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Topic: Rational Result (Read 2341 times) 

Sir Col
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Rational Result
« on: Jun 22^{nd}, 2003, 1:26pm » 
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Prove that there explicity exists irrational numbers, a and b, such that a^{b} = r is rational. It's certainly not a new problem and if you've not see it before, please feel free to produce some proofs. As I don't want to cloud anyone's creativity, I won't post the few 'textbook' proofs that I've seen – as I have some issues with them. I'd love to see what people make of this problem...


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NickH
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Re: Rational Result
« Reply #1 on: Jun 22^{nd}, 2003, 3:04pm » 
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I've seen this before, so I won't spoil others' fun by giving perhaps the most elementary and elegant choices for a and b. Below is a solution, without proof, to a^{a} = r, where r is rational and a is not only irrational, but transcendental! ::r = 2; a is the positive real solution to a^{a} = 2. For proof: (i) Show that a cannot be rational; (ii) To show that a is transcendental, use the GelfondSchneider Theorem ::

« Last Edit: Jun 22^{nd}, 2003, 3:06pm by NickH » 
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BNC
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Re: Rational Result
« Reply #2 on: Jun 22^{nd}, 2003, 11:28pm » 
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How about a=e, b=ln(2), r=2? PS: I havn't seen this before, but I assume that an example is good enough as a "proof". I also didn't prove that my choise for a and b are irrational, but I think I did see a proof of that somewhere.

« Last Edit: Jun 22^{nd}, 2003, 11:31pm by BNC » 
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Sir Col
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Re: Rational Result
« Reply #3 on: Jun 23^{rd}, 2003, 10:33am » 
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An excellent choice, BNC, and an explicit example certainly is sufficient. In fact you could generalise your result, If r is a rational number, then ln(r) is irrational, e is irrational, and e^{ln(r)}=r is rational. It's only my opinion, but the issue that I have with this result is that the problem (the way that I have phrased it) requests two explicitly irrational numbers. There is always a minimum level that a proof need sensibly reduce to, but I believe that the statement, "ln(r) is irrational, e is irrational," is not sufficiently elementary for the proof to be valid. It's certainly not difficult to annex the proof with these lemmas... If r is rational, let us assume that ln(r)=a/b, then r=e^{a/b}, therefore r^{b}=e^{a}. This is a contradiction, as the LHS is rational and the RHS is irrational. Hence ln(r) must be irrational. All it would take now it to show that e is, at least, irrational. But can we find a simple elementary solution to the problem?


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Leo Broukhis
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Re: Rational Result
« Reply #4 on: Jun 23^{rd}, 2003, 2:55pm » 
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on Jun 23^{rd}, 2003, 10:33am, Sir Col wrote: All it would take now it to show that e is, at least, irrational. But can we find a simple elementary solution to the problem? 
 As e is sum{i:0..Inf} 1/i!, assuming e is some a/b and trying to find any factorisation of b will fail. Therefore e is irrational. Proving that it is transcendental is harder.


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Sir Col
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Re: Rational Result
« Reply #5 on: Jun 23^{rd}, 2003, 4:42pm » 
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I think that you would need to qualify two things: Show that, (i) e can be obtained from the sum to infinity, e=1+1/2!+1/3!+... (ii) e=a/b cannot be factored, as I do not think this is trivial. I'll share two other proofs: 1. Consider sqr(2)^{sqr(2)}. If it is rational, then we have proved the result, as sqr(2) is irrational (trivial proof) and we have an irrational number raised to an irrational power to give a rational result. If it is irrational, then (sqr(2)^{sqr(2)})^{sqr(2)}=sqr(2)^{2}=2, which is rational, we have, again, shown that an irrational number raised to an irrational power to give a rational result. The problem: without the aid of the GelfondScheider Theorem we are still none the wise as to whether or not sqr(2)^{sqr(2)} is rational or irrational. I believe this proof fails, because it does not explicity give two irrational numbers without utilising a complex tool from the higher mathematics. 2. Let x be an irrational number and y be any rational number. Consider log_{x}(y). Assume that log_{x}(y)=a/b, then y=x^{a/b}, therefore, y^{b}=x^{a}. We have a contradiction as LHS is rational and RHS is irrational, hence log_{x}(y) is irrational. Now we can see that x^{logx(y)}=y and we have completed the proof. Any comments?


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SWF
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Re: Rational Result
« Reply #6 on: Jun 23^{rd}, 2003, 10:58pm » 
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I was about to post exactly along the lines of proof 1, but I see you already posted the solution to your own riddle only one day after the question. What exactly, does "explicitly exists" mean in the original question? Is that the same as "Give two irrational numbers a and b such that a^{b} is rational, and prove that this is the case" ? In proof 2, x^{a} is not necessarily irrational.


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Sir Col
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Re: Rational Result
« Reply #7 on: Jun 24^{th}, 2003, 12:46am » 
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on Jun 23^{rd}, 2003, 10:58pm, SWF wrote:I was about to post exactly along the lines of proof 1, but I see you already posted the solution to your own riddle only one day after the question. 
 I'm sorry about that, SWF. I only posted it as I felt that it is not a valid proof... Quote:What exactly, does "explicitly exists" mean in the original question? 
 ...which was my objection to that particular proof. If the problem requested the existence of an irrational number raised to an irrational power to give a rational result then the nonconstructuve proof given would work. When I added the phrase "explicity exists", I meant to solve the problem with two values, a and b, that can each be shown to be irrational. This particular puzzle has been bothering me for some time now and I haven't yet found a satisfactory solution, which is why I posted it here. I realised after starting the thread that I was running the risk allowing someone to post a proof so that I would appear all smug by saying, "Ah, but..." Thatwas not my intention. I feel very humbled by the expertise of thinkers here, so I would appreciate any insights or thoughts on this problem. Quote:In proof 2, x^{a} is not necessarily irrational. 
 I agree, which is why I said, "Any comments?" (I should have used a wink smilie). It can be modified by changing the nature of x... Let x be a transcendental number and y be any rational number. Consider log_{x}(y). Assume that log_{x}(y)=a/b, then y=x^{a/b}, therefore, y^{b}=x^{a}. We have a contradiction as LHS is algebraic and RHS is transcendental, hence log_{x}(y) cannot be rational. Now we can see that x^{logx(y)}=y and we have completed the proof. The question is whether this proof is any more acceptable than the proof involving e? Can we just state that x is a nonspecific transcendental?

« Last Edit: Jun 24^{th}, 2003, 1:28am by Sir Col » 
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towr
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Re: Rational Result
« Reply #8 on: Jun 24^{th}, 2003, 2:30am » 
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I'm still unsure what the question really is, are we asked to proof forall r rat(r) => thereare a,b such that irat(a), irat(b), a^b=r, or are we to proof thereare a,b,r such that rat(r), irat(a), irat(b), a^b=r In other words, do we need one example, or do we need to show that for every r at least one combination of a and b exists.


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Sir Col
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Re: Rational Result
« Reply #9 on: Jun 24^{th}, 2003, 4:32am » 
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The latter; that there exists, explicitly, two irrational numbers, a and b, for which a^{b}=r is rational.


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SWF
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Re: Rational Result
« Reply #10 on: Jun 24^{th}, 2003, 5:17pm » 
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Is this considered explicit: Let b=a, and solve a^{a}=2 for a. If rational a=p/q which implies that p^{p}=q^{p}2^{q} The number of factors of 2 in p^{p} must be a multiple of p, so for some integer N>=0, q=N*p. Therefore, a=p/q=1/N, which means 1=N*2^{N}. There is obviously no integer N that satisfies this, and a is irrational.


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Sir Col
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Re: Rational Result
« Reply #11 on: Jun 25^{th}, 2003, 6:46am » 
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Now that is scary! I was discussing exactly this result with someone last night. I have two problems with it: (i) although not stated clearly in the original problem, you have not found two different irrational numbers, and, (ii) you have not explicity stated the value of a. In fact, is it possible to find an analytical solution to equations of that type? Perhaps we need to refine the problem (I know, I am becoming quite annoying, as I keep moving the goalposts). State the values of two different irrational numbers, a and b, such that a^{b} = r is rational.


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SWF
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Re: Rational Result
« Reply #12 on: Jun 25^{th}, 2003, 5:21pm » 
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Excuse me, NickH. Rereading this thread, I see you already mentioned what I suggested yesterday. Looking up that GS theorem, it looks like at least one of the two irrational numbers will be transcendental. That could make it difficult to find simple expressions for the irrationals that are also easy to prove. One pair that is simple, but not easy to prove irrational is: a=2^{sqrt(2)} and b=sqrt(2). However, that the expression for a is irrational does not seem surprising. It is easy get two different irrationals from the easy to prove a^{a}=2 solution. Just square it to get a^{2*a}=4. The same value of a is obtained but b is now different, equal to 2*a. An explicit, although infinite, expression for a is (with ^ denoting raising to a power): a=2^(2^(2^(2^( ... ^(2^(2^(2^(1/2)))) ... )))) That gives approximately 1.55961 for a and 3.11922 for b.


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ThudnBlunder
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Re: Rational Result
« Reply #13 on: Jun 25^{th}, 2003, 6:11pm » 
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Quote:In fact, is it possible to find an analytical solution to equations of that type? 
 Yes it is, in terms of Lambert's Wfunction.


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Sir Col
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Re: Rational Result
« Reply #14 on: Jun 26^{th}, 2003, 12:52am » 
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T&B, could you shed some light on how this is achieved? I've looked it up, but there isn't much on specifics. It seems to suggest that it is an invented system to express the solution to an otherwise insoluable (analytically speaking) equation; can we actually express the solution in terms of elementary/computable functions?


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TenaliRaman
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Re: Rational Result
« Reply #15 on: Jun 26^{th}, 2003, 5:16am » 
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i think Lambert W function can be thought of as elementary function (just like sin(x) or cos(x)) and it is certainly computable with the aid of computers (since it has a series expansion).


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ThudnBlunder
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Re: Rational Result
« Reply #16 on: Jun 26^{th}, 2003, 11:31am » 
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on Jun 26^{th}, 2003, 12:52am, Sir Col wrote:T&B, could you shed some light on how this is achieved? I've looked it up, but there isn't much on specifics. It seems to suggest that it is an invented system to express the solution to an otherwise insoluable (analytically speaking) equation; can we actually express the solution in terms of elementary/computable functions? 
 Sorry, Sir Col, I don't know the theory offhand. But here is an excellent free downloadable program for investigating this and other strangelybehaved functions. http://www.peda.com/grafeq/gallery/rogue/xx_exponential.html As to your second question, I once had an amusing disagreement on Usenet's sci.math with a couple of professors about this very point: QUESTIONER: I am asked to solve the equation x^3=3^x. I want to know how to find the root 2.4780526802883 without using numerical methods. Besides, I think there may be many unreal roots (infinitely many?) but I cannot find them yet. T&B: There is NO analytical method for finding the solution x = 2.4780526802883.... PERSON A: That's what I guessed after fiddling with the equation for a while. But (pardon my naiveté) how can you know for certain that there's no analytical method? I don't doubt you're right, but I'd like to know why. T&B: Because solving the equation is equivalent to solving a socalled transcendental equation ( e.g. x = lnx OR x = e^x OR x = a^x). It is not possible (in general), by the very definition of a transcendental number. And there's no reason why we should naively believe (as I used to) that it should be possible . PROFESSOR BOB: Bzzzt! Wrong. Thank you for playing. Look up Lambert's W function. It gives an analytical answer. Next time you might try investigating the problem before spouting misinformation. T&B: I am familiar with Lambert's W function, thank you very much. But I do not consider reams of tabulated results to be an analytic solution. Tomorrow I'm going to invent T&B's Qfunction and call it the analytic solution to y = tanQxtanhQx. You will then be able to quote it in your diatribes. (Bob? Bob? Isn't that what you call a guy with no arms or legs, in a swimming pool?) PROFESSOR DAVID: Hi T&B. I really don't understand what you're objecting to. While I certainly would "not consider reams of tabulated results to be an analytic solution" either, using the Lambert W relation, all real solutions of 3^x = x^3 are given neatly by the expression W(ln(3)/3)/(ln(3)/3) where both the 0 (i.e. principal) and the 1 branch are to be used. [In the case of the 1 branch, by an identity, the solution happens to simplify to 3.] I don't see any "reams of tabulated results" above and would say that it does indeed represent an analytic solution. T&B: Thanks, David. I don't mind being put right by a perfect gentleman like yourself. The original questioner asked if there was a nonnumerical method to solve 3^x = x^3. It's all very well saying that the solution is W(ln(3)/3)/(ln(3)/3) but how would Maple, say, evaluate this expression? By using numerical methods, right? So, the next time I have a transcendental equation and I need an analytical solution, can I just invent an XYZ function, and Bob's your uncle (so to speak)? PROFESSOR BOB: Your objections make no sense (no flame intended). By what you say there is no such thing as an analytical solution to ANY transcendental equation. Even solving (say) 1/5 = sin(x) for x as arcsin(1/5) is not an analytical expression by your objection because to evaluate arcsin(1/5) requires a numerical method. Evaluation of transcendental functions, whether closed form or not ALWAYS requires numerical methods, except in special cases e.g. arcsin(1/2). Why does one NEED to evaluate W(ln(3)/3)/(ln(3)/3) if you don't need to evaluate arcsin(1/5)? Once you need to evaluate EITHER, you then need numerical methods. What is an "analytical expression" depends on the breadth of knowledge of the perceiver. Would you consider a solution in terms of confluent hypergeometric functions to be closed form? What matters [to most people; not to mathematicians] is whether there is a solution in terms of *elementary* functions because these are the functions they *know*. One never sees erf(x), si(x), ei(x), psi(x), gamma(x), zeta(x), airy(x), bessel_0(x) etc. until one gets beyond 1st year calc. The continued fraction 1 + 1/(2+1/(3+1/(4+1/(5...) can be expressed in 'closed form' as the ratio of two Bessel functions. Most mathematicians would accept this as 'closed form'.

« Last Edit: Mar 26^{th}, 2011, 9:33am by ThudnBlunder » 
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TenaliRaman
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Re: Rational Result
« Reply #17 on: Jun 26^{th}, 2003, 12:02pm » 
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Well Said,Bravo And yes ,a guy with no arms or legs, in a swimming pool is called Bob and ofcourse there are millions and billions of various other interpretations


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Sir Col
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Re: Rational Result
« Reply #18 on: Jun 26^{th}, 2003, 12:07pm » 
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Thanks for your post, T&B; a most amusing runin, indeed. on Jun 26^{th}, 2003, 11:31am, THUDandBLUNDER wrote:What is an "analytical expression" depends on the breadth of knowledge of the perceiver. 
 I find this point fascinating and you're probably going to certify me as bonkers, but... the more mathematics I learn, the more I feel unhappy with 'solutions'. To cite a simple case: the positive solution of x^{2}=2 is x=sqr(2), but what is sqr(2)? In one context it is the hideous apparition that exists geometrically as the diagonal length of the unit square, but is entirely incommensurable. Whereas I appreciate the use of these numbers in intermediate steps in manipulating equations, I have objections to it as a solution in its own right. How is tan^{1}(3/4) the smallest angle in 345 triangle? Compact and closed, maybe, but surely it is no more the solution to x^{2}–2=0 is x^{2}=2? I have total empathy with the sort of person who says, "God made the natural numbers; all else is the work of man." Does anyone else have similar reservations, or have I totally lost the plot?


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Icarus
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Re: Rational Result
« Reply #19 on: Jun 26^{th}, 2003, 6:19pm » 
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The point of "solving an equation" is to put it in a form where you can obtain more information about the variable than was obvious from the original. A study of the properties of the W "function" allows you to get a handle on how many solutions there are to x^{3} = 3^{x}, and give you a convenient way to approximate the values. Whereas looking at the equation itself just gives you headaches, and any numerical method schemes based directly on the equation are unlikely to provide you all solutions with any confidence. To demand that x=sqr(2) is not really a solution to x^{2} = 2, but rather another way of saying the same thing (given x>=0 of course), is to some extent valid. But this sentiment ignores the fact for evaluation of x, x=sqr(2) is an easier starting off point than x^{2} = 2. Quote:"God made the natural numbers; all else is the work of man." 
 Alas  Kroeneker was wrong! The natural numbers are just as little or as much the invention of man as the rest of mathematics. They too are definable in terms of "simpler" concepts.


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wowbagger
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Re: Rational Result
« Reply #20 on: Jun 27^{th}, 2003, 2:20am » 
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I know Kronecker is supposed to have said something like: "Die natürlichen Zahlen sind vom lieben Gott geschaffen, alles andere in der Mathematik ist nur Menschenwerk." And Dedekind allegedly said: "Die natürlichen Zahlen sind freie Schöpfungen des menschlichen Geistes."


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Icarus
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Re: Rational Result
« Reply #21 on: Jun 27^{th}, 2003, 3:51pm » 
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Arrgghh! Don't tell me you're going to force me to dig out my GermanEnglish dictionary! Sorry about forgetting the c in Kroenecker (I'm used to seeing oe for "oumlaut", so I won't apologize for that). I suppose it's disrespectful to mispell his name. But then again, he is the reason there is no Nobel Prize for Mathematics, so don't expect me to be too sorry!


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Sir Col
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Re: Rational Result
« Reply #22 on: Jun 27^{th}, 2003, 4:13pm » 
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Actually, I believe that the story is a myth, as Alfred Nobel was never married. What did Dedekind mean by, "The natural numbers are the free creation of the human mind"? And, Wowbagger, what is the best way to translate 'Geistes'?


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wowbagger
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Re: Rational Result
« Reply #23 on: Jun 30^{th}, 2003, 6:18am » 
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on Jun 27^{th}, 2003, 3:51pm, Icarus wrote:Arrgghh! Don't tell me you're going to force me to dig out my GermanEnglish dictionary! 
 No, I won't force you. The less you understand me, the less you can correct me! Ok, it's not me who said these German words. Wait! I could switch to German when it comes to the dubious parts of future posts! Quote:Sorry about forgetting the c in Kroenecker (I'm used to seeing oe for "oumlaut", so I won't apologize for that). I suppose it's disrespectful to mispell his name. But then again, he is the reason there is no Nobel Prize for Mathematics, so don't expect me to be too sorry! 
 I have to admit I'm confused. The only way I've ever seen his name written in German is "Kronecker" (plain o, no umlaut). However, googling around for a few minutes produced these "alternatives": Kroeneker (seems to be widespread in English writing, also in Spanish and Portuguese), Kröneker (English, French), Krönecker (English), Kroenecker (English). BTW, it's "Kronecker" in Mathematica. on Jun 27^{th}, 2003, 4:13pm, Sir Col wrote:What did Dedekind mean by, "The natural numbers are the free creation of the human mind"? And, Wowbagger, what is the best way to translate 'Geistes'? 
 Well, I don't really know what he meant by it. Maybe he didn't like the reference to some transcendental power. I'd have translated "der menschliche Geist" as "the human mind" as well, I suppose. Or did your question aim at the genitive?

« Last Edit: Jun 30^{th}, 2003, 6:26am by wowbagger » 
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Sir Col
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Re: Rational Result
« Reply #24 on: Jun 30^{th}, 2003, 9:05am » 
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on Jun 30^{th}, 2003, 6:18am, wowbagger wrote:I'd have translated "der menschliche Geist" as "the human mind" as well, I suppose. Or did your question aim at the genitive? 
 I wasn't sure if 'Geistes' was best translated as mind or spirit.


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