\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{} 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{} 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