What is i?

[NotLKH]

This just in! snakes says the time honored definition of i has changed.
bugz says i is defined as: i² = -1, where i <> (-1)^(1/2)

[bugzpodder]

lol infinity is not a value. they are simiply mathematical ideas. there are mathematical definitions of i and infinity. definition of i is: i^2=-1, and definition of infinity is (by my Colin's math dictionary): not FINITE; having a size or absolute value that is greater than any natural number (or growing out of all bounds). too bad my dictionary says i=sqrt(-1) also. but i believe that the definition of i is i^2=-1.


***saying i=sqrt(-1) is saying that sqrt(-1) exists and that simiply we invented i for no reason!!***

[bugzpodder]

snakeeyes make a very good point.

it's not a real number; hence, we can't tell you what i equals.

the form

i^2=-1

has no errors in it; you can square i and -1 is a real number. But you cannot take a real value from Sqrt(-1), so the former is the preferred method.

[DavidHooper]

bugz is wrong. NotLKH, you are, as always, correct (I've never known you to make a mistake in all the time I've been here ). Look here for the proof of the pudding.

Bugz, I don't know where you get your information from but, no offence, you need slow down before you hit the Post button and make sure you don't make simple errors.

[bugzpodder]

lol, whats wrong?? enlighten me! instead of telling me whats wrong, you just tells me i am wrong and lecture me on how i should post my msgs lol. btw your argument is pretty weak. NotLKH maybe right before and he may take Fourier Series Analysis and so on, but I sure as hell know what I am talking about, since i would never go gabbling on something i have no idea of. and i sure don't need someone to tell me to shut up! i know what these website says. if you are willing to share your expertise i am willing to listen, but last thing i want to hear is for ppl to tell me to shut up. btw i know what these amateur website says (and i don't see the proof, all it says is i=sqrt(-1)). did you went to the university of toronto page i listed? do you have a reply for both snakeeyes' and my points? and what "simple mistake" did i make?

[Destined Soul]

Oi, people. Calm down. I have to admit, BugZ is right DavidHooper in that the link you gave doesn't really prove it. However, BugZ is wrong in assuming that his/her answer is the only possible answer.

After a lot of searching on the net, I've found multiple sources that have either of the definitions. I think it's more a matter of taste as to how you "define i" I also believe that if you handle your math correctly, both should result in the same result.

Not to disrespect you BugZ, but you tend to complain that people don't prove their side or prove you wrong when you in fact don't do it yourself.

That said, here's a question for you all: where would you find the actual definition that determines i? How can you show that one definition is the "right definition?" I can define a = 2, whereas someone else might say a = 3. So who's right?

Destined

[Masterbandit666]

The definition I have always gone by is i² = -1, since it has never gone me wrong. Someone made a miniature thesis about how the other definition can lead to screw-ups in a thread I read about 30 minutes ago, but I can't remember who. I agree totally with him/her.

[NotLKH]

You do have to admit, saying i² = -1 is
not a definition of i. It only defines i².

But, lets use i² = -1, since this is a valid property of i.
Certainly, if it is a valid equation, then you can raise both sides to the (1/2) power.

Thus:
i² = -1 transforms to
(i²)^(1/2) = (-1)^(1/2)
Which simplifies to
i = (-1)^(1/2)

Now, if you say we can't do that, then i² = -1 can't be mathematically valid.

Furthur, how can it be valid to use i^n when n is odd, if i^1 cannot be represented directly?

BTW, Your Website seems to be pretty old, and its almost dead.
I wonder what the updated site will say.

BTBTW, Why don't you agree with your dictionary? Which one is it anyways?

And, David, I could be wrong, but my track record isn't as
spotless as you claim, however, I do appreciate the perception!



-Lou

[bugzpodder]

Just saying the other person is wrong, like what David Hooper did was no way to go about things. i agree that it is hard to let go of something you've known for years. but hey, u are right, they are really don't make that much of a difference if you do your math right. but back to the point, i believe the definition of i^2=-1. and i can offer NotLKH an explaination of his attempt to say that i=sqrt(-1)

NotLKH, sqrt(x) says the domain of x is that x>=0.

this is how i've understand the complex number theory. i itself is an imaginary number. its relation with the real number set is that i^2=-1. since we developed the imaginary number theories eons after the real number theories, it would make sense to define i based on the real numbers, wouldn't you say. since you can't legally take a square root of a negative number, so instead of saying sqrt(-1)=i, we say i^2=-1. all complex numbers can be represented as:
rcos(t)+r*i*sin(t), because every complex number on the comple plan can right in terms of coordinates (rcos(t),rsin(t)) on the complex plane (where the x-axis becomes the Re[al] axis and the y axis becomes the Im[aginary] axis), or simiply a polar coordinate of (r,t), where r is the maginitude (distance from center) and t is the angle it make with the center.

now as to your problem NotLKH,

-1=cos(180+360k)+isin(180+360k), k is an element of Integers

using demoivre's theorem
the (cos180+isin180)^(1/2)=cos(90+180k)+isin(90+180k)
which yields i and -i

my point is to use demoivre's theorem since square roots and cube roots should not be used as part of complex equations.

[snakeeyes1000]

all imaginary numbers are based on i, so is it not logical to make i a real number? Saying i=Sqrt(-1) is a circular definition, because

Sqrt(-1)

is an imaginary number, and all imaginary numbers are based on i. The other form doen't have the same definition.

[Jimw]

Hi,
I would like to support Bugz' suggestion that since functions ought to return single values, the description of i that says
i^2 = -1 is better than a pseudo-functional prescription of i as sqrt(-1). Here is why. Sqrt is not a single-valued function on the real line, even if you restrict the domain to nonnegative numbers.
That is not really a problem, because mathematicians have defined something called the "principal square root" which, as Bugz points out again is the nonnegative square root in this case.
But for functions with complex domains, the notion of "principle square root" is more problematic, as you can imagine. What is the principle square root of -1? Is it i or -i? Both qualify, but we do not have the convenient notion of positivity for complex numbers, as we do for Reals.
In this case, we hace to resort to "Riemann Surfaces" to define such things as roots and logs on the Complex plane.
There is another, more geometric, definition of i that you might like to think about. Complex numbers can be thought of as mappings from the plane to the plane that first expand it by a positive factor and then rotate it. i is a rotation by 90 degrees in this view. All such geometric maps can be represented by 2x2 matrices of the form:
[a -b]
[b a] for a and b real.
Such a matrix "represents" a+bi. THis geometric interpretation may be the simplest one. Notice that the matrix representing i:
[0 -1]
[1 0]
has the property that its "square" is:
[-1 0]
[0 -1]
which is the matrix for real number -1. But also notice that the matrix has a real, concrete interpretation. It is the matrix of the rotation of the plane through 90 degrees.

Why go to all this trouble to define i ? Well, that is another story with its roots deep on 16th Century algebra. As Bugz points out, DeMoivre's Theorem gives an excellent interpretation of "imaginary numbers". I would like to add a famous formula by Euler that Richard Feynman felt was the "deepest equation in mathematics" when he was a teenager. I'm inclined to agree. It goes:

e^(i*pi) + 1 = 0
It is certainly one of the most beautiful equations I have ever seen.

My name is Jim White, I would like to invite you to visit the Mathwright Library at http://www.mathwright.com and explore the properties of imaginary numbers and of other more abstract numbers for yourselves by reading a book I wrote called: Cardano. You will not only read, but you may experiment with complex numbers, and will learn a new and nifty way to solve cubic equations. The book is free, so jump in. There is another excellent book on DeMoivre's Theorem written by one of my colleagues, and several others on matrices, etc. They are all Active Math Books. That means they let you ask your own questions, and try to answer them.

I built the Library and Bugz will tell you that I am always happy to talk with students who have questions about mathematics (as this overlong reply will show!)

I know that I have not closed this fascinating debate, but I hope that I have contributed a little more food for thought. Carry on!

[Destined Soul]

Interesting. I'll have to have a look at those books later. Somehow, I think this discussion could go on forever - largely because there is no right answer. I think that defining i exactly (and only one way) is similar to the debates of dividing by zero and that of infinity; there is no answer. They just are what they are. Imaginary, nothing, or infinite. (respectively.)

Destined

[bugzpodder]

I built the Library and Bugz will tell you that I am always happy to talk with students who have questions about mathematics

Yes, and I would also recommend his site since these books are very well worth reading, since they are arranged from beginner to advanced. I guarentee that any one of you will learn something out of these books.

[sql_lall]

very interesting.

Just one thing bugging me. (no pun intended)
Bugz, you said two things:
1)

which yields i and -i

2)

you can say i^2=-1, but not i = sqrt(-1) -generalisation

Does this mean that i = -i?? as i^2 = -1, AND -i^2 = -1
=> i = -i

[bugzpodder]

lol its just like the squares thing. if the squres of the two numbers are equal, does it mean that they are equal? 2^2=4, (-2)^2=4 does it mean that 2=-2? no. but we know the absolute value of the two numbers are equal.

[Masterbandit666]

In reply to sql_lall, 4² = 16, (-4)² = 16 but 4 does not = -4...

[sql_lall]

Good, but not right. There is one MAJOR difference:
2 is not DEFINED by: 2^2 = 4
same as 4 isn't DEFINED by 4^2 = 16

You said:
i is defined by i^2 = -1
=> i = ANY NUMBER that satisfies this.
(-i) satisfies this,
=> i = -i

see the diffenece??

[bugzpodder]

lol i see what you mean. so tell me, how is 2 defined and I'll answer your question.

[beachbum]

Lou, you are obviously not asking the right question... it should read:


Who am i? Grammar is terrible in the U.S these days

[NotLKH]

Originally posted by beachbum
Lou, you are obviously not asking the right question... it should read:


Who am i? Grammar is terrible in the U.S these days

How do you know Grammar?
She's not only terrible, she's Rotten! Haveing been dead and Buried here for decades!!!





-Lou

[DavidHooper]

Hi Bugz,
I'm gonna pick a few holes then apologise later ok?!

1. I did not "lecture" you, I suggested using the Shift key for capitals.
2. I did not tell you to "shut up". (Twice).
3. MathWorld is not an amateur site.

No offence was, however, intended in any of my posts - apologies retrospectively. So no hard feelings ok?

[bugzpodder]

yea i want to apologize when i tend to get somewhat emotional when debating. so did you find any holes?

[NotLKH]

Originally posted by Jimw blaa is meant to represent gems of knowledge that are not pertinant to my focus, and in no way is meant to be a slur on the knowledgability of Jimw, nor is it meant in any disrespectful manner.
blaaa[a -b]
[b a] for a and b real.
Such a matrix "represents" a+bi. THis geometric interpretation may be the simplest one. Notice that the matrix representing i:
[0 -1]
[1 0]
blaaa

You point out that sqrt(-1) is problematical, since assigning + or - to such a number has no anchor.

However, the same can be said where the definition of i is that which satisfies the equation i^2 = -1. {See bugz's site}.

Now, my beef isn't the property i^2 = -1. My beef is the insistance that i^2 = -1 WHILE i <> (-1)^(1/2).

How can this be?
Certainly, if we say Variable B, if it is equal to radical(-1), then that satisfies the property statement for i. BUT bugz says b CAnnot be equal to i, even though it satisfies i's property of i^2 = -1, because i <> radical(-1), and therefore anything designated equal to radical(-1) cannot be i. {Even though we are allowed to have radical(negative number) in equations. }

So, doesn't [(-1)^(1/2)]² = -1 or not?

BTW, Using matrix notation to define i is just saying i = i, since the two positions represent the real portion, and the scalar value of the imaginary portion multiplied by i. Defining i with complex notation is circular logic, and is meaning less, IMHO.




-Lou

[bugzpodder]

what is IMHO?

anywayz, my reasoning for i<>sqrt(-1) is that sqrt(-1) does not exist (in your words, we are not allowed to have radical (negative number) in equations) b/c the domain of sqrt(x) says x>=0. if you want to evaluate the sqrt of a negative number, you should use the demoivre's theorem instead.

[NotLKH]

Ehhh, yes, it isn't calculateable. But then, in a seperate fashion, neither is pi.

But, radical(-n) occurs during calculations. BUT, we could assign the variable Garp = radical(-n), and work with Garp from then on, right?

No one is sayin "Work it out", since its impossible.

Its an assignment, nothing more.

[bugzpodder]

well, i want to make a point here:

i am saying that the definition of i is i^2=-1. it just simiply means that the definition of i is not i=sqrt(-1)

also, since sqrt(-1) is imaginary (since i know that its not real), your definition is saying imaginary=imaginary, which is retorical.

however, in i^2=-1. RHS is a real value. so a connection between real and imaginary set can be easily made.

i=sqrt(-1) maybe true but i^2=-1 is a better definition for i. how is that?

btw pi is an irrational number, you can still approximate its value to a certain degree. sqrt(-1) is uncalculatable and un-approximatable (in terms of real numbers) b/c it doesnt exist (at least in the real plane, you can argue that it is an imaginary number or that it doesn't exist at all)

[NotLKH]

Originally posted by bugzpodder
well, i want to make a point here:

i am saying that the definition of i is i^2=-1. it just simiply means that the definition of i is not i=sqrt(-1)

also, since sqrt(-1) is imaginary (since i know that its not real), your definition is saying imaginary=imaginary, which is retorical.

however, in i^2=-1. RHS is a real value. so a connection between real and imaginary set can be easily made.

i=sqrt(-1) maybe true but i^2=-1 is a better definition for i. how is that?

btw pi is an irrational number, you can still approximate its value to a certain degree. sqrt(-1) is uncalculatable and un-approximatable (in terms of real numbers) b/c it doesnt exist (at least in the real plane, you can argue that it is an imaginary number or that it doesn't exist at all)

Point #1)
But also, since ^2 is a function, you are saying there exists a number that can be squared that returns -1. But, Nothing exists that can be squared that retuns -1. so YOU to, are arguing the same.


Point #2, prologue)
However, Yes, Radical (-1) again, is impossible, imaginary, call it whatever you want. However, we still can create the notation radical(-1), nothing prevents us from writing it, nothing prevents the quadratic equation from returning components that are multiples of it. Why balk at saying, "Whenever you see radical(-n), you can note it as i*Radical(n)"?

Point #2)
What you are saying is, "Whenever you see Radical(m), if radical(m) squared returns a negative number, then you can note it as i*radical(-m)"

Epilogue)

Thats pretty much a roundabout way of doing it, but thats what your "definition" requires.

BTW, don't forget to comment on point #1.

[Destined Soul]

also, since sqrt(-1) is imaginary (since i know that its not real), your definition is saying imaginary=imaginary, which is retorical.

Just thought I'd comment on this. You're confusing his statement on i in that he's defining i = sqrt(-1), whereas you're trying to say that, although sqrt(-1) is an imaginary number, it is not the definition. You seem to be getting your logic a little bit confused.

however, in i^2=-1. RHS is a real value. so a connection between real and imaginary set can be easily made.

This can be equally argued for the other side. Here, you define a pure real on the RHS, and some not-quite-imaginary number on the left. If I define it as i = sqrt(-1) (or i = (-1)^(1/2)), you have a pure imaginary on the LHS and a not-quite-real number on the right. Your LHS does have an imaginary number IN it, but this one does have a real number IN the RHS.

Well, you get the idea of what I'm arguing. Both definitions have a method of translating one set to the other.

Destined

[bugzpodder]

lol i know what you are saying.

You're confusing his statement on i in that he's defining i = sqrt(-1), whereas you're trying to say that, although sqrt(-1) is an imaginary number, it is not the definition.

actually, i am just saying why i^2=-1 might be a better definition than i=sqrt(-1)

the only way to say that there is a hole (maybe thats all it takes?) in the ladder one is:
i*i=sqrt(-1)*sqrt(-1)=sqrt(-1*-1)=sqrt(1)=1

for your point one: i is something new that once being squared, would get you -1.

Well JimW made the following comment, he has a good point at the end which I highlighted below:

Now about your question about i. The specification
that i should be a
"number" such that i^2 = -1 does not by itself
"define" the number i. To
define it, you have to build a new number system.
There are two ways to do
that, algebraically or geometrically. And there are
at least 2 algebraic
ways to do it. The first is this. You could simply
declare that the set of
2x2 matrices that I described to you of the form:
[a -b]
[b a] with a and b real is the number system (the
complex numbers!) and
to prove that it has all the usual properties except
that it is
2-dimensional, unordered, and has this special
"number"
[0 -1]
[1 0] that we call i. Thus there is no need to
"define" i. Instead, you
define a whole system of numbers (containing the reals
as the diagonal
matrices) which includes i.

The second algebraic way is the way it was actually
done. I know you know
about this because I saw at the Forum a contribution
you made on Cardano's
Method (Cardano or Cardan, not Cardino). This
involves polynomials. Take
any polynomial of degree greater than 1 with real
coefficients that cannot
be factored into a product of lower degree polynomials
with real
coefficients. As it happens, such a polynomial must
have degree two (This
is the so-called Fundamental Theorem of Algebra). The
polynomial x^2+1 will
do nicely but there are many others. There is a
well-defined algebraic
procedure for constructing a new number system,
containing the Real numbers,
and also containing a solution to this polynomial.
Any way you do it, you
will get the complex numbers, by the way. But the
point is that this is
constructive too. You actually build that number
system and demonstrate
that it is one.

This latter procedure was explored by mathematicians
Abel and especially
Evariste Galois, who actually discovered it, but was
killed in a duel at the
age of 20 or so. A real tragedy, because with that
work, Evariste
established the foundations for all of Modern Algebra.

Now the Geometric Way is entirely different. Picture
the Euclidean Plane.
Now consider those mappings of the plane to itself
that leave the origin
fixed and preserve orientation and take any figures to
"similar" figures.
These will be compositions of rotations about the
origin and expansions at
the origin. These mappings of the plane to itself
form a number system.
That is not so difficult to prove. So you can "add"
two of them. And you
can "multiply" two of them by composing the mappings.
ny such mapping
(except 0) has an inverse, so you can also divide.
Etc. etc. In this
number system, the 90 degree rotation is what we call
i. Why? If you do it
twice (square it) you get a 180 degree rotation, which
is tantamount to
multiplying by "-1"

I prefer the latter (non-constructive) way. But
notice that it is never
any more a matter of "defining" i. You must "define"
an entire system of
numbers to discover i. I do not think that the
question you asked is the
right one. One does not define i the way one
"defines" pi or e. Those
numbers already exist in the reals and can then be
determined uniquely by
their properties. If you have built a system of
numbers as I described,
then you could "name" i, but the equation x^2+1 = 0
would not specify it
uniquely as you point out. Because of the basic
symmetry of the complex
numbers that is associated with conjugation, there is
no way to distinguish
a priori between i and -i. You have both. The best
you can do is "pick"
one, and call it i.

[NotLKH]

Let me quote my last post here. Bugz, I think you missed this.
Care to comment?

OOPS, you asked "What is IMHO",
Thats easy. In My Humble Opinion.


-Lou

Originally posted by NotLKH


Point #1)
But also, since ^2 is a function, you are saying there exists a number that can be squared that returns -1. But, Nothing exists that can be squared that retuns -1. so YOU to, are arguing the same.


Point #2, prologue)
However, Yes, Radical (-1) again, is impossible, imaginary, call it whatever you want. However, we still can create the notation radical(-1), nothing prevents us from writing it, nothing prevents the quadratic equation from returning components that are multiples of it. Why balk at saying, "Whenever you see radical(-n), you can note it as i*Radical(n)"?

Point #2)
What you are saying is, "Whenever you see Radical(m), if radical(m) squared returns a negative number, then you can note it as i*radical(-m)"

Epilogue)

Thats pretty much a roundabout way of doing it, but thats what your "definition" requires.

BTW, don't forget to comment on point #1.

[bugzpodder]

lol i didn't miss ur post: for your point one: i is something new that once being squared, would get you -1.

point 2 is a good point.

[Dreamlax]

Does it really matter how it's defined? We know it's properties (when squared it's equal to -1). One thing I find strange is that the complex number:
2 - i
is a LOT smaller than
1000000 - 999999i

but when both are squared, you can clearly see the large difference.

But hey, I don't complain, I just take it in.

[bugzpodder]

lol this is new, i've never heard how to tell from two complex numbers which one is greater and which one is smaller. how could you tell that? by its magnitude (sqrt(a^2+b^2))?

[Dreamlax]

Hmmm, maybe it looks like something my teacher tried to start... he told me if the modulus of a certain complex number is higher than another, then the number can be described as 'larger'.

I suppose the modulus of i and -i are identical, but are obviously different numbers (same goes for 1 and -1 I suppose).... now that I think about it... it is all bollocks....

Thanks Bugz (no proof needed)!

[bugzpodder]

Well the modulus of a complex number is the magnitude of a complex number, lol. i did a search and come up with nothing mainly because i don't know what to search for, like i tried "order of complex numbers". but i came up with something pretty interesting to read if you are into complex numbers:

http://www.math.tau.ac.il/~baizen/A_M2.PDF

[NotLKH]

Originally posted by bugzpodder
lol this is new, i've never heard how to tell from two complex numbers which one is greater and which one is smaller. how could you tell that? by its magnitude (sqrt(a^2+b^2))?

I think so.

Since complex numbers are vectorial in nature, it would seem natural to use their absolute magnitude to determine if one is larger than the other.

Of course, I'd skip the sqrt function, if all we are doing is
determining size comparisons in a series of complex numbers.

[bugzpodder]

To NotLKH

I've consulted with my teacher, and basically offered me this reply to your two points:

i is a symbol that is introduced such that i^2=-1. i quote

i is not a number

Also, as you pointed out, radical(-1) does not exists. We are prevented from writing it because it is mathematically impossible. the eaisest way to solve for complex roots of a quadratic is "Whenever you see radical(-n), you can note it as i*Radical(n)", n>0. Now, if we want to solve the following quadratic using the real way:

x^2-2x+2=0

(x-1)^2=-1
now you write -1 as cos180+isin180
and use demoivre's theroem to get:
x-1=cos90+isin90 or x-1=cos270+isin270
so
x-1=i or -i
x=i+1 or -i+1

to sql_lall:
-i is simiply -1*i. a complex number is a real number plus another real number times i. i is a symbol that upon being squared, gives you -1. just like pi is the ratio of the area of a circle to the area of a square with the side length of the circle's radius.

[Dreamlax]

I'm lost... I dunno much about the concept of complex numbers... When I first heard "Complex numbers are a set of numbers outside real numbers" I immediately thought they had two decimal places or something stupid like that.

Can someone introduce me to 'radical' ?

[NotLKH]

Hate to quible, but:

Originally posted by bugzpodder

i is a symbol that is introduced such that i^2=-1. i quote Also, as you pointed out, radical(-1) does not exists. We are prevented from writing it because it is mathematically impossible.

Are you sure we are prevented from writing it because it is mathematically impossible?
radical(-1)
radical(-1)
radical(-1)
radical(-1)
radical(-1)
radical(-1)

Hmmm, I'm able to write it. You must mean it we are prevented from calculating it because it is mathematically impossible.

Originally posted by bugzpodder

the eaisest way to solve for complex roots of a quadratic is "Whenever you see radical(-n), you can note it as i*Radical(n)", n>0. Now, if we want to solve the following quadratic using the real way:

x^2-2x+2=0

(x-1)^2=-1
now you write -1 as cos180+isin180
and use demoivre's theroem to get:
x-1=cos90+isin90 or x-1=cos270+isin270
so
x-1=i or -i
x=i+1 or -i+1

So you are saying that, since (-b ⁺_- [b²-4ac]^(1/2))(/2a) can and does produce radical(-n) expressions, then we should stop useing it, and resort to untangling the order 2 equation by hand to get to the point where we can use demoivre's theorem?

And, since any plug in equation to generate roots for order 3 and order 4 polynomials inherently have the same "danger", we shouldn't use those either?

All this side stepping, wild and crazy guy dancing around just to keep from seeing radical(-n) is exhausting!
Its also similar to superstition, and the lenghths some people will go to evade black cats.



-Lou

[bugzpodder]

you missed my point.

back to the sqrt(-1)

there are two roots to sqrt(-1) -- i and -i. sqrt(x), where x<-1 is not well defined, thus the use of demoivre's theorem. sqrt(x) here is still easy. what if its 100th-root(x) -- x^(1/100)? where x=-1? can you figure out all the roots at once? no! then you have to use demoivre's theorem. so what i am saying is use it on all cases, whether u can figure them out or not.

now unlike real numbers, complex numbers does not have positive or negative. so how do you know which one you want?

the way complex numbers is defined, if you choose -i to be your i, you would still have the same number system, with a rotation of numbers in the complex plane. hope this explains better of my position.