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does it make sense to add a real to a complex given that addition binary operation is only defined for set of complex numbers OR real numbers

also a related question: how can exponential $e^x$ which is all to do with banking and money have anything to do with sin and cos which they teach you is to do with triangle

Michael Hardy
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kimm
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    This is two different questions. The answer to the second question is long and involved, and you really should split it off into another Stack Exchange question. Stack Exchange questions are meant to be precisely that - individual questions - so that anyone searching in future only has to search for one specific thing. – Patrick Stevens Aug 25 '15 at 17:29
  • related for second question – Blex Aug 25 '15 at 17:49
  • The main reason that trig gets involved is because of De Moivre. Basically, if you multiply a complex number by $\cos A+i\sin A$, then the result is the original number rotated around the origin by an angle of $A$. – Akiva Weinberger Aug 25 '15 at 17:55
  • I take your second question to be what this is all about. My answer posted below may be the best really short answer that can be given. I could write a longer one if I had some assurance that you'd be willing to follow a longer one carefully. ${}\qquad{}$ – Michael Hardy Aug 25 '15 at 17:56
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    In fact, you don't even need to get to complex numbers to find a (slight) connection. Notice that the function $(-1)^n$ oscillates between $-1$ and $1$, just like a trig function does, despite being an exponential function. – Akiva Weinberger Aug 25 '15 at 17:58
  • @PatrickStevens : You say the answer to the second question is long and involved. What do you think of my short answer to it posted below. Yes, it's long and involved, and I could write a long and involved answer, but I have gradually developed an ability to write things like the shorter answer below. ${}\qquad{}$ – Michael Hardy Aug 25 '15 at 17:58
  • Saying "$e^x$ which is all to do with banking and money" is a lot like saying "decimal fractions, which are all to do with banking and money." It's possible for someone to have learned only this one application of either of these things, but they both have many, many more applications. – David K Aug 25 '15 at 18:14
  • @MichaelHardy I'm having trouble putting myself in a position of someone who doesn't already understand the point. I don't know whether it's helpful or not :P – Patrick Stevens Aug 25 '15 at 18:17
  • @PatrickStevens : I'm taking the "banking and money" comment to mean it's about exponential growth. One mathematical concept that came from banking and money is negative numbers. I don't know whether the study of exponential growth originated in that field as well. ${}\qquad{}$ – Michael Hardy Aug 25 '15 at 18:30

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The real numbers inherit complex addition. Indeed, the inclusion map $i : \mathbb{R} \to \mathbb{C}$ by $x \mapsto x + 0i$ allows you to view any real number as a complex number in a canonical way. That is, when you add a real to a complex, you are effectively using complex addition.

So, to answer the question directly: if you are being really mathematically precise (or pedantic), it doesn't make sense to add a real to a complex, any more than it makes sense to add an apple to a banana. However, in any context where you don't have to be totally precise about these things, it does make sense, because we implicitly identify each real number with its associated complex number.

For example: $1_{\mathbb{R}} + (1+i)_{\mathbb{C}}$ very strictly doesn't make sense. However, we identify $1_{\mathbb{R}}$ with $(1+0i)_{\mathbb{C}}$, and so we may identify the original sum with $(1+0i)_{\mathbb{C}} +_{\mathbb{C}} (1+i)_{\mathbb{C}} = (2+i)_{\mathbb{C}}$. There are very few instances where a mathematician would not do this "view a real as a complex" automatically and without thinking about what they were doing.

All this is to say that the reals are isomorphic to a subfield of the complexes, and we freely identify elements of the field $\mathbb{R}$ with their corresponding elements of the field $\mathbb{C}$ (where we pick the correspondence in the obvious way).

The exponential function is to do with lots and lots of different things, and is very much worth asking in a different question (if the answer isn't already on Math.SE).

Patrick Stevens
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  • I think the question is a bot more nuanced than you give it credit for. A real number is not the same as its image under the inclusion map. I would say the inclusion map allows you to map any real number to a complex number (in a canonical way). – TonyK Aug 25 '15 at 17:30
  • so the answer to the first question is strictitly speaking no? it makes no sense. we can only 'add' complex to complex or real to real but by your inclusion map, we can also in some sense add real to complex – kimm Aug 25 '15 at 17:31
  • @TonyK I thought I was quite careful to specify that a real wasn't the same thing as a complex - "allows you to view", "effectively using". If it's unclear, I'll edit that. – Patrick Stevens Aug 25 '15 at 17:31
  • so tonyk, what do you mean its not the same? what you mean by same? – kimm Aug 25 '15 at 17:32
  • Seems clear to me. I don't think we should obscure the basic intuition of the mapping with too many caveats. – Brian Tung Aug 25 '15 at 17:33
  • @kimm: I've hopefully clarified that in my second paragraph. – Patrick Stevens Aug 25 '15 at 17:33
  • so what esle is there to say about this? is the group sructure within the reals or the field structure on the reals, included in some sense within the complexes? the inclusion is not only a subset but it also inherits the structures no? – kimm Aug 25 '15 at 17:36
  • @kimm: added fourth paragraph to say that the reals form a subfield of the complexes, with the inherited structure. – Patrick Stevens Aug 25 '15 at 17:40
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    This pretty much holds true for every inclusion map up the chain, since when we are all rigorous we define rational numbers as sets of equivalence classes for the field of factions of the integers, but then we use the canonical embedding of the integers into the field of fractions, etc. – Alan Aug 25 '15 at 17:47
  • @Alan: the field of factions sounds really scary ;) I thought we were beyond the days of warring tribes! – Patrick Stevens Aug 25 '15 at 17:49
  • You have covered my objection nicely in your third paragraph. (But then you go on to say that the reals "form a subfield" of the complexes. And we're back to square one!) – TonyK Aug 25 '15 at 17:58
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    @TonyK You can define everything so that you have real inclusions $\mathbb{N}\subset \mathbb{Z}\subset \mathbb{Q}\subset \mathbb{R}\subset \mathbb{C}$. Then $\mathbb{R}$ is literally a subfield of $\mathbb{C}$. And adding a real number and a complex number strictly makes sense, since real numbers are just complex numbers with some property. – Daniel Fischer Aug 25 '15 at 18:01
  • @DanielFischer: Yes, but nobody does it that way, surely? – TonyK Aug 25 '15 at 18:02
  • @Tony: In my experience, people who are not algebraists will often see things as Daniel describes, since it makes it easier to just work inside $\Bbb C$. – Asaf Karagila Aug 25 '15 at 18:03
  • Patrick, is there a reason you write $(1+0i)\Bbb C$ and not $1\Bbb C$? Also, is the $0$ there $0_\Bbb C$ or $0_\Bbb R$ when looking at $\Bbb C$ as an $\Bbb R$-module? – Asaf Karagila Aug 25 '15 at 18:04
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    @TonyK Usually not when one constructs the things. But it's really convenient to have inclusions, so one often assumes them. If a pedant comes along, one can say one defined $\tilde{\mathbb{N}},\tilde{\mathbb{Z}}$ etc. and calls $\mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R}$ the images of the tilde-things under the canonical inclusions into $\mathbb{C}$. – Daniel Fischer Aug 25 '15 at 18:08
  • @Daniel: Well, exactly. – TonyK Aug 25 '15 at 18:09
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    @TonyK: Actually, I would say that virtually every practicing mathematician does exactly that. The time when people would be forced to take more care is, for example, writing computer code with a carefully typed language, that has two different types for real numbers and for complex numbers. But outside of such situations where linguistic precision is truly necessary, people take the same shortcuts with mathematical language that they take with ordinary everyday language. – Lee Mosher Aug 25 '15 at 18:09
  • @LeeMosher: There is a difference between seeing things that way and defining them that way. – TonyK Aug 25 '15 at 18:10
  • @AsafKaragila: I did that so as to make it even clearer that it was literally a complex number, rather than just "a way of viewing a real number". The zero is intended to be $0_{\mathbb{R}}$, viewing as an $\mathbb{R}$-module. – Patrick Stevens Aug 25 '15 at 18:12
  • @TonyK I've changed "the reals form a subfield of" to "the reals are isomorphic to a subfield of". – Patrick Stevens Aug 25 '15 at 18:13
  • @PatrickStevens: Thank you! – TonyK Aug 25 '15 at 18:15
  • This answer is all about the first question. I thought the second question was what this is really about. ${}\qquad{}$ – Michael Hardy Aug 25 '15 at 18:23
  • @MichaelHardy I consider the two questions to be substantially different. I answered the first one in detail, and I think you'll agree that my answer to the first question is not closely related to the second question at all. – Patrick Stevens Aug 25 '15 at 18:30
  • so what about if you take the number 2 from the set of naturals and then from the set of integers. are they same or different numbers? because from what i understand of the comments made, you can add in the negative numbers to the naturals to get integers yet when it comes to complex numbers, we dont do that? we dont add in imaginary numbers to the reals do we? we just define complex numbes from scracth which makes the reals to the complexes. seem like less of an inclusion as the naturals do to the integers. – kimm Aug 25 '15 at 20:38
  • @kimm They include in exactly the same way: the monoid of naturals includes into the group of integers. Strictly they are "different types", but for all practical purposes we can regard them as the same type (that is, integers). We do add in imaginary numbers to the reals: that's how to make the complexes. (It does depend on your definition of "complexes" - there are several equivalent ones, like a particular $\mathbb{R}$-module over the basis ${1, i }$, or the algebraic closure of $\mathbb{R}$, or whatever.) – Patrick Stevens Aug 25 '15 at 21:18
  • @Tony: By the way, often times in analysis related books, they will define $\Bbb N$ as the smallest inductive subset of $\Bbb R$; and $\Bbb Z$ as the additive-inverse closure of $\Bbb N$ in $\Bbb R$; and the rational numbers as the multiplicative-inverse closure of $\Bbb Z$ in $\Bbb R$. In that case, the inclusion is actual factual inclusion rather than "canonically embeds into". – Asaf Karagila Aug 25 '15 at 21:25
  • @AsafKaragila: I have never seen that approach, so thanks for the info. It doesn't get us to $\mathbb C$, though! – TonyK Aug 25 '15 at 21:39
  • @Tony: True, but I suspect this is because the set theory and "basic constructions" happen usually when the reader is new to mathematics, and analysis is limited to $\Bbb R$. By the time we move to $\Bbb C$ these things are chalked aside. In any case you repeat them entirely over $\Bbb C$. – Asaf Karagila Aug 25 '15 at 21:40
  • @TonyK: simply say that $\mathbb{R}$ is the subfield of $\mathbb{C}$ with zero imaginary part, surely? – Patrick Stevens Aug 25 '15 at 21:41
  • @AsafKaragila: And $\mathbb H$. And $\mathbb O$. The trouble with that approach is that you need to know where you're going to end up before you start $-$ if you're ever going to be using octonions, then you'd better define $\mathbb N$ as a subset of $\mathbb O$, hadn't you? – TonyK Aug 25 '15 at 21:45
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    @Tony: That is correct. And you can just do everything about a similar construction over the surreals, just to make sure that you catch every field, ring, or whatever. :-) but in all seriousness, this is about usefulness. If you want to teach someone basic real analysis, insisting that $\Bbb N$ is not a subset of $\Bbb R$ may end up being more confusing than not; and I just gave you a definition of $\Bbb N$ which gives you a subset of $\Bbb R$ explicitly. I don't recall octonions or sedenions popping up in real analysis, sorry. – Asaf Karagila Aug 25 '15 at 21:48
  • it would be nice to be able to understand this. i am lost now – kimm Aug 25 '15 at 22:00
  • @kimm Can we move this to a chat? https://chat.stackexchange.com/rooms/27404/is-2-both-natural-and-integer – Patrick Stevens Aug 25 '15 at 22:06
  • @kimm: See what you've started! The point is this: The natural way to define $\mathbb N,\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ is as a sequence of ever more sophisticated constructions, wherein each member of the sequence has a canonical embedding into its successor. In this view, $\mathbb R$ is not a subfield of $\mathbb C$. But once you have done all this, it just becomes convenient to treat the various members as if they were all subfields or subrings of $\mathbb C$. This is a kludge, if you like, but it can always be justified when required. – TonyK Aug 25 '15 at 22:12
  • @patrick it seems i cant chat until i have 20 points. – kimm Aug 25 '15 at 22:17
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(Every real number is a complex number, e.g. $3=3+0i$, so adding a complex number and a real number is just adding two complex numbers.)

With sine and cosine you should think in this context of circles rather than triangles. A first step toward understanding this is to realize this:

  • Multiplying by $i$ means rotating $90^\circ$ counterclockwise. Doing that repeatedly means going around in circles.

  • Multiplying by $2$ repeatedly means growing progressively faster.

Thus if $x$ is an increasing real quantity then

  • $x\mapsto e^x$ grows progressively faster, and

  • $x\mapsto e^{ix}$ goes around in circles.

(A complete answer to this question would be long and involved.)

Michael Hardy
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  • Repeatedly multiplying by $i$ sounds more like $i^x$. I think you should clarify the connection to $e^{ix}$. – ipsec Aug 25 '15 at 18:44
  • @ipsec good point – kimm Aug 25 '15 at 20:59
  • @ipsec : The connection to $x\mapsto e^{ix}$ is similar to the connection between $x\mapsto e^x$ and $x\mapsto 2^x$. This is related to the reason of just what is "natural" about $e$, and that is about instantaneous growth rates, i.e. derivatives. So I said "a first step toward understanding this", and it is a first step. I also said a complete answer would be longer. However, I do show a connection between exponential growth (as in earning interest) and circular motion, showing that they're not so unrelated as one might guess. ${}\qquad{}$ – Michael Hardy Aug 25 '15 at 21:48
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When addition binary operation is defined on complex numbers, a complex number and a real number can be added giving a complex number. This is because a real number is actually a complex number whose imaginary part is 0. You can express a complex number as Z= a + bi where a is the real part and b is the imaginary part.

But, when this binary operation is defined on real numbers, we cannot add a real number and a complex number. This is simply because we cannot give a binary operation an argument (operand) which is not present in the set on which it is defined. You are defining it on real numbers and then trying to add a complex number.

For the e part, you should read this: http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/

An Analogy: For the first part of your question. Let's assume I have 4 apples and 3 bananas in the kitchen, and 5 apples and 1 banana in the drawer. Some mathematician comes to me and asks two questions:

1. How many bananas do you have?

2. How many fruits do you have?

For the fruits you can add bananas and apples. But for the bananas you add bananas only. Here the operator 'How many' behaves like addition. But the context where it is asked is different. The mathematician was talking about bananas in the first question and about fruits in the second one.

Shashi Dwivedi
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So as has been said, the answer is in the strictest technical sense sense no, for essentially type-checking reasons, but in many cases, it is reasonable to convert a real number to the corresponding complex number (with imaginary part 0) so that it may be added to complex numbers.

One thing to keep in mind, though, is that while the notions of real numbers, algebraic numbers, rational numbers, integers, and complex numbers all extend our basic notion of natural numbers, they don't always form a strict hierarchy. That is, there are some situations in which it makes sense to use complex numbers, but only those where both the "real" and "complex" parts are integers, or where it makes sense to use real numbers, but only positive ones.

According to the definitions of $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$, there is of course a hierarchy, but many problems require using subsets of these sets (e.g. $\mathbb{R^+}$) that defy that hierarchy.

Getting back to your original question, then, What matters most in determining whether it's OK to add real and complex numbers together is whether or not that makes sense in the context of the problem you're working on. Ask yourself, "Are these two numbers representing the same kind of thing or idea?" If so, then the real number is probably best thought of as a complex number with real part 0 in which case they may be added.

Gabriel Burns
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You can use also use Taylor series to show that exponential and sinusoid functions are some how related in the set of complex numbers. $$e^{x}=\sum_{n=1}^{\infty}\frac{x^{n}}{n!}$$

Thus, $$e^{ix} = \sum_{n=1}^\infty\frac{i^n x^n}{n!}$$

Due to the property of $i^{n}$ we can see the Taylor series of sine and cosine functions: $$e^{ix}=(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots+\frac{x^{2n}}{2n!}-\cdots)+i(x-\frac{x^{3}}{3}+\frac{x^{5}}{5!}-\cdots+\frac{x^{2n-1}}{(2n-1)!}-\cdots)$$ This is recognized as $$e^{ix}=\cos(x)+i\sin(x)$$ This proof really opened my eyes, I think this might also help you!

Michael Hardy
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Socre
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