Difference between revisions of "1971 AHSME Problems/Problem 6"
Coolmath34 (talk | contribs) (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<...") |
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<math>\textbf{(B) }\ast\text{ is associative over }S</math> | <math>\textbf{(B) }\ast\text{ is associative over }S</math> | ||
− | <cmath>a \ast (b \ast c) = a \ast 2bc = | + | <cmath>a \ast (b \ast c) = a \ast 2bc = 2a2bc = 4abc</cmath> |
− | <cmath>(a \ast b) \ast c = 2ab \ast c = | + | <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> | ||
− | <cmath>a \ast 1 | + | <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> | ||
− | + | Statement D is true. | |
− | |||
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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 22:40, 25 July 2024
Problem
Let be the symbol denoting the binary operation on the set of all non-zero real numbers as follows: For any two numbers and of , . Then the one of the following statements which is not true, is
Solution
Statement A is true.
Statement B is true.
Statement C is true.
Statement D is true.
Since the identity is , not 1, we can see that statement E is false.
The answer is
-edited by coolmath34 -fixed by DoctorSeventeen