Relations

Question

My explanation I have provided, Please rectify the mistake where I am going wrong :)

Comments

Please correct me where I am going wrong , Thank you :)

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Ans 27?

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@ srestha They have provided ans as option ( D)

Will you explain the approach please.

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i hope this images helps

 

Answer 1

Option d is right

 

Comments

Why 3^n(n-1)/2 ?

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@ abhishekmehta4u

Thank a lot !

Can you please explain :

1. # neither reflexive nor irreflexive

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@gauravkc Hope this will help.

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Thanks :)

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@abhishekmehta4u

Is this a correct approach for "neither...nor " question I asked ?  Like, 1 (a,a) pair must be rejected so that it will not be reflexive, and 1 should be kept so that it wont he irreflexive. Way of  doing this is " 1st term or 2nd term " as shown in fig. Please correct me if I am wrong .

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@Headshot..

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@ abhishekmehta4u

Wow...

Very nice explanation... cleared my most of the concepts.

I solved remaining problems because of your help...

Great help.

Thank you so much :)

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@ abhishekmehta4u

Will you please check my  method, I guess it is working too (I  calculated for n= 4 ) so that I ll get to know it is valid way or not.

P.S : Last term is 2^1 insted of 2^(n-n) .

I realized its 2^[n-(n-1)] .

Please ans if possible, it will be really helpful. Thanks :)

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can you  explain little bit more . i can not understand ur logic??

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@ abhishekmehta4u

formula for antisymmetric relation $3^{\frac{n^{2}-n}{2}}.2^{n}$ where $2^{n}$ is for reflexive relation

or we can also say antisymmetric relation without reflexive relation is asymmetric

right?

Isnot then answer incorrect?

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Antisymmetric relation without reflexive relation is asymmetric.

No its not correct.

S={1,2,3}

R={(1,2)(2,3)(1,1)}--->it is antisymm ant not reflexive nor asymmetric.

 

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@ abhishekmehta4u

I thought over it again, but I missed a point.. The image below is what I want to convey, now I  corrected it but it is not giving the answer.. Will you analyze it once and see if its somewhere close..not as a solution but approach and tell where my concept going wrong. 

After satisfying both the conditions , i think it doesnt matter whether we select one pair or (+) two pairs or (+) and so on...

Thats why there are additions in above expression.

Still I am not sure whether I am conveying it in right way or not but  have a look , it will surely help me .

Thank you :)

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@ abhishekmehta4u

Will it be fine if i say ...

" Antisymmetric with irreflexivity is assymetric "

?

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Antisymmetric with irreflexivity is assymetric

yes true. because definition of irreflexive is

A relation  on a set  is irreflexive provided that no element is related to itself; in other words,  for no  in 

So, not reflexive different from irreflexive 

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But if the set has numbers from 1 to n (Say)

For a relation to be reflexive, all (x,x) for x=1..n should be in relation.

Any one of them missing in relation will make it irreflexive.

However, in asymmetric, there should be no (x,x) relation. So I guess we cannot say that.

Please correct me if I'm wrong.

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but definition of irreflexive is No $\left ( x,x \right )$ element will be present there

Any $\left ( x,x \right )$ is present then it is not irreflexive and nor even reflxive

right?

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Oops, my mistake.

Sets who contain at least all (x,x) pairs are reflexive

Sets who do not contain any (x,x) pair are irreflexive

Sets who have few (x,x) pairs are neither reflexive nor irreflexive