\problemdata     
{relations-5}           
{partial orders, minimal, maximal elements}              
{Vellemen 4.4.6}              
{SO2 ps3} 		
{SO2 Radhika}        	


\begin{problem} Let $B$ be the set of integers greater than 1, and let $R$ be
the divisibility relation on the set $B$ (which means that $aRb$ iff
$ak=b$ for some natural number $k$).

\begin{problemparts}
\problempart Does the poset $(B,R)$ have any minimal elements? If so
what are they?

\solution{Yes, all prime numbers are minimal elements of the poset $(B,R)$ since no numbers divide them. Prime numbers divide all other numbers in their chains and can thus be considered the ``head'' of the chain.  }

\problempart Does poset $(B,R)$ have a smallest element? If so, what
is it?  

\solution{There is no unique minimal element or ``smallest'' element.  Intuitively, since there are separate chains each with its own minimal element (``head''), there is no ``smallest'' element. For example, consider the numbers 2 and 3.  2 divides all multiples of 2 and 3 divides all multiples of 3, thus 2 and 3 are the heads of their respective chains.  But 2 and 3 are incomparable, so there is no least element. }

\problempart Does poset $(B,R)$ have a maximal element?  \solution{No,
for any $n$, we can create a larger number, namely $2n$, so the poset
$(B,R)$ does not have a maximal element.  Note that a maximal element here is a number which is divisible by all the numbers in the chain (and can thus be considered as the ``tail'' of the chain).  Since we can always make a larger number in the chain, no such maximal element exists.  

}

\problempart Let $C$ be the natural numbers (includes zero). What are
the maximal and minimal element of the poset $(C,R)$.  

\solution{Now $C$ is the natural numbers, which include 0 and 1.  Thus for $C$, the minimal element is 1 since 1 divides all natural numbers.  The maximal element is 0 since no numbers divide 0.}

\end{problemparts}
\end{problem}
