Symmetrical, Transitional and Antisymmetrical question! If $R \subseteq A \times A$ is it true that $R$ is symmetrical since $xRy$ then $yRx$ ? 
I have written that this is also antisymmetrical if both $x\leq y$ and $y\leq x$ if $x =y$
How does I then make $R \subseteq A \times A$ transitive?
 A: You are either confused, or we are running against a language barrier (both are certainly possible).
It is not true that simply by virtue of being a subset of $A\times A$, a subset $R$ will be "symmetrical."
Rather: we define $R$ to be "symmetrical" if and only if for every $x$ and $y$ in $A$, if $(x,y)\in R$ (that is, if $xRy$), then $(y,x)\in R$ (that is, $yRx$).
An example of a symmetric relation on $\mathbb{R}$ is
$$R = \{ (a,b)\in\mathbb{R}\times\mathbb{R} \mid |a|=|b|\}$$
since, if $aRb$, then $|a|=|b|$, hence $|b|=|a|$, hence $bRa$.
An example of a non-symmetric relation on $\mathbb{R}$ is
$$S = \{(a,b)\in\mathbb{R}\times\mathbb{R} \mid a\geq 0\}.$$
It is not symmetric, because $(1,-1)\in S$, but $(-1,1)\notin S$.
We define $R$ to be "antisymmetric" if and only if for every $x$ and $y$ in $A$, if $xRy$ and $yRx$ are both true, then $x=y$. 
(You have the implication reversed).
In the example $R$ above, $R$ is not antisymmetric, because $(1,-1)\in R$, $(-1,1)\in R$, but $-1\neq 1$.
However, an example of an antisymmetric relation on $\mathbb{R}$ is:
$$T = \{(a,b)\in\mathbb{R}\times\mathbb{R}\mid a\leq b\}.$$
This is antisymmetric because if $(a,b)\in T$ and $(b,a)\in T$, that means that $a\leq b$ and $b\leq a$, so then we conclude that $a=b$.
Finally, a relation $R\subseteq A\times A$ is said to be "transitive" if and only if for every $a,b,c\in A$, if $aRb$ and $bRc$ both hold, then $aRc$ holds.
Both $R$, $S$, and $T$ above are transitive. To see an example of a relation on $\mathbb{R}$ that is not transitive, let
$$U = \{(a,b)\in\mathbb{R}\times\mathbb{R}\mid a\neq b\}.$$
Then $0U1$ holds, and $1U0$ holds, but $0U0$ does not hold. So $U$ is not transitive.
