**Puzzle 60.** Yes. Suppose we have \\(x \le y\\) and \\(x' \le y'\\). We know all of the variables are nonnegative, so we can freely multiply our variables into these inequalities to obtain \\(x \cdot x' \le y \cdot x'\\) and \\(y \cdot x' \le y \cdot y'\\); by transitivity, we have \\(x \cdot x' \le y \cdot y'\\).

**Puzzle 61.** No, because Puzzle 62 says one of Puzzle 60 and Puzzle 61 must have "no" as its answer. ;) Without appealing to outside information, we can merely demonstrate that \\(-1 \le 1\\) and \\(-2 \le 1\\) but \\(-1 \cdot -2 = 2 \not\le 1 = 1 \cdot 1\\).

**Puzzle 62.** Define \\(x \preceq y\\) by \\(\lvert x \rvert \le \lvert y \rvert\\). Then if \\(x \preceq y\\) and \\(x' \preceq y'\\), we know \\(\lvert x \rvert \le \lvert y \rvert\\) and \\(\lvert x' \rvert \le \lvert y' \rvert\\); these are all positive quantities as before, so by the same process we find \\(\lvert x \cdot x' \rvert = \lvert x \rvert \cdot \lvert x' \rvert \le \lvert y \rvert \cdot \lvert y' \rvert = \lvert y \cdot y' \rvert\\). Therefore, \\(x \cdot x' \preceq y \cdot y'\\), as desired.

We do get two units out of this (\\(-1\\) and \\(1\\)), but I think that's okay as long as we pick one and stick to it. They're equivalent in the sense that \\(-1 \preceq 1\\) and \\(1 \preceq -1\\).

**Puzzle 63.** The structures \\(\langle \mathcal{P}(X), \subseteq, \cup, \emptyset \rangle\\) and \\(\langle \mathcal{P}(X), \subseteq, \cap, X \rangle\\) are monoidal preorders. More generally, any lattice possesses two (dual) monoidal structures.

**Puzzle 64.** We can always apply either the discrete or codiscrete preorders to make a monoid into a monoidal preorder.

**Puzzle 65.** We can always apply the discrete preorder to make a monoid into a monoidal poset.

**Puzzle 66.** This is surprisingly hard to think through. I've got a weak hunch that the poset below cannot be made into a monoidal poset, but I'm still working through it. The poset can't be a lattice, per Puzzle 63, and this is the smallest non-semilattice non-trivial-order at my disposal.

![](https://i.imgur.com/Km2PosD.png)

**Puzzle 61.** No, because Puzzle 62 says one of Puzzle 60 and Puzzle 61 must have "no" as its answer. ;) Without appealing to outside information, we can merely demonstrate that \\(-1 \le 1\\) and \\(-2 \le 1\\) but \\(-1 \cdot -2 = 2 \not\le 1 = 1 \cdot 1\\).

**Puzzle 62.** Define \\(x \preceq y\\) by \\(\lvert x \rvert \le \lvert y \rvert\\). Then if \\(x \preceq y\\) and \\(x' \preceq y'\\), we know \\(\lvert x \rvert \le \lvert y \rvert\\) and \\(\lvert x' \rvert \le \lvert y' \rvert\\); these are all positive quantities as before, so by the same process we find \\(\lvert x \cdot x' \rvert = \lvert x \rvert \cdot \lvert x' \rvert \le \lvert y \rvert \cdot \lvert y' \rvert = \lvert y \cdot y' \rvert\\). Therefore, \\(x \cdot x' \preceq y \cdot y'\\), as desired.

We do get two units out of this (\\(-1\\) and \\(1\\)), but I think that's okay as long as we pick one and stick to it. They're equivalent in the sense that \\(-1 \preceq 1\\) and \\(1 \preceq -1\\).

**Puzzle 63.** The structures \\(\langle \mathcal{P}(X), \subseteq, \cup, \emptyset \rangle\\) and \\(\langle \mathcal{P}(X), \subseteq, \cap, X \rangle\\) are monoidal preorders. More generally, any lattice possesses two (dual) monoidal structures.

**Puzzle 64.** We can always apply either the discrete or codiscrete preorders to make a monoid into a monoidal preorder.

**Puzzle 65.** We can always apply the discrete preorder to make a monoid into a monoidal poset.

**Puzzle 66.** This is surprisingly hard to think through. I've got a weak hunch that the poset below cannot be made into a monoidal poset, but I'm still working through it. The poset can't be a lattice, per Puzzle 63, and this is the smallest non-semilattice non-trivial-order at my disposal.

![](https://i.imgur.com/Km2PosD.png)