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I'd like to know why the divides relation on the set of positive integers antisymmetric. The book says $a|b$ and $b|a$ then $a=b$. But I think if a|b and b not divides a for example $1|2$ but not $2|1$.

amWhy
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user32104
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1 Answers1

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The relation is antisymmetric if and only if for every $a, b$ in the set,

IF $(a\mid b$ AND $b\mid a)$, then it must follow that $a = b$.

If it's NOT true that both $a\mid b$ AND $b\mid a$, then it's perfectly consistent to have $a \neq b$. Indeed, the only time $a \mid b$ AND $b\mid a$ is exactly when $a = b$, since then we have $a \mid b \iff a \mid a = \text{true for all a}$. Hence the relation is antisymmetric.

Antisymmetry here doesn't mean that it must hold that $a \mid b$ and $b\mid a$. It is true that to be symmetric, the relation must be such that if $a \mid b$, then $b\mid a$, too. So clearly, this relation is NOT symmetric.

But since $a \mid b$ and $ b\mid a$ is true if and only if $a = b$, then the relation satisfies the property of being ANTI-symmetric.

amWhy
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  • But I think it's false that a|b and b|a ,right? – user32104 Dec 13 '13 at 14:21
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    It's true if and only if $a = b$. Take $a = b = 2$. Then $2\mid 2$ and $2\mid 2$. If $a\neq b$, then it may be that $a\mid b$, but not $b\mid a$, or else $b\mid a$ but not $a\mid b$, or else neither one divides the other. – amWhy Dec 13 '13 at 14:25
  • Thank you but as I know if p then q is not equal if q then p – user32104 Dec 13 '13 at 15:08
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    Antisymmetric: Let $p: a\mid b; \land ; b\mid a$. Let $q: a = b$. Then a relation is antisymmetric if and only if $p \rightarrow q$. If $p$ happens to be false, that automatically makes $p\rightarrow q$ true, regardless of whether q is true or false, (hence in this case, true means antisymmetric). Recall that an implication is true whenever $p$ is false. – amWhy Dec 13 '13 at 15:13
  • @amWhy: Needs another TU +1 – Amzoti Dec 14 '13 at 01:17