# 1971 AHSME Problems/Problem 6

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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.

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

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

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

By process of elimination, we can see that statement D is false.

The answer is $\textbf{(D) }\text{Every element of }S\text{ has an inverse for }\ast\qquad .$

-edited by coolmath34