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Topic: e^pi = 1 (Read 864 times) |
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Icarus
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[smiley=e.gif][smiley=suppi.gif] = 1.
Proof:
Euler's formula tells us that
[smiley=e.gif][smiley=sup2.gif][smiley=suppi.gif][smiley=supi.gif] = [smiley=c.gif][smiley=o.gif][smiley=s.gif] 2[pi] + [smiley=i.gif] [smiley=s.gif][smiley=i.gif][smiley=n.gif] 2[pi] = 1
so [smiley=e.gif][smiley=sup2.gif][smiley=suppi.gif] = ([smiley=e.gif][smiley=sup2.gif][smiley=suppi.gif][smiley=supi.gif])[smiley=supminus.gif][smiley=supi.gif] = 1[smiley=supminus.gif][smiley=supi.gif] = 1 (since 1 raised to any power is still 1).
[smiley=e.gif][smiley=suppi.gif] = [sqrt][smiley=e.gif][smiley=sup2.gif][smiley=suppi.gif] = [sqrt]1 = 1.
QED
Since the Gelfond-Schneider Theorem shows that [smiley=e.gif][smiley=suppi.gif] is a transcendental number, it follows that 1 is transcendental, and thus is not an integer as people have generally believed.
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| « Last Edit: Aug 26th, 2003, 5:27pm by Icarus » |
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Sir Col
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Re: e^pi = 1
« Reply #1 on: Aug 26th, 2003, 6:52pm » |
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You're very naughty, Icarus.
Perhaps I could present one along similar lines...
e[pi]i = cos[pi] + i sin[pi] = –1.
As e2[pi]i = 1 (see previous post), we can raise both sides to the power 1/2, (e2[pi]i)1/2 = 11/2 [bigto] e[pi]i = 1, but we have just shown that e[pi]i = –1.
[therefore] 1 = –1, and multiplying both sides by x leads to the generalisation that x = –x; in other words, all positive numbers are exactly equal to their negative counterpart. Of course, we already knew that. 
It's an interesting general question that you pose, Icarus. I suspect that many problems are often misplaced, with regards to difficulty. I'm never quite sure where to post problems myself, as too many standards have been set with medium/hard problems being posted in the easy section. The boundaries are certainly not clearly defined – a little like the terms in the proof.
By the way, Medium got my vote, as in the grand scheme of things it's not Hard, but it certainly couldn't be described as Easy, either.
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| « Last Edit: Aug 27th, 2003, 8:13am by Sir Col » |
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towr
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Re: e^pi = 1
« Reply #2 on: Aug 27th, 2003, 2:26am » |
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Part of those proofs are along the same lines as proving -1 = 1 by squaring both sides and noticing they're equal..
[sqrt]1 = 1 [vee] [sqrt]1=-1, you can't simply pick the one that suits you..
Of course [smiley=e.gif][smiley=suppi.gif] isn't -1 either.. 
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| « Last Edit: Aug 27th, 2003, 2:27am by towr » |
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Sameer
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Re: e^pi = 1
« Reply #3 on: Aug 27th, 2003, 7:01am » |
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Aren't we supposed to use De Moivre's therom here to find roots ... this was like saying that if 1 and 3 are roots of equation f(x) = 0 then 1 = 3 ?
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TenaliRaman
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Re: e^pi = 1
« Reply #4 on: Aug 27th, 2003, 11:21am » |
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i always knew e^pi = 1.But i voted for medium as this would have allowed me to post something in the medium section 
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Icarus
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Re: e^pi = 1
« Reply #5 on: Aug 27th, 2003, 3:29pm » |
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on Aug 27th, 2003, 2:26am, towr wrote:Part of those proofs are along the same lines as proving -1 = 1 by squaring both sides and noticing they're equal..
[sqrt]1 = 1 [vee] [sqrt]1=-1, you can't simply pick the one that suits you.. |
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This is the case for Sir Col's variation, but for mine, I am justified in picking the positive square root because [smiley=e.gif][smiley=suppi.gif] > 0.
Quote:| Of course [smiley=e.gif][smiley=suppi.gif] isn't -1 either.. |
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on Aug 26th, 2003, 6:52pm, Sir Col wrote:| It's an interesting general question that you pose, Icarus. I suspect that many problems are often misplaced, with regards to difficulty. I'm never quite sure where to post problems myself, as too many standards have been set with medium/hard problems being posted in the easy section. The boundaries are certainly not clearly defined – a little like the terms in the proof. |
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That's why I decided to add the poll. It's easy for me (and would be even if I wasn't the one to come up with it), but I have specialized background that makes it so. So I decided to put it where I thought it should go, but let others have a say. If significantly more people think it should be somewhere else, I'll move it.
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Sir Col
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Re: e^pi = 1
« Reply #6 on: Aug 27th, 2003, 7:09pm » |
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Okay, I'll take the bite, Icarus...
on Aug 26th, 2003, 5:03pm, Icarus wrote:| Since the Gelfond-Schneider Theorem shows that [smiley=e.gif][smiley=suppi.gif] is a transcendental number, it follows that 1 is transcendental, and thus is not an integer as people have generally believed. |
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The Gelfond-Schneider Theorem states that [alpha][beta] is transcendental if [alpha][ne]0,1 and [beta] are algebraic. Neither e nor [pi] are algebraic, so the GS Theorem does not apply.
Quote:| (since 1 raised to any power is still 1). |
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Really? 11/3 = 1, true, but 11/3 = -(1[pm]i[sqrt]3)/2, also. 
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| « Last Edit: Aug 27th, 2003, 7:12pm by Sir Col » |
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Icarus
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Re: e^pi = 1
« Reply #7 on: Aug 27th, 2003, 7:55pm » |
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The Gelfond-Schneider theorem does apply. You just have to be canny about it. It was proved by Gelfond and Schneider independently for the express purpose of showing the [smiley=e.gif][smiley=suppi.gif] is transcendental. (This was one of Hilbert's problems.)
But that was not part of this puzzle. It was simply a side comment to show a curious consequence of this startling result.
(-1)2 = 1, but 11/2 = 1, not -1. This is a straightforward consequence of the definition of exponentiation of Real numbers.
But to settle any doubts:
1 = [smiley=e.gif][sup0]. (Any base will do.) So,
1 [smiley=supminus.gif][supi] = ([smiley=e.gif][sup0])[smiley=supminus.gif][supi] = [smiley=e.gif][sup0][smiley=suplp.gif][smiley=supminus.gif][supi][smiley=suprp.gif] = [smiley=e.gif][sup0] = 1.
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towr
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Re: e^pi = 1
« Reply #8 on: Aug 28th, 2003, 2:33am » |
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on Aug 26th, 2003, 5:03pm, Icarus wrote:| [smiley=e.gif][smiley=suppi.gif] = 1. |
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[smiley=e.gif] > 1 and [smiley=pi.gif] > 1 thus [smiley=e.gif][smiley=suppi.gif] > 1
so we've got [smiley=e.gif][smiley=suppi.gif] = 1 and [smiley=e.gif][smiley=suppi.gif] > 1
clearly the universe should now explode
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Sir Col
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Re: e^pi = 1
« Reply #9 on: Aug 28th, 2003, 7:53am » |
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Okay, I'm quite happy with, e2k[pi][smiley=i.gif] = cos(2k[pi]) + [smiley=i.gif] sin(2k[pi]) = 1.
Of course, when k=0, everything is perfect: e0 = 1.
For k=1, I'll accept that, e2[pi][smiley=i.gif] = 1.
However, I do not accept that, e2[pi] = 1; clearly it is not true.
So it seems that raising both sides to the power of –[smiley=i.gif] causes a problem. I don't know why, but I suspect that is where the problem lies.
By the way, Icarus, do you know how we use the GS Theorem to show that e[pi] is transcendental?
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| « Last Edit: Aug 28th, 2003, 7:55am by Sir Col » |
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Sameer
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Re: e^pi = 1
« Reply #10 on: Aug 28th, 2003, 11:52am » |
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Assume the statement is correct:
i.e.
e[pi] = 1
Taking Natural logarithm on both sides
ln e[pi] = ln 1
[pi] ln e = 0
[pi] = 0
Hmm the egyptians definitely miscalculated the value of [pi] 
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James Fingas
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Re: e^pi = 1
« Reply #11 on: Aug 28th, 2003, 12:45pm » |
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on Aug 28th, 2003, 7:53am, Sir Col wrote:Okay, I'm quite happy with, e2k[pi][smiley=i.gif] = cos(2k[pi]) + [smiley=i.gif] sin(2k[pi]) = 1.
Of course, when k=0, everything is perfect: e0 = 1.
For k=1, I'll accept that, e2[pi][smiley=i.gif] = 1.
However, I do not accept that, e2[pi] = 1; clearly it is not true.
So it seems that raising both sides to the power of –[smiley=i.gif] causes a problem. I don't know why, but I suspect that is where the problem lies. |
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For the first line, we could say that:
e2k[pi][smiley=i.gif] = e2m[pi][smiley=i.gif] [forall] k,m [in] [smiley=bbi.gif]
but it is evident that
e2k[pi] [ne] e2m[pi] [forall] k,m [in] [smiley=bbi.gif] unless k=m
This encompasses the more specific case we're dealing with here.
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Sameer
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Re: e^pi = 1
« Reply #12 on: Aug 28th, 2003, 3:12pm » |
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on Aug 28th, 2003, 12:45pm, James Fingas wrote:
For the first line, we could say that:
e2k[pi][smiley=i.gif] = e2m[pi][smiley=i.gif] [forall] k,m [in] [smiley=bbi.gif]
but it is evident that
e2k[pi] [ne] e2m[pi] [forall] k,m [in] [smiley=bbi.gif] unless k=m
This encompasses the more specific case we're dealing with here. |
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I guess k is a multiple of m would be better.
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"Obvious" is the most dangerous word in mathematics.
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Sir Col
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Re: e^pi = 1
« Reply #13 on: Aug 28th, 2003, 3:48pm » |
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I think you've cracked it, James. The multi-value function, e2k[pi][smiley=i.gif] = 1, only holds for complex numbers; we cannot apply the same identity to real numbers. Raising both sides to the power [smiley=i.gif], converts it to an equation that is exclusively real and, consequently, single valued.
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Icarus
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Re: e^pi = 1
« Reply #14 on: Aug 28th, 2003, 4:18pm » |
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Close, but not quite correct.
Yes, I do know how to use GS to show that [smiley=e.gif][smiley=suppi.gif] is transcendental. But I am going to wait until someone has correctly stated what the problem is with the proof.
Hint 1: The nasty thing about the "proof" is that every statement by itself is true with the appropriate interpretation.
Hint 2: My previous post was a deliberate obfuscation.
Hint 3: Now why would I not want to show the GS thing until after the puzzle is solved?
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| « Last Edit: Aug 28th, 2003, 4:22pm by Icarus » |
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James Fingas
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Re: e^pi = 1
« Reply #15 on: Aug 29th, 2003, 7:16am » |
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The fundamental problem is that e2[pi][smiley=i.gif] = 1 is true, but when you raise each side to the power of -[smiley=i.gif], you are implicitly doing the following:
a = b
eln(a) = eln(b)
e-[smiley=i.gif]ln(a) = e-[smiley=i.gif]ln(b)
This may break down if you use two different branches of the natural logarithm function. So the rule you've broken is: if you take the logarithm of a function, you must use the same branch of the logarithm everywhere.
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Icarus
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Re: e^pi = 1
« Reply #16 on: Aug 29th, 2003, 3:58pm » |
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That isn't how I would of described it, but it amounts to the same thing.
Like the logarithm, exponentiation is not single-valued for complex exponents. Real exponentiation suffers from this as well, for rational exponents, but in that case there is a "natural choice" - one that is favored by the mathematics itself, not just the mathematicians: the positive real value.
The same is not true for complex exponents. With the exception of base [smiley=e.gif], where the multiple values collapse, complex exponents give infinitely many values, and none is "natural". So we get around this inconvenient fact by declaring complex exponentiation a "multi-valued" function. And such expressions can represent any of the values, depending on which "branch" of the exponential function you choose.
So 1[supminus][supi] = [smiley=e.gif][sup2][smiley=suppi.gif] is true for one branch of the exponential function.
1[supminus][supi] = 1 is also true for a branch. But they are true for different branches. Therefore it is incorrect to string them together as I did.
Sir Col: By one branch of the exponential function, (-1)[supminus][supi] = [smiley=e.gif][smiley=suppi.gif]. The expression on the left satisfies the requirements of the Gelfond-Schneider Theorem, so all of its possible values must be transcendental, including [smiley=e.gif][smiley=suppi.gif].
Last I knew, another of the numbers Hilbert asked about, [pi][smiley=supe.gif], is still unknown.
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SWF
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Re: e^pi = 1
« Reply #17 on: Aug 30th, 2003, 3:28pm » |
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The Mathworld description of G-S Theorem does not mention that it applies when [beta] is non-rational and algebraic (allowing complex values)- only irrational and algebraic. I found some other references that say it permits non-rational values. Not that Mathworld's description is wrong, just that its description of the theorem is incomplete.
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