Difference between revisions of "Problem 39"

(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...")
 
m (Proposed for deletion)
 
Line 2: Line 2:
  
 
<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>
 +
 +
{{delete|title is too vague}}

Latest 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:

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

This article has been proposed for deletion. The reason given is: title is too vague.

Sysops: Before deleting this article, please check the article discussion pages and history.