Question 993909
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ R\ =\ \{x, y\ \in \mathbb{R}\ |\ x\ +\ y\ =\ 0\}]


Which is the same thing as


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \{x, y\ \in \mathbb{R}\ |\ x\ =\ a,\ y\ =\ -a\}]


So if *[tex \Large <a, b>\ \in\ R] then *[tex \Large b\ =\ -a], and therefore *[tex \Large <a, a>\ \not\in R] unless *[tex \Large a\ =\ 0].


Therefore *[tex \Large R] is neither reflexive or irreflexive.


But since *[tex \Large a] can be negative, *[tex \Large <a, -a> \in\ R\ \right\ <-a,a>\ \in\ R]


Therefore, R is symmetric.


Since *[tex \Large a\ \not=\ -a] except when *[tex \Large a\ =\ 0], *[tex \Large R] is not asymmetric.


Since if *[tex \Large <a, b> \in\ R] then *[tex \Large b\ =\ -a], so if *[tex \Large <b, c> \in\ R] then *[tex \Large c\ =\ a].


Therefore R is transitive.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \