river_rat Posted July 4, 2007 Share Posted July 4, 2007 Addition is a GROUP operation. On the reals yes but not on the naturals so your point is? Link to comment Share on other sites More sharing options...

the tree Posted July 6, 2007 Share Posted July 6, 2007 From the definition of a group, each member must have an inverse. If you'd read what I was responding to you might have made the connection. Link to comment Share on other sites More sharing options...

shadowacct Posted July 6, 2007 Share Posted July 6, 2007 You need to make sure you're calculating it properly. To calculate [math]y= x^{x^{x^x}}[/math], you calculate [math]x_1 = x^x[/math], then [math]x_2 = x^{x_1}[/math], then [math]y = x^{x_2}[/math] (i.e. you start from the top and work down). If evaluation is like Dave suggests, then the answer is the square root of 2, sqrt(2). The power-repeating function can be written as .....^(x^(x^(x^(x^x)))). Now start evaluating from the right and compute the series z0 = x = sqrt(2) z1 = sqrt(2) ^ z0 z2 = sqrt(2) ^ z1 z3 = sqrt(2) ^ z2 z4 = sqrt(2) ^ z3 z5 = .... etc. You'll see that this converges to 2. It is easy to prove this. I'll leave that as an exercise to the reader. So, the solution is x = 2^(1/2) = sqrt(2). For general N in the range from 1 to e (2.7181...), the solution of the equation ...^(x^(x^(x^(x^x))))...) = N can be written as x = N^(1/N). For N larger than e, I'm not sure whether this equation has a solution, I have strong doubts on that. I did some further analysis. I think that beyond e (2.71828...), there is a branch-point for x as function of N. The solution breaks down into a set of complex solutions. No real solution exists for N > e. For N < 1, the solution, posted in my previous post remains valid. The solution of that type is valid for the interval <0, e]. Link to comment Share on other sites More sharing options...

blue_cristal Posted July 7, 2007 Author Share Posted July 7, 2007 Interesting analysis, shadowacct ! Link to comment Share on other sites More sharing options...

river_rat Posted July 9, 2007 Share Posted July 9, 2007 From the definition of a group, each member must have an inverse. If you'd read what I was responding to you might have made the connection. Ok, reread what you replied to again and i still can't see your connection. Link to comment Share on other sites More sharing options...

Tom Mattson Posted July 9, 2007 Share Posted July 9, 2007 Addition is a GROUP operation. No, addition doesn't "belong" to any particular algebraic structure. Addition is a binary operation, and the definition of a binary operation is made prior to the definition of a group, monoid, or even semigroup. And what is the definition of a binary operation, you might ask? A binary operation b on a set S is a map b:SxS-->S. That's it! There is nothing more implied by this. Not associativity, not an identity element, and not inverses. Of course, mathematicians find it perverse to use the "+" symbol to indicate a noncommutative binary operation, so they almost universally never do it. But that's just an aesthetic choice. Link to comment Share on other sites More sharing options...

the tree Posted July 10, 2007 Share Posted July 10, 2007 Ok, reread what you replied to again and i still can't see your connection.Whether or not infinity has an inverse is relevant, in the context of group operations, because groups require inverses. Tom fair enough, in all the talk of careless statements I guess I made one myself. Although IIRC associativity, identity element et cetera et cetera are axioms of addition if not of binary operations generally, no? Link to comment Share on other sites More sharing options...

river_rat Posted July 10, 2007 Share Posted July 10, 2007 But you have not explained why we are limited to group operations. Lets change the story a bit, we needed a way of talking about [math]\infty[/math] and addition on the naturals for this whole setup to work for the problem at hand. Now addition is not a group operation on the set of natural numbers but we have a perfectly legitimate semigroup operation which extends addition to the set [math]\mathbb{N} \cup \{ \infty \}[/math]. Just treat the added point as a zero under addition (i.e. [math] x + \infty = \infty + x = \infty[/math] [math]\forall x \in \mathbb{N} \cup \{ \infty \}[/math]) Link to comment Share on other sites More sharing options...

Tom Mattson Posted July 10, 2007 Share Posted July 10, 2007 Although IIRC associativity, identity element et cetera et cetera are axioms of addition if not of binary operations generally, no? No, they aren't. As river_rat has already pointed out, addition is defined on the naturals, which don't include zero or negative numbers (so no additive identity or inverses). The definition of a binary operation b on a set S implies closure of S under b, but it does not imply not any of the group axioms for S. Link to comment Share on other sites More sharing options...

Country Boy Posted July 10, 2007 Share Posted July 10, 2007 Take [math]\mathbb{R}[/math] and adjoin two points to it, namely [math]+\infty[/math] and [math]-\infty[/math], denote this new set by [math]\left[-\infty, +\infty\right][/math]. Define addition as follows : [math] a + b = \begin{cases} a + b & \text{ if } a, b \in \mathbb{R} \\ +\infty & \text{ if } a = +\infty \text{ or } \left( a \in \mathbb{R} \text{ and } b = +\infty \right) \\ -\infty & \text{ if } a = -\infty\ \text{or } \left( a \in \mathbb{R} \text{ and } b = -\infty \right) \end{cases}[/math] This operation is closed and associative so we have a semigroup with both [math]+\infty[/math] and [math]-\infty[/math] acting as zeros on [math]\mathbb{R}[/math] under addition and is an extension of the normal addition on [math]\mathbb{R}[/math]. In fact both added points are left-zeros of the entire space The trick here is to not demand that addition be commutative for all possible values, only for real values (else define a local semigroup and work in the two point compactification of [math]\mathbb{R}[/math] if you want to keep the commutativity.) For our original problem though we only actually need to add one point to [math]\mathbb{R}[/math] so the issue is actually a moot point. You HAVEN'T defined [math]\infty+ (-\infty)[/infty][/math]so this is not a well defined operation. Link to comment Share on other sites More sharing options...

timo Posted July 10, 2007 Share Posted July 10, 2007 Sure that's defined by above rules: [math] \infty + \, -\infty = \infty, \ -\infty + \infty = -\infty[/math]. To what extent this definition makes sense is another issue and probably only decidable in a given context (e.g. you now will run into problems applying this to limits of series or functions). But to come back to my original statement being "don't write infinity minus one equals inifnity": I see this kind of proven as an analysis of the statement did indeed cause confusion as plainly demonstrated - not even the two definitions of the people who said "sure, no problem to include the infinities" do match. In fact, I've additionally asked a friend of mine who did his diploma thesis in a related field (set theory) and he spontaneously gave me a sketch of a 3rd way of including the infinities. EDIT:@Tom: [smart-ass]Afaik, closure is a group-axiom.[/smart-ass] Link to comment Share on other sites More sharing options...

Tom Mattson Posted July 11, 2007 Share Posted July 11, 2007 EDIT:@Tom: [smart-ass]Afaik, closure is a group-axiom.[/smart-ass] That's what I thought too, until last fall. I took a graduate course in algebra, and the professor defined a group as a set with binary operation that satisfies the axioms for associativity, identity element, and inverses. That day I was the smart-ass, and I "reminded" him of closure. He rolled his eyes and said that that comes with "binary operation". He said that without closure the operation is just a "partial operation". But whatever, you can certainly list closure a second time if you want. The wiki page on Groups does just that. Link to comment Share on other sites More sharing options...

timo Posted July 11, 2007 Share Posted July 11, 2007 Hm, might be formulation-dependent, i.e. depending on what you demand from the operation in the first place. However, my algebra book seems to agree with your professor, so listing closure explicitely as a group axiom might indeed be uncommon. Wikipedia imho doesn't count as a reliable source. Articles on relatively "everyday topics" like this one (or alternatively topics mentioned in popular science books" do, according to my experiences with the german Wikipedia, have the tendency to be written or at least heavily influenced by interested pupils or undergraduate students at best, which often lack the overview over the topic and only know the particular formulation that they learned in their classes. Link to comment Share on other sites More sharing options...

doG Posted July 11, 2007 Share Posted July 11, 2007 Wikipedia imho doesn't count as a reliable source. Is this better? Groups Link to comment Share on other sites More sharing options...

Xerxes Posted July 11, 2007 Share Posted July 11, 2007 Tom, I can't convince myself of the truth of this. So, let's call a set with a binary operation an algebraic set. Grant me that, if A is an algebraic set, then any subset B of A "inherits" the operation from A. Now consider the algebraic set Z; this is an abelian group under the operation +,by the closure axiom. The subset P of Z, the primes, also admits of the same operation +, but P is most definitely not closed under + and is therefore not a group. You surely must have closure as an additional axiom? I cannot see in what sense the + on Z is said to be a "complete" operation, whereas the + on P (it's subset) is partial. Did your prof give any more info? Link to comment Share on other sites More sharing options...

Tom Mattson Posted July 11, 2007 Share Posted July 11, 2007 You surely must have closure as an additional axiom? Not for groups, but for subgroups. If you take the definition of binary operation that I was taught, then later on down the line we have to have the following theorem. Let <G,*> be a group and let H be a nonempty subset G. <H,*> is a subgroup of <G,*> iff it is closed both under * and the taking of inverses. I cannot see in what sense the + on Z is said to be a "complete" operation, whereas the + on P (it's subset) is partial. Did your prof give any more info? No, but I think he pretty much said it all with what I quoted. + on Z is a map from ZxZ-->Z, and it's not a map from PxP-->P. There's the difference right there. Link to comment Share on other sites More sharing options...

Xerxes Posted July 11, 2007 Share Posted July 11, 2007 Not for groups, but for subgroups. Umm, well, I still don't see your point. Every group G has at least one subgroup, that is G itself, by the definition. How does this comment work here? + on Z is a map from ZxZ-->Z, and it's not a map from PxP-->P. There's the difference right there.But this is a property of P, not the operation +. Sorry to sound assertive, but; +: Z × Z → Z defines an abelian group (let's say) whereas +: P × P -/-> P, rather +: P × P → Z, hence P is not closed under +, and therefore P is not a group, therefore not a subgroup of Z. But for certain sure, P is a proper algebraic subset of the algebraic set Z. Are you suggesting there are two "varieties" of +? PS (by edit): In case there may be hard feeling here, I am merely suggesting that what you called your "smart-ass" question was well motivated, and, unless your prof could give a better explanation, (s)he was wrong to "roll their eyes", I think. Link to comment Share on other sites More sharing options...

Tom Mattson Posted July 11, 2007 Share Posted July 11, 2007 Umm, well, I still don't see your point. Every group G has at least one subgroup, that is G itself, by the definition. How does this comment work here? If H=G itself, then the requirement that H be closed under * is redundant. It's only not redundant if H is a proper subset of G. But this is a property of P, not the operation +. It's a property of + acting on P. P is just a set, and set theory knows nothing of +. Sorry to sound assertive, but; +: Z × Z → Z defines an abelian group (let's say) whereas +: P × P -/-> P, rather +: P × P → Z, hence P is not closed under +, and therefore P is not a group, therefore not a subgroup of Z. Right, because + is not a binary operation on P. So in the formalism I learned, we don't even get to the group axioms. <P,+> isn't eligible for consideration as a group. But for certain sure, P is a proper algebraic subset of the algebraic set Z. Are you suggesting there are two "varieties" of +? No. I'm suggesting that + is a binary operation on some sets, and it is not a binary operation on some other sets. Edited to add: How do you define binary operation, Xerxes? The definition I cited is the only one I've ever seen. Edited to add again: Xerxes: Are you suggesting there are two "varieties" of +? Tom: No. Actually, come to think of it, maybe I am! In the spring I took a second algebra course, and in that course the professor was very specific about what he meant by a function. To define a function, one must specify the domain, the codomain (a superset of the range), and a rule for assigning each element of the former set to a unique element of the latter set. He maintained the point of view that changing any of these three items changes the function. For instance... [math]f:\mathbb{R} \rightarrow [0,1][/math], given by [math]f(x)=\sin(x)[/math] is not the same function as [math]g:\mathbb{Q} \rightarrow [0,1][/math], given by [math]g(x)=\sin(x)[/math] So if we look at + as a function, then under this interpretation of functions [math]+_{\mathbb{Z}}:\mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Z}[/math] is not the same as [math]+_P:P\times P\rightarrow P[/math]. hmmmmm....(scratches head) Link to comment Share on other sites More sharing options...

Xerxes Posted July 11, 2007 Share Posted July 11, 2007 It's a property of + acting on P. P is just a set, and set theory knows nothing of +.Umm. So, if P is "just a set", and set theory knows nothing of +, how can you then assert that It's a property of + acting on P.? Recall I suggested the notion of an "algebraic set", i.e one with a binary operation.Right, because + is not a binary operation on P. Oh? 3 +5 is undefined? Surely not. <P,+> isn't eligible for consideration as a group.I completely agree, but not for the reasons you gave. Binary operations, one might say, are insensitive to the space they operate on. No. I'm suggesting that + is a binary operation on some sets, and it is not a binary operation on some other sets.True, startling as it may appear. How do you define binary operation, Xerxes? The definition I cited is the only one I've ever seen.As you did; +: X × X → Y, ×: Y × Y → Z, where possibly X = Y, Y = Z, I'm reasonably sure all other imaginable operations can be derived from these. Link to comment Share on other sites More sharing options...

Tom Mattson Posted July 11, 2007 Share Posted July 11, 2007 Umm. So, if P is "just a set", and set theory knows nothing of +, how can you then assert that [edit]"It's a property of + acting on P"[/edit]? Because operations can be defined on sets? I really don't see why my remarks are so bewildering. If you're only talking about some set P, then surely the only "properties" of P that exist without introducing an operation such as + are set theoretic properties such as [math]P\subset\mathbb{Z}[/math], no? But no matter, because as you point out... Recall I suggested the notion of an "algebraic set", i.e one with a binary operation. By the definition of a binary operation that I cited, P is not an algebraic set, because as I said, "Right, because + is not a binary operation on P, "to which you responded... Oh? 3 +5 is undefined? Surely not. Who said that? Not me. I completely agree, but not for the reasons you gave. Binary operations, one might say, are insensitive to the space they operate on. Yes and then again, one mightn't say it. The definition I cited is certainly sensitive to the set S. As you did; +: X × X → Y, ×: Y × Y → Z, where possibly X = Y, Y = Z, I'm reasonably sure all other imaginable operations can be derived from these. That is not how I defined a binary operation. I defined it as a map [math]b:S\times S \rightarrow S[/math]. My definition necessarily entails closure of S under b, while yours doesn't. Link to comment Share on other sites More sharing options...

Xerxes Posted July 12, 2007 Share Posted July 12, 2007 So if we look at + as a function, then under this interpretation of functions [math]+_{\mathbb{Z}}:\mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Z}[/math] is not the same as [math]+_P:P\times P\rightarrow P[/math]. I just saw this edit. I must be truly dim, but consider the following. You will agree that each p, q in P is integer, right? So let (p,q) be an element in P × P with [math]+_P:P\times P\rightarrow \mathbb{Z}[/math]. (Actually you wrote [math]+_P:P\times P\rightarrow P[/math] which is probably a typo, no?). Let [math]+_P(p,q) = x \in \mathbb{Z}[/math]. Now consider [math]+_{\mathbb{Z}}:\mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Z}[/math]. If p and q are integer, then (p, q) is in Z × Z with [math]+_{\mathbb{Z}}(p,q) = x [/math] again. Being bold I'll say that, if two operations give the same output for the same input, then they must coincide. Am I mad? Link to comment Share on other sites More sharing options...

river_rat Posted July 12, 2007 Share Posted July 12, 2007 To what extent this definition makes sense is another issue and probably only decidable in a given context (e.g. you now will run into problems applying this to limits of series or functions). Ah, but that is a different question. I only stated i could extend the algebraic operation of addition to a larger set that includes some ideal points. To start talking about limits you must have introduced a topology. To talk about limits and addition it would be nice if addition was continuous with respect to this topology. Now if you want a "nice" topological extension of the reals i would have to suggest the Stone-Cech compactification where addition can be extended by the universal property of that compactification. Sadly I think we only get an operator which is left continuous but we are still better equipped to talk about limits here. But to come back to my original statement being "don't write infinity minus one equals inifnity": I see this kind of proven as an analysis of the statement did indeed cause confusion as plainly demonstrated - not even the two definitions of the people who said "sure, no problem to include the infinities" do match. In fact, I've additionally asked a friend of mine who did his diploma thesis in a related field (set theory) and he spontaneously gave me a sketch of a 3rd way of including the infinities. Well share the sketch Link to comment Share on other sites More sharing options...

sciencenoob Posted October 6, 2007 Share Posted October 6, 2007 Starting again with the equation, x^x^x^…x = 2. If you isolate the first x then the remaining sequence of exponents remains infinite: (x) ^ ( x^x^x^…x) = 2 Therefore the second part, between parenteses, would still be equal 2 because it is still a infinite sequence ( ∞ - 1 = ∞ ). So by replacing the second part by 2 you get : (x) ^ 2 = 2 Therefore: x = sqrt(2). Interesting riddle, but unfortunately I think this proof has fallacies, simply because you happen to work with the magic number '2' that it seems to work, and hence a more generalized solution following the same processes as you undertook will not yield the same conclusion. For example: let (X)^X^X^X^X^X^X^X... = A ----- (1) if (X)^(X^X^X^X^X^X^X^X...) = A Therefore using Eq(1) (X)^(Eq(1))=A Then (X)^A=A Therefore X=A^(1/A) <= is your generalized form But If you tried this with say A=4, then the iteration breaks down, where Eq(1) does not make sense in that if A=4 then X=4^0.25, then Eq(1), A=/=X^X^X^X^X^X^X... Infact, if A=4, then you will get X^X^X... = 4, given that X=4^0.25 when you have only powered it 4 additonal times, and that any more powers on top of that will start going above 4. i.e.: given your solution that X=A^(1/A), and that A=4 Then: X=4^(1/4)=4^0.25 Then X^X^X^X^X^X^X... = 4 is true only when: (4^0.25)^(4^0.25)^(4^0.25)^(4^0.25)^(4^0.25)=4 <= EXACTLY such that: (4^0.25)^(4^0.25)^(4^0.25)^(4^0.25)^(4^0.25)^(4^0.25)=7.1 Hence I think theres something wrong with this 'proof' and that it seems to work only because its a 2? Please correct me if im wrong Link to comment Share on other sites More sharing options...

Fred56 Posted October 7, 2007 Share Posted October 7, 2007 What happens if x is infinite? Link to comment Share on other sites More sharing options...

sciencenoob Posted November 12, 2007 Share Posted November 12, 2007 bump.. look at my solution suggesting a problem in the original 'answer', that it only applies simply due to the magic of the number 2 in this situation instead of being a real answer. Link to comment Share on other sites More sharing options...

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