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Difference between revisions of "1971 AHSME Problems/Problem 6"

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\textbf{(B) }\ast\text{ is associative over }S\qquad \\
 
\textbf{(B) }\ast\text{ is associative over }S\qquad \\
 
\textbf{(C) }\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad  
 
\textbf{(C) }\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad  
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\textbf{(D) }\text{Every element of }S\text{ has an inverse for }\ast\qquad  
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\textbf{(D) }\text{Every element of }S\text{ has an inverse for }\ast\qquad \\
 
\textbf{(E) }\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S    </math>
 
\textbf{(E) }\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S    </math>
  
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Since the identity is <math>\frac{1}{2}</math>, not 1, we can see that statement E is false.
 
Since the identity is <math>\frac{1}{2}</math>, not 1, we can see that statement E is false.
  
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The answer is <math>\textbf{(E)}</math>
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Hence, the answer is <math>\boxed{\textbf{(E)}}</math>.
  
 
-edited by coolmath34
 
-edited by coolmath34
 
-fixed by DoctorSeventeen
 
-fixed by DoctorSeventeen
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== See Also ==
 +
{{AHSME 35p box|year=1971|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 09:51, 1 August 2024

Problem

Let $\ast$ be the symbol denoting the binary operation on the set $S$ of all non-zero real numbers as follows: For any two numbers $a$ and $b$ of $S$, $a\ast b=2ab$. Then the one of the following statements which is not true, is

$\textbf{(A) }\ast\text{ is commutative over }S \qquad \textbf{(B) }\ast\text{ is associative over }S\qquad \\ \textbf{(C) }\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad \textbf{(D) }\text{Every element of }S\text{ has an inverse for }\ast\qquad \\ \textbf{(E) }\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S$

Solution

$\textbf{(A) }\ast\text{ is commutative over }S$ \[a \ast b = b \ast a = 2ab\] Statement A is true.


$\textbf{(B) }\ast\text{ is associative over }S$ \[a \ast (b \ast c) = a \ast 2bc = 2a2bc = 4abc\] \[(a \ast b) \ast c = 2ab \ast c = 2(2ab)(c) = 4abc\] Statement B is true.


$\textbf{(C) }\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad$ \[a \ast \frac{1}{2} = a\] Statement C is true.

$\textbf{(D) }\text{From the previous answer choice, we know }\frac{1}{2}\text{ is the identity element for }\ast\text{ in }S.\qquad$ $\text{For inverses to exist, they must evaluate to the identity under the operator }\ast.\text{ Thus, } a \ast b = \frac{1}{2}\text{, which leads to } b = \frac{1}{4a}$

Statement D is true.


$\textbf{(E) }\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S$ \[\frac{1}{2a} \ast a = 1\]

Since the identity is $\frac{1}{2}$, not 1, we can see that statement E is false.

Hence, the answer is $\boxed{\textbf{(E)}}$.

-edited by coolmath34 -fixed by DoctorSeventeen

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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
All AHSME Problems and Solutions

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