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 | ||
| − | \textbf{(D) }\text{Every element of }S\text{ has an inverse for }\ast\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 </math> | \textbf{(E) }\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S </math> | ||
| Line 18: | Line 18: | ||
<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. | |
| − | |||
| − | |||
| − | + | <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. | ||
| + | |||
| + | Hence, the answer is <math>\boxed{\textbf{(E)}}</math>. | ||
-edited by coolmath34 | -edited by coolmath34 | ||
| + | -fixed by DoctorSeventeen | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 35p box|year=1971|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:51, 1 August 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
![\[a \ast b = b \ast a = 2ab\]](http://latex.artofproblemsolving.com/d/a/9/da98493b596a2b5f5e10d448fd4e2b3cbd2e3d0e.png)
Statement A is true.
![\[a \ast (b \ast c) = a \ast 2bc = 2a2bc = 4abc\]](http://latex.artofproblemsolving.com/4/7/a/47af5cba180e658900cb7d4484520f0144ee7be1.png)
![\[(a \ast b) \ast c = 2ab \ast c = 2(2ab)(c) = 4abc\]](http://latex.artofproblemsolving.com/1/1/b/11b920aa2324dbf366b81d71d884af14baf225d7.png)
Statement B is true.
![\[a \ast \frac{1}{2} = a\]](http://latex.artofproblemsolving.com/0/d/f/0dff967d0cda711f1bec225e5f939fe2b30d071c.png)
Statement C is true.
Statement D is true.
![\[\frac{1}{2a} \ast a = 1\]](http://latex.artofproblemsolving.com/a/c/4/ac497d0a34eb3576f69b01ee9f168d6372c9f381.png)
Since the identity is 
, not 1, we can see that statement E is false.
Hence, the answer is 
.
-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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
