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If multiplication is just repeated addition, then how can be $i\cdot i=-1$?

The question was posted on AoPS Facebook page a few weeks ago. I think this is a question worth thinking. And I want to hear from Math SE community members about it.

Oshawott
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    If multiplication is just repeated addition, how can you make sense of $\pi\cdot (-\sqrt3)$? I suggest you start there, and only begin tackling complex numbers after you understand the real numbers. – Arthur Jul 11 '21 at 17:57
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    Multiplication by natural numbers is like repeated addition: multiplication by $i$ is like $90^\circ$ rotation in the complex plane – J. W. Tanner Jul 11 '21 at 17:58
  • Compare to: if exponentiation is just repeated multiplication, then how is $8^{1/3}=2$ or $e^{i \pi} = -1$. Some "natural" operations are extended to real and complex numbers (or arbitrary rings/fields) in a way that makes sense mathematically, but the elementary definitions do not necessarily hold everywhere. – dxiv Jul 11 '21 at 18:07
  • Time flies like an arrow. Fruit flies like a banana. – Steven Alexis Gregory Jul 12 '21 at 02:17

1 Answers1

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Multiplication isn't just repeated addition. "Repeated addition" tells you how to take $n \cdot a$ where $n \in \mathbb{N}$, but it cannot get you any further than that.

What does it mean to "repeatedly add" something to itself $-1$ times? You have to extend the definition to include "repeated subtraction". This gets you as far as $n \cdot a$ where $n \in \mathbb{Z}$.

What does it mean to repeatedly add something to itself $1/2$ times? It doesn't necessarily mean anything. This is where the scheme starts to really break down, but you can salvage it by saying that $(1/2) \cdot a$ should be the unique value $z$ such that $2 \cdot z = a$. This gets you as far as $q \cdot a$ where $q \in \mathbb{Q}$.

But what does it mean to add something to itself $\sqrt{2}$ times? $\sqrt{2}$ is not rational, so we don't really have a nice answer in terms of repeated addition. The way we compute $\sqrt{2} \cdot a$ is to find a sequence $q_1, q_2, ...$ of rational numbers which converges to $\sqrt{2}$, and then define $\sqrt{2} \cdot a = \lim\limits_{n \to \infty} q_n \cdot a$. This gets you as far as $x \cdot a$ where $x \in \mathbb{R}$.

But what does it mean to add something to itself $i$ times? At this point, we don't even try to come up with a connection to repeated addition. We just define multiplication to work the way we want it to, so that $i \cdot i = -1$ by the definition of multiplication.

Notice that at every stage, we have to redefine multiplication. There are many different multiplication operators: one each for $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$, plus multiplication defined for countless other sets.

A general theory of how multiplication "should work" is the theory of rings, which states that multiplication is just an operation that follows some rules such as $1 \cdot x = x$ and $(a + b) \cdot c = (a \cdot c) + (b \cdot c)$. Ring theory is typically covered in the first or second semester of an abstract algebra curriculum.

Mark Saving
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