From b243177be957de36ed30982d952aa622f06aa7cc Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Sun, 20 Feb 2022 14:29:31 -0500
Subject: howard homeworks and all the mastery mailings

---
 gupta/hw5.tex | 8 +++-----
 1 file changed, 3 insertions(+), 5 deletions(-)

(limited to 'gupta')

diff --git a/gupta/hw5.tex b/gupta/hw5.tex
index bb1bd43..8e908d7 100644
--- a/gupta/hw5.tex
+++ b/gupta/hw5.tex
@@ -164,11 +164,9 @@ $$(A\times B)\cup(C\times D) = \{(1,1),(2,2)\} \neq
 b.$
 
 \answer
-We will show this by contrapositive.
-Let $a\neq b.$ We will show that $a\nmid b$ or $b\nmid a.$
+{\bf Disproof.}
 
-WLOG, $a > b.$
-Immediately, $a\nmid b.$
+$2\mid -2$ and $-2\mid 2,$ but $-2\neq 2.$
 \endanswer
 
 \problem{30.} There exist integers $a$ and $b$ for which $42a + 7b = 1.$
@@ -192,7 +190,7 @@ B.$
 Let $A = \{1\},$ $B = \{2\},$ and $X = A\cup B.$
 Immediately, $X\subseteq A\cup B.$
 And then, $2\in X,$ but $2\not\in A,$ so $X\not\subseteq A.$
-Also, $1\in X,$ but $1\not\in B,$ so $X\not\in B.$
+Also, $1\in X,$ but $1\not\in B,$ so $X\not\subseteq B.$
 \endanswer
 
 \section{Problem not from the textbok}
-- 
cgit