Definition of totally ordered monoid?

Question

Suppose I have the monoid $(\mathbb{N},\times)$.

It is my understanding that for the relation $\le$ to form a total order on the above monoid, the following must be true:

$$a\le b\iff (\forall c\in\mathbb{N})\ a\times c\le b\times c$$

If that is the case, then I could plug in $10$ for $a$, $5$ for $b$ and $0$ in for $c$ and arrive at a contradiction:

$$\begin{align} 10\le5&\iff \ 10\times 0\le 5\times 0\\ F&\iff \ 0\le 0\\ F&\iff \ T\\ \end{align}$$

However if instead the definition was (if vs iff):

$$a\le b\implies (\forall c\in\mathbb{N})\ a\times c\le b\times c$$

Then the contradiction I showed goes away (the right hand side becomes vacuously true). So which is the right definition? Or really, which is the more useful definition?

Answer 1

There is a logical error in your argument: In $$a\le b\iff (\forall c\in\mathbb{N})\ a\times c\le b\times c$$ you cannot "plug in" something for $c$, as the $\forall$ is not at the top level.

Consider that:

• "I am 12 feet tall iff for every person $x$, $x$ is a mathematician" is true, because I am not 12 feet tall and there are people who are not mathematicians.
• "I am 12 feet tall iff Jean-Pierre Serre is a mathematician" is false, because I am not 12 feet tall but Jean-Pierre Serre is a mathematician. So "plugging in" Serre for $x$ is not valid.

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For a monoid $(M,\times)$, which definition you use doesn't matter, as the two definitions you propose are actually equivalent. This is because the implication $$a\le b \Longleftarrow (\forall c\in M)\ a\times c\le b\times c$$ is always true: assuming $(\forall c\in M)\ a\times c \le b\times c$ we can plug in $c = 1$ to obtain $a \le b$.

For a general magma, the $\Longrightarrow$ definition is probably the "right" one and as far as I can find, it is the only one that is used. See for example https://en.wikipedia.org/wiki/Ordered_semigroup