1957 AHSME Problems/Problem 3

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Problem 3

The simplest form of $1 - \frac{1}{1 + \frac{a}{1 - a}}$ is:

$\textbf{(A)}\ {a}\text{ if }{a\not= 0} \qquad \textbf{(B)}\ 1\qquad  \textbf{(C)}\ {a}\text{ if }{a\not=-1}\qquad \textbf{(D)}\ {1-a}\text{ with not restriction on }{a}\qquad \textbf{(E)}\ {a}\text{ if }{a\not= 1}$

Solution

We have $1 - \frac{1}{1 + \frac{a}{1 - a}} = 1 - \frac{1}{\frac{1}{1-a}} = 1 - \frac{1-a}{1} = a$ for almost all $a$. However, the first step is invalid when $a=1$, and each step is valid otherwise, so the answer is (E).

See Also

1957 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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