Difference between revisions of "GATE2010 q4"

(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...")
 
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<b>(A) </b>A Group
 
<b>(A) </b>A Group
 
 
−
(B) <A Ring  
+
(B) A Ring  
 
 
−
(C) <An integral domain
+
(C) An integral domain
 
 
 
(D) A field
 
(D) A field
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
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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>
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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>
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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>
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<li>Associativity : multiplication operation is anyway associative.</li>
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<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  
+
<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>
−
So S satisfies all 4 properties of group, so it is a group. Infact 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.  
 
  property.  
 
<br>
 
<br>
−
So option <b>(A)</b> is correct.
+
So, option <b>(A)</b> is correct.
  
 
{{Template:FBD}}
 
{{Template:FBD}}

Revision as of 21:33, 14 April 2014

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

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.

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.




blog comments powered by Disqus