1971 AHSME Problems/Problem 6

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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1971 AHSME Problems/Problem 6

Problem

Solution