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Difference between revisions of "1958 AHSME Problems/Problem 8"

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== Problem ==
 
== Problem ==
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Which of these four numbers <math> \sqrt{\pi^2},\,\sqrt[3]{.8},\,\sqrt[4]{.00016},\,\sqrt[3]{-1}\cdot \sqrt{(.09)^{-1}}</math>, is (are) rational:
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Which of these five numbers <math> \sqrt{\pi^2},\,\sqrt[3]{.8},\,\sqrt[4]{.00016},\,\sqrt[3]{-1}\cdot \sqrt{(.09)^{-1}}</math>, is (are) rational:
  
 
<math> \textbf{(A)}\ \text{none}\qquad  
 
<math> \textbf{(A)}\ \text{none}\qquad  
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{{AHSME box|year=1958|num-b=7|num-a=9}}
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{{AHSME 50p box|year=1958|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:00, 29 June 2017

Problem

Which of these five numbers $\sqrt{\pi^2},\,\sqrt[3]{.8},\,\sqrt[4]{.00016},\,\sqrt[3]{-1}\cdot \sqrt{(.09)^{-1}}$, is (are) rational:

$\textbf{(A)}\ \text{none}\qquad \textbf{(B)}\ \text{all}\qquad \textbf{(C)}\ \text{the first and fourth}\qquad \textbf{(D)}\ \text{only the fourth}\qquad \textbf{(E)}\ \text{only the first}$

Solution

$\sqrt{\pi^2}=\pi$ is not rational, so eliminate choices $(B)$, $(C)$, and $(E)$. Note that $(-1)^3 = (-1)$, so $\sqrt[3]{-1}=-1$ is rational. We have found one rational number, so we can eliminate choice $(A)$. The answer is $\boxed{ \text{(D)}}$.


See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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