Transitive Closure of Relations on S: Solutions

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Homework Statement



Let S = {1, 2, 3, 4}. For each of the following relationson S give its transitive
closure.

(a) {(1, 1), (3, 4)}
(b) {(1, 2), (4, 4), (2, 1), (4, 3), (2, 3)}
(c) {(1, 1), (2, 2), (3, 3), (4, 4), (4, 1)}
(d) {(1, 3), (3, 2), (2, 4), (4, 1)}

Homework Equations


N/A?


The Attempt at a Solution


I honestly have absolutely no idea what I am supposed to be doing. Any push in the right direction would be appreciated. Not asking for an answer here.
 
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What's the definition of transitive closure for a relation?
 
I have absolutely no idea. I got some help from a classmate and I think I have it somewhat figured out?

for a, there is none
for b, I found (1,1) (2,2) and (1,3)
for c, there is also none since anything you can find is already there..
d still working on.

Just wondering now how do I format this into an answer? Do I just do {(1,1),(2,2),(1,3)} and so on?
 
How can you give the transitive closure of a relation without knowing what it is?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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