Query_Relation

[sourav neogi]

Let A be {a,b,c}. Let the relation R be {(b,c),(a,b),(c,c)}.

1.Is R Transative?

2.Can a relation have same ordered pair more than once?

Thanks

Sourav.

[Anand Iyer]

It's not transitive.  (a,b) and (b, c) are pairs in the relation, but not (a,c)

Relations are sets.  No set can have one item appear more than once.  So, relations also can't.

[sourav neogi]

Thanks.  

Let A be {a,b,c}. Let the relation R be AxA - {(b,b),(c,b)}. 

Is this relation transitive.

Thanks

Sourav. 

[Anand Iyer]

yes, it's transitive.

--

Cheers,

Anand

[Boss Annapillai]

Dear Anand,

R be AxA - {(b,b),(c,b)} 

Could you pl explain how this is transitive?

Thanks in advance,

[Anand Iyer]

Better than me explaining, watch this doubt-clearing session from IIT-M.  I've set it to the right time.  Watch it until the end of the question.

https://youtu.be/7ZGkJBGef-E?t=1513

If you're not clear, please reply..

--

Cheers,

Anand

[Boss Annapillai]

Sorry Anand,

I am unable to follow the hypothesis logic mentioned in the doubt clearing session.

Could you please explain?

[Malabika Guha Mustafi]

@ Boss Annapillai

 Transitive

Let us consider the set A as given below.

A  =  {a, b, c}

Let R be a transitive relation defined on the set A. 

Then, R  =  { (a, b), (b, c), (a, c)}

That is,

If "a" is related to "b" and "b" is related to "c", then "a" has to be related to "c".  

Important Note :

For a particular ordered pair in R, if we have (a, b) and we don't have (b, c), then we don't have to check transitive for that ordered pair. 

So, we have to check transitively, only if we find both (a, b) and (b, c) in R. 

Examples:

1)Let A  =  {1, 2, 3} and R be a relation defined on set A as

R  = {(1, 1), (2, 2), (3, 3), (1, 2)}

 The relation is transitive.

Note : For the two ordered pairs (2, 2) and (3, 3), we don't find the pair (b, c). So, we don't have to check the condition for those ordered pairs.  

2)Let A  =  {1, 2, 3} and R be a relation defined on set A as

R  = {(1, 1), (2, 2), (1, 2), (2, 1)}

so the relation is transitive.

3) Let A  =  {1, 2, 3} and R be a relation defined on set A as

R  = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}

In the table above, for the ordered pair (1, 2), we have both (a, b) and (b, c). But, we don't find (a, c). 

So, the relation is not transitive.

[Boss Annapillai]

Thanks Malabika,

But I was still not convinced and I googled the question verbatim and came up with this resource of UCSB.  

Chk out Qn.2 in Quiz 2

https://sites.cs.ucsb.edu/~pconrad/cs40/13W/hwk/cs40_W13_IC14.pdf

[Malabika Guha Mustafi]

I deleted my previous responses since it may create confusion.

@ Anand , You are right.

The relation is transitive .

@ Boss Annapillai,

Trying to explain Sir's Talking about the hypothesis.

In transitivity, We need not to search that the elements of transitivity(b,c) and final(a,c) should be present for a particular pair (a, b). If it is not there still it 'll be transitive.

But if (a,b) and (b,c), both the pairs are there then definitely we need to see that (a,c) will be there.

So, for transitivity, missing elements doesn't a problem unless it contradicts the definition.

R be AxA - {(b,b),(c,b)} 

here B is realed to b and b is related to c.

(c,b) is there. 

So. It is transitive.

@ Anand 

Thank you so much for your constant persuation and rectify my perception.

Thanks again.

[Anand Iyer]

Sure.  You’re welcome malabika

  I’m sort of relieved that I don’t have to try and understand some corner case around this😀

--

Cheers,

Anand

[Malabika Guha Mustafi]

:) With connotation logic and handling of outliers , no corner case is appreciated in Maths at least. With the problems, we are already cornered.