Why the divides relation on the set of positive integers antisymmetric
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 $2|1$.
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$\begingroup$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.
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