Why does the imaginary number $i$ satisfy $i\times 0=0$?

Question

Why does the imaginary number $i$ satisfy $i\times 0=0$? I mean, we don't really know what $i$ is. How could we be sure about that? I think there's a reason behind why mathematicians decided that.

Answer 1

Ignore this answer if you've never heard of matrix multiplication. Better yet, learn matrix multiplication and then read this answer!

The answer to your question depends on what definition of complex numbers you're using.

For example, I like to think of the complex numbers as matrices of the form $$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} $$ where $a$ and $b$ are real numbers. This allows us to define two complex numbers \begin{align*} \mathbf 0 &= \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix} & i &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \end{align*} Hence we have the identity $$ \mathbf 0\cdot i = \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0\cdot 0+0\cdot 1 & 0\cdot(-1)+0\cdot 0\\ 0\cdot 0+0\cdot 1 & 0\cdot (-1)+0\cdot 0 \end{pmatrix} = \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix} = \mathbf0 $$

One of the nice things about defining the complex numbers this way is that we avoid the confusing equation $i=\sqrt{-1}$. We also don't have to resort to using silly words like "imaginary."

Note, however, that we do have $$ i^2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} $$ So, if we define the complex number $$ \mathbf{1}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ then we recover the formula $$ i^2=-\mathbf{1} $$ To convince yourself that our definition of the complex number $\mathbf1$ is not arbitrary, note that $\mathbf 1$ enjoys the property $$ \mathbf 1 \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} $$ This is remarkably similar to the celebrated equation $1\cdot a=a$. It's also worth noting that the complex number $\mathbf0$ satisfies $$ \mathbf0+ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} 0&0\\0&0 \end{pmatrix}+ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} 0+a & 0-b \\ 0+b & 0+a \end{pmatrix} = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} $$ which is similar to our usual equation $0+a=a$.

The point here is that complex-arithmetic "feels" like ordinary arithmetic but is indeed different. As @Hurkyl points out, algebraic gadgets that behave like this are called rings.

Answer 2

The complex numbers are defined to satisfy all of the usual ring axioms: i.e. all of the usual identities involving $0,1,+,-,\times,\div$ hold for complex numbers. $0z=0$ is one of those identities.

Answer 3

Just like real numbers, we can write: $0z = (0+0)z = 0z + 0z$. Canceling like terms on each side allows us to see that $0 = 0z$. Thus, this is really a property of zero itself.

Also, as others have mentioned, complex numbers (including imaginaries) are fairly well understood. However, one important nuance is that @iHubble's definition quickly leads to contradictions (such as using properties of square roots to show that $-1 = i^2 = \sqrt{-1}\sqrt{-1} = \sqrt{(-1)^2} = 1$). It is best to instead always define $i$ as a number such that $i^2 = -1$.

Answer 4

Technically, $0\,i\not\in\mathbb{R}$; instead, $0\,i\in\mathbb{C}$. Thus, being (overly?) pedantic, $0\times i\ne0$; instead $0\times i=0+0\,i$. However, being the zero element of $\mathbb{C}$, we usually abbreviate $0+0\,i$ as $0$, not meaning an element of $\mathbb{R}$ but an element of the subset of $\mathbb{C}$ that is homeomorphic to $\mathbb{R}$, that is $\{x+0\,i:x\in\mathbb{R}\}$. With this abbreviation, we do have $0\times i=0$

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This comment by AlexB says

The thing is that when one defines the integers, they come with a canonical embedding of $\mathbb{N}$. Similarly, the rationals come with a canonical embedding of $\mathbb{Z}$ and so on. So it does make sense to speak of subsets. Indeed, it becomes so cumbersome to distinguish between genuine subsets and images under an embedding that one fairly quickly drops this distinction in practice, unless one really has to thing about the foundations.

The question "Why does the imaginary number $i$ satisfy $i\times0=0$?" seems to me to be fairly foundational, so I think that my approach above seems appropriate.

Once we have established why $i\times0=0$, then we can move on as Arturo Magidin says in this answer:

So even though they are actually very different sets, we have copies of each sitting inside the "next one", copies that respect all the structures we are interested in, so we can still think of them as being "subsets".