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Consider the binary relation:

$S= \left\{\left(x, y\right) \mid y=x+1 \text{ and } x, y \in \left\{0, 1, 2\right\} \right\}$

The reflexive transitive closure is $S$ is

  1. $\left\{\left(x, y\right) \mid y >x \text{ and } x, y \in \left\{0, 1, 2\right\} \right\}$ 

  2. $\left\{\left(x, y\right) \mid y \geq x \text{ and } x, y \in \left\{0, 1, 2\right\} \right\}$ 

  3. $\left\{\left(x, y\right) \mid y < x \text{ and } x, y \in \left\{0, 1, 2\right\} \right\}$ 

  4. $\left\{\left(x, y\right) \mid y \leq x \text{ and } x, y \in \left\{0, 1, 2\right\} \right\}$

asked in Set Theory & Algebra by Veteran (56k points)   | 229 views
@Habibkhan, can u explain this problem in detail. Thanks in advance.
Reflexive closure is taken to add the reflexive property and same for transitive as well..

Now for reflexivity we should have (x,x) ordered pair..which is not possible if y = x + 1..So the modification required is y = x need also be added..

For transitive property we require here if y >= x and z >= y implies z >= x as original condition is not transitive clearly bcoz if y = x + 1 and z = y + 1 this does not imply z = x + 1 as z = x + 2..So we need to redefine it as y > x instead of y = x + 1 and for reflexitvity y = x is also to be mentioned..

So y >= x is the appropriate closure..Hence B) option is correct.
What is asked to do in question..??
simple, just take the reflexive closure by adding the diagonal elements and then take the transitive closure and verify with the options.

S={(0,1),(1,2)}

reflexive closure of S={(0,0), (1,1), (2,2,) ,(0,1),(1,2)}

Now take transitive closure={(0,0), (1,1), (2,2,) ,(0,1),(1,2),(0,2)}

clearly x<=y.

2 Answers

+7 votes
Best answer
Option b. Transitive means, x is related to all greater y (as every x is related to x + 1) and reflexive means x is related to x.
answered by Loyal (2.9k points)  
selected by
+2 votes

Relation contains = { (0,1) , (1,2) }

When said Reflexive Transitive then apply first Transitive closure and then Reflexive closure.

After Transitive = { (0,1) (1,2) (0,2) }

After Reflexive ={ (0,1) (1,2) (0,2) (0,0) (1,1) (2,2) }

So B is ans.

Some Important Note: Wiki

The transitive closure of a binary relation R on a set X is the smallest relation on X that contains R and is transitive.

For example, if X is a set of airports and x R y means "there is a direct flight from airport x to airport y" (for x and y in X), then the transitive closure of R on X is the relation R+such that x R+ y means "it is possible to fly from x to y in one or more flights". Informally, the transitive closure gives you the set of all places you can get to from any starting place.

More formally, the transitive closure of a binary relation R on a set X is the transitive relation R+ on set X such that R+ contains R and R+ is minimal . If the binary relation itself is transitive, then the transitive closure is that same binary relation; otherwise, the transitive closure is a different relation.

answered by Veteran (21.2k points)  
edited ago by
@Gabbar  Best one :)

Thank u for explaing
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