From 7c2398bcf5c5933cf841c454f7e95d2ecc64de6d Mon Sep 17 00:00:00 2001 From: Mark Date: Fri, 6 Oct 2023 14:12:42 -0700 Subject: [PATCH 1/2] Added a problem --- Advanced/Intro to Proofs/main.tex | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/Advanced/Intro to Proofs/main.tex b/Advanced/Intro to Proofs/main.tex index ba22a3e..8e8779a 100755 --- a/Advanced/Intro to Proofs/main.tex +++ b/Advanced/Intro to Proofs/main.tex @@ -101,4 +101,26 @@ \end{itemize} + \vfill + \pagebreak + + + \problem{} + Let $f$ be a function from a set $X$ to a set $Y$. We say $f$ is \textit{injective} if $f(x) = f(y) \implies x = y$. \par + We say $f$ is \textit{surjective} if for all $y \in Y$ there exists an $x \in X$ so that $f(x) = y$. \par + Let $A, B, C$ be sets, and let $f: A \to B$, $g: B \to C$ be functions. Let $h = g \circ f$. + + \vspace{2mm} + \begin{itemize} + \item Show that if $h$ is injective, $f$ must be injective and $g$ may not be injective. + \item Show that if $h$ is surjective, $g$ must be surjective and $f$ may not be surjective. + \end{itemize} + + \vfill + \pagebreak + + \problem{} + + + \end{document} \ No newline at end of file From f8d66bdc226fcb18e37eb02bad64283de1e1b9a3 Mon Sep 17 00:00:00 2001 From: Mark Date: Fri, 6 Oct 2023 14:17:34 -0700 Subject: [PATCH 2/2] Added a problem --- Advanced/Intro to Proofs/main.tex | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/Advanced/Intro to Proofs/main.tex b/Advanced/Intro to Proofs/main.tex index c6d05ea..126d59b 100755 --- a/Advanced/Intro to Proofs/main.tex +++ b/Advanced/Intro to Proofs/main.tex @@ -16,6 +16,10 @@ \maketitle + + + + \problem{} We say an integer $x$ is \textit{even} if $x = 2k$ for some $k \in \mathbb{Z}$. We say $x$ is \textit{odd} if $x = 2k + 1$ for some $k \in \mathbb{Z}$. \par @@ -53,6 +57,9 @@ + + + \problem{} Let $r \in \mathbb{R}$. We say $r$ is \textit{rational} if there exist $p, q \in \mathbb{Z}, q \neq 0$ so that $r = \frac{a}{b}$ @@ -82,6 +89,22 @@ + + + + + \problem{} + Show that there are infinitely may primes. \par + You may use the fact that every integer has a prime factorization. + + + \vfill + \pagebreak + + + + + \problem{} For a set $X$, define its \textit{diagonal} as $\text{D}(X) = \{ (x, x) \in X \times X ~\bigl|~ x \in X \}$.