ISRO-DEC2017-2

Question

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

• A. Reflexive, Not Symmetric, Transitive
• B. Not Reflexive, Not Anti-symmetric, Transitive
• C. Reflexive, Anti-symmetric, Transitive
• D. Not Reflexive, Not Anti-symmetric, Not Transitive

Comments

no it is not like that.

answer will be none of these

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bdw answer given is b which is wrong

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divide by 0 is not a problem for transitivity

It only effects on reflexivity property

Answer 1

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.

Comments

@Rahul More

eg : 4D8 here 4 divides 8 and quotient is 8/4= 2 i.e. an integer

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For the Anti-Symmetric (-1,1) and (1,-1)

By this example firstly it could not hold Anti-Symmetric Property ?

Question is asked for Division ? Can anyone explain this please ?

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In case of transitivity,if aDb and bDc then aDc

For eg - but 4D8 and 8D2 are not valid since if we divide 2 by 8 then answer is a decimal which is not an integer.

It will not reach to the 3rd quarter of the transitivity proof.

Cn anybody explain.

Answer 2

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.