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CS 6505 Readiness Assessment - Problem 2
Author
Laura Hamilton
Last Updated
8 anni fa
License
Creative Commons CC BY 4.0
Abstract
Problem #2
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Problem #2
\documentclass[10pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb, graphicx, multicol, array}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newenvironment{problem}[2][Problem]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}\hskip \labelsep {\bfseries #2.}]}{\end{trivlist}}
\begin{document}
\title{Readiness Assessment}
\author{Laura Hamilton\\
Course number: CS 6505}
\maketitle
\begin{problem}{2(a)}
Give the negation of: For all $x \in A$, there exists $y \in B$ such that $x + y = -1$.
\end{problem}
\begin{proof}[Solution]
There exists $x \in A$ such that for all y in B, ${x + y \neq -1}$
\end{proof}
\begin{problem}{2(b)}
Give the contrapositive of: "If $y \in A$, then $y \in A \cap B$."
\end{problem}
\begin{proof}[Solution]
If $y \notin A \cap B$, then $y \notin A$.
\end{proof}
\begin{problem}{2(c)}
Is the statement in (b) true?
\end{problem}
\begin{proof}[Solution]
No, the statement in (b) is false. Let A be the set of integers, B be the set of even integers, and y be 1. Y is an integer, so it is in A. But it is odd, so it is not in B. Because the statement is false, its contrapositive must also be false.
\end{proof}
\end{document}