Ratchit

To the Ineffable All,
There seems to be some confusion about "i", which is sometimes wrongly called the square root of -1. The square root of -1 is a numeric constant, specifically 1/_90deg in polar form. "i" is correctly defined as a operator in correct math texts. 5i means take 5 and rotate it 90 degrees CCW. 2+5i defines a duplex number that has a real part and a orthogonal part. There is confusion because of bad notation. i^2 is not an exponential operation. It really means an operation that rotates 1 twice by 90 degrees. A better notation would be something like i(2). Now we get correct answers by treating "i" like a constant, but that is due to conformal similarity, not because the concept is correct. So again, Sqrt(-1) is not "i". It is 1/_90 in polar form or i(1) in rectangular form. Ratch

3iMaJ

Lets define i in a rigorous manner so we can be done with all of this nonsense.

Given an ordered pair (a,b), lets define two operations:
(a,b) + (c,d) = (a + b,c +d)
(a,b)*(c,d) = (ac - bd,bc + ad)

We will cleverly call our operations addition and multiplication. For our newly defined operations we would like them to satisfy a few properties, namely:

Closure under both + and *.
Both + and * are associative.
Both + and * are communitive.
The operation * is distributive over +.
Existance of additive identity.
Existance of multiplicitive identity.
Existance of additive inverses.
Existance of multiplicitve inverses.

If our operations have all of these properties then we've defined a field. A field is a communitive ring. A ring is simply a algebraic structure that contains two operations, in this case multiplication and addition. A communitive ring (or a field) is simply a ring where multiplication is communitive. You might note that the matrix operations do not form a field, as their multiplication is not communitive (generally speaking). As well as the problem with the existance of an inverse in all cases. An example of a field is the set of real numbers.

(a,b)*(1,0) = (a,b) = (1,0)(a,b) = (a,b): Our multiplicitive identity
(a,b) + (0,0) = (a,b) = (0,0) + (a,b): Our additive identity
(a,b)*(c,d) = (1,0)
(ac - bd, bc + ad) = (1,0)
ac - bd = 1
ad + bc = 0
Solving for c and d we get (c,d) = (a/(a^2 + b^2), -b/(a^2 + b^2))
which is our multiplicitive inverse.
(a,b) + (c,d) = (0,0)
a + c = 0
b + d = 0
c = -a
d = -b
(a,b) + (-a,-b) = (0,0): Our additive inverse

These properties are obviously hold and I will not prove them.

Closure under both + and *.
Both + and * are associative.
Both + and * are communitive.
The operation * is distributive over +.

So, where does i fit in?

This is the definition of i.

(a,b) = a(1,0) + b(0,1)
let (0,1) = i
(0,1)*(0,1) = (-1,0)
i^2 = -1

Since (a,b) uniquely specifies a + ib, the complex numbers are a one-to-one correspondence to the complex plane.

A question often arises what are complex numbers useful for? Since real numbers are a subset of complex numbers there aren't solutions to all equations with out invoking the more general complex number, namely

x^2 + 1 = 0, has no real number solution, but as we've just seen in the definition of i, the solution to that equation i.

To recap:

The "imaginary" number i is a consequence of how we define the operations + and * in our new field we're calling the complex numbers. A complex number is simply a number that has operations + and * that satisfy several properties that were listed earlier.

The unit i arises when we write (a,b) = a(1,0) + b(0,1). We can write it in this form because of the properties of our field. The interesting piece of this equation is (0,1), what exactly is (0,1)? It is (0,1)*(0,1) = i^2 = -1. So we can see where i = sqrt(-1) comes from.

Its VERY important to note, and this was the point made in the previous post is that the sqrt(.) function is only defined for real x >= 0. Using sqrt(-1) to manipulate numbers can result in false results namely:

-1 = i*i = sqrt(-1)*sqrt(-1) = sqrt(1) = 1. This is incorrect because the square root function IS NOT defined for x < 0, which is the reason for the bad result. To avoid this mistake simply do the following:

i = i*sqrt(1), so -1 = i*i = i*i*sqrt(1)*sqrt(1) = -1. Crisis averted.

Ratchit

3iMaJ,

I don't know why you are proving binary properties of associativity, communitivity, and distributivity on the unary operator i.

I don't believe so. Anyone who has worked in engineering knows how useful duplex numbers are.

Actually, there is no "imaginary" number. That is a misnomer. The number most folks call imaginary is actually a number that is orthogonal to the real number taken as the reference. For instance, in electrical engineering, the energy in AC circuits within the magnetic and electrostatic fields represented by "imaginary" numbers is every bit as real as the energy dissipated by the resistive component.

Well, every negative number has two square roots. I already showed that -1 has a square root of plus or minus 1/_90.

Well, I say that that isn't so for the above reason.

Crisis that never was. Ratch

3iMaJ

I was deriving complex numbers based on two operators. And I'd like to point out that i is not an operator. The operators of the field are + and *. i is simply (0,1) its a number within the complex set, not an operator.

You'll also note that I put "imaginary" in quotations just as I've done here. That is commonly what they're called, even though it is technically incorrect.

Let me be very specific here:

The square root operator is ONLY defined for real numbers >= 0. Now let me show you why you can use them on complex numbers.

Reminding you (0,1)(0,1) = i^2 = -1.
sqrt(-16) = sqrt(-1)*sqrt(16) = i*4. HOWEVER sqrt(-16) w/o separating it into sqrt(-1)*sqrt(16) is not defined for the square root function.

The mathematical definition of the square root is that its the inverse function of f(x) = x^2 for all real x >= 0. In other words the square root function is a mapping of non-negative real numbers to R+ union 0. The key there is non-negative.


Quit with all the orthogonal nonsense, of course a purely real number is orthogonal to a "imaginary" by construction.

(a,b) is our ordered pair construction, so (a,0)*(0,b) = (0,0). Which is by definition orthogonal. This true because of how we constructed our field, and not for any other reason.

(0,1) is 1<90 because of the mapping of the complex numbers to the complex plane. Your orthogonality isn't a consequence of the mapping like you seem to claim, but rather a consequence of the construction of the field.

Ratchit

3iMaJ,

I disagree, as do many math textbooks that define i to be a operator. Yes, + and * are binary operators too, but that does not keep i from being a unary operator.

We both agree on that. I think they should be called orthogonal numbers.


You are treating i as a constant, which I don't agree with.

The Sqrt(-16) is calculated, not defined. I would not say that 2+2 is defined to be 4. It is calculated to be 4. I would, however, say that 0! can be defined to be 1 because its calculation is indefinite.

You can map it anywhich way you want, but why restrict yourself. I know that the Sqrt(-4) is either -2/_90 or 2/_90. How? Because when I multiply each of those number by themselves, I get -4. That is the basic definition of a square root.

A orthogonal number is what it is. And it is not the mapping that makes it orthogonal. The orthogonal number is the dog that wags the tail (map).

The tail wagging the dog again.

The orthogonal number itself determines its properties, not the mapping. Ratch

tkbits

Not this twat (tiresome argument) again.

Just because i can be interpreted as an operator doesn't mean it is the ONLY correct interpretation. For example, i can simply be a marker for the second number in an ordered pair <real-part, imaginary-part>. Then we can define operations on ordered pairs that are consistent with how we calculate with complex numbers.

I'm very aware from algebra theory that a value can be reinterpreted as an operator. So even if we define i to be a value (which the majority of us prefer to do), it can be "reinterpreted", just as desired, to be an "operator". We do this in linear algebra, where some matrices are algebraically shown as operators (also known as "transformations").

There does not seem to be much advantage to treating i as an operator, because you still need to redefine addition as a vector addition (rather than a simple numeric addition). Then i might just as well represent the unit vector in the "i" direction, and the real number attached to it can be its scaling factor.

For all its assumed "naturalness" because it embeds rotation in the "meaning" of i, it doesn't show how rotation occurs when you're multiplying arbitrary complex numbers. For that, the polar form of complex numbers is a much better choice.

3iMaJ

Horray for a reasonable person!!

I quit arguing because I saw it to be pointless.

Ratchit

tkbits,

"Not this twat again."

Are you going into ad hominem mode now?

"Just because i can be interpreted as an operator doesn't mean it is the ONLY correct interpretation."

No matter how it is interpreted, a duplex number is a real plus an orthogonal number.

"So even if we define i to be a value (which the majority of us prefer to do), it can be "reinterpreted", just as desired, to be an "operator"."

If it were not for its conformal similarity, you would not be able to do that. i is rather unique in that respect.

"We do this in linear algebra, where some matrices are algebraically shown as operators (also known as "transformations")."

And is a matrix not a operator that specifies predetermined multiplications and additions with another matrix or number?

"There does not seem to be much advantage to treating i as an operator, because you still need to redefine addition as a vector addition (rather than a simple numeric addition). Then i might just as well represent the unit vector in the "i" direction, and the real number attached to it can be its scaling factor."

Like said before, you get correct answers treating i as a constant because of its conformal similarity. But conceptionally, 5i is not i+i+i+i+i, it is 5 rotated by 90 degrees.

"For all its assumed "naturalness" because it embeds rotation in the "meaning" of i, it doesn't show how rotation occurs when you're multiplying arbitrary complex numbers. For that, the polar form of complex numbers is a much better choice."

Sure, everyone knows that the rectangular form is better for addition/subtraction and the polar form it better for multiplication/division/exponentiation.

3iMaJ,

"Horray for a reasonable person!!"

You mean, someone who agrees with you?

"I quit arguing because I saw it to be pointless."

And you could not refute it. Ratchit

__________________
Hopelessly pedantic

tkbits

So I misused a term to refer to the argument.

As for "conformal similarity", i is not unique. Any fixed rotation can be viewed as a value. It need not be the same as the arithmetic numbers we are all so familiar with.

As for 5i being a rotation of 5, a number cannot be rotated. You can rotate a vector of length 5 pointing in the "positive real" direction, which appears to be what you are saying.

I can also insist that i is a vector that is of length 1 pointing in the "positive imaginary" direction. Then I can insist that 5i represents i + i + i + i + i under vector addition, or better, that it represents a magnification of i by a factor of 5. In the latter case, 5 is the operator, not i.

What's tiresome is the insistence that rotation is the "one true way" of viewing i.

Ratchit

tkbits,
"So I misused a term to refer to the argument."

OK.

"As for "conformal similarity", i is not unique. Any fixed rotation can be viewed as a value. It need not be the same as the arithmetic numbers we are all so familiar with."

Well, if the "fixed" rotation were say, 50 degrees, you could not treat the rotation as a value and get correct answers. A 90 degree rotation keeps repeating itself on the boundries of each quadrant, thereby bestowing i its conformal properties.

"As for 5i being a rotation of 5, a number cannot be rotated. You can rotate a vector of length 5 pointing in the "positive real" direction, which appears to be what you are saying."

Any real or natural number can be represented as a vector (magnitude and direction) along a reference axis. The magnitude is not rotated, the direction is.

"I can also insist that i is a vector that is of length 1 pointing in the "positive imaginary" direction. Then I can insist that 5i represents i + i + i + i + i under vector addition, or better, that it represents a magnification of i by a factor of 5. In the latter case, 5 is the operator, not i."

No, then multiplication would be the binary operator with 5 and i being the operands.

"What's tiresome is the insistence that rotation is the "one true way" of viewing i."

Rotation is the operative path that brings a real number to an orthogonal existence. It is a simple and elegant way of conceptualizing it. Ratchit

__________________
Hopelessly pedantic

tkbits

My only request is that you do not argue it to be the "one true (correct) way". It may or may not be the best introduction to the vector properties of complex numbers.

Any further discussion ought to be on another thread.

Ratchit

tkbits,

"My only request is that you do not argue it to be the 'one true (correct) way'. "

If there is another way, I am willing to look at it. But I will not compromise myself to say that i is a constant equal to the sqrt(-1).

"It may or may not be the best introduction to the vector properties of complex numbers."

I think it is the best way, but anyone is free to learn about duplex numbers any which way they want.

"Any further discussion ought to be on another thread."

Why? Keep the continuity of the discussion intact. Ratchit

__________________
Hopelessly pedantic

3iMaJ

In the field with operators + and * that I derived earlier, i was a constant, where i = (0,1)*(0,1) = -1. I never said that i was equal to sqrt(-1), generally some liberties are taken and it can be treated as such. But in the field that I derived, under the operators that I defined, i was a constant and i^2 = -1, that is the definition of i in my field, a constant. Now if you want i to be an operator, define it to be one, set up and show what properties it has and where it fits into a field of your choosing. But i as a rotational operator is not the only way to view i, it can be viewed as a constant depending on how you setup your field.

Ratchit

3iMaJ,

"But i as a rotational operator is not the only way to view i, it can be viewed as a constant depending on how you setup your field."

Because i has its conformal similarity, you can use it as a constant to get correct answers in calculations, and even define a field as you have done above. Heck, I even use it that way myself to do duplex number calculations. But underlying all the proofs and conclusions is the unique property of a 90 degree rotation. You would not be able to do what you did for any other rotation. As far as I can see, you have proved that i can be used as a constant to get correct answers, but that was known already. Ratchit

__________________
Hopelessly pedantic

3iMaJ

History suggests that the reason was used as the sqrt(-1) so the following error could be avoided:

-1 = i^2 = sqrt(-1)*sqrt(-1) = sqrt(1) = 1.

Now I have to admit I know where you're coming from. The supposition that the inner product of i with any real number is zero, ie is orthogonal to the real set. Which is consistant with what tkbits was saying about vector notation, where a vector r = xax + yay + zaz where ax, ay, and az are all unit vectors pointing in their respective directly, which is why r can be written in the way it was shown above, which is also what makes i special in that sense, that a complex number can be written z = a + ib, since i is orthogonal to the real number set.

I claim i to be a constant, you claim it to be a rotational operator, but in the end we're both correct. i is a special constant that is orthgonal to the real number set (which seems to be what you're saying), but it in my eyes i is just (0,1) which is in the complex number set, a constant, albeit one with a special property. But this property was a result of how the my field was defined, and it just happens to have the (necessary orthogonal property) otherwise my field would cease to be a field.

History tells a different story where i was simply invented to solve equations that do not have real number solutions. This was at a time when negative numbers were on weak ground, the existance of i wasn't justified until 150 years later, when a geometric interpretation of i was invented, this is the interpretation you choose to use. But either method is fine as they can be derived independantly. So we can go on arguing until we're blue in the face, but in the end we're both correct, and both supported by firm mathematics and history.