# Am I correct about the transitive closure of this relation?

Transitive closure given by relation $$A$$ is?

$$A = \{ (1,2) , (2,3) , (3,4) , (3,1), (4,3)\}$$

We have $$(1,2)$$ and $$(2,3)$$ gives $$(1,3)$$

moreover, $$(1,3)$$ and $$(3,1)$$ gives $$(1,1)$$,

$$(1,3)$$ and $$(3,4)$$ gives $$(1,4)$$

We have $$(2,3)$$ and $$(3,4)$$ gives $$(2,4)$$, $$(2,4)$$ and $$(4,3)$$ gives $$(2,3)$$

We have $$(2,3)$$ and $$(3,1)$$ gives $$(2,1)$$

$$(2,1)$$ and $$(1,2)$$ gives $$(2,2)$$

We have $$(3,4)$$ and $$(4,3)$$ gives $$(3,3)$$

We have $$(3,1)$$ and $$(1,2)$$ gives $$(3,2)$$

We have $$(4,3)$$ and $$(3,1)$$ gives $$(4,1)$$

We have $$(4,3)$$ and $$(3,4)$$ gives $$(4,4)$$ ,

So Transitive closure given by relation $$A$$ is given by

$$\{ (1,1),(1,2) ,(1,3),(1,4) ,(2,1),(2,2),(2,3) ,(2,4), (3,1) , (3,2),(3,3),(3,4), (4,1),(4,2),(4,3),(4,4)\}$$

Now my question is, am I correct? My friend said I was incorrect but I do not know what is wrong with my answer. If anyone can point out why I'm wrong, it would be appreciated.

You showed that $(2,3)$ is an element of the transitive closure of $A,$ but this is not necessary, since $(2,3)\in A.$ You didn't show that $(4,2)$ is an element of the transitive closure of $A,$ but it is.
• Yes. Remove the redundant justification, and add the missing justification. Can you see why $(4,2)$ is in the transitive closure of $A$? Apr 6, 2017 at 23:50
• Just keep doing the same kind of thing you've been doing. There are multiple ways to get there. There are $3$ ways, to be exact. Apr 6, 2017 at 23:55