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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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.
Brian Fitzpatrick
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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 $i^2 = -1$.
PepeOfMath
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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. – Bill Dubuque Aug 05 '14 at 13:10
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1@BillDubuque: I think he means it is a property of $0$ in any ring, in this case, $\mathbb{C}$. – robjohn Aug 05 '14 at 22:15
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@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) – Bill Dubuque Aug 05 '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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2I 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?) – user1729 Aug 06 '14 at 08:22
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1@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. – robjohn Aug 06 '14 at 12:17
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1Another 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 $\mathbb{R}$. – robjohn Aug 07 '14 at 00:52
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2There 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." – anon Aug 07 '14 at 01:38
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@blue: yes, I agree. However, the question "why does $i\times0=0$?" seems pretty "low level", so I thought that being pedantic was appropriate. – robjohn Aug 07 '14 at 02:21
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@blue: I have added some quotes that support what I am saying. I am not trying to say that we need always to make the distinction between $\mathbb{R}$ and the canonical subspace of $\mathbb{C}$, but when asking a question like "why does $i\times0=0$?" I think it is a good idea. – robjohn Aug 07 '14 at 02:44
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@Rob $ $ If $,\Bbb R$ is disjoint from $\Bbb C,$ and $,0\in\Bbb R,\ i\in\Bbb C,,$ then what do you mean by the "product" $,0,i,$ when you write $,0,i\not\in\Bbb R,,$ i.e. what type of "product" is denoted? $ $ It cannot denote the multiplication operation of a ring since the operands are from disjoint rings. – Bill Dubuque Aug 07 '14 at 03:15
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@BillDubuque: $\mathbb{C}$ is a vector space over $\mathbb{R}$ with basis ${(1,0),(0,1)}$ and an additional vector space product defined by $(a,b)\times(c,d)=(ac-bd,ad+bc)$ (the field product). So it makes sense to multiply by a scalar $0\cdot(0,1)=(0,0)$ and by a vector $(0,0)\times(0,1)=(0,0)$. – robjohn Aug 07 '14 at 03:42
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That's what I figured. So the next question is, do you think that the product in the OP's question denotes a scalar product vs. ring multiplication? If so, why? – Bill Dubuque Aug 07 '14 at 03:45
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That's a good question. I had assumed that he intended $0\in\mathbb{R}$, not $0+0,i\in\mathbb{C}$; otherwise; I wouldn't see a problem arising. – robjohn Aug 07 '14 at 03:49
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@Rob What if the question was instead about $,0+i,?$ How would you handle that given that the summands are from disjoint rings. – Bill Dubuque Aug 07 '14 at 03:50
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@BillDubuque: The only problem would arise if we were considering $0\in\mathbb{R}$. Then we would need to "promote" $0$ to $(0,0)$ via the canonical embedding. It turns out that with products, whether we multiply by scalar or the canonical embedding, we get the same thing. – robjohn Aug 07 '14 at 04:00
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@robjohn Thank you for this answer! Along these lines, shouldn't we really describe Euler's equation as $e^{i\pi} = -1 + 0i$ ? Is it as pretty if we don't neglect that the RHS is in the complex plane and say that $Re(e^{i\pi}) = -1$? – Shape Mar 04 '17 at 18:33
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@Shape: if you want to be overly pedantic, then yes, I guess we could write $e^{i\pi}=1+0i$. However, we usually omit zero and write $1=1+0i\in\mathbb{C}$; usually, we only write $1+0i$ if we need to emphasize that $1\in\mathbb{C}$. Since we have the complex expression $e^{i\pi}$, it is pretty clear that we are dealing with the complex plane. – robjohn Mar 04 '17 at 18:46
C. Which brings us to the problem with this question, which could be that, apparently, the OP believes that the numberiexists per se, separately of everything else, while (as you know)iis simply one element of a precisely defined set. Ultimately, we do not know whatCis to the OP and the question only makes sense if the OP explains this (the trouble being that, as soon as one has a precise definition ofCat hand, the problem disappears--but let us forget this point for now). – Did Jul 30 '14 at 10:19