Determine whether the relations are symmetric or partially ordered, or totally ordered? Determine whether the relations $R$ on the real numbers
$x, y$ given below are symmetric or partially ordered. Are they totally ordered?
$x \geq y$
 A: Hint: 
Use the definitions of symmetric, partially ordered and totally ordered to say whether the real numbers and $\ge$ meet the definition or there is a counter example.
For example, for a symmetric relation, you would need $x \ge y$ if and only if $y \ge x$.  You should be able to find a counterexample.
A: Pay attention to the quantifiers in the definition. Symmetry means: for all $x,y$ the implication $x\leq y\Rightarrow y\leq x$ holds. 
You can't say (as you do in the comments) that the relation is symmetric for certain elements; a relation is either symmetric or not. 
If you think it is symmetric, try to prove this. If not, you should find a counter-example. The latter means you should find certain elements $x,y\in\mathbb{R}$ such that $x\leq y$ holds, but $y\leq x$ does not hold.
Edit: There is only one relation: $R=\{(x,y)\in\mathbb{R}\times \mathbb{R}: x\leq y\}$. You seem to be under the impression that elements of $R$ are (also) called relations. This is not true. To say that $(x,y)\in R$ means that $x$ is related to $y$, which in the current case means that $x$ is not larger than $y$. Stated this way, symmetry of $R$ means that the implication $(x,y)\in R\Rightarrow (y,x)\in R$ holds for all pairs $(x,y)\in\mathbb{R}\times \mathbb{R}$. Or: is true that whenever $x$ is not larger than $y$, then it follows that also $y$ cannot be larger than $x$? Obviously, if $x$ and $y$ happen to be equal, then this holds. But what if $x\neq y$?
