Art of Problem Solving

Problem 39: Difference between revisions

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Created page with "To satisfy the equation <math>\frac{a+b}{a}=\frac{b}{a+b}</math>, <math>a</math> and <math>b</math> must be: <math>\textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text..."
 
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<math>\textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text{both real but not rational}\qquad\textbf{(C)}\ \text{both not real}\qquad</math> <math>\textbf{(D)}\ \text{one real, one not real}\qquad\textbf{(E)}\ \text{one real, one not real or both not real}</math>
<math>\textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text{both real but not rational}\qquad\textbf{(C)}\ \text{both not real}\qquad</math> <math>\textbf{(D)}\ \text{one real, one not real}\qquad\textbf{(E)}\ \text{one real, one not real or both not real}</math>
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Revision as of 17:03, 10 December 2024

To satisfy the equation $\frac{a+b}{a}=\frac{b}{a+b}$, $a$ and $b$ must be:

$\textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text{both real but not rational}\qquad\textbf{(C)}\ \text{both not real}\qquad$ $\textbf{(D)}\ \text{one real, one not real}\qquad\textbf{(E)}\ \text{one real, one not real or both not real}$


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