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| − | Consider the set S = $\{1, ω, ω^2\}$, where $ | + | Consider the set S = $\{1, ω, ω^2\}$, where $\omega$ and $\omega^2$ are cube roots of unity. If $*$ |
| − | denotes the multiplication operation, the structure (S, *) forms | + | denotes the multiplication operation, the structure $(S, *)$ forms |
| − | + | '''(A) A Group''' | |
| − | (B) | + | (B) A Ring |
| − | (C) | + | (C) An integral domain |
(D) A field | (D) A field | ||
==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
| − | + | We can directly answer this question as "A Group", because other three options require two operations over structure, | |
| − | but let us see whether (S, *) satisfies group properties or not. | + | but let us see whether $(S, *)$ satisfies group properties or not. |
<ul> | <ul> | ||
| − | <li>Closure : If we multiply any two elements of S, we get one of three elements of S, so S is closed over *.</li> | + | <li>Closure: If we multiply any two elements of $S$, we get one of three elements of $S$, so $S$ is closed over $*$.</li> |
| − | <li>Associativity : multiplication operation is anyway associative.</li> | + | <li>Associativity: multiplication operation is anyway associative.</li> |
| − | <li>Identity element : 1 is identity element of S.</li> | + | <li>Identity element: $1$ is identity element of $S$.</li> |
| − | <li>Inverse element : inverse of 1 is 1 because 1 * 1 = 1, inverse of $ω$ is $ω^2$, because | + | <li>Inverse element: inverse of $1$ is $1$ because $1 * 1 = 1$, inverse of $ω$ is $ω^2$, because |
$ω * ω^2 = 1$. Also inverse of $ω^2$ is $ω$, because $ω^2 * ω = 1$</li> | $ω * ω^2 = 1$. Also inverse of $ω^2$ is $ω$, because $ω^2 * ω = 1$</li> | ||
</ul> | </ul> | ||
| − | + | Thus, $S$ satisfies all $4$ properties of group, so it is a group. In fact, $S$ is an abelian group, because it also satisfies commutative | |
property. | property. | ||
<br> | <br> | ||
| − | So option <b>(A)</b> is correct. | + | So, option <b>(A)</b> is correct. |
{{Template:FBD}} | {{Template:FBD}} | ||
| − | + | ||
[[Category: GATE2010]] | [[Category: GATE2010]] | ||
| − | [[Category: | + | [[Category: Set Theory & Algebra questions from GATE]] |
Consider the set S = $\{1, ω, ω^2\}$, where $\omega$ and $\omega^2$ are cube roots of unity. If $*$ denotes the multiplication operation, the structure $(S, *)$ forms
(A) A Group
(B) A Ring
(C) An integral domain
(D) A field
We can directly answer this question as "A Group", because other three options require two operations over structure, but let us see whether $(S, *)$ satisfies group properties or not.
Thus, $S$ satisfies all $4$ properties of group, so it is a group. In fact, $S$ is an abelian group, because it also satisfies commutative
property.
So, option (A) is correct.
Consider the set S = $\{1, ω, ω^2\}$, where $ω$ and $ω^2$ are cube roots of unity. If * denotes the multiplication operation, the structure (S, *) forms
(A) A Group
(B) <A Ring
(C) <An integral domain
(D) A field
We can directly answer this question as "A Group", because other three options require two operations over structure,
but let us see whether (S, *) satisfies group properties or not.
So S satisfies all 4 properties of group, so it is a group. Infact S is an abelian group, because it also satisfies commutative
property.
So option (A) is correct.
GATECSE