For any real number a and positive integer k, define
${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$
What is
${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$?
$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 199$
We expand both the numerator and the denominator.
\begin{align*} \binom{-\frac{1}{2}}{100}\div\binom{\frac{1}{2}}{100} &= \frac{   \dfrac{     (-\frac{1}{2})     (-\frac{1}{2} - 1)     (-\frac{1}{2} - 2)     \cdots     (-\frac{1}{2} - (100 - 1))   }{\cancel{(100)(99)\cdots(1)}} }{   \dfrac{     (\frac{1}{2})     (\frac{1}{2} - 1)     (\frac{1}{2} - 2)     \cdots     (\frac{1}{2} - (100 - 1))   }{\cancel{(100)(99)\cdots(1)}} } \\ &= \frac{     (-\frac{1}{2})     (-\frac{1}{2} - 1)     (-\frac{1}{2} - 2)     \cdots     (-\frac{1}{2} - 99) }{     (\frac{1}{2})     (\frac{1}{2} - 1)     (\frac{1}{2} - 2)     \cdots     (\frac{1}{2} - 99) } \end{align*}
Now, note that $-\frac{1}{2}-1=\frac{1}{2}-2$, $-\frac{1}{2}-2=\frac{1}{2}-3$, etc.; in essence, $-\frac{1}{2}-n=\frac{1}{2}-(n+1)$. We can then simplify the numerator and cancel like terms.
\begin{align*} \frac{     (-\frac{1}{2})     (-\frac{1}{2} - 1)     (-\frac{1}{2} - 2)     \cdots     (-\frac{1}{2} - 99) }{     (\frac{1}{2})     (\frac{1}{2} - 1)     (\frac{1}{2} - 2)     \cdots     (\frac{1}{2} - 99) } &= \frac{     \cancel{(\frac{1}{2} - 1)}     \cancel{(\frac{1}{2} - 2)}     \cancel{(\frac{1}{2} - 3)}     \cdots     (\frac{1}{2} - 100) }{     (\frac{1}{2})     \cancel{(\frac{1}{2} - 1)}     \cancel{(\frac{1}{2} - 2)}     \cdots     \cancel{(\frac{1}{2} - 99)} } \\ &= \frac{\frac{1}{2}-100}{\frac{1}{2}} \\ &= \frac{-\frac{199}{2}}{\frac{1}{2}} \\ &= \boxed{\textbf{(A) } -199.} \end{align*}