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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 $-1 \ne 1$

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  • $\begingroup$ I'm confused, what does $aR2,b$ mean in the title? $\endgroup$ – Eevee Trainer May 12 at 7:57
  • $\begingroup$ sorry its a R b $\endgroup$ – fiksx May 12 at 7:58
  • $\begingroup$ Okay. What's the statement $b \iff a|b$ supposed to mean? I feel like you're trying to say $aRb \iff a|b$ and just made some slight typos, but I want to be sure. $\endgroup$ – Eevee Trainer May 12 at 7:58
  • $\begingroup$ im sorry its typo, i fix it $\endgroup$ – fiksx May 12 at 7:59
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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.)

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  • $\begingroup$ @cansomeonehelpmeout Okay I derped - the reason why I said that is that I looked on the antisymmetry page on Wikipedia because I wanted to be sure on the definition. Conveniently, it had the divisibility relation being antisymmetric there ... but on $\Bbb N$, which I overlooked. You're right, and that's my bad. $\endgroup$ – Eevee Trainer May 12 at 8:13
  • $\begingroup$ thanks but how to proof that it is symmetric ?i dont want to find counter example by trial and error, cause its hard , if i can proof it symmetric to give insight of the counterexample it would be great $\endgroup$ – fiksx May 12 at 8:24
  • $\begingroup$ It's not symmetric. Since $3|6$ for example, but $6$ does not divide $3$, it can't be. If $R$ is symmetric, then $xRy \iff yRx$. The preceding example shows that doesn't hold here. $\endgroup$ – Eevee Trainer May 12 at 8:25
  • $\begingroup$ but it in Z, 6|3 give rational number so it must be not in Z? $\endgroup$ – fiksx May 12 at 8:26
  • $\begingroup$ The definition of $a|b$ is that there exists a $k \in \Bbb Z$ such that $b = ak$. The stipulation that $k$ is an integer is important. If $6|3$ then the $k$ here would be $1/2$ which is not an integer. $\endgroup$ – Eevee Trainer May 12 at 8:27
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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 $4\not\mid 2$.

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  • $\begingroup$ is this has relation with the proof of not symmetry and not antisymmetry? $\endgroup$ – fiksx May 12 at 8:46

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