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A={1,2,3}, find ordered pairs on A which satisfy
1) R1 is transitive
2) R2 is non-symmetric and non-antisymmetric

At first, I thought 1) is a simple question, and the result is {<1,2>,<2,3>,<1,3>}.
However, the answer tells I am wrong.

As for 2), I am not sure if or not the ordered pairs exist on A?

Any help will be greatly appreciated!

Really thanks for Scott's help.

1) The answer for the first question is {<1,2>,<2,1>,<1,1>,<2,2>,<3,3>}
However,I don't know why the answer is above.

Find a non-symmetric relation on A: {<1,2>}
Find a non-antisymmetric relation on A: {<1,3>,<3,1>}
so R2 is {<1,2>,<1,3>,<3,1>}

I am not sure if it is right, because {<1,3>,<3,1>} is symmetric.

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As I said, there are many correct answers to the first question; $\{\langle 1,2\rangle,\langle 2,1\rangle,\langle 1,1\rangle,\langle 2,2\rangle,\langle 3,3\rangle\}$ is certainly one of them. $\{\langle 1,2\rangle,\langle 1,3\rangle,\langle 3,1\rangle$ is both non-symmetric, because it includes $\langle 1,2\rangle$ but not $\langle 2,1\rangle$, and non-antisymmetric, because it includes both $\langle 1,3\rangle$ and $\langle 3,1\rangle$; if it were antisymmetric, it would have to include at most one of those two pairs. (Sorry to be so slow; I didn’t realize that you’d updated it.) – Brian M. Scott Apr 8 '12 at 6:32
up vote 1 down vote accepted

Your first answer is not wrong: $\{\langle 1,2\rangle,\langle 2,3\rangle,\langle 1,3\rangle\}$ is a perfectly good transitive relation on $A=\{1,2,3\}$. There are many others, including all of the partial orders that you found in the other problem, not to mention the total relation $A\times A$, the set of all nine possible ordered pairs of elements of $A$.

For the second question, you need to find a relation $R_2$ on $A$ that is neither symmetric nor antisymmetric. Given two ordered pairs $\langle a,b\rangle$ and $\langle b,a\rangle$, a symmetric relation must either include both of them or neither of them. If a relation includes just one of a such a pair, it is necessarily not symmetric. Thus, you can make $R_2$ non-symmetric by including $\langle 1,2\rangle$ but not $\langle 2,1\rangle$.

A relation is antisymmetric if it always contains exactly one of two reversed pairs; if for some $a$ and $b$ it contains both $\langle a,b\rangle$ and $\langle b,a\rangle$ or neither of $\langle a,b\rangle$ and $\langle b,a\rangle$, it cannot be antisymmetric. Thus, you want to make sure that there’s some pair $a,b\in A$ such that $R_2$ contains both $\langle a,b\rangle$ and $\langle b,a\rangle$ or neither of $\langle a,b\rangle$ and $\langle b,a\rangle$. You can’t do it with $1$ and $2$, since we’ve already decided to put just one of $\langle 1,2\rangle$ and $\langle 2,1\rangle$ into $R_2$; can you do it with $1$ and $3$, say?

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