GATE2010 q4 - Revision history

Starry Starr at 00:42, 7 May 2015

2015-05-07T00:42:16Z

Arjun Suresh at 11:35, 15 July 2014

2014-07-15T11:35:09Z

Arjun Suresh at 22:47, 13 July 2014

2014-07-13T22:47:27Z

Arjun Suresh at 14:25, 29 June 2014

2014-06-29T14:25:39Z

Arjun Suresh at 12:28, 29 June 2014

2014-06-29T12:28:30Z

Arjun Suresh at 12:27, 29 June 2014

2014-06-29T12:27:47Z

Arjun Suresh at 07:11, 15 April 2014

2014-04-15T07:11:20Z

Arjun Suresh at 21:33, 14 April 2014

2014-04-14T21:33:24Z

Arjun Suresh at 21:33, 14 April 2014

2014-04-14T21:33:00Z

Arjun Suresh: Created page with "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, *) for..."

2014-04-14T21:31:33Z

Created page with "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, *) for..."

New page

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
==={{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.
<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>Associativity : multiplication operation is anyway associative.</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
$ω * ω^2 = 1$. Also inverse of $ω^2$ is $ω$, because $ω^2 * ω = 1$</li>
</ul>
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.

{{Template:FBD}}
[[Category:Discrete Mathematics]]
[[Category: GATE2010]]
[[Category: Previous year GATE questions]]

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@@ Line 1: / Line 1: @@
-Consider the set S = $\{1, &omega;, &omega;^2\}$, where $\omega$ and $\omega^2$ are cube roots of unity. If *
+Consider the set S = $\{1, &omega;, &omega;^2\}$, 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'''

@@ Line 1: / Line 1: @@
-Consider the set S = $\{1, &omega;, &omega;^2\}$, where $&omega;$ and $&omega;^2$ are cube roots of unity. If *
+Consider the set S = $\{1, &omega;, &omega;^2\}$, where $\omega$ and $\omega^2$ are cube roots of unity. If *
 	denotes the multiplication operation, the structure (S, *) forms
@@ Line 11: / Line 11: @@
 ==={{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>
-			 <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>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 $&omega;$ is $&omega;^2$, because
+			 <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>
 		 </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
+		 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.
 		 <br>
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-[[Category:Discrete Mathematics]]
 [[Category: GATE2010]]
 [[Category: Discrete Mathematics questions]]

@@ Line 2: / Line 2: @@
 	denotes the multiplication operation, the structure (S, *) forms
-		<b>(A) </b>A Group
+		'''(A) A Group'''
 		(B) A Ring