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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.

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    $\begingroup$ We've understood complex numbers for a few centuries now; the meaning of mathematical word "imaginary" hasn't had any relation to the meaning of the English word "imaginary" for a long time. $\endgroup$ – Hurkyl Jul 30 '14 at 5:11
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    $\begingroup$ How is closing this helpful? It is clear what the question is, and it is a good un'! It is also not obvious to me how to prove $0\times i=0$ given simply that $i^2=-1$. It is a theorem, not something we have decided. Thus, an answer will prove this theorem! (As, for example, user13157 has done.) $\endgroup$ – user1729 Jul 30 '14 at 10:07
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    $\begingroup$ @Did Hmm, I don't know. How do you define $\mathbb{C}$ if not as "the real numbers with an element $i$ such that $i^2=-1$"? (Implicitly, we assume brackets work as we like.) Then the result follows according to user13157's proof (and I remember people struggling with the analogous proof in a first course on rings - I do not think "trivial" is really the best word!). $\endgroup$ – user1729 Jul 30 '14 at 10:23
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    $\begingroup$ @Did I have three relevant books on my desk: Calculus: A complete Course by Robert A. Adams, Engineering Mathematics by K.A.Stroud, and Fundamentals of University Mathematics by McGregor, Nimmo and Stothers. The first two, Adams and Stroud, go for the "$i^2=-1$ and complex numbers have the form $ai+b$, $a, b\in\mathbb{R}$" definition, while McGregor, Nimmo and Stothers start with the complex plane and work backwards. So...I would think that the "extension" was quite common. No? $\endgroup$ – user1729 Jul 30 '14 at 10:52
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    $\begingroup$ @robjohn If you want to learn more about the general way in which complex numbers were intorduced to Did, look up the "third" book I mention in my comments above, Fundamentals of University Mathematics by McGregor, Nimmo and Stothers. The main advantage of this method, as I see it, is that you start with the complex plane. So it is more geometric and less algebraic. $\endgroup$ – user1729 Aug 6 '14 at 8:29
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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.

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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 $i^2 = -1$.

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  • $\begingroup$ It's not a "property of zero itself". Rather, it is a property of any algebraic structure (e.g. ring) which satisfies the laws that you employed in your proof, e.g. the distributive law, and additive cancellation law. $\endgroup$ – Bill Dubuque Aug 5 '14 at 13:10
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    $\begingroup$ @BillDubuque: I think he means it is a property of $0$ in any ring, in this case, $\mathbb{C}$. $\endgroup$ – robjohn Aug 5 '14 at 22:15
  • $\begingroup$ @robjohn Possibly. The point of my comment was to help clarify that, i.e. to whittle it down to the algebraic essence of the matter (which may not be so clear to readers who have not had much experience with abstract algebra) $\endgroup$ – Bill Dubuque Aug 5 '14 at 22:25
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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.

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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 $0\times i=0$


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".

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  • $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn Aug 6 '14 at 5:08
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    $\begingroup$ I didn't downvote, but I am not convinced by your pedantry. (Although it certainly made me think!) Are you also claiming that $i^2=-1\not\in\mathbb{R}$? (Mainly I am not convinced because, if we adopt the language of rings which seems to be the fashion in these answers, $0i$ is contained in the subring $\mathbb{R}$, $0i\in\mathbb{R}$. No?) $\endgroup$ – user1729 Aug 6 '14 at 8:22
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    $\begingroup$ @user1729: $\mathbb{R}$ is one-dimensional. $\{x+0\,i:x\in\mathbb{R}\}$ is a one-dimensional subspace of $\mathbb{C}$. They are isomorphic, but not the same. It is possible that after $\mathbb{C}$ is introduced, $\mathbb{R}$ could be redefined to be this one-dimensional subspace of $\mathbb{C}$, but this feels circular to me. $\endgroup$ – robjohn Aug 6 '14 at 12:17
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    $\begingroup$ Another downvote without comment? Consider this: is $(0,0)\in\mathbb{R}$? How can $\mathbb{R}$ be a subset of $\mathbb{C}$ when we can't even define $\mathbb{C}$ without $\mathbb{R}$ so that either $\mathbb{C}=\{x+iy:x,y\in\mathbb{R}\}$ or $\mathbb{C}=\{(x,y):x,y\in\mathbb{R}\}$. There seems to be a confusion between $\mathbb{R}$ and the subspace of $\mathbb{C}$ that is homeomorphic to $\mathbb{R}$. $\endgroup$ – robjohn Aug 7 '14 at 0:52
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    $\begingroup$ There is a well-established tradition in mathematics of identification that permeates it at all levels. This practice is endlessly convenient and enlightening. E.g. we identify things with other things in order to get $\Bbb N\subset\Bbb Z\subset\Bbb Q\subset\Bbb R\subset\Bbb C$. It's true that on a machine-language level, pedantry is useful for rigorous constructions, but after the constructions are done or trivial or subconscious and we've moved on, it is no longer useful but distracting and gets in the way of intuitively seeing the big picture and what's "morally correct." $\endgroup$ – blue Aug 7 '14 at 1:38

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