Lecture 21  Monoidal Preorders
Now let's look at a mathematical approach to resource theories. As I've mentioned, resource theories let us tackle questions like these:
 Given what I have, is it possible to get what I want?
 Given what I have, how much will it cost to get what I want?
 Given what I have, how long will it take to get what I want?
 Given what I have, what is the set of ways to get what I want?
Our first approach will only tackle question 1. Given \(y\), we will only ask is it possible to get \(x\). This is a yesorno question, unlike questions 24, which are more complicated. If the answer is yes we will write \(x \le y\).
So, for now our resources will form a "preorder", as defined in Lecture 3.
Definition. A preorder is a set \(X\) equipped with a relation \(\le\) obeying:

reflexivity: \(x \le x\) for all \(x \in X\).

transitivity \(x \le y\) and \(y \le z\) imply \(x \le z\) for all \(x,y,z \in X\).
All this makes sense. Given \(x\) you can get \(x\). And if you can get \(x\) from \(y\) and get \(y\) from \(z\) then you can get \(x\) from \(z\).
What's new is that we can also combine resources. In chemistry we denote this with a plus sign: if we have a molecule of \(\text{H}_2\text{O}\) and a molecule of \(\text{CO}_2\) we say we have \(\text{H}_2\text{O} + \text{CO}_2\). We can use almost any symbol we want; Fong and Spivak use \(\otimes\) so I'll often use that. We pronounce this symbol "tensor". Don't worry about why: it's a long story, but you can live a long and happy life without knowing it.
It turns out that when you have a way to combine things, you also want a special thing that acts like "nothing". When you combine \(x\) with nothing, you get \(x\). We'll call this special thing \(I\).
Definition. A monoid is a set \(X\) equipped with:
 a binary operation \(\otimes : X \times X \to X\)
 an element \( I \in X \)
such that these laws hold:

the associative law: \( (x \otimes y) \otimes z = x \otimes (y \otimes z) \) for all \(x,y,z \in X\)

the left and right unit laws: \(I \otimes x = x = x \otimes I\) for all \(x \in X\).
You know lots of monoids. In mathematics, monoids rule the world! I could talk about them endlessly, but today we need to combine the monoids and preorders:
Definition. A monoidal preorder is a set \(X\) with a relation \(\le\) making it into a preorder, an operation \(\otimes : X \times X \to X\) and element \(I \in X\) making it into a monoid, and obeying:
$$ x \le x' \textrm{ and } y \le y' \textrm{ imply } x \otimes y \le x' \otimes y' .$$
This last condition should make sense: if you can turn an egg into a fried egg and turn a piece of bread into a piece of toast, you can turn an egg and a piece of bread into a fried egg and a piece of toast!
You know lots of monoidal preorders, too! Many of your favorite number systems are monoidal preorders:

The set \(\mathbb{R}\) of real numbers with the usual \(\le\), the binary operation \(+: \mathbb{R} \times \mathbb{R} \to \mathbb{R} \) and the element \(0 \in \mathbb{R}\) is a monoidal preorder.

Same for the set \(\mathbb{Q}\) of rational numbers.

Same for the set \(\mathbb{Z}\) of integers.

Same for the set \(\mathbb{N}\) of natural numbers.
Money is an important resource: outside of mathematics, money rules the world. We combine money by addition, and we often use these different number systems to keep track of money. In fact it was bankers who invented negative numbers, to keep track of debts! The idea of a "negative resource" was very radical: it took mathematicians over a century to get used to it.
But sometimes we combine numbers by multiplication. Can we get monoidal preorders this way?
Puzzle 60. Is the set \(\mathbb{N}\) with the usual \(\le\), the binary operation \(\cdot : \mathbb{N} \times \mathbb{N} \to \mathbb{N}\) and the element \(1 \in \mathbb{N}\) a monoidal preorder?
Puzzle 61. Is the set \(\mathbb{R}\) with the usual \(\le\), the binary operation \(\cdot : \mathbb{R} \times \mathbb{R} \to \mathbb{R}\) and the element \(1 \in \mathbb{R}\) a monoidal preorder?
Puzzle 62. One of the questions above has the answer "no". What's the least destructive way to "fix" this example and get a monoidal preorder?
Puzzle 63. Find more examples of monoidal preorders.
Puzzle 64. Are there monoids that cannot be given a relation \(\le\) making them into monoidal preorders?
Puzzle 65. A monoidal poset is a monoidal preorder that is also a poset, meaning
$$ x \le y \textrm{ and } y \le x \textrm{ imply } x = y $$
for all \(x ,y \in X\). Are there monoids that cannot be given any relation \(\le\) making them into monoidal posets?
Puzzle 66. Are there posets that cannot be given any operation \(\otimes\) and element \(I\) making them into monoidal posets?
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