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| author | Holden Rohrer <hr@hrhr.dev> | 2022-02-20 14:29:31 -0500 |
|---|---|---|
| committer | Holden Rohrer <hr@hrhr.dev> | 2022-02-20 14:29:31 -0500 |
| commit | b243177be957de36ed30982d952aa622f06aa7cc (patch) | |
| tree | a49ab238a61894afa0b1118ca87ea24f42ddfd9b /gupta | |
| parent | 6d54a14e80dfdadfc4ee08989744442fbe02e6c0 (diff) | |
howard homeworks and all the mastery mailings
Diffstat (limited to 'gupta')
| -rw-r--r-- | gupta/hw5.tex | 8 |
1 files changed, 3 insertions, 5 deletions
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} |