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		<id>http://helmut.knaust.info/mediawiki/index.php?title=24178:_HW5&amp;diff=2132&amp;oldid=prev</id>
		<title>HelmutKnaust: Created page with &quot;For a natural number $n$, let ${\cal D}_n$ denote the set of the divisors of $n$. For example, ${\cal D}_{42}=\{1,2,3,6,7,14,21,42\}$ and ${\cal D}_{12}=\{1,2,3,4,6,12\}$. For...&quot;</title>
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				<updated>2017-04-03T19:10:25Z</updated>
		
		<summary type="html">&lt;p&gt;Created page with &amp;quot;For a natural number $n$, let ${\cal D}_n$ denote the set of the divisors of $n$. For example, ${\cal D}_{42}=\{1,2,3,6,7,14,21,42\}$ and ${\cal D}_{12}=\{1,2,3,4,6,12\}$. For...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;For a natural number $n$, let ${\cal D}_n$ denote the set of the divisors of $n$. For example, ${\cal D}_{42}=\{1,2,3,6,7,14,21,42\}$ and ${\cal D}_{12}=\{1,2,3,4,6,12\}$. For $m,n\in\mathbb{N}$ let $m\sqcap n$ denote the greatest common divisor of $n$ and $m$, and $m\sqcup n$ their least common multiple. For instance $6\sqcap 4=2$ and $6\sqcup 4=12$. It turns out that ${\cal D}_{42}$ with these two operations $\sqcap$ and $\sqcup$ forms a Boolean Algebra, while ${\cal D}_{12}$ does '''not'''.&lt;br /&gt;
&lt;br /&gt;
'''Problem 21.'''&lt;br /&gt;
# Verify Boolean Algebra Law 7 for ${\cal D}_{42}$. &lt;br /&gt;
# Show that ${\cal D}_{12}$ does not form a Boolean Algebra. &lt;br /&gt;
# Conjecture for which values of $n$ the set ${\cal D}_{n}$ forms a Boolean Algebra.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
'''Problem 22.'''  Let $R$ and $S$ be two relations on $\mathbb{R}$: $R=\{(x,y)\in\mathbb{R}\times\mathbb{R}\ |\ y&amp;lt;x^2\}$ and $S=\{(x,y)\in\mathbb{R}\times\mathbb{R}\ |\ y=2x-1\}$. Find $S\circ R$ and $R\circ S$.&lt;br /&gt;
&lt;br /&gt;
'''Problem 23.''' Let $R$ be a relation from $A$ to $B$. For an element $b\in B$ define the set $R_b:=\{a\in A\ |\ (a,b)\in R\}$. Show \[\bigcup_{b\in B} R_b=\mbox{Dom}\, R.\]&lt;br /&gt;
&lt;br /&gt;
'''Problem 24.''' Define a relation $S$ on $\mathbb{R}$ as follows: $a\,S\,b$ if $a-b$ is irrational. Prove or disprove: $S$ is (a) reflexive, (b) symmetric, (c) transitive. &lt;br /&gt;
&lt;br /&gt;
'''Problem 25.''' Let $R$ be a relation on $A$.&lt;br /&gt;
# Show that $R$ is reflexive iff $I_A\subseteq R$. Here $I_A$ denotes the identity relation on $A$: $I_A=\{(a,a) \ |\ a\in A\}$. &lt;br /&gt;
# Show that $R$ is symmetric iff $R=R^{-1}$.&lt;br /&gt;
# Show that $R$ is transitive iff $R\circ R\subseteq R$.&lt;/div&gt;</summary>
		<author><name>HelmutKnaust</name></author>	</entry>

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