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Consider again the preorder \( ( \mathbb{R} , \le) \).

Someone proposes 1 as a monoidal unit and \(*\) (usual multiplication) as a monoidal product. But an expert walks by and says “that won’t work.”

Figure out why, or prove the expert wrong!

**Example 2.4**.
There is a well-known preorder structure, denoted \( \le \), on the set \( \mathbb{R} \) of real numbers; e.g. \( −5 \le 2 \) .

**Definition 2.2**.
A symmetric monoidal structure on a preorder \( (X, \le) \) consists of two constituents:

- an element \(I \in X\), called the monoidal unit, and
- a function \( \otimes : X \times X \rightarrow X \), called the monoidal product.

These constituents must satisfy the following properties:

**(a)**for all \(x_1 , x_2 , y_1 , y_2 \in X, \text{ if } x_1 \le y_1 \text{ and } x_2 \le y_2 , \text{ then } x_1 \otimes x_2 \le y_1 \otimes y_2 \)**(b)**for all \(x \in X\), the equations \(I \otimes x = x\) and \(x \otimes I = x\) hold**(c)**for all \(x, y, z \in X\), the equation \((x \otimes y) \otimes z = x \otimes (y \otimes z) \) holds, and**(d)**for all \(x, y \in X\), the equivalence \(x \otimes y \cong y \otimes x\) holds.

A preorder equipped with a symmetric monoidal structure, \( (X, \le, I, \otimes) \), is called a symmetric monoidal preorder.

## Comments

Puzzle FE-1Would \( I = 1 \) and \( \otimes = * \) with the preorder \( (\mathbb{R}^+, \le) \) form a symmetric monoidal preorder?`**Puzzle FE-1** Would \\( I = 1 \\) and \\( \otimes = \* \\) with the preorder \\( (\mathbb{R}^+, \le) \\) form a symmetric monoidal preorder?`

Let \(x_1, x_2, y_1, y_2\) be \(-2, -3, 2, 1\), respectively. We have \( −2 \le 2 \) and \(-3 \le 1\). However, \((-2) ∗ (-3) \) is greater than \( 2 ∗ 1 \): thus, condition 1 of Definition 2.1 is not satisfied. This is similar to Puzzle 61. I'm tempted to say that, if we consider positive reals, the answer is yes -- I did not find any counterexample yet.

`Let \\(x_1, x_2, y_1, y_2\\) be \\(-2, -3, 2, 1\\), respectively. We have \\( −2 \le 2 \\) and \\(-3 \le 1\\). However, \\((-2) ∗ (-3) \\) is greater than \\( 2 ∗ 1 \\): thus, condition 1 of Definition 2.1 is not satisfied. This is similar to Puzzle 61. I'm tempted to say that, if we consider positive reals, the answer is yes -- I did not find any counterexample yet.`