Let N be the set of all natural num...
Question
Let N be the set of all natural numbers and let R be a relation in N, defined by R={(a,b)}:a is a factor of b} Then, show that R is reflexive and transitive but not
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Let ( N ) be the set of all natural numbers and let ( R ) be a relation in ( N ), defined by ( R={(a, b)}: a ) is a factor of ( b} ) Then, show that ( R ) is reflexive and transitive but not symmetric.

11th - 12th Class
Maths
Solution
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Let N be the set of all natural numbers and let R be a relation in N, defined by R={(a,b)}:a is a factor of b} Then, show that R is reflexive and transitive but not symmetric.
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Here, ( R ) satisfies the following properties:
(i) Reflexivity Let ( a ) be an arbitrary element of ( N ).
Then, clearly, ( a ) is a factor of ( a ).
( therefore(a, a) in R forall a in N )
So, ( R ) is reflexive.
(ii) Transitivity Let ( a, b, c in N ) such that ( (a, b) in R ) and ( (b, c) in R )
Now, ( (a, b) in R ) and ( (b, c) in R )
( Rightarrow quad(a ) is a factor of ( b) ) and ( (b ) is a factor of ( c) )
( Rightarrow quad b=a d ) and ( c=b e ) for some ( d, e in N )
( Rightarrow quad c=(a d) e=a(d e) )
[by associative law]
( Rightarrow a ) is a factor of ( c )
( Rightarrow(a, c) in R )
( therefore(a, b) in R ) and ( (b, c) in R Rightarrow(a, c) in R )
Hence, ( R ) is transitive.
(iii) Nonsymmetry Clearly, 2 and 6 are natural numbers and 2 is a factor of ( 6 . )
( therefore(2,6) in R )
But, 6 is not a factor of 2 .
( therefore(6,2) otin R )
Thus, (2,6)( in R ) and (6,2)( otin R )
Hence, ( R ) is not symmetric.

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