Reflexive and symmetric can be proved as $|x|+|x|=|x+x|$ hence reflexive and $|y|+|x|=|y+x|$ hence symmetric but how transitive?
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3 Answers
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Hint: $|x|+|y|=|x+y|$ is true if $x$ and $y$ have the same sign or one of them is $0$.
Thus, all positive numbers are related to each other, and all negative numbers are related to each other, and $0$ is related to everything ...
answered Oct 28, 2015 at 20:23
hmakholm left over Monicahmakholm left over Monica
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1$\begingroup$ So it would be an equivalence relation on $\mathbb R \setminus \{0\}$. And it would be just a very uncommon way of describing the fibres of the sign function... $\endgroup$– MooSOct 28, 2015 at 20:29
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It is not transitive. So no.
$1$ is related to $0$ but also $0$ is related to $-1$. Is it true that $1$ and $-1$ are related?
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Hint: Question is not exactly defined. Relation can be equivalence, if you restrict the base set (from complex or real).
