Ken's formula for inverses:

> The inverse operation takes \\(x\\) to \\(-x\\), \\(-x\\) to \\(x\\), and the identity 0 to itself.

works better than Keith's:

> However, \\(-x \mapsto x\\), \\(x \mapsto x\\), and \\(0 \mapsto 0\\) I think will work.

You're on the right track, Ken. But the case of building the integers from the natural numbers fails to illustrate some of the strange things that can happen. So try applying your procedure to the Boolean monoid! This is the monoid with two elements 0 and 1, and addition defined as follows:

\[ 0 + 0 = 0, \qquad 0 + 1 = 1, \qquad 1 + 0 = 1, \qquad 1 + 1 = 1 .\]

In the end everything will work out beautifully, and you're on the road to success, but there is a road bump.

> The inverse operation takes \\(x\\) to \\(-x\\), \\(-x\\) to \\(x\\), and the identity 0 to itself.

works better than Keith's:

> However, \\(-x \mapsto x\\), \\(x \mapsto x\\), and \\(0 \mapsto 0\\) I think will work.

You're on the right track, Ken. But the case of building the integers from the natural numbers fails to illustrate some of the strange things that can happen. So try applying your procedure to the Boolean monoid! This is the monoid with two elements 0 and 1, and addition defined as follows:

\[ 0 + 0 = 0, \qquad 0 + 1 = 1, \qquad 1 + 0 = 1, \qquad 1 + 1 = 1 .\]

In the end everything will work out beautifully, and you're on the road to success, but there is a road bump.