Is this relation symmetric? $R={(1,1),(0,0)}\ $ R={(1,1),(0,0)}
A relation is symmetric if xRy and yRx but can it be x=y?
 A: 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.
A: 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 $\color{orange}{\small{\text{y}}}R\color{purple}{\large{\text{X}}}$.
