Ok, here I'll give it a go.

> (a) Consider the category where the objects are natural numbers and where there is a unique morphism from m to n if m divides n. Given two numbers, for example 42 and 27, what is their product? What is the name of this binary operation?

Given two objects A, B in a category, a product of A and B consists of another object, call it \\(A \times B\\), along with two morphisms \\(\pi_1: A \times B \rightarrow A\\), and \\(\pi_2: A \times B \rightarrow B\\), which satisfies a certain universal property in the category.

But before talking about the universal property, let's instantiate this first part of the definition to the category at hand.

Here, object A = m, and object B = n.

We have to now posit what the object \\(A \times B\\) is, and provide the two morphisms \\(\pi_1: A \times B \rightarrow A\\) and \\(\pi_2: A \times B \rightarrow B\\).

@FabricioOlivetti posited that \\(A \times B\\) is the greatest common divisor of m and n, gcd(m,n). Indeed it is, as we are now proving.

Claim: \\(A \times B = m \times n = gcd(m,n)\\)

(Note, don't be thrown off by the odd appearance of the above equation. Here the symbol \\(\times\\) is being used in a generic categorical sense, which in the case is being used to describe an operation that is not multiplication.)

Now we must produce \\(\pi_1: m \times n \rightarrow m\\) and \\(\pi_2: m \times n \rightarrow n\\).

But because this category is "thin," which means that there is at most one arrow between any two objects, there could only be one possible choice for \\(pi_1\\) and \\(\pi_2\\). Let's start with \\(\pi_1: gcd(m,n) \rightarrow m\\). Does any such arrow exist going from gcd(m,n) to m? Well yes, because gcd(m,n) divides m! So that is what \\(\pi_1\\) is. Similarly, \\(\pi_2\\) is the unique arrow going from gcd(m,n) to n.

> (a) Consider the category where the objects are natural numbers and where there is a unique morphism from m to n if m divides n. Given two numbers, for example 42 and 27, what is their product? What is the name of this binary operation?

Given two objects A, B in a category, a product of A and B consists of another object, call it \\(A \times B\\), along with two morphisms \\(\pi_1: A \times B \rightarrow A\\), and \\(\pi_2: A \times B \rightarrow B\\), which satisfies a certain universal property in the category.

But before talking about the universal property, let's instantiate this first part of the definition to the category at hand.

Here, object A = m, and object B = n.

We have to now posit what the object \\(A \times B\\) is, and provide the two morphisms \\(\pi_1: A \times B \rightarrow A\\) and \\(\pi_2: A \times B \rightarrow B\\).

@FabricioOlivetti posited that \\(A \times B\\) is the greatest common divisor of m and n, gcd(m,n). Indeed it is, as we are now proving.

Claim: \\(A \times B = m \times n = gcd(m,n)\\)

(Note, don't be thrown off by the odd appearance of the above equation. Here the symbol \\(\times\\) is being used in a generic categorical sense, which in the case is being used to describe an operation that is not multiplication.)

Now we must produce \\(\pi_1: m \times n \rightarrow m\\) and \\(\pi_2: m \times n \rightarrow n\\).

But because this category is "thin," which means that there is at most one arrow between any two objects, there could only be one possible choice for \\(pi_1\\) and \\(\pi_2\\). Let's start with \\(\pi_1: gcd(m,n) \rightarrow m\\). Does any such arrow exist going from gcd(m,n) to m? Well yes, because gcd(m,n) divides m! So that is what \\(\pi_1\\) is. Similarly, \\(\pi_2\\) is the unique arrow going from gcd(m,n) to n.