Do you believe in √(-1)?
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  Do you believe in √(-1)?
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Question: Do you believe in √(-1)?
#1
Yes
 
#2
No
 
#3
I'm agnostic
 
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Total Voters: 42

Author Topic: Do you believe in √(-1)?  (Read 5471 times)
Storebought
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« Reply #25 on: June 09, 2009, 12:45:34 AM »
« edited: June 09, 2009, 12:48:18 AM by Storebought »


Hamilton's greatest achievement. Too bad we quit teaching and using them in favor of vector and matrix formalism. Matrices always work as a representation, but the beauty of simple elements like quaternions directly shows symmetry.

After lecturing for a day or so about quaternions, the professor just admitted, "Physicists struggle, mightily, to find a proper use for these things."

Math as given through matrix representatives is simply easier to follow. But even with finite rotations, which are well handled by matrices, the math gets to be too much for us poor humble chemists. Just give us the character table for the rotation and be done with it.
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Queen Mum Inks.LWC
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« Reply #26 on: June 09, 2009, 01:54:20 AM »

Of course.
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« Reply #27 on: March 08, 2021, 03:50:22 AM »
« Edited: March 08, 2021, 02:18:14 PM by Spring is coming ⛅🌱🍃🌸🐑🐰🐣 »

No.  Imaginary numbers are, by definition, imaginary, i.e. products of the imagination.  Therefore, they are not real; if they were, they would be real numbers.  Since the square root of -1 is clearly not a real number but an imaginary number, then by its very definition it cannot exist.

The name has little to do with its meaning and more to do with its history. That would be like saying a strange quark has less reality than an electron because of its fanciful name.


Imaginary numbers were therefore called lateral numbers after their discovery.
They are nowadays called "imaginary" because they are too complex for too many people to understand... 🤓

What still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯
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True Federalist (진정한 연방 주의자)
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« Reply #28 on: March 08, 2021, 09:00:21 AM »

https://www.quantamagazine.org/imaginary-numbers-may-be-essential-for-describing-reality-20210303/
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muon2
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« Reply #29 on: March 08, 2021, 02:01:58 PM »

No.  Imaginary numbers are, by definition, imaginary, i.e. products of the imagination.  Therefore, they are not real; if they were, they would be real numbers.  Since the square root of -1 is clearly not a real number but an imaginary number, then by its very definition it cannot exist.

The name has little to do with its meaning and more to do with its history. That would be like saying a strange quark has less reality than an electron because of its fanciful name.


Imaginary numbers were therefore called lateral numbers after their discovery.
They are nowadays called "imaginary" because they are too complex for too many people to understand... 🤓

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯

Why would the result seem so unusual? Is it because there is more than one value in the answer?

The natural log is the inverse of the exponential, and is often used to find the result of a real as well as a complex exponential. Since complex exponentials are equivalent to certain trigonometric functions, the complex logarithm behaves in part like an inverse trig function. Inverse trig functions with real arguments have multiple values for the answer, offset by 2pi. Similarly offsets by 2pi appear in the complex logarithm, so it may have multiple values. Hence complex exponentiation may also have multiple values as in this case.
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« Reply #30 on: March 08, 2021, 02:35:13 PM »

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯
Why would the result seem so unusual? Is it because there is more than one value in the answer?

No, because the result is a real number. The proof of that is very easy to understand, but it's hard to visualize why the result comes into being.
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vitoNova
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« Reply #31 on: March 08, 2021, 08:06:08 PM »

I saw some crazy sh**t on YouTube how irrational/imaginary numbers were the natural byproducts of trying to explain the universe in post-Euclidean, post-Newtonian terms.  

Damn. No wonder you STEM people give birth to autistic babies.  Way above my head, bruh.   Gimme the cliff notes.  
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muon2
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« Reply #32 on: March 08, 2021, 08:14:36 PM »

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯
Why would the result seem so unusual? Is it because there is more than one value in the answer?

No, because the result is a real number. The proof of that is very easy to understand, but it's hard to visualize why the result comes into being.


But it is defined such that exponentiation results in a real number - square it and you get -1, fourth power gives +1. Why would other powers, including i, also being real be a surprise?
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« Reply #33 on: March 08, 2021, 08:36:46 PM »

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯
Why would the result seem so unusual? Is it because there is more than one value in the answer?

No, because the result is a real number. The proof of that is very easy to understand, but it's hard to visualize why the result comes into being.


But it is defined such that exponentiation results in a real number - square it and you get -1, fourth power gives +1. Why would other powers, including i, also being real be a surprise?

I was talking about i i.
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Antonio the Sixth
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« Reply #34 on: March 08, 2021, 11:08:10 PM »

It is a fully defined mathematical construct, so it's just as real (even if not "Real") as any other number.
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muon2
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« Reply #35 on: March 09, 2021, 12:29:30 AM »
« Edited: March 09, 2021, 12:35:03 AM by muon2 »

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯
Why would the result seem so unusual? Is it because there is more than one value in the answer?

No, because the result is a real number. The proof of that is very easy to understand, but it's hard to visualize why the result comes into being.


But it is defined such that exponentiation results in a real number - square it and you get -1, fourth power gives +1. Why would other powers, including i, also being real be a surprise?

I was talking about i i.


As am I. My point is that i z is real for a number of values of z, and is defined based on its exponential to a specific real value of z = 2. Why is is surprising that there are many other values of z, including complex values such as i, that also give real results?

If the problem is one of visualization, then I have to start by asking how you visualize i 2. If it is only as an algebraic formula, I can see where that is hard to extend. If instead you visualize points in a plane that can shift and rotate then it is easier to see how some can end up on the real axis.
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Benjamin Frank
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« Reply #36 on: March 12, 2021, 07:04:34 AM »

According to a calculus instructor I had, imaginary numbers are essential for electrical engineering.  So, if imaginary numbers weren't real, all the computers that this site is on and that you use to access it wouldn't exist either.


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Former President tack50
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« Reply #37 on: March 12, 2021, 12:19:56 PM »

According to a calculus instructor I had, imaginary numbers are essential for electrical engineering.  So, if imaginary numbers weren't real, all the computers that this site is on and that you use to access it wouldn't exist either.


As someone who had to take a small electrical engineering class in college yes, this is correct and imaginarny numbers have plenty of applications in electricity and electronics.

Even if it is not the most obvious application, imaginary numbers can and definitely do exist in the "real world"
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« Reply #38 on: March 12, 2021, 12:34:08 PM »

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯
Why would the result seem so unusual? Is it because there is more than one value in the answer?

No, because the result is a real number. The proof of that is very easy to understand, but it's hard to visualize why the result comes into being.


But it is defined such that exponentiation results in a real number - square it and you get -1, fourth power gives +1. Why would other powers, including i, also being real be a surprise?

I was talking about i i.


As am I. My point is that i z is real for a number of values of z, and is defined based on its exponential to a specific real value of z = 2. Why is is surprising that there are many other values of z, including complex values such as i, that also give real results?

If the problem is one of visualization, then I have to start by asking how you visualize i 2. If it is only as an algebraic formula, I can see where that is hard to extend. If instead you visualize points in a plane that can shift and rotate then it is easier to see how some can end up on the real axis.

It's just too hard to understand for me what "to the power of i" even means.
Is there even a real number x that, exponentiated with itself, delivers the same result as ii = xx?
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muon2
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« Reply #39 on: March 12, 2021, 05:11:47 PM »

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯
Why would the result seem so unusual? Is it because there is more than one value in the answer?

No, because the result is a real number. The proof of that is very easy to understand, but it's hard to visualize why the result comes into being.


But it is defined such that exponentiation results in a real number - square it and you get -1, fourth power gives +1. Why would other powers, including i, also being real be a surprise?

I was talking about i i.


As am I. My point is that i z is real for a number of values of z, and is defined based on its exponential to a specific real value of z = 2. Why is is surprising that there are many other values of z, including complex values such as i, that also give real results?

If the problem is one of visualization, then I have to start by asking how you visualize i 2. If it is only as an algebraic formula, I can see where that is hard to extend. If instead you visualize points in a plane that can shift and rotate then it is easier to see how some can end up on the real axis.

It's just too hard to understand for me what "to the power of i" even means.
Is there even a real number x that, exponentiated with itself, delivers the same result as ii = xx?

Philosophically it should be just as hard to understand what it means "to the power of π (pi)". Exponentiation by integers is natural in that 53 means 5 x 5 x 5. Exponentiation by fractions can be inferred from the product rule: xaxb = xa+b so 51/2 means the square root because 51/2 51/2 = 51 = 5. But if a or b is an irrational number, then it isn't at all clear what it means in an exponent.

Exponentiation to an irrational power can be defined one of two ways. It can be found as the limit of powers to an infinite sequence of rational numbers that converge to the real number. That's fine if you grasp infinite convergent sequences, but philosophers have had problems with that going back to Zeno's paradoxes. The alternative is to recognize that exponentiation is the inverse of a logarithm, so you can find 5π = eπ ln 5 by using a table of natural logarithms and inverses. But from your statement, it seems this doesn't cause you any philosophical problems.

So if exponentiation and logarithms are inverse and you accept Euler's formula from knowledge of the infinite series of the trig functions, then there should be no philosophical problem with treating 5i as one would 5π. e i ln 5 = cos(ln 5) + i sin(ln 5). Here the table of natural logs is augmented by tables of sine and cosine, and the knowledge that a complex exponential can produce a complex value.

To answer your last question, the principal value of ii is approximately 0.2079. With a little calculus it can be shown that the lowest real value for x>0 is at (1/e)(1/e) = 0.6922 and xx is not generally defined with real number values for x<0. So there is no real value x that solves xx = ii.
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