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a divides b is not a partial ordering on the set Z of integers since a | b and b | a need not imply a = b.

This is a line from Schaum's outline of discrete mathematics ( chapter 2 Relations pg. 33)

I understood that it is a POSET for set of Natural numbers $N$. But, it even holds the necessary conditions to be a POSET on set of integers $Z$.

My approach:

  1. It is reflexive as $a|a\ \forall a \in Z $.
  2. It is anti-symmetric as $a|b \ , b|a \implies a=b \forall a,b \in Z$.
  3. It is transitive too.

Then why is it mentioned that it is not a POSET on $Z$ ?

Asaf Karagila
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taurus05
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    Two of the conditions remain true. One fails. Here's a slightly obscure hint: $(-1)^2=1$. – Jamie Radcliffe Aug 11 '21 at 17:08
  • @JamieRadcliffe Thanks for the quick response. This is really helpful because $1|-1$ and $-1|1$ but 1 != -1. Hence not antisymmetric and so not a poset on Z. Thank you! – taurus05 Aug 11 '21 at 17:10
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    In general, $-n\mid n$ and vice versa – J. W. Tanner Aug 11 '21 at 17:12
  • @J.W.Tanner, yes sir indeed. If it was asked to check if Is a divides b a POSET on set of non negative integers $Z+$, it would've been false, right? As it is not reflexive when a is 0. – taurus05 Aug 11 '21 at 17:15
  • @taurus05 That's actually a poset, there's no such problem with $0$, it will be simply the greatest element: we have $n|0$ for every nonnegative integer $n$, while $0|n\implies n=0$. – Berci Aug 11 '21 at 18:22
  • @Berci Thanks for your valuable feedback. – taurus05 Aug 11 '21 at 19:46
  • Apart from this, i also found that $0 \in Z$ and $0|0\ is\ undefined$. Hence, it is not reflexive too. Is this right? – taurus05 Aug 13 '21 at 07:51

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