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ISRO-DEC2017-2

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Consider the set of integers $I.$ Let $D$ denote "divides with an integer quotient" (e.g. $4D8$ but not $4D7$). Then $D$ is

  1. Reflexive, Not Symmetric, Transitive
  2. Not Reflexive, Not Anti-symmetric, Transitive
  3. Reflexive, Anti-symmetric, Transitive
  4. Not Reflexive, Not Anti-symmetric, Not Transitive
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For reflexivity for all $x \in I, R(x,x)$ but this is violated for $x = 0$. So, given relation is not reflexive.

For symmetry, for all $x,y \in I, R(x,y) \implies R(y,x).$ This is violated for $x = 2, y = 1$.

For anti-symmetry, for all $x,y \in I,  (R(x,y) \wedge R(y,x)) \implies x = y$. We have $R(-1, 1)$ and $R(1,-1)$ so $R$ is not anti-symmetric.

For transitivity, for all $x,y,z \in I, R(x,y) \wedge R(y,z) \implies R(x,z).$ As per the rule of division this holds.

So, option B is the answer.
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Consider the set of integers I

As soon as you read this in a relations question, your mind must automatically click on testing negatives and 0.

  1. Reflexive? No. Because (0,0) doesn't exist.
     
  2. Symmetric? No. Because if 4D8, then 8D4 doesn't exist.
     
  3. Anti-Symmetric? No. Because 10D-10, and -10D10 both exist.
     
  4. Transitive? Yes. If aDb and bDc then aDc. This is elementary stuff.
    Still, give this a read.
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