Page 2 of 4

Re: 2^?

Posted: Wed Apr 03, 2013 9:35 am
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
1.246868988?

Re: 2^?

Posted: Wed Apr 03, 2013 9:50 am
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.

Re: 2^?

Posted: Wed Apr 03, 2013 9:59 am
by iansjack
CWood wrote:pi-th root of 2 (2^(1/pi))
That's not narrowing it down very much.

Re: 2^?

Posted: Wed Apr 03, 2013 10:10 am
by Mikemk
I think a nice one is 2^(X in base 1)

Re: 2^?

Posted: Wed Apr 03, 2013 11:09 am
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.

Re: 2^?

Posted: Wed Apr 03, 2013 11:16 am
by Gigasoft
m12 wrote:
iansjack wrote:-i
Wouldn't that be 1?
No, it's equal to cos(ln(2)) - sin(ln(2))i.

Re: 2^?

Posted: Wed Apr 03, 2013 11:38 am
by Mikemk
how do you all know all this?

Re: 2^?

Posted: Wed Apr 03, 2013 11:46 am
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.

Re: 2^?

Posted: Wed Apr 03, 2013 12:02 pm
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.
- XKCD 1047 (Approximations)

Re: 2^?

Posted: Wed Apr 03, 2013 12:07 pm
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.

Re: 2^?

Posted: Wed Apr 03, 2013 12:09 pm
by iansjack
m12 wrote:how do you all know all this?
Six years at university studying complex numbers helps a little.

Re: 2^?

Posted: Wed Apr 03, 2013 12:45 pm
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

Re: 2^?

Posted: Wed Apr 03, 2013 12:53 pm
by VolTeK
m12 wrote:how do you all know all this?

When your in high school, you'll understand :roll:

Re: 2^?

Posted: Wed Apr 03, 2013 12:57 pm
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.

Re: 2^?

Posted: Wed Apr 03, 2013 1:13 pm
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).