2^?

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Mikemk
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Re: 2^?

Post by Mikemk »

CWood wrote:pi-th root of 2 (2^(1/pi))

I'm sure there will be some of you who get the reference
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iansjack
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Re: 2^?

Post by iansjack »

i is defined as follows:

i^2 = -1
That's obviously not so. A definition of a number that applies to two numbers cannot reasonably be called a definition, can it? A definition can't be ambiguous like that. That is a property of i, not a definition of it.

i is the complex number (0,1). Simple.
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Re: 2^?

Post by iansjack »

CWood wrote:pi-th root of 2 (2^(1/pi))
That's not narrowing it down very much.
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Re: 2^?

Post by Mikemk »

I think a nice one is 2^(X in base 1)
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Re: 2^?

Post by Gigasoft »

iansjack wrote:
i is defined as follows:

i^2 = -1
That's obviously not so. A definition of a number that applies to two numbers cannot reasonably be called a definition, can it? A definition can't be ambiguous like that. That is a property of i, not a definition of it.

i is the complex number (0,1). Simple.
The property i^2 = -1 does in fact completely define i, as long as you don't read things into it that aren't there (such as "the solution of the equation i^2 = -1", or "the complex number with the property i^2 = -1", or "every number that has the property i^2 = -1"). The fact that the introduction of i also implies another number with the same properties, does not make the definition of i invalid. The statement "i is the complex number (0,1)" follows from the definition of complex numbers, so if we redefine i by substituting -i, i is still equal to the complex number (0,1), and nothing changes.
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Re: 2^?

Post by Gigasoft »

m12 wrote:
iansjack wrote:-i
Wouldn't that be 1?
No, it's equal to cos(ln(2)) - sin(ln(2))i.
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Re: 2^?

Post by Mikemk »

how do you all know all this?
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Re: 2^?

Post by Gigasoft »

AJ wrote:Hi,
Griwes wrote:i^2 = -1

No other definition of i is correct
i^6 = -1?

Cheers,
Adam
No, i^6 = -1 is consistent with i, -i^3 and i^5 being distinct entities, while i^2 = -1 is not.
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Re: 2^?

Post by CWood »

iansjack wrote:
CWood wrote:pi-th root of 2 (2^(1/pi))
That's not narrowing it down very much.
if you ever find yourself raising log(anything)^e or taking the pi-th root of anything, set down the marker and back away from the whiteboard; something has gone horribly wrong.
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iansjack
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Re: 2^?

Post by iansjack »

Gigasoft wrote:
iansjack wrote:
i is defined as follows:

i^2 = -1
That's obviously not so. A definition of a number that applies to two numbers cannot reasonably be called a definition, can it? A definition can't be ambiguous like that. That is a property of i, not a definition of it.

i is the complex number (0,1). Simple.
The property i^2 = -1 does in fact completely define i, as long as you don't read things into it that aren't there (such as "the solution of the equation i^2 = -1", or "the complex number with the property i^2 = -1", or "every number that has the property i^2 = -1"). The fact that the introduction of i also implies another number with the same properties, does not make the definition of i invalid. The statement "i is the complex number (0,1)" follows from the definition of complex numbers, so if we redefine i by substituting -i, i is still equal to the complex number (0,1), and nothing changes.
I'm sorry, but you are still incorrect. You are confusing a property of a number with something that uniquely defines it. You might as well define 3 by saying it is a prime factor of 12. That is certainly a property of 3, but it doesn't define it.

Mathematicians define i as being the tuple (0, 1) in the complex plane - at least that's the way it was when I was doing my thesis in Complex Analysis.
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Re: 2^?

Post by iansjack »

m12 wrote:how do you all know all this?
Six years at university studying complex numbers helps a little.
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Re: 2^?

Post by Griwes »

iansjack wrote:
Gigasoft wrote:
iansjack wrote:That's obviously not so. A definition of a number that applies to two numbers cannot reasonably be called a definition, can it? A definition can't be ambiguous like that. That is a property of i, not a definition of it.

i is the complex number (0,1). Simple.
The property i^2 = -1 does in fact completely define i, as long as you don't read things into it that aren't there (such as "the solution of the equation i^2 = -1", or "the complex number with the property i^2 = -1", or "every number that has the property i^2 = -1"). The fact that the introduction of i also implies another number with the same properties, does not make the definition of i invalid. The statement "i is the complex number (0,1)" follows from the definition of complex numbers, so if we redefine i by substituting -i, i is still equal to the complex number (0,1), and nothing changes.
I'm sorry, but you are still incorrect. You are confusing a property of a number with something that uniquely defines it. You might as well define 3 by saying it is a prime factor of 12. That is certainly a property of 3, but it doesn't define it.

Mathematicians define i as being the tuple (0, 1) in the complex plane - at least that's the way it was when I was doing my thesis in Complex Analysis.
Although I rarely rely on wikipedia when talking about math, I want to quote it here, as it is consistent with what I was taught at uni:
The imaginary number i is defined solely by the property that its square is −1:

i^2 = -1

With i defined this way, it follows directly from algebra that i and −i are both square roots of −1.
http://en.wikipedia.org/wiki/Imaginary_unit#Definition
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Re: 2^?

Post by VolTeK »

m12 wrote:how do you all know all this?

When your in high school, you'll understand :roll:
Last edited by VolTeK on Wed Apr 03, 2013 12:59 pm, edited 1 time in total.
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iansjack
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Re: 2^?

Post by iansjack »

Griwes wrote: Although I rarely rely on wikipedia when talking about math....
A very sensible policy that I wholeheartedly agree with. According to that "definition" the tuple (0, -1) is i. It isn't.
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Re: 2^?

Post by trinopoty »

if you ever find yourself raising log(anything)^e or taking the pi-th root of anything, set down the marker and back away from the whiteboard; something has gone horribly wrong.
Are you talking about (log(anything))^e or log((anything)^e).
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