R={(1,1),(0,0)}
A relation is symmetric if xRy and yRx but can it be x=y?
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1Yes. The relation is also transitive and if the relation is treated as being with domain ${0,1}$ is also reflexive and so an equivalence relation... the trivial equivalence relation where something is "equivalent" to another if and only if they are equal to eachother. – JMoravitz Nov 17 '20 at 20:19
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While not a direct duplicate, my answer to Can a relation be both symmetric and antisymmetric or neither? where I describe the graphical interpretation of symmetry answers your question. – JMoravitz Nov 17 '20 at 20:23
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@JMoravitz its also antisymmetric too right? and it's transitive because it doesn't violate the definition of transitive relation? – Light Lamps Nov 17 '20 at 20:23
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Yes to both, and I refer you again to the linked question. – JMoravitz Nov 17 '20 at 20:24
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A relation $ R $ defined at a set $ E $ is said to be symetric if, each time an element $ a $ is in relation with an element $ b$, then the element $ b $ is also in relation with $ a $.
$$(\forall a,b\in E)\; \Bigl(a\;R\; b\;\implies\;b\;R\;a\Bigr)$$
In your case, clearly, the relation $ R $ is symetric.
The relation is antisymetric if
$$(\forall a,b \in E)$$ $$\Bigl(a \;R\; b \;\wedge\;b\;R\;a\implies a=b\Bigr)$$
The negation is $$(\exists a,b \in E)\ :$$ $$(\;a\;R\;b)\;\wedge\; (b\;R\;a)\;\wedge\;(a\ne b)$$
This is not the case. Your relation is also antisymetric.
It is transitive iff $$(\forall a,b,c\in E)$$ $$(a\;R\;b)\wedge (b\;R\;c)\implies a\;R\;c$$
The negation is $$(\exists a,b,c\in E)\;:$$ $$(a\;R\;b) \wedge (b\;R\;c)\;\wedge (a\;not\;R\;c)$$
this is not the case, so it is transitive.
hamam_Abdallah
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It symmetric.
If $\color{green}0R\color{red}0$ then $\color{red}0R\color{green}0$.
And if $\color{green}1R\color{red}1$ then $\color{red}1R\color{green}1$
And as those are the only two possible cases of $\color{green}x R\color{red}y$ possible, If we ever have $\color{green}x R\color{red}y$ it will always be the case we also have $\color{red}y R\color{green}x$.
Please let me know if you hare red/green colorblind. If so consider this explanation:
If $\color{purple}{\large{\text{ZERO}}}R\color{orange}{\small{\text{zero}}}$ then $\color{orange}{\small{\text{zero}}}R\color{purple}{\large{\text{ZERO}}}$.
And if $\color{purple}{\large{\text{ONE}}}R\color{orange}{\small{\text{one}}}$ then $\color{orange}{\small{\text{one}}}R\color{purple}{\large{\text{ONE}}}$
And as those are the only two possible cases of $\color{purple}{\large{\text{X}}}R\color{orange}{\small{\text{y}}}$ possible, If we ever have $\color{purple}{\large{\text{X}}}R\color{orange}{\small{\text{y}}}$ it will always be the case we also have $\color{orange}{\small{\text{y}}}R\color{purple}{\large{\text{X}}}$.
fleablood
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@hamam_Abdallah I don't understand your comment. If $0R0$ then $0R0$. Nothing wrong with that tautology. – fleablood Nov 18 '20 at 00:13