \lceil x \rceil$, we must have that $\lfloor x \rfloor = x = \lceil x \rceil$. \par
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$\lfloor x \rfloor$ is an integer, so $x$ must be an integer.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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We don't always need to prove each direction of an iff statement separately. \par
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\begin{itemize}[itemsep = 1mm]
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\item Convince yourself that we can \say{chain} iffs together: \par
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If we show that $A \iff B \iff C \iff D$, do we know that $A \iff D$?
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\item Does this still work if $A \iff B \implies C \iff D$?
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\item Show that $x^2 - 6x - 6 = 3 \iff x = 3$ by building a chain of iffs. \par
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\hint{You remember how to factor quadratics, right?}
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\end{itemize}
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\begin{solution}
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Does this still work if $A \iff B \implies C \iff D$? \par
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Of course not. $D \notimplies A$ since $C \notimplies B$.
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We can only conclude that $A \implies D$.
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\linehack{}
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$x^2 - 6x - 6 = 3$ \par
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$\iff x^2 - 6x - 9 = 0$ \par
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$\iff (x-3)^2 = 0$ \par
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$\iff x-3 = 0$ \par
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$\iff x = 0$
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Note that this is a well-defined argument. Every step is an iff statement we can rigorously justify.
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We're not hand-wavily \say{rearranging} one equation into another,
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we're building a chain of implications that eventually bring us to our result.
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This is the logic behind most algebraic proofs.
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\end{solution}
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\vfill
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\problem{}
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Another trick you may find useful is the \say{implication cycle.} \par
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Convince yourself that if $A \implies B \implies C \implies D \implies A$, \par
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we can conclude that $A$, $B$, $C$, and $D$ are equivalent. \par
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\note{$A \iff B$ means that $A$ and $B$ are equivalent. See \ref{iffdef}.}
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\vfill
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\problem{Bonus}
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Use an implication cycle to show that the following definitions of a \textit{squarefree integer} are equivalent.
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%\hint{Show that $A \implies B \implies C \implies D \implies A$}
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\begin{enumerate}
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\item $n^2$ does not divide $q$ for any $n \in \mathbb{Z}^+$, $n \neq 1$
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\item $p^2$ does not divide $q$ for any prime $p$
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\item $q$ is a product of distinct primes
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\item $q ~|~ n^k \implies q ~|~ n$ for all $n, k \in \mathbb{Z}^+$
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\end{enumerate}
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\begin{solution}
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Assume $q$ has a square factor, so that $q = an^2$ for some $a, n \in \mathbb{Z}^+$ \par
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By D, we know that $q ~|~ (an)^2 \implies q ~|~ an$ \par
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But $q ~|~ an \implies an^2 ~|~ an$ \par
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$\implies n = 1$
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\vspace{2mm}
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So, $q$ cannot have a square factor that isn`t 1.
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\end{solution}
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\vfill
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\pagebreak
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Often enough, proving a statement is simply a matter of \say{definition chasing,}
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where we expand the symbols used in the statement we're proving, and then do a bit of
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rearranging to arrive at the result we want.
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\definition{}
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Let $n, x \in \mathbb{Z}$. \par
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We say \say{$n$ divides $x$} if $x = kn$ for some $k \in \mathbb{Z}$
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\definition{}
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Let $a, b \in \mathbb{Z}$ and $n \in \mathbb{Z}^+$. \par
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We say \say{$a$ is congruent to $b$ modulo $n$} (and write $a \equiv_{n} b$) if $n$ divides $a - b$. \par
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\definition{}
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Let $a, b \in \mathbb{Z}$. We define $a ~\%~ b$ as the remainder of $a \div b$.
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\problem{}
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Let $a, b, n$ be positive integers. \par
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Show that $a + b ~\equiv_n (a ~\%~ n) + (b ~\%~ n)$
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\begin{solution}
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$a + b ~\equiv_n (a ~\%~ n) + (b ~\%~ n)$ \par
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\vspace{2mm}
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...can be rewritten as... \par
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$n$ divides $a + b - (a ~\%~ n) - (b ~\%~ n)$ \par
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\vspace{2mm}
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...which can be rearranged to... \par
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$n$ divides $(a - (a ~\%~ n)) + (b - (b ~\%~ n))$
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\vspace{2mm}
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...which is clearly true, if you think about the meaning of \say{$n$ divides $x$} and \say{$a ~\%~ b$}.
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\end{solution}
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\vfill
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\pagebreak
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63
Misc/Proof Techniques/parts/1 contradiction.tex
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Misc/Proof Techniques/parts/1 contradiction.tex
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\section{Proofs by Contradiction}
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\definition{}
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A very common proof technique is \textit{proof by contradiction}.
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It works as follows:
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\vspace{2mm}
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Say we want to prove a statement $P$. Assume that $P$ is false, and show that this implies a false statement.
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In other words, we show that $P$ can't \textit{not} be true. \par
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If it's false, we either get a known impossibility ($1 = 2$, pigs fly, et cetera), \par
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or we find that (not $P$) $\implies$ (not (not $P$)), which is a contradiction in itself.
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\problem{}
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Show that the set of integers has no maximum using a proof by contradiction.
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\begin{solution}
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Assume there is a maximal integer $x$. \par
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$x + 1$ is also an integer. \par
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$x + 1$ is larger than $x$, which contradicts our original assumption!
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\vspace{2mm}
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This is a \textit{proof by infinite descent}, a special type of proof by contradiction.\par
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Such proofs have the following structure:
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\begin{itemize}
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\item Assume there is a smallest (or largest) object with property $X$.
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\item Show that we have an even smaller object that has the same property $X$.
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\end{itemize}
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\end{solution}
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\vfill
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\definition{}
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We say a number $x \in \mathbb{R}$ is \textit{rational} if we can write $x$ as $\nicefrac{p}{q}$, \par
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where $p, q$ are integers with no common factors.
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\problem{}
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Show that $\sqrt{2}$ is irrational. \par
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\hint{Start by chasing definitions. \say{Not irrational} $=$ \say{rational.}}
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\begin{solution}
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Suppose $\sqrt{2}$ is rational. Then, there exist $p, q$ so that $\sqrt{2} = \frac{p}{q}$.
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\vspace{2mm}
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Multiply by $q$ and square to find that $2q^2 = p^2$, which implies that $p^2$ is even. \par
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This then implies that $p$ is even, \par
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which implies that $p^2$ is divisible by 4, \par
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which implies that $q^2$ is divisible by 2, \par
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and thus we see that $q$ is also even.
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\vspace{2mm}
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$p$ and $q$ are both even, so they cannot be coprime. \par
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Therefore, we cannot write $\sqrt{2}$ as $\nicefrac{p}{q}$ for comprime $p, q$, \par
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and $\sqrt{2}$ is therefore irrational.
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\end{solution}
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\vfill
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\pagebreak
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243
Misc/Proof Techniques/parts/2 induction.tex
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243
Misc/Proof Techniques/parts/2 induction.tex
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\section{Proofs by Induction}
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\definition{}
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The last proof technique we'll discuss in this handout is \textit{induction.} \par
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This is particularly useful when we have a \say{countable} variable, usually an integer. \par
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Let's say we're proving a statement $A$ for all positive integers $n$. \par
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We'll write the special case \say{$A$ holds for $n$} as $A_n$.
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\vspace{2mm}
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A proof by induction consists of two parts: a \textit{base case} and a \textit{inductive step}. \par
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The base case is usually fairly simple: we show that our statement holds for $n = 0$. \par
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In other words, the base case shows that $A_0$ is true.
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The inductive step is a bit more confusing: we show that if our statement holds for
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$n$, it must hold for $n = 1$. In other words, we show that $A_n \implies A_{n + 1}$.
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\vspace{2mm}
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In this way, we build an infinite implication chain: \par
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The base case proves that $A_0$. By the inductive step,
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$A_0 \implies A_1$, $A_1 \implies A_2$, and so on. \par
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We can thus conclude that $A_n$ is true for all $n \in \{0, 1, 2, 3, ...\}$
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\problem{}
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Proof by induction will make a bit more sense with an example. \par
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Read and understand the following proof.
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\begin{examplesolution}
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Show that $1 + 2 + ... + n = \frac{n(n+1)}{2}$
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\linehack{}
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\textbf{Base case:} $n = 1$ \par
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Substitute $n = 1$ into the hypothesis:
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$1 \qe \frac{1(1 + 1)}{2}$ \par
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but we have that
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$\frac{1(1 + 1)}{2} = \frac{2}{2} = 1$,
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so this is of course true.
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\vspace{2mm}
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\textbf{Inductive step:} \par
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Now, we assume our hypothesis is true for $n$, \par
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and show it is true for $n + 1$.
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\vspace{2mm}
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Write the hypothesis for $n + 1$:
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$$
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1 + 2 + ... + n + (n + 1) \qe \frac{(n+1)(n+2)}{2}
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$$
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We know that $1 + 2 + ... + n = \frac{n(n+1)}{2}$, so:
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$$
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1 + 2 + ... + n + (n + 1) = \frac{n(n+1)}{2} + (n+1)
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$$
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now we can complete the proof with some algebra:
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$$
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\frac{n(n+1)}{2} + (n+1) = \frac{n(n+1) + 2(n+1)}{2} = \frac{(n + 2)(n+1)}{2}
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$$
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So, we've shown that the $n^\text{th}$ case implies the $(n+1)^\text{th}$ case.\par
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We're therefore done, the hypothesis is true for $n \in \{1, 2, 3, ...\}$
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\end{examplesolution}
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\vfill
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\pagebreak
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\problem{}
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Why do we need a base case when constructing a proof by induction? \par
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\hint{Try to prove that $n = n + 1$ for all $n \in \mathbb{Z}^+$}
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\begin{solution}
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Consider the following example: \par
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Say we want to prove that $n = n + 1$ for all $n \in \mathbb{Z}^+$.
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\vspace{2mm}
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\textbf{Inductive step:}
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Assume our hypothesis is true for $n$ (that is $n = n + 1$). Is this true for $n + 1$? \par
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Adding 1 to both sides, we get $n+1=n+1+1$, which means that $(n+1)=(n+1)+1$,
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which is the statement we wanted to prove. We've thus completed the inductive step!
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\vspace{2mm}
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Our problem is as follows: we've shown that $A_1 \implies A_2 \implies ...$,
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but we have no reason to believe that $A_1$ is true. If it was, our hypothesis
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would be correct---but since it isn't, this is not a complete proof.
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\end{solution}
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\vfill
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\problem{}
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Show that $1^2 + 2^2 + 3^3 + ... + n^2 = \frac{1}{6}(n)(n+1)(2n+1)$.
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\begin{solution}
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\textbf{Base case:}\par
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$1^2 = \frac{1}{6}(1)(1+1)(2 + 1) = 1$, which is true.
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\vspace{2mm}
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\textbf{Induction:}\par
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Assume $1^2 + ... + n^2$ satisfies the equation above. \par
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$$
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1^2 + 2^2 + ... + n^2 + (n+1)^2 =
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\frac{(n)(n+1)(2n+1)}{6} + (n + 1)^2
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$$
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which is equal to
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$$
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\frac{(n)(n+1)(2n+1) + 6(n+1)^2}{6}
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$$
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now expand and factor to get
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$$
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\frac{(n+1)(n+2)(2(n+1)+1)}{6}
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$$
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\end{solution}
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\vfill
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\problem{}
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Show that $2^n - 1$ is divisible by 3 for all odd $n$ \par
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\hint{If $n$ is odd, the next odd number is $n + 2$.}
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\begin{solution}
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\textbf{Base case:} \par
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$2^2 - 1 = 3$, which is divisible by 3..
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\vspace{2mm}
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\textbf{Induction:}\par
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Assume $2^n - 1$ is divisible by 3. \par
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$2^{(n+2)} - 1 = 4(2^n) - 1 = 4(2^n - 1) + 3$ \par
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By our induction hypothesis, $(2^n - 1)$ has a factor of 3. \par
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Therefore, $4(2^n - 1) + 3$ must also have a factor of 3.
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\end{solution}
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\vfill
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\pagebreak
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\definition{}
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As you may already know, \say{n choose k} is defined as follows: \par
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$$
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\binom{n}{k} = \frac{n!}{k!(n-k)!}
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$$
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This counts the number of ways to choose $k$ things from a set of $n$,
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disregarding the order of the chosen items.
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\theorem{Pascal's Identity}
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The binomial coefficient defined above satisfies the following equality:
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$$
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\binom{n+1}{k} = \binom{n}{k-1} \times \binom{n}{k}
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$$
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\problem{}<binomsum>
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Using induction, show that
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$$
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\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n} = 2^n.
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$$
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||||
|
||||
|
||||
\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
|
||||
|
||||
\makeatletter
|
||||
\@makeORMCbox{tmpbox}{Alternative Proof}{ogrape!10!white}{ogrape}
|
||||
\makeatother
|
||||
|
||||
\begin{tmpbox}
|
||||
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{tmpbox}
|
||||
|
||||
|
||||
\pagebreak
|
||||
Reference in New Issue
Block a user