on the set of all integer,For all $a, b ∈ Z, a R b,$ $ ⇔ a | b, $ is R antisymmetric? on the set of all integer,For all $a, b ∈ Z, a R b,$ $⇔ a | b, $ is  R antisymmetric? 
the answer is symmetric  
but i dont know how to prove it and how to find the counter example
$a,b \in Z$ 
$ka=b$ , $cb=a$ 
$(kc)b=b$ 
$kc=1$
but since all integer either $k=c=1$ ,  or $k=c=-1$ 
but this weird if k=c=1 then b=a, im not prove this symmetric 
is this correct??
but how to find counter example?
for symmetric i need to find $a,b \in Z$ and $b,a \in Z$ that
$a \ne b$
is the counter example 1|-1  and -1|1 ? 
$(1,-1) , (-1,1) \in Z$ such that $-1 \ne 1$
 A: So recall the symmetry property of relations: if $R$ is symmetric, then $aRb \iff bRa$. For this scenario, then, $R$ is symmetric $b|a \iff a|b$. This obviously doesn't hold for quite a few numbers. For example,  $3|6$ but $6 \not \mid 3$.
$1,-1$ do not provide a counterexample as $1|-1$ and $-1|1$. This follows as $1 = (-1)(-1)$ ($1$ is an integer multiple of $-1$, meeting the definition of divisibility), and $-1 = 1(-1)$. Of course as noted above there are other counterexamples so in the end it's somewhat moot.
Not sure why you have the answer given as symmetric though.
Interestingly enough this counterexample is not without merit - this gives us a case for $R$ not being an antisymmetric relation either. For a relation $R$ to be antisymmetric, if $xRy$ and $yRx$, then $x=y$. However, we have $1R(-1)$ and $(-1)R1$ yet $1 \ne -1$. So $R$ is not antisymmetric.
(Thanks to cansomeonehelpmeout for pointing out a dumb oversight of mine in the comments.)
A: Well, if $a=kb$ and $b=la$, then $(kl)b = b$.
If $b\ne 0$, the shortening rule holds: $ab=cb\Rightarrow a=c$. (In general, this holds for all nonzero-divisors $b$ in a commutative ring $R$. But the ring of integers has no zero-divisors.)
Here $kl=1$. But the only invertible elements in the ring of integers are $\pm 1$.
Thus $k=l=1$ or $k=l=-1$. Hence, $a=b$ or $a=-b$.
In view of antisymmetry, $a\mid b\wedge b\mid a\Rightarrow a=b$, this is wrong for integers, since $2\mid -2 \wedge -2\mid 2$, but $2\not= -2$. This shows the proof above.
In view of symmetry, $a\mid b\Rightarrow b\mid a$ is also wrong, since $2\mid 4$ but $4\not\mid 2$.
