It is known that

  1. If you multiply more than 1 natural numbers together, it doesn't matter which order you put them in.
  2. If you multiply more than 2 natural numbers together, then it makes no difference how you bracket them.

If we are dealing with 2 natural numbers in 1., then it is the commutative law of multiplication. If we are dealing with 3 natural numbers in 2., then it is the associative law of multiplication. In other cases, 1. and 2. can be proven using the commutative and associative laws of multiplication and by induction on the natural numbers.

I came up with a non-abstract argument for 1., and would like some help in coming up with a non-abstract argument for 2. as well.

A non-abstract argument for 1. (not limited to only 2 natural numbers, can be 3, 4, natural numbers etc)

Let $a,b \in \mathbb{N}$. I excluded $0$ from $\mathbb{N}$. I view $a\cdot b$ as having $b$ times of $a$. That is, $a\cdot b = a + a + \dots + a$, where there are $b$ many $a$'s in the sum.

For example, I would view $2\cdot 5\cdot 8\cdot 3$ as starting with $2$ objects, I take $5$ groups of them, then take $8$ groups of the $2\cdot 5$ objects, and finally take $3$ groups of $2\cdot 5\cdot 8$.

According to 1., for example, $2\cdot 5\cdot 8\cdot 3 = 8\cdot 3\cdot 2\cdot 5$.

I'll start with any side of the equation and show it is equal to the other.

Starting with $8\cdot 3\cdot 2\cdot 5$, I could think $8$ as 8 groups of some objects, rather than $8$ objects. To do so, I first distribute each $1$ of the $8$ objects into separate group, thus having $1$ object in a group, and a total of $8$ groups.

Then because $3$ comes after $8$ in the expression $2\cdot5\cdot8\cdot3$, I will just take $3$ groups of $8$. Now, I have $3$ groups of $8$, whereby each $8$ is a group containing $1$ object. ($8\cdot3$)

Then $2$ comes before $8$ in the expression $2\cdot5\cdot8\cdot3$, so I will need to have $2$ objects in each of the $8$ groups, instead of $1$ object. This is the same as multiplying $8\cdot 3$ by $2$, thus having $8\cdot3\cdot2$.

Lastly, $5$ is between $2$ and $8$. I just need to multiply the $2$ objects in each of the $8$ groups by $5$, to form $5$ groups of $2$ objects within each of the $8$ groups. And this is the same as multiplying $8\cdot 3\cdot 2$ by $5$.

Thus $2\cdot 5\cdot 8\cdot 3 = 8\cdot 3\cdot 2\cdot 5$.

This non-abstract argument is similar to the interpretation of stacking objects in a configuration of a cube. First place $a$ objects in a row, then $a\cdot b$ means have $b$ rows of $a$, thus forming $a$ rows and $b$ columns, then $a\cdot b \cdot c$ by stacking one on top of the other, thus having $c$ "height". And to count the total number of objects, we can look at either the rows, columns, or "heights" first.

I would like some help to come up with a non-abstract argument for 2. as well. Thanks.


1 Answer 1


Interesting question! Firstly, a couple of things to point out.

You say that

From the abstract point of view, 1. and 2. can be proven using the fact that the operation multiplication is both commutative and associative.

This is not quite right. 1 and 2 are, as you have written them, the definitions of commutativity and associativity.

Now, you go on to show us your "non-abstract" argument for 1. What you really mean is a proof. That's absolutely fine language to use. To be clear, what you are attempting to prove is that multiplication is both commutative and associative on the natural numbers $\mathbb{N}$.

The proof that you provide however, needs some work. What you've got there is an example where you shows how your argument should work in practise. To be precise, what you're aiming to show is:

For any $a,b \in \mathbb{N}, a \cdot b = b \cdot a $.

You have only shown us one case. You should bash your head against the wall for a while trying to figure out how you can show this in general (i.e. for any $a$ and $b$). It'll be bad for the walls, but it'll be worth it.

Now, to answer your question, what you are looking to show is that

For any $a,b, c \in \mathbb{N}, (a \cdot b) \cdot c = a \cdot (b \cdot c)$.

I won't write a proof out for you, because that would ruin it a bit for you, but I think that I can provide some hints:

  1. In your first attempt at a proof for 1., you have the right idea but it's only in one case. In words, you are showing that "$b$ lots of $a$ is the same as $a$ lots of $b$". Using similar words, for 2., you'd like to show that $c$ lots of $(a \cdot b)$ is the same as $a$ lots of $(b \cdot c)$.

  2. In order to show what I suggest above, you may need to "expand brackets" more than once.

  3. It is definitely worth trying to iron 1. out properly before you attempt 2.

  • $\begingroup$ I have edited the question slightly. $\endgroup$
    – yh05
    Mar 23, 2019 at 3:09

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