how to prove that a relation is antisymmetric? I have this question that I didn't know how to prove it and need your help.
$R$ is a transitive and not reflexive relation on $A$. Prove that $R$ is antisymmetric.
I tried to apply the definition of the transitive and reflexive and antisymmetric but I didn't get any result.
and thank you in advance for your help 
 A: Do you mean "irreflexive" instead of "not reflexive"?


*

*A relation is reflexive if $\forall x\in A, xRx$.

*It's irreflexive if $\forall x\in A, \neg (xRx)$.

*However, if it's not reflexive, you only know that $\exists x  \in A, \neg(xRx)$.

I ask, because the result you have to prove is wrong.
Take a relation on the set $A=\{x,y,z\}$ with
$\neg (xRx), \neg(yRx),\neg(zRx),xRy, yRy,zRy,xRz,yRz,zRz$
That is, for $(a,b) \in A^2$, $aRb$ is true whenever $b\neq x$.
You may represent $R$ by the following table
$$\begin{matrix}
  & x & y & z \\
x & F & T & T \\
y & F & T & T \\
z & F & T & T \\
\end{matrix}$$
Then $R$ is transitive, but is neither reflexive nor irreflexive, and is not antisymmetric since you have $yRz$ and $zRy$ but not $y=z$.
It's transitive because for $(a,b,c)\in A^3$, if $aRb$ and $bRc$, then necessarily $c\neq x$, so $aRc$ is certainly true.

However, if you assume that $R$ is irreflexive, you can conclude, since by transitivity you have that if $aRb$ and $bRa$, then $aRa$, which is always false, hence one of $aRb$ or $bRa$ is always false, and the implication 
$aRb \wedge bRa \implies a=b$ is thus vacuously true.
