KJW

But one-time pad keys are not public keys.

No, according to Birkhoff's theorem, the spacetime inside a non-rotating spherical shell of matter is flat.

What I do understand is that Cantor's theorem is about sets, their elements, and their subsets, as well as things that are represented by sets, their elements, and their subsets, and also things that represent sets, their elements, and their subsets. My vision is not limited to an 1891 paper. Infinite sequences of binary symbols can be placed into one-to-one correspondence with subsets of the natural numbers. As such, it doesn't really matter if one considers them as sequences or subsets. I actually avoided considering the sequences as real numbers due to a complication that I did not wish to discuss. Cantor was a mathematician, not a snake oil salesman. What Cantor demonstrated was a mathematical truth concerning infinite sets. He wasn't "selling" anything. It is disappointing that you didn't see any value in the mathematics I presented about non-convergent infinite series. You started a topic that was ultimately about the infinite, but you failed to accept that the infinite is more complicated than the common simplistic understanding of it. I do wish to conclude with the following: Consider the negative of the derivative: [math]-\dfrac{d}{dx} \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} = -\displaystyle \sum_{k=0}^{\infty} (-1)^{k} \dfrac{dx^{k}}{dx} = \displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k\,x^{k-1}[/math] [math]= -\dfrac{d}{dx} \dfrac{1}{1 + x} = \dfrac{1}{(1 + x)^2}[/math] For [math]x = 1[/math]: [math]\displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k = \eta(-1) = \dfrac{1}{4}[/math] [math]\zeta(z) = \dfrac{1}{1 - 2^{1-z}}\>\eta(z)[/math] [math]\zeta(-1) = -\dfrac{1}{3}\>\eta(-1) = -\dfrac{1}{12}[/math] Note that I have evaluated [math]\zeta(-1)[/math], not simply obtained its value from elsewhere. [math]\dfrac{d}{dx} \left(x\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k}\right) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} \dfrac{dx^{k+1}}{dx} = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)x^{k}[/math] [math]= \dfrac{d}{dx} \dfrac{x}{1 + x} = \dfrac{1}{1 + x} - \dfrac{x}{(1 + x)^2} = \dfrac{1 + x}{(1 + x)^2} - \dfrac{x}{(1 + x)^2} = \dfrac{1}{(1 + x)^2}[/math] This is the same as above. However, by multiplying by [math]x[/math] before differentiation, the exponents of [math]x[/math] in the series remains unchanged, leading to coefficients that are powers of [math]k + 1[/math] (equivalent to powers of [math]k[/math]) rather than factorials when the operation is iterated. [math]\dfrac{d}{dx} \left(x\displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1) x^{k}\right) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^2 x^{k}[/math] [math]= \dfrac{d}{dx} \dfrac{x}{(1 + x)^2} = \dfrac{1}{(1 + x)^2} - \dfrac{2x}{(1 + x)^3} = \dfrac{1 + x}{(1 + x)^3} - \dfrac{2x}{(1 + x)^3} = \dfrac{1 - x}{(1 + x)^3}[/math] [math]\dfrac{1}{1 - 2^{3}} \dfrac{1 - x}{(1 + x)^3} = \zeta(-2) = 0[/math] for [math]x = 1[/math] [math]\dfrac{d}{dx} \left(x\displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^2 x^{k}\right) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^3 x^{k}[/math] [math]= \dfrac{d}{dx} \dfrac{x - x^2}{(1 + x)^3} = \dfrac{1 - 2x}{(1 + x)^3} - \dfrac{3x - 3x^2}{(1 + x)^4} = \dfrac{1 - x - 2x^2}{(1 + x)^4} - \dfrac{3x - 3x^2}{(1 + x)^4} = \dfrac{1 - 4x + x^2}{(1 + x)^4}[/math] [math]\dfrac{1}{1 - 2^{4}} \dfrac{1 - 4x + x^2}{(1 + x)^4} = \zeta(-3) = \dfrac{1}{120}[/math] for [math]x = 1[/math] [math]\dfrac{1}{1 - 2^{n+1}} \overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n}\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \dfrac{1}{1 - 2^{n+1}} \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^n x^{k}[/math] [math]= \dfrac{1}{1 - 2^{n+1}} \overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n} \>\>\dfrac{1}{1 + x} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \zeta(-n)[/math] for [math]x = 1[/math]

deleted

[math]\overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n}\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^n x^{k} = \eta(-n)[/math] for [math]x = 1[/math]

[math]\zeta(z) = \displaystyle \sum_{k=1}^{\infty} k^{-z}[/math] [math]\eta(z) = \displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k^{-z} = \displaystyle \sum_{k=1}^{\infty} (2k-1)^{-z} - \displaystyle \sum_{k=1}^{\infty} (2k)^{-z} = \displaystyle \sum_{k=1}^{\infty} k^{-z} - 2 \displaystyle \sum_{k=1}^{\infty} (2k)^{-z}[/math] [math]= \displaystyle \sum_{k=1}^{\infty} k^{-z} - 2^{1-z} \displaystyle \sum_{k=1}^{\infty} k^{-z} = (1 - 2^{1-z}) \displaystyle \sum_{k=1}^{\infty} k^{-z}[/math] [math]= (1 - 2^{1-z})\>\zeta(z)[/math] For [math]z = -1[/math], [math]\eta(-1) = \displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k = 1 - 2 + 3 - 4 + 5 - 6\>+\>...\>= -3\>\zeta(-1) = \dfrac{1}{4}[/math] Although the series [math]1 + 1 + 1 + 1 + 1 + 1\>+\>...\>[/math] can't be determined as the geometric series, it can be determined from the Riemann zeta function: [math]\zeta(0) = \displaystyle \sum_{k=1}^{\infty} k^0 = 1 + 1 + 1 + 1 + 1 + 1\>+\>...\>= -\dfrac{1}{2}[/math] Note that this could also have been obtained from: [math]\zeta(z) = \dfrac{1}{1 - 2^{1-z}}\>\eta(z)[/math] For [math]z = 0[/math], [math]\zeta(0) = -\eta(0) = -\dfrac{1}{2}[/math], where [math]\eta(0) = 1 - 1 + 1 - 1 + 1 - 1\>+\>...\>= \dfrac{1}{2}[/math] had already been obtained in a previous post.

Given that the above infinite series are alternating, one might assume that their values are some sort of average of the terms. But the non-alternating series: [math]\displaystyle \sum_{k=0}^{\infty} x^k[/math] where [math]x > 1[/math] dispels that notion: [math]\displaystyle \sum_{k=0}^{\infty} x^k = \displaystyle \sum_{k=0}^{m-1} x^k + \displaystyle \sum_{k=m}^{\infty} x^k =\dfrac{x^m - 1}{x - 1} + x^m \displaystyle \sum_{k=0}^{\infty} x^k[/math] [math]\displaystyle \sum_{k=0}^{\infty} x^k - \dfrac{x^m - 1}{x - 1} = x^m \displaystyle \sum_{k=0}^{\infty} x^k[/math] Note that removing initial terms from the infinite series yields the same infinite series multiplied by some value, and that removing more initial terms yields an even larger multiple of the infinite series. This behaviour of the infinite series is clearly different to the behaviour of any corresponding finite partial series. Also note that if [math]x = 10[/math], then the removed partial sum has a decimal value of the form [math]11111 \cdots 11111[/math] which is [math]\dfrac{1}{9}[/math] of [math]99999 \cdots 99999[/math], hence the general form for base-[math]x[/math] numbers. Continuing: [math](x^m - 1) \displaystyle \sum_{k=0}^{\infty} x^k = -\dfrac{x^m - 1}{x - 1}[/math] [math]\displaystyle \sum_{k=0}^{\infty} x^k = \dfrac{-1}{x - 1} < 0[/math] for [math]x > 1[/math] All the partial sums of the infinite series are positive-valued, yet the infinite series itself is negative-valued. In the case of [math]x = 2[/math], the value of the infinite series is [math]-1[/math]. But treated as an infinite sum, the value of the infinite series can be expressed in binary as: [math]\cdots 1111111111[/math] This is the same value that was excluded by Cantor's diagonal argument in an earlier post for the mapping of the set of all finite subsets of the natural numbers to the natural numbers. In other words, the sum of the infinite series goes beyond [math]\aleph_0[/math]. However, the binary expression for the infinite sum also corresponds to the two's complement representation of [math]-1[/math] used in computing, albeit extended leftward to infinite bits.

[math]\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk} = \dfrac{1}{1 + x^{n}} \\ y = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk}[/math]

But the correct method leads to the correct result. The correct method is the method I have shown above. In the case of [math]1 - 1 + 1 - 1 + 1 - 1\>+\>...\>[/math], the correct method is the method you labelled as "self referential", leading to a value of [math]\dfrac{1}{2}[/math] (in the above formula, [math]x = 1[/math], [math]n =[/math] whatever). It's important to note that no attempt was made to sum the series. For [math]|x| \ge 1[/math], the series does not converge. Instead, the series was assumed to have a value, which was then determined algebraically as shown above. The correctness of the value is the result of being the only value that is consistent with the definition of the series. Whether or not a non-convergent series can be regarded as having a value is a matter of debate, but if it does have a value, then that value must be the value obtained by the method I have shown above. In the case of [math]1 + 1 + 1 + 1 + 1 + 1\>+\>...\>[/math], the above formula with [math]x = -1[/math], [math]n = 1[/math] does not produce a finite result. That is, it is not just the series that fails to converge, but the corresponding function is also unbounded.

The above proof is incomplete: Proof: Let [math]y = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk}[/math] [math]y = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} + \displaystyle \sum_{k=m}^{\infty} (-1)^{k} x^{nk}[/math] [math]y = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} + (-1)^{m} x^{nm} \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk}[/math] [math]y = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} + (-1)^{m} x^{nm}\>y[/math] [math]y\>(1 - (-1)^{m} x^{nm}) = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math] [math]y \left(1 + \displaystyle \sum_{k=1}^{m-1} (-1)^{k} x^{nk} - \displaystyle \sum_{k=1}^{m-1} (-1)^{k} x^{nk} - (-1)^{m} x^{nm}\right) = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math] [math]y\left(\displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} - \displaystyle \sum_{k=1}^{m} (-1)^{k} x^{nk}\right) = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math] [math]y\left(\displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} + x^{n} \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}\right) = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math] [math]y\>(1 + x^{n}) \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math] [math]y = \dfrac{\displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}}{(1 + x^{n}) \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}}[/math] [math]y = \dfrac{1}{1 + x^{n}}[/math] Therefore: [math]\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk} = \dfrac{1}{1 + x^{n}}[/math] Q.E.D. Two things to note about this proof is that it generalises the previous result to other powers of [math]x[/math], and that it splits the series at an arbitrary location, thus establishing the consistency of the result with respect to the location of the split (the result is independent of [math]m[/math]).

Proof: Let [math]X = 1 − 4 + 16 − 64 + 256 − 1024\>+\>...[/math] [math]X - 1 = −4 + 16 − 64 + 256 − 1024 + 4096\>-\>...[/math] [math]= −4\>(1 − 4 + 16 − 64 + 256 − 1024\>+\>...)[/math] [math]= −4\,X[/math] [math]5\,X - 1 = 0[/math] [math]X = \dfrac{1}{5}[/math] Q.E.D. In general: Proof: Let [math]y = 1 − x^2 + x^4 − x^6 + x^8 − x^{10}\>+\>...[/math] [math]y - 1 = −x^2 + x^4 − x^6 + x^8 − x^{10} + x^{12}\>-\>...[/math] [math]= −x^2\>(1 − x^2 + x^4 − x^6 + x^8 − x^{10}\>+\>...)[/math] [math]= −x^2\>y[/math] [math](1 + x^2)\>y - 1 = 0[/math] [math]y = \dfrac{1}{1 + x^2}[/math] Q.E.D.

I see it as a contradiction of terms. You can't approach something that is out of reach. Here is an analogy. Did you continue to read what I said after "The term "approaching infinity" is often used in connection to limits."? You seem to have put in a lot of effort to argue against a phrase that is intended to be intuitive. Why can't you approach something that is out of reach? The notion of limits is about approaching something but not actually reaching it.

I think ii and vii are eliminations, the clue being "hot concentrated ethanoic potassium hydroxide solution".

Do you know where the value [math]-\dfrac{1}{12}[/math] comes from? It is the value of [math]\zeta(-1)[/math], where [math]\zeta(z)[/math] is the Riemann zeta function over the complex number field. The Riemann zeta function is defined as: [math]\zeta(z) \buildrel\rm def\over= \dfrac{1}{1^z} + \dfrac{1}{2^z} + \dfrac{1}{3^z} + \dfrac{1}{4^z} + \dfrac{1}{5^z} \> + \> ...[/math] Substituting [math]z = -1[/math] gives the series [math]1 + 2 + 3 + 4 + 5 \> + \> ... \>[/math], but that's not how [math]\zeta(-1) = -\dfrac{1}{12}[/math] is calculated. Calculating [math]\zeta(-1) = -\dfrac{1}{12}[/math] involves a process called "analytic continuation" where the domain of a function is extended as the domain of a continuous function beyond where it would otherwise fail to converge. The series converges for real values [math]> 1[/math], but there is a "pole" at [math]z = 1[/math], and analytic continuation to negative real values from real values [math]> 1[/math] involves going around the pole in the complex number domain. In the case of the Casimir effect, one uses the identity: [math]\zeta(-3) = 1 + 8 + 27 + 64 + 125 \> + \> ... \> = \dfrac{1}{120}[/math] A simple example of a function that is continuous over the entire real number domain, yet its infinite series only converges within a limited domain of the real numbers is: [math]\dfrac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + x^8 - x^{10} \> + \> ... [/math] Although the function is well-defined for [math]|x| \ge 1[/math], the infinite series fails to converge for [math]|x| \ge 1[/math]. But note that for [math]-1 < x < 1[/math], the infinite series converges to the same value as the function. Therefore, one could say that: [math]1 - 4 + 16 - 64 + 256 - 1024 \> + \> ... \> = \dfrac{1}{5}[/math]

[math]f(x)\; {\buildrel\rm def\over=} \;x+1[/math] [math]\buildrel\rm def\over=[/math] [math]\buildrel def \over=[/math]

Did you get your Physics PhD from a home schooling environment or something? There seems to be some rather glaring holes in your physics education... Yeah, I thought that. William.Walker39's argument seems too much like a n00b mistake for it to have come from a PhD physicist.

Again?? You previously asked me a similar question which I answered. The answer to this question is that [math]\aleph_0[/math] is not an element of [math]\mathbb {N}[/math]. [math]\mathbb {N}[/math] contains only finite numbers and [math]\aleph_0[/math] is not a finite number. We've already discussed this, so I suggest you go back to that discussion. The term "approaching infinity" is often used in connection to limits. Suppose a1, a2, …, an, … is a sequence of real numbers. When the limit of the sequence exists, the real number L is the limit of this sequence if and only if for every real number ε > 0, there exists a natural number N such that for all n > N, we have |an − L| < ε. The common notation: [math]\displaystyle \lim _{n \to \infty} a_{n}=L[/math] is read as: "The limit of an as n approaches infinity equals L" or "The limit as n approaches infinity of an equals L". But the formal definition itself doesn't mention "infinity". Indeed, it seems to me that the whole notion of limits is about dealing with the infinite and infinitesimal in a way that remains firmly in the realm of the finite.

[math]\displaystyle \lim _{n \to \infty} a_{n}=L[/math] [math]\lim _{n \to \infty} a_{n}=L[/math]

It concerns the motivation of the author. There have been real world cases of scientists who manipulate data to support their theories. In the justice system, motive is important. There is one important difference that you are overlooking: the mathematical theorem is completely transparent. There is no reliance on trust. I can see from the theorem itself that the motivation of the author is irrelevant, there is no data to manipulate, and this is not the justice system. I disagree. Cantor's theorem is about the existence of different transfinite numbers, based on the mappings of sets of numbers. Have you ever seen the following identity?: [math]1 + 2 + 3 + 4\> +\> ...\> = -\dfrac{1}{12}[/math] This is fairly well-known because it is recognised as one the most bizarre identities in mathematics. It emerges as a result of analytic continuation of the Riemann zeta function. But surely such apparent absurdity has no relevance to physical reality? Actually, a related equally bizarre identity manifests itself in the Casimir effect. I find this quite extraordinary, to be honest. What is its value? The value is [math]\aleph_0[/math]. Why do you think its value should be something else? Where have I said that it does? I think I have indicated that "infinite" is a property. But you also have to recognise that there are different properties that lead to different infinities. In particular, the property of being able to be placed into one-to-one correspondence with the natural numbers, and the property of not being able to be placed into one-to-one correspondence with the natural numbers are two different properties leading to two different infinities.

A fundamental problem with GR is that it is for the most part a local theory. That is, it deals with the various fields at individual points in spacetime. When one changes the coordinate system, the numerical values of the various components of the fields at the various points change according to how the coordinate system changed at those points. And the change in the components at one point is largely independent of the change in the components at another point due to the general nature of coordinate transformations. Thus, for many types of fields, it is impossible to create a total over a region of spacetime, quite simply because changing the coordinate system changes the total in a way that GR does not allow. And because there is no preferred coordinate system in GR, there is no way to say which value of the total over a region of spacetime is the correct value. So, it is impossible to have a total energy-momentum over the entire universe. Furthermore, regions in spacetime are four-dimensional, whereas we tend to think of three-dimensional regions and totals over three-dimensional regions, which also depend on the particular coordinate system.

I heard a joke a few days ago: Women regard men who know a second language as more attractive... unless it's Klingon.

[math] \documentclass{article} \begin{document} Evaluate the sum $\displaystyle\sum\limits_{i=0}^n i^3$. \end{document} [/math] [math]Evaluate the sum $\displaystyle\sum\limits_{i=0}^n i^3$.[/math]

That the union of two disjoint sets, both of which have a cardinality of [math]\aleph_0[/math], has a cardinality of [math]\aleph_0[/math] is logical because we are not dealing with finite sets. The logic of finite sets does not apply to infinite sets. Infinite sets have a logic of their own. If we have two disjoint sets A = {a1, a2, ...,an, ...} and B = {b1, b2, ...,bn, ...}, both of which can be placed into one-to-one correspondence with [math]\mathbb {N}[/math]: [math]1 \longleftrightarrow a_1[/math] [math]2 \longleftrightarrow a_2[/math] [math]...[/math] [math]n \longleftrightarrow a_n[/math] [math]...[/math] and: [math]1 \longleftrightarrow b_1[/math] [math]2 \longleftrightarrow b_2[/math] [math]...[/math] [math]n \longleftrightarrow b_n[/math] [math]...[/math] then we can interleave these two lists to form a list that is also a one-to-one correspondence with [math]\mathbb {N}[/math]: [math]1 \longleftrightarrow a_1[/math] [math]2 \longleftrightarrow b_1[/math] [math]3 \longleftrightarrow a_2[/math] [math]4 \longleftrightarrow b_2[/math] [math]...[/math] [math]2n-1 \longleftrightarrow a_n[/math] [math]2n \longleftrightarrow b_n[/math] [math]...[/math]

So, you have reduced the notion of a soul to mere physiology. The heart obviously exists, but what connection does it have to religious belief? You say the heart moves our body. Are you sure you're not talking about the brain, the nervous system, or muscles?

Earlier in this thread I said that I was "unwilling to debate over a translation of the precise wording of an 1891 paper". I should also have included discussions about the man. The fact of the matter is that the validity of a mathematical theorem is independent of who originated it or how it is expressed in particular writings. Thus, I accept the diagonal argument purely on the basis of its logic, and for you to convince me otherwise, your argument would also have to be purely on the basis of its logic. I don't know about that. It turns out that the mapping from the group of integers [math]\mathbb {Z}[/math] to the group of even integers [math]2\mathbb {Z}[/math] is an endomorphism. This means that properties of the odd numbers map to properties of double of the odd numbers under the endomorphism. And given that this mapping is one-to-one, it is unlikely that the integers [math]\mathbb {Z}[/math] are any richer than the even integers [math]2\mathbb {Z}[/math]. Leopold Kronecker was a subscriber to the philosophy of finitism and was thus in the minority among mathematicians. I am a subscriber to the philosophy of formalism, and regard mathematics as not being rigorous unless it can be derived from first principles (axioms) by mechanical symbol manipulation. The meanings of the symbols are abstract, not necessarily connected to reality, and manifest in the process of manipulation. I do not ask if notions represented by the symbols exist in reality. However, I ask you whether our description of reality is in any way adversely affected by the concept of the transfinite numbers, or the notion that an infinite set can be placed into one-to-one correspondence with proper subsets of that set. In other words, I challenge your notion of existence in mathematics. I don't think any of the reasoning I put forward in this thread even remotely suggests a thinking of a large number as closer to infinity than a small one. I would say the opposite. For example, in the one-to-one mapping of [math]n \longleftrightarrow 2^n[/math], for large [math]n[/math], [math]2^n[/math] is a lot larger than [math]n[/math], yet there was no suggestion that [math]2^n[/math] was any closer to the "end of the list" than [math]n[/math]. How is this E=2D? The mapping: [math]1 \longleftrightarrow 2[/math] [math]2 \longleftrightarrow 4[/math] [math]3 \longleftrightarrow 6[/math] [math]4 \longleftrightarrow 8[/math] [math]...[/math] [math]n \longleftrightarrow 2n[/math] [math]...[/math] is just as true a one-to-one correspondence as the mapping [math]2n-1 \longleftrightarrow 2n[/math]. The cardinality is [math]\aleph_0[/math]. Banach-Tarski paradox