what is wrong with this?

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.

Horray for a reasonable person!!

I quit arguing because I saw it to be pointless.

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

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.

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

Ratchit said:

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

Click to expand...

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.

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

Ratchit said:

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

Click to expand...

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.

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

Ratchit said:

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

Click to expand...

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.

3iMaJ,

The important thing is that the calculatons using i result in the same answer no matter which of the two ways we each expounded. Ratchit.