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Consider the set S = $\{1, ω, ω^2\}$, where $ω$ and $ω^2$ are cube roots of unity. If *
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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'''
 
'''(A) A Group'''
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==={{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,
 
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>
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<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>
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<li>Identity element: $1$ is identity element of $S$.</li>
<li>Inverse element: inverse of 1 is 1 because 1 * 1 = 1, inverse of $&omega;$ is $&omega;^2$, because  
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<li>Inverse element: inverse of $1$ is $1$ because $1 * 1 = 1$, inverse of $&omega;$ is $&omega;^2$, because  
 
$&omega; * &omega;^2 = 1$. Also inverse of $&omega;^2$ is $&omega;$, because $&omega;^2 * &omega; = 1$</li>
 
$&omega; * &omega;^2 = 1$. Also inverse of $&omega;^2$ is $&omega;$, because $&omega;^2 * &omega; = 1$</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
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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>
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{{Template:FBD}}
 
{{Template:FBD}}
[[Category:Discrete Mathematics]]
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[[Category: GATE2010]]
 
[[Category: GATE2010]]
[[Category: Discrete Mathematics questions]]
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[[Category: Set Theory & Algebra questions from GATE]]

Latest revision as of 00:42, 7 May 2015

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

Solution by Happy Mittal

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.

  • Closure: If we multiply any two elements of $S$, we get one of three elements of $S$, so $S$ is closed over $*$.
  • Associativity: multiplication operation is anyway associative.
  • Identity element: $1$ is identity element of $S$.
  • 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 $ω^2 * ω = 1$

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.




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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

Solution by Happy Mittal[edit]

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.




blog comments powered by Disqus