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

(Created page with "==Problem== Let <math>\ast</math> be the symbol denoting the binary operation on the set <math>S</math> of all non-zero real numbers as follows: For any two numbers <math>a<...")
 
m (→‎Solution)
Line 18: Line 18:
  
 
<math>\textbf{(B) }\ast\text{ is associative over }S</math>
 
<math>\textbf{(B) }\ast\text{ is associative over }S</math>
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<cmath>a \ast (b \ast c) = a \ast 2bc = 2abc</cmath>
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<cmath>a \ast (b \ast c) = a \ast 2bc = 2a2bc = 4abc</cmath>
−
<cmath>(a \ast b) \ast c = 2ab \ast c = 2abc</cmath>
+
<cmath>(a \ast b) \ast c = 2ab \ast c = 2(2ab)(c) = 4abc</cmath>
 
Statement B is true.
 
Statement B is true.
  
  
 
<math>\textbf{(C) }\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad </math>
 
<math>\textbf{(C) }\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad </math>
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<cmath>a \ast 1/2 = a</cmath>
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<cmath>a \ast \frac{1}{2} = a</cmath>
 
Statement C is true.
 
Statement C is true.
  
 +
<math>\textbf{(D) }\text{From the previous answer choice, we know }\frac{1}{2}\text{ is the identity element for }\ast\text{ in }S.\qquad </math>
 +
<math>\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}</math>
  
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<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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Statement D is true.
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<cmath> 1/2a \ast a = 1</cmath>
 
  
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By process of elimination, we can see that statement D is false.
 
  
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The answer is <math>\textbf{(D) }\text{Every element of }S\text{ has an inverse for }\ast\qquad .</math>
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<math>\textbf{(E) }\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S </math>
 +
<cmath> \frac{1}{2a} \ast a = 1</cmath>
 +
 
 +
Since the identity is <math>\frac{1}{2}</math>, not 1, we can see that statement E is false.
 +
 
 +
The answer is <math>\textbf{(E)}</math>
  
 
-edited by coolmath34
 
-edited by coolmath34
 +
-fixed by DoctorSeventeen

Revision as of 23:40, 25 July 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.

The answer is $\textbf{(E)}$

-edited by coolmath34 -fixed by DoctorSeventeen

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