# Intuitive justification of associative law

It has been previously asked how one can see that multiplication of real numbers is associative. The answer given there is this:

Let's use the following analogy for the multiplication case; suppose we have a block of wood in the corner of the room, dimensions $a \times b \times c$. It doesn't matter whether you lift the side with dimension $a \times b$ up $c$ or if you take a wood-line $a$ and have it fill out the area $b \times c$; it will fill the same volume. You can extend this to a single negative number if you "drill through walls" with a sort of wood-debt; multiple negatives will require a kind of checkerboard pattern for the corner of the room.

I understand why $a\times (b\times c)=(a\times b)\times c$ for nonnegative $a, b, c$. Now if I want to see why this holds in general (where $a, b, c$ are any – possibly negative – real numbers), do I have to go through all of the 8 combinations of algebraic signs that the numbers $a, b, c$ could have? I mean a case analysis like this:

1. $a, b$, and $c$ are all nonnegative
2. $a$ is negative and $b,c$ nonnegative
3. $a$ and $b$ are negative and $c$ nonnegative
4. all of $a, b, c$ are negative

Or is there an easier way to see why $a\times (b\times c)=(a\times b)\times c$ holds in general? An intuition?

• Do you accept general, or at least pairwise, commutativity of multiplication? Are you willing to replace $-8$ with $-1 \times 8$? Nov 5, 2016 at 16:03
• @EricTowers: yes. And now?
– user384011
Nov 5, 2016 at 17:34
• Commutativity is obvious. Associativity is an easy corollary. Nov 5, 2016 at 20:17

Since you accept commutativity, you can reduce to four cases:

First, eliminate a triviality: If any of the multiplicands is $0$, the product is $0$. Therefore, assume none of the multiplicands is $0$. (We can skip this and modify the four cases below to accommodate the change, but then they are harder for a reader to follow. (Replace each "positive" with "nonnegative" and each "negative" with "nonpositive".) Since confusing the reader is not the goal, ...)

By commutativity, we have $$a \times b \times c = b \times c \times a = c \times a \times b$$ and we may assume that all the negative factors are written before all the positive factors. This reduces the problem to four cases:

1. Suppose $a$, $b$, and $c$ are all positive. This case is associative by previous work.
2. Suppose $a$ is negative and both $b$ and $c$ are positive. ...
3. Suppose $a$ and $b$ are negative and $c$ is positive. ...
4. Suppose $a$, $b$, and $c$ are all negative. ...

You can save yourself half of those cases by first showing arbitrary associativity among $-1$ and all positive numbers: $$(-1 \times a) \times b = -1 \times a \times b = -1 \times (a \times b) \text{.}$$ (You can motivate this with the "drilled out wood" analogy from your quoted material.)

Then case 4 above is the same as case 2:
$$a \times (-1 \times (-b)) \times (-1 \times (-c)) = a \times -b \times (-1 \times -1) \times -c = a \times -b \times -c$$ and case 3 is the same as case 1: $$(-1 \times (-a)) \times (-1 \times (-b)) \times c = -a \times (-1 \times -1) \times -b \times c = -a \times -b \times c$$

This only leaves case 2 to work out.

Given your intuition from the block example, the only intuition left that you need for negative numbers is that the sign of a measurement is a product of interpretation. For instance 500 dollars in debt is a savings of -500 dollars. A 500 pound force in one direction is a -500 pound force in the other direction. Thus the magnitude of the measurement should not be affected by the sign.