[math-fun] Re: favorite theorem
Let me say, only half in jest, that my favorite theorem in math is the classification theorem for finite sets, i.e., the fact that finite sets are classified up to isomorphism by the counting numbers (including zero)!
And I'll respond, half-seriously: How would you prove that?
I wouldn't call it a theorem in the theorem-proof sense. If a proof is defined (epistemo-sociologically) as a text that makes you believe and/or understand a proposition more than you did before you read the proof (leaving aside the question of what "believe" and "understand" mean), then I don't think this theorem *has* a proof! You can undoubtedly set up a formal system in which this claim is proved by induction, but nobody who didn't already accept the claim would be able to understand the formal system, let alone follow the proof. Instead, I am using the word "theorem" to mean any mathematical fact that makes you say "Well, will you look at that!" And this proposition is a real looker. We all encountered it when we were kids and noticed that a set doesn't change its cardinality when its elements are shifted around, and this agreeable invariance probably fuels many kids' enthusiasm for counting (even though their knowledge of this invariance is intuitive and most of them couldn't begin to articulate it as a general proposition). My serious point is that some mathematical facts may simultaneously be so profound and yet so commonplace that we don't recognize them as "theorems" as such. But I think I want to change my mind about my favorite mathematical fact. How about the fact that there are infinitely many counting numbers? The realization that you can keep on counting forever and never run out of numbers (even if you run out of names for them) was such a mind-blowing experience for my young self that I retained absolutely, positively NO memory of the experience. :-) But my mind remained permanently blown, so there was nothing for me to do with my life but become a mathematician... Is it a "theorem"? Well, I doubt that there are any formal bases for mathematics in which the axiom of infinity can be proved. Either you see it, or you don't. And once you've seen it, you can try to doubt it, but it's really hard to do. Jim
Other criteria I would use are the surprisingness of the theorem (imagine Euler's first reaction!), its vast implications, its relating previously unrelated but commonplace things (all the REALLY important numbers), its elegance, its compactness. On these scores, e^(i pi)+1=0 is unbeatable. Steve Gray ----- Original Message ----- From: "James Propp" <propp@math.wisc.edu> To: <math-fun@mailman.xmission.com> Sent: Friday, April 28, 2006 2:51 PM Subject: [math-fun] Re: favorite theorem
Let x=0, y=Pi. Then e^(x+iy)=e^0(cos(Pi) + i sin(Pi)) = 1(-1+0) = -1. ----- Original Message ----- From: "Emeric Deutsch" <deutsch@duke.poly.edu> To: "Steve Gray" <stevebg@adelphia.net> Cc: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, April 28, 2006 3:28 PM Subject: Re: [math-fun] Re: favorite theorem
On Fri, 28 Apr 2006, Steve Gray wrote:
... On these scores, e^(i pi)+1=0 is unbeatable.
I am sure that I am missing here something. What is the definition of e^(x+iy) ? Isn't it e^(x+iy)=e^x (cos y + i sin y) ?
Emeric
On Fri, 28 Apr 2006, Steve Gray wrote:
Let x=0, y=Pi. Then e^(x+iy)=e^0(cos(Pi) + i sin(Pi)) = 1(-1+0) = -1.
This is my point. If e^(x+iy)=e^x (cos y + i sin y) is the definition WE have chosen, then e^(i pi)=-1 is a simple consequence thereof. However, if we say that e^(x+iy)=e^x (cos y + i sin y) is the ONLY definition such that the "new" function e^(x+iy) satisfies this and this, then indeed, e^(i pi) =-1 is a remarkable fact. Emeric
----- Original Message ----- From: "Emeric Deutsch" <deutsch@duke.poly.edu> To: "Steve Gray" <stevebg@adelphia.net> Cc: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, April 28, 2006 3:28 PM Subject: Re: [math-fun] Re: favorite theorem
On Fri, 28 Apr 2006, Steve Gray wrote:
... On these scores, e^(i pi)+1=0 is unbeatable.
I am sure that I am missing here something. What is the definition of e^(x+iy) ? Isn't it e^(x+iy)=e^x (cos y + i sin y) ?
Emeric
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The wonder of "e^(i pi) = -1" is from discovery, not from definition. - Scott
On Fri, 28 Apr 2006, Steve Gray wrote:
Let x=0, y=Pi. Then e^(x+iy)=e^0(cos(Pi) + i sin(Pi)) = 1(-1+0) = -1.
This is my point. If e^(x+iy)=e^x (cos y + i sin y) is the definition WE have chosen, then e^(i pi)=-1 is a simple consequence thereof.
However, if we say that e^(x+iy)=e^x (cos y + i sin y) is the ONLY definition such that the "new" function e^(x+iy) satisfies this and this, then indeed, e^(i pi) =-1 is a remarkable fact.
Emeric
----- Original Message ----- From: "Emeric Deutsch" <deutsch@duke.poly.edu> To: "Steve Gray" <stevebg@adelphia.net> Cc: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, April 28, 2006 3:28 PM Subject: Re: [math-fun] Re: favorite theorem
On Fri, 28 Apr 2006, Steve Gray wrote:
... On these scores, e^(i pi)+1=0 is unbeatable.
I am sure that I am missing here something. What is the definition of e^(x+iy) ? Isn't it e^(x+iy)=e^x (cos y + i sin y) ?
Emeric
The identity is compelled by the several infinite series (derived independently), the complex plane, derivatives, etc. It couldn't be otherwise. ----- Original Message ----- From: "Emeric Deutsch" <deutsch@duke.poly.edu> To: "Steve Gray" <stevebg@adelphia.net> Cc: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, April 28, 2006 4:14 PM Subject: Re: [math-fun] Re: favorite theorem
On Fri, 28 Apr 2006, Steve Gray wrote:
Let x=0, y=Pi. Then e^(x+iy)=e^0(cos(Pi) + i sin(Pi)) = 1(-1+0) = -1.
This is my point. If e^(x+iy)=e^x (cos y + i sin y) is the definition WE have chosen, then e^(i pi)=-1 is a simple consequence thereof.
However, if we say that e^(x+iy)=e^x (cos y + i sin y) is the ONLY definition such that the "new" function e^(x+iy) satisfies this and this, then indeed, e^(i pi) =-1 is a remarkable fact.
Emeric
On Friday 28 April 2006 23:28, Emeric Deutsch wrote:
On Fri, 28 Apr 2006, Steve Gray wrote:
... On these scores, e^(i pi)+1=0 is unbeatable.
I am sure that I am missing here something. What is the definition of e^(x+iy) ? Isn't it e^(x+iy)=e^x (cos y + i sin y) ?
Hardly the only definition. You could, e.g., work along these lines: - log x = integral from 1 to x of 1/t dt [x real, >0] - exp x = inverse function of log - find the Taylor series for exp - plug in complex numbers as well as real ones - behold, exp (i pi) + 1 = 0. Or these: - exp is defined to be anything (C -> C) satisfying exp 0 = 1 and exp' = exp - this turns out to be unique - and to satisfy exp (i pi) + 1 = 0. What's really startling isn't so much any single result as the fact that many different approaches all lead to the *same* results. It's not surprising that exp(x+y) = exp(x) exp(y) if you start by defining e and then define exp x as e^x; it's very surprising if you start from the Taylor series. It's not surprising that exp (i pi) = -1 if you start from the definition Emeric wrote above; it's very surprising if you start from the differential equation. It's not surprising that exp' = exp if you start from the Taylor series; it's very surprising if you start with e and define exp(x) = e^x. The real surprise is that all these things fit together so neatly. Which, more generally, is one reason why I find this sort of favourite-theorem question very difficult. Sure, Goedel's theorem is lovely, but an order of magnitude lovelier is the whole constellation of ideas that surrounds it: incompleteness, undecidability, equivalence of formal systems to computing systems, diagonal arguments, ... . Likewise for most of the plausible candidates for "best theorem"; they're gems, but they only really sparkle properly when their brilliance is reflected and re^n-reflected off all the other gems located nearby in theorem-space. So which one do you pick to receive the credit? -- g
I would normally define e^z = sum_{n=0}^{infinity} z^n/n!. Then e^(pi i) = -1 is a non-trivial result. Some nominees for my favorite theorem: * The order of the element divides the order of the group. * Unique factorization of integers. * The intermediate value theorem. As you can see, I tend to favor simple but fundamental theorems. e^(pi i) = -1 is neat, but doesn't lead to anything. Franklin T. Adams-Watters -----Original Message----- From: Emeric Deutsch deutsch@duke.poly.edu On Fri, 28 Apr 2006, Steve Gray wrote:
... On these scores, e^(i pi)+1=0 is unbeatable.
I am sure that I am missing here something. What is the definition of e^(x+iy) ? Isn't it e^(x+iy)=e^x (cos y + i sin y) ? Emeric _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com
Since no one has yet mentioned them, I am compelled to cite the (distributional) idea of the Dirac delta function - the connection between the discrete and the continuous. Scott Beaver James Propp wrote:
Let me say, only half in jest, that my favorite theorem in math is the classification theorem for finite sets, i.e., the fact that finite sets are classified up to isomorphism by the counting numbers (including zero)!
And I'll respond, half-seriously: How would you prove that?
I wouldn't call it a theorem in the theorem-proof sense. If a proof is defined (epistemo-sociologically) as a text that makes you believe and/or understand a proposition more than you did before you read the proof (leaving aside the question of what "believe" and "understand" mean), then I don't think this theorem *has* a proof! You can undoubtedly set up a formal system in which this claim is proved by induction, but nobody who didn't already accept the claim would be able to understand the formal system, let alone follow the proof.
Instead, I am using the word "theorem" to mean any mathematical fact that makes you say "Well, will you look at that!"
And this proposition is a real looker. We all encountered it when we were kids and noticed that a set doesn't change its cardinality when its elements are shifted around, and this agreeable invariance probably fuels many kids' enthusiasm for counting (even though their knowledge of this invariance is intuitive and most of them couldn't begin to articulate it as a general proposition).
My serious point is that some mathematical facts may simultaneously be so profound and yet so commonplace that we don't recognize them as "theorems" as such.
But I think I want to change my mind about my favorite mathematical fact. How about the fact that there are infinitely many counting numbers? The realization that you can keep on counting forever and never run out of numbers (even if you run out of names for them) was such a mind-blowing experience for my young self that I retained absolutely, positively NO memory of the experience. :-) But my mind remained permanently blown, so there was nothing for me to do with my life but become a mathematician...
Is it a "theorem"? Well, I doubt that there are any formal bases for mathematics in which the axiom of infinity can be proved. Either you see it, or you don't. And once you've seen it, you can try to doubt it, but it's really hard to do.
Jim
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A theorem can be beautiful in many different ways. For me, one of the most striking is when the proof becomes almost trivial provided you look at the problem in the right way. Example; Fermat's "Little Theorem" (high school version). If p is a prime and a is not divisible by p then a^(p-1)-1 is divisible by p. You try various a's and p's and sure enough-. Why should this be? But n years later in grad school you see it's a very special case of the simply proved but elegant Lagrange's Theorem* that the order of a subgroup divides the order of the group (key idea, coset!). I think one's fondness for a theorem depends on how and when one first encountered it, the so called "mathematical experience'. Among my other pets are Pappus's hexagon theorem and the already mentioned fact that a prime is the sum of two squares if and only if it is of the form 4n+1. DG * Does anyone know the reference to Lagrange? I thought groups weren't even invented until much later
Several people have mentioned prime = sum of 2 squares theorem as a favorite theorem. It is indeed a fun theorem, but I think that the elegant way to see this is in the context of Gaussian integers. From this perspective, it isn't the theorem itself that is that interesting, but the fact that it is a clue to the really cool concept of the Gaussian integers. At 01:13 AM 4/30/2006, David Gale wrote:
A theorem can be beautiful in many different ways. For me, one of the most striking is when the proof becomes almost trivial provided you look at the problem in the right way. Example; Fermat's "Little Theorem" (high school version). If p is a prime and a is not divisible by p then a^(p-1)-1 is divisible by p. You try various a's and p's and sure enough-. Why should this be? But n years later in grad school you see it's a very special case of the simply proved but elegant Lagrange's Theorem* that the order of a subgroup divides the order of the group (key idea, coset!).
I think one's fondness for a theorem depends on how and when one first encountered it, the so called "mathematical experience'. Among my other pets are Pappus's hexagon theorem and the already mentioned fact that a prime is the sum of two squares if and only if it is of the form 4n+1. DG
* Does anyone know the reference to Lagrange? I thought groups weren't even invented until much later
I think you're mistaken in dismissing this as a "theorem". Once when I TA'd in a course "math for elementary school teachers" taught by Leon Henkin, he went over a proof of this theorem---I thought it was pretty revealing, (although I have to say it wasn't very appropriate for the future elementary school teachers, who didn't understand the point.) But for me, after going through the proof and trying to explain it to my students, it made a lot of sense why there is something that needs proof and that can be proven, our early indoctrination to the contrary notwithstanding. In fact, there's a good proof reducing it to on more primitive intuition than counting. It's very close to the standard proof that finite-dimensional vector spaces over a given field are classified by their dimension---which few people think is obvious without proof! I.e., when you count a finite set in two different orders, why do you get the same number? You can transform one order of counting to the other step-by-step using a simple exchange. I kind of like the point of view that proofs as a dialectical construct, as in Lakatos' book "Proofs and their refutations", rather than part of a formal system (which in practice they definitely are not). Many things people accept for a time, or for ever, until they are challenged, when their beliefs can be shaken. Did you ever tabulate a big set of data by hand and find the column sums and the row sums did not reconcile? At times like that, my belief in the commutative law for addition can become shaken --- thinking back through why sums, or counts, should reconcile is a non-obvious task. Bill On Apr 28, 2006, at 5:51 PM, James Propp wrote:
Let me say, only half in jest, that my favorite theorem in math is the classification theorem for finite sets, i.e., the fact that finite sets are classified up to isomorphism by the counting numbers (including zero)!
And I'll respond, half-seriously: How would you prove that?
I wouldn't call it a theorem in the theorem-proof sense. If a proof is defined (epistemo-sociologically) as a text that makes you believe and/or understand a proposition more than you did before you read the proof (leaving aside the question of what "believe" and "understand" mean), then I don't think this theorem *has* a proof! You can undoubtedly set up a formal system in which this claim is proved by induction, but nobody who didn't already accept the claim would be able to understand the formal system, let alone follow the proof.
Instead, I am using the word "theorem" to mean any mathematical fact that makes you say "Well, will you look at that!"
And this proposition is a real looker. We all encountered it when we were kids and noticed that a set doesn't change its cardinality when its elements are shifted around, and this agreeable invariance probably fuels many kids' enthusiasm for counting (even though their knowledge of this invariance is intuitive and most of them couldn't begin to articulate it as a general proposition).
My serious point is that some mathematical facts may simultaneously be so profound and yet so commonplace that we don't recognize them as "theorems" as such.
But I think I want to change my mind about my favorite mathematical fact. How about the fact that there are infinitely many counting numbers? The realization that you can keep on counting forever and never run out of numbers (even if you run out of names for them) was such a mind-blowing experience for my young self that I retained absolutely, positively NO memory of the experience. :-) But my mind remained permanently blown, so there was nothing for me to do with my life but become a mathematician...
Is it a "theorem"? Well, I doubt that there are any formal bases for mathematics in which the axiom of infinity can be proved. Either you see it, or you don't. And once you've seen it, you can try to doubt it, but it's really hard to do.
Jim
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participants (10)
-
Bill Thurston -
David Gale -
Emeric Deutsch -
franktaw＠netscape.net -
Gareth McCaughan -
Henry Baker -
James Propp -
Scott Beaver -
Scott Huddleston -
Steve Gray