relation on A that is symmetric and transitive but not reflexive

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Homework Statement


Let A = {1,2,3,4}. Give an example of a relation on A that is symmetric and transitive, but not reflexive.

Homework Equations


Symmetric: if aRb then bRa
Transitive: if aRb and bRc then aRc
Reflexive: aRa for all a in A

The Attempt at a Solution


{(1,2),(2,1),(1,1)} It's symmetric because 1R2 and 2R1. Not reflexive because (2,2)...(4,4) are not elements and transitive because 1R2 and 2R1 so 1R1. Yet, this one got marked wrong on my homework. I'm going to assume my teacher is right, and I'm wrong. Can anyone find my mistake?
 
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2R1 and 1R2 implies 2R2 if your relation is transitive. But (2,2) isn't in A. So your teacher is right.
 


Oh wow, I can't believe I missed that! Thanks for that!
 
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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