Talk:Imaginary number

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is this correct?

"An imaginary number bi can be added to a real number a to form a complex number of the form a + bi, where the real numbers a and b are called, respectively, the real part and the imaginary part of the complex number." Should this be "An imaginary number bi can be added to a real number a to form a complex number of the form a + bi, where the real numbers a and bi are called, respectively, the real part and the imaginary part of the complex number"? --Richardson mcphillips (talk) 16:09, 14 February 2015 (UTC)

Most sources call b (not bi) the imaginary part of the complex number a+bi. See, for instance [1]. I have added the source ([2]). - DVdm (talk) 16:17, 14 February 2015 (UTC)
It is correct. "bi" is not a "real" number but an imaginary number. "b" is a real number and since it's multiplied by "i" we call "b" the imaginary part. Keep in mind that the actual form is (a * 1 + b * i) where "a" and "b" are real numbers. "b" is the coefficient of the imaginary part just like "a" is the coefficient of the real part. If you describe it as a point on the complex plane "a" would be the amount on the real axis and "b" is the amount on the imaginary axis. --Bunny99s (talk) 05:46, 15 December 2016 (UTC)
Slight quibble: you say: ""b" is the coefficient of the imaginary part just like "a" is the coefficient of the real part", but it should be more like: ""b" is a coefficient, called the imaginary part, just like "a" is a coefficient, called the real part". - DVdm (talk) 07:48, 15 December 2016 (UTC)

Multi-valued function

@DVdm: On the multivalued function page, the very first example it gives is the square root. Besides, the fallacy shown in the example is wrong: if you notice the reference cited, the order of equations is reversed. The error in this equation is actually in the last step, unlike the reference, where the error is in a different step. Kingsindian   06:32, 17 May 2016 (UTC)

Yes, it says that "the square roots (plural) of 4 are in the set {+2,−2}". It does not say that "the square root of 4 is the set {+2,−2}". Note that the square root (singular) of 4 is +2, aka the principal root. When we write √x, we immediately have a single value. Otherwise no equation involving square roots would make sense. - DVdm (talk) 06:57, 17 May 2016 (UTC)
@DVdm: Firstly, I suggest you simply read the reference, where it says that sqrt(-1) has two possible values. Secondly, the text as written (last sentence) says that sqrt(-1) is either meaningless or two-valued in this context. Thirdly, as I point out, the equation is reversed in the reference cited, which makes nonsense of the justification cited, because it is pointing to an error in the wrong step.

To clarify, the fallacy in the reference cited is, reading from left to right:

${\displaystyle 1={\sqrt {1}}={\sqrt {(-1)(-1)}}={\sqrt {-1}}{\sqrt {-1}}=i^{2}=-1}$

The error, as the reference says, is in the second last step, because -1 can take the value both i and -i

While the fallacy in the article is:

${\displaystyle -1=i^{2}={\sqrt {-1}}{\sqrt {-1}}={\sqrt {(-1)(-1)}}={\sqrt {1}}=1}$

where the error is in the last step.

(I made a typo in my edit as well, btw. I was trying to fix it before reversion) Kingsindian   07:07, 17 May 2016 (UTC)

The cited source https://books.google.com/books?id=zNvvoFEzP8IC&pg=PA37 clearly defines ${\displaystyle {\sqrt {-1}}\equiv i}$, not ${\displaystyle {\sqrt {-1}}=\pm i}$. - DVdm (talk) 07:21, 17 May 2016 (UTC)
@DVdm: Are you looking at page 47, where the solution to the fallacy is presented? Kingsindian   07:24, 17 May 2016 (UTC)
Ah. Sorry, I hadn't seen the "solution" at page 47 (https://books.google.com/books?id=zNvvoFEzP8IC&pg=PA47) before. In that case you are correct, but that implies that (at least i.m.o.) this is a very bad, old-fashioned source. Revised in 2006, good grief. - DVdm (talk) 07:56, 17 May 2016 (UTC)
Do note that top of page 47 says that ${\displaystyle {\sqrt {1}}=1}$ is correct, so your claim above that the "error is in the last step" must be wrong. Pretty bad source, no? - DVdm (talk) 08:10, 17 May 2016 (UTC)
@DVdm: No, as I pointed out, the order of equations is reversed in the reference. I have made an attempt to reverse the order and follow the source. See if this makes sense. Kingsindian   08:21, 17 May 2016 (UTC)
Yes, I agree that you more or less follow the source, but I think that the source sucks, and that the current text does not make sense at all. It almost hurts.
We could simply delete the whole example - there is nothing special about imaginary numbers here. The main point is the multivalued square root, since there are similar fallacies without imaginary numbers. Kingsindian   12:24, 17 May 2016 (UTC)
Yes, I fully agree with that. Afaiac, dump it. By the way, I like your username—my favourite way-back opening. - DVdm (talk) 13:06, 17 May 2016 (UTC)
I have deleted the section. Anyone who disagrees is free to revert/discuss etc. Kingsindian   13:15, 17 May 2016 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── FWIW, I found a source that backs the original reasoning: see first paragraph of https://books.google.com/books?id=PflwJdPhBlEC&pg=PA12, which is in line with Square root#Properties and uses ("For all non-negative real numbers x and y, ${\displaystyle {\sqrt {xy}}={\sqrt {x}}{\sqrt {y}}}$"), and of course without mentioning "multi-valued functions". - DVdm (talk) 08:14, 18 May 2016 (UTC)

@DVdm: (Ping does not work unless you add another signature in the same diff). Note that the section clarifies the terminology used: ${\displaystyle {\sqrt {x}}}$ is actually the principal value of the square root. With this terminology, the statement that sqrt(xy) = sqrt(x)sqrt(y) for non-negative numbers is of course correct. But the root of this fallacy is still the multi-valued square root. As mentioned above, one can create a similar paradox/fallacy without using negative numbers or imaginary numbers at all. I think the (now deleted) section is misleading and simply useless. Kingsindian   02:10, 19 May 2016 (UTC)
It's like you say indeed, square root (in words} is multi-valued (althoug "the square root" is not), but ${\displaystyle {\sqrt {x}}}$ always denotes the principle value, and therefore is a proper (i.e. mono-valued) function. If that was not the case, the radical sign would be absolutely useless. Some would say, but hey, it is multi-valujed, because clearly ${\displaystyle {\sqrt {x^{2}}}=\pm x}$. Yes, but the RHS still denotes one single value only: ${\displaystyle {\sqrt {x^{2}}}=+x{\text{ for }}x\geq 0{\text{ and }}{\sqrt {x^{2}}}=-x{\text{ for }}x\leq 0}$ . I guess that this is why some think that sqrt() is multi-valued.
Heh... can you show me such a similar paradox/fallacy without using negative numbers or imaginary numbers at all? Very curious . - DVdm (talk) 06:33, 19 May 2016 (UTC)
I shouldn't have said "without using negative numbers at all", since some intermediate steps do use them. But one can find examples without imaginary numbers. The first fallacy on page 37 in the original source is one such example. As the solution on page 45 says, both the first and the second example are basically the same fallacy. There is another example on page 45. Kingsindian   10:12, 19 May 2016 (UTC)
I really don't agree with your interpretaton.
• The error in the first fallacy on page 37 is that from the opening line it does not follow that ${\displaystyle 1+cos(x)=1+\left(1-sin^{2}(x)\right)^{1/2}}$. It follows that ${\displaystyle 1+cos(x)=1\pm \left(1-sin^{2}(x)\right)^{1/2}}$. The value x=π satisfies the other alternative with the minus sign.
• The error in the second example on page 45 is that the original equation is squared (even twice!), and therefore extra invalid solutions can be induced: 4 is not a solution of the original equation and not even of the result of the first squaring. 4 is induced after the second squaring.
These examples are clearly not related to some ambiguity in the meaning of ${\displaystyle {\sqrt {x}}}$, because for positive x the thing ${\displaystyle {\sqrt {x}}}$ is defined as always positive. If that would not be the case, every book of algebra (and, by extension, of engineering) in the world would be essentially useless. - DVdm (talk) 11:06, 19 May 2016 (UTC)
It is of course correct that when one uses the radical symbol, it usually means the principal value of the square root. Also, by convention ${\displaystyle {\sqrt {-1}}=i}$. One can of course "solve" this fallacy by simply saying that ${\displaystyle {\sqrt {xy}}={\sqrt {x}}{\sqrt {y}}}$ only holds when at least one of x and y is positive. But that would be rather ad hoc; the reason why the identity does not hold for the case where x and y are both negative is because of the multi-valued square root. Coming back to the original issue: I think there's very little value in trying to explain this fallacy on this page - it doesn't really have much to do with imaginary numbers per se. Kingsindian   10:54, 21 May 2016 (UTC)
I still can't agree. Neither ${\displaystyle {\sqrt {xy}}}$ or ${\displaystyle {\sqrt {x}}{\sqrt {y}}}$ are ambiguous. For every x and y both expressions can be unambiguously calculated. Whether they are equal, simply depends on the relationship between x and y. Challenge: find me a counter-example: an x and y making it impossible to unambiguously calculate the (single-valued) expressions. I bet you can't .
I.m.o. the biggest problem is writing things like ${\displaystyle {\sqrt {-1}}}$ to begin with. There's a very good reason why the symbol i was introduced in the first place. Never write ${\displaystyle {\sqrt {-7}}}$, but in stead write ${\displaystyle i{\sqrt {7}}}$ and there can be no confusion.
I still think that the original section was very on topic in this article, but with the original explanation—not with the (i.m.o. very erroneous) multi-valued explanation of the original source. The section was sort of a warning that using expressions with negative numbers under the radical sign is dangerous and to be avoided. It could be properly sourced now with this new source—my 2 cents of course. - DVdm (talk) 11:46, 21 May 2016 (UTC)
What exactly does the source say? You mentioned that it says ${\displaystyle {\sqrt {xy}}={\sqrt {x}}{\sqrt {y}}}$ for nonnegative x and y, but what does it say about negative x and/or negative y? Does it specifically address this fallacy? If not, it would be WP:SYNTH to suggest a connection which was not made in the source. Kingsindian   01:51, 22 May 2016 (UTC)
You can see what it exacly says here, first parapgraph:
Complex numbers obay many of the obvious rules, e.g. [...]. But you have to be careful. For example, if a and b can both only be positive, then ${\displaystyle {\sqrt {ab}}={\sqrt {a}}{\sqrt {b}}}$. But if we allow negative numbers, too, this rule fails, e.g., ${\displaystyle {\sqrt {(-4)(-9)}}={\sqrt {36}}=6\neq {\sqrt {-4}}{\sqrt {-9}}=(2i)(3i)=6i^{2}=-6}$. Euler was confused on this very point in his 1770 Algebra.
Note the usage of "e.g.": it's just an example where it does not work. I think, without being guilty of synthing, we could reverse the order of the equalities in the left hand side of the inequality to
${\displaystyle 6={\sqrt {36}}={\sqrt {(-4)(-9)}}\neq {\sqrt {-4}}{\sqrt {-9}}=(2i)(3i)=6i^{2}=-6}$,
and then replace both 4 and 9 to 1, to find:
${\displaystyle 1={\sqrt {1}}={\sqrt {(-1)(-1)}}\neq {\sqrt {-1}}{\sqrt {-1}}=(i)(i)=i^{2}=-1}$.
Looks okay to me, and certainly sufficiently interesting to keep it in the article. - DVdm (talk) 09:02, 22 May 2016 (UTC)
The link goes to the old source, so I'm guessing you made a mistake in the link. However, assuming that whatever you wrote is the same as the source, it is indeed close to the listed fallacy. That said, I have the same feeling as before: the fallacy is really about square roots, not imaginary numbers. However, if you want to add it to the article, I won't object. Kingsindian   09:32, 22 May 2016 (UTC)
Yes, I already had updated the link. You probably missed it through an edit conflict. Indeed, the fallacy is really about square roots. That's why the section was there, to sort of warn the reader never to work with imaginary numbers expressed as square roots of negative numbers. - DVdm (talk) 09:43, 22 May 2016 (UTC)
Ok, I have added a little section ([3]), upon which I think we can agree. I kept your other amendments to the article. - DVdm (talk) 10:35, 22 May 2016 (UTC)

Semi-protected edit request on 24 May 2016

Change the following line:

${\displaystyle 6={\sqrt {36}}={\sqrt {(-4)(-9)}}\neq {\sqrt {-4}}{\sqrt {-9}}=(4i)(9i)=36i^{2}=-6.}$

to:

${\displaystyle 6={\sqrt {36}}={\sqrt {(-4)(-9)}}\neq {\sqrt {-4}}{\sqrt {-9}}=(2i)(3i)=6i^{2}=-6.}$

Because (4i)(9i) is not equal to sqrt(-4)sqrt(-9). And 36 i^2 does not equal -6.

Instead sqrt(-4)*sqrt(-9) = (2i)(3i) = 6 i^2 = -6.

74.66.89.121 (talk) 01:01, 24 May 2016 (UTC)

Done Kingsindian   01:09, 24 May 2016 (UTC)
Oops . - DVdm (talk) 06:39, 24 May 2016 (UTC)

Mathematical operations

Should a section on how exponentiation and square rooting is applied to these kinds of numbers be added? Gluons12 talk 20:12, 1 June 2016 (UTC).

This seems to be already covered at Complex number#Exponentiation (where we just need to make b=0), and at Square root#Square roots of negative and complex numbers. Probably not really needed here. - DVdm (talk) 20:36, 1 June 2016 (UTC)

ALL "NON-REAL" NUMBERS ARE IMAGINARY--a + bi as well as bi--SO SAYS THE DICTIONARY!

EVERY MATH TEXTBOOK I've ever read has said that "imaginary numbers" are complex numbers a + bi such that b is not zero. That is, all complex numbers other than real numbers (a) are imaginary--not just bi, which is called pure imaginary. This is also what Merriam Webster's Collegiate Dictionary, Eleventh Edition (published 2014!) says--and this is a 1,600+-page dictionary with terms ranging from tech-math like Fourier series to dirty words like fuck. Definition of imaginary number as follows (page 620):

"A complex number (as 2 + 3i) in which the coefficient of the imaginary unit is not zero--called also imaginary; compare PURE IMAGINARY"

This is a dictionary that has a lot of stuff not found in most dictionaries--and very technical-minded to boot. Besides, the idea that, within the set of complex numbers, the set of imaginary numbers represents the full complement of the set of real numbers is consistent with the other English meanings of the word "imaginary"--anything not real. If real numbers are a and imaginary numbers are only bi, what the hell are a + bi numbers called? Think about it.

I recommend that this article be rewritten, and a category/article created for "pure imaginary numbers". RobertGustafson (talk) 06:31, 15 April 2017 (UTC)

Different authors use different terminology. Your point is described in the last sentence of the lead: Some authors use the term pure imaginary number to denote what is called here an imaginary number, and imaginary number to denote any complex number with non-zero imaginary part.. Kingsindian   06:37, 15 April 2017 (UTC)