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Question: For \(n,m\in N\),\(n|m\) means that \(n\) is factor of \(m\), the relation \(|\) is A. Reflexive a...

For n,mNn,m\in N,nmn|m means that nn is factor of mm, the relation | is
A. Reflexive and symmetric
B. Transitive and symmetric
C. Reflexive, transitive and symmetric
D. Reflexive, transitive and not symmetric

Explanation

Solution

Hint: Here we have been given that nn is a factor of mm. So use (a,a)RaA(a,a)\in R\forall a\in A, a,b,cX:(aRbbRc)aRc\forall a,b,c\in X:(aRb\wedge bRc)\Rightarrow aRc and (a,b)R(b,a)R(a,b)\in R\Rightarrow (b,a)\in R. You will get the answer.

Complete step-by-step answer:

In maths, a binary relation RR across a set XX is reflexive if each element of set XX is related or linked to itself. In terms of relations, this can be defined with (a,a)RaX(a,a)\in R\forall a\in X or IRI\in R where I is the identity relation on A. It has a reflexive property and is said to hold reflexivity. Symmetry, transitivity and reflexivity are the three properties representing equivalence relations.
In relation and functions, a reflexive relation is the one in which every element maps to itself. For example, let us consider a set A=1,2A=\\{1,2\\}. Now here the reflexive relation will be R=(1,1),(2,2),(1,2),(2.1)R=\\{(1,1),(2,2),(1,2),(2.1)\\}. Hence, a relation is reflexive if:
(a,a)RaA(a,a)\in R\forall a\in A
where aa is the element, AA is the set and RR is the relation.
A binary relation RR over a set XX is transitive if whenever an element aa is related to an element bb, and bb is in turn related to an element cc, then a is also related to cc.
a,b,cX:(aRbbRc)aRc\forall a,b,c\in X:(aRb\wedge bRc)\Rightarrow aRc
On the other hand, "is the mother of" is not a transitive relation, because if Alice is the mother of Brenda, and Brenda is the mother of Claire, then Alice is not the mother of Claire. What is more, it is anti transitive: Alice can never be the mother of Claire.
Let AA be a set in which the relation RR defined. Then RR is said to be a symmetric relation, if (a,b)R(b,a)R(a,b)\in R\Rightarrow (b,a)\in R, that is, aRbbRaaRb\Rightarrow bRa for all (a,b)R(a,b)\in R.
Consider, for example, the set AA of natural numbers. If a relation AA be defined by (x+y)''(x+y)'', then this relation is symmetric in AA, for a+b=5b+a=5a+b=5\Rightarrow b+a=5.
But in the set AA of natural numbers if the relation RR be defined as ‘xx is a divisor of yy’, then the relation RR is not symmetric as 3R93R9 does not imply 9R39R3; for, 33 divides 99 but 99 does not divide 33.
For a symmetric relation RR, R1=R{{R}^{-1}}=R.

Since nn is a factor of nn, since every natural number is a factor of itself so the relation is reflexive.
If nn is a factor of mm and mm is a factor of pp, then nn is surely a factor of pp, so the relation is transitive.
If however nn is a factor of mm, mm is not necessarily a factor of nn so the relation is not symmetric.
Hence, the answer is option D.

Note: Read the question carefully. You should know the concept of reflexive, transitive and symmetric. Also, you should know the basics of these and their properties. You must also know the types of properties.