uncool

Does anyone know of the mathematical formulation of the postulates of special (and perhaps later, general) relativity? I want to see what I can work out on my own, just from what I've learned about relativity. Thanks!

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There is no "room infinity." That is the entire point of the "Hilbert Hotel" paradox. No last room, nothing. None at all.

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How do you know that 2 is an nth power? And how do you know that its powers generate half of N(q)?

And where do you use d?

And what is the problem with the final form that you have written?

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I would say that Numb3rs isn't that good because it tries to overcomplicate the math. An example - the episode in which the person is attacking the electrical stations - the guy keeps on talking about the intersection of each of the sets, where all he really needs to say is whatever each of the stations covers. It keeps on doing stuff like that which is really annoying to me...as a math person.

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I have to agree with the other guys; what you wrote is word salad. Can you please try to simplify what you are saying so that we can understand it?

 

What "inertia phenomena" are you talking about?

=Uncool-

Each integer - individually - has finitely many digits. The set of all integers, on the other hand, has infinitely many digits.

 

And what your proof showed was that for all placeholders, there is an integer that makes that placeholder nonzero. Or, to state it like above, that the set of all integers uses each digit placeholder.

=Uncool-

Once again, you are mixing up quantifiers. You have proven that for all digits, there is a natural number such that that digit is nonzero. However, you have not proven that there is a natural number such that each digit is nonzero. You can have an infinite set using a finite possible set of digits - take, for example, the set 0, S(0) = 1, S(S(0)), etc., where S denotes the successor function. This clearly is the infinite set that is the natural numbers, but only uses a total of 4 symbols - S, (, 0, ).

 

Remember, there is a huge difference between the integers being infinite, and an individual integer being infinite.

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Actually, there happens to be another Peano axiom that your thing happens to violate - S(a) =/= 0 for any a.

Reason: Consider the "infinite number" ...999, which is all 9s. The successor of this number is ...000, which is exactly 0.

Also, in your system, it can be proven that ordering does not work.

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The new number need only have as many digits as there are numbers in the list, in this case 4. Also, it would be 1100, not 1110.

Notice the ... - that means an infinite list. However, you are right in the 1100 thing.

 

Fine, then. They are all finite but have a number of digits that keeps getting bigger and bigger without end, and the number itself gets bigger and bigger without end.

Only as you change the number - any number by itself has a finite number of digits that stays constant.

 

 

Do you have a link for that?

http://en.wikipedia.org/wiki/Natural_number. Also, every mathematician uses it in that way - that is, that every natural number can be obtained by adding 1 to itself a finite number of times.

 

Eh? So then [math] \sum_{n=0}^{\infty}\frac{1}{2^n} = 2[/math] shows that 2 is irrational?

Where did I say that? All I'm saying is that there is no need to have infinite sums to find them.

 

Then the sum [math]\sum_{n=0}^{\infty}1[/math] is finite. Since that sum is usually considered to be infinite, then perhaps finite = infinite according to your proof? You say that "clearly" adding 1 to any finite number results in another finite number, but it is the "clear" things that are hardest to prove

How is that sum finite? My proof deals with numbers that are expressed as a finite list of digits in binary form. That number cannot be expressed in that form and therefore has nothing to do with the proof.

 

And usually it is the clearest things that can be most easily proven. That is why they are usually clear.

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Hm, but consider the binary numbers,

0001

0010

0011

0100

...

1100 is not on the list

 

Then you can take the diagonal from the top right to the bottom left, and invert each of the digits, to get a different number not on the list. Would this not show that the natural numbers (in binary) are just as uncountable?

No, it does not. The reason that the proof works is that it shows ANY enumeration does not work. You have just shown an enumeration. I, on the other hand, say that the enumeration

1

10

11

100

... is an enumeration of the binary integers. If you try the diagonal argument, the result you get (which should be something like ....11110) is not a natural binary number, as all binary natural numbers are defined to have a finite number of nonzero entries.

 

 

Is countable by definition different from uncountable?

Yes - in fact, the class of all countable sets is disjoint from the class of uncountable sets. S being countable means that there is a surjective function from the natural numbers to S - that is, there is an enumeration that covers all elements of S (but there might be repeats). Uncountable means that the set S does not have such a function.

 

 

 

But does that make it finite? There is no maximum number of digits a natural number can have.

There is no maximum number of digits that every natural number can have - but each number itself has to be finite.

 

 

 

I think it was clear what he meant, about its decimal expansion

I will agree here.

 

 

 

Depends on what you mean by countable. Finite sets are not isomorphic to N.

That is correct.

 

 

Yes, but what about the 3 at the start? In my example, the decimal expansion keeps going toward the left. Yes, that is much less elegant because it diverges instead of converging, but it has no end on that side.

Which means that it is not a natural number. Natural numbers, by definition, are finite.

 

 

 

Yes, what is troubling about this is that it is just as impossible to say what an infinitely long natural number "starts" with as it is to say what an irrational number "ends" with.

There is no such thing as an infinitely long natural number. All natural numbers are finite in length when expressed in binary.

 

It should be writable as an infinite sum though. Just like an irrational number is written as an infinite sum, but you can't actually write it as a decimal expansion either.

You can express an irrational number as an infinite sum - because that is the definition of an infinite sum. It's that the partial sum converges to the irrational number. However, there is no such definition that will work for numbers expanding on the left.

 

 

 

This is probably what is wrong with my argument. However, I cannot see how there can be an infinite number of numbers, each with a different number of digits, without the number of digits also being infinite. Which of the Peano axioms wouldn't hold with infinitely many digits?

First, every digit has an element of N such that that digit is nonzero. However, that does not mean that there is an element of N such that every digit (or even infinitely many digits) are nonzero. And the induction axiom is not true, if you allow "infinitely many" digits. For example, let phi(n) be the expression "n is finite in length." Clearly, phi(0) is true. Also, clearly phi(n) -> phi(n+1). Thereore, phi should be true for all n - but clearly, your thing is not finite in length.

 

Secondly, Cantors diagonalisation argument is very general, it doesn't require a certain method of enumeration, it only posits that one exists and goes on to show that given any enumeration an element can be produced that is in the reals (although the argument is easily generalised to lots of different uncountable sets) but will not be produced by the enumeration.

 

What if you do the diagonalization argument with the natural numbers?

You can't - as I showed above.

 

 

 

Countably infinite?

The natural numbers are infinite, and yet can be enumerated.

 

 

 

I think that all irrational numbers are also calculated by taking a limit to infinity. This does seem to be different in the naturals and reals though. Need to think a bit more on this one.

Not all irrational numbers have to be calculated in that way - algebraic numbers can usually be left alone. However, if you want to learn a lot about the algebraic numbers, then it is nice to be able to relate them to rationals by using limits of sequences.

 

 

 

I did a proof by negation. So I assumed that there was a maximum finite number of digits n for the naturals, then showed that there would be a number with n+1 digits. Then the assumption is false, meaning that there is not a maximum finite number of digits for [math] n \in \mathbb{N}[/math]

You are correct - there are an infinite number of digits. However, for any natural number, only finitely many of them can be nonzero. That is, what you proved is that for all digits, there exists a number n such that that digit is nonzero. However, that does not prove what you are trying to prove, which is that there exists a number n such that every digit (or even infinitely many of them) is nonzero. You have switched the "for all" quantifier and the existence quantifier.

 

Is there anything wrong with saying there are an infinite number of zeros to the left of any finite number?

No, there is not. However, there is no natural, or real, number with infinitely many nonzero entries to the left of the decimal point.

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The way I would always do it is to find the actual thing and then use that.

 

This can be seen as a "displaced" geometric series, and in some way that should be able to prove everything. That is:

x_n = x_(n-1)*10 - 3. We know that for x_1 = 7, x_1^2 = 49.

 

x_n^2 = x_(n-1)^2*100-60*x_(n-1)+9 = [4]_(n-1)4[8]_(n-1)9 - 60*[6]_(n-2)7 + 9, which should all work out.

=Uncool-

For a wheel going downhill the friction should be in the same direction as the motion, i.e your tutor is correct. If the wheel was going uphill then the direction of the frictional force would be opposite to the direction of motion.

 

Shouldn't it be preventing the motion, i.e. going against the motion?

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Well, my first objection so far is that you are assuming that a is a generator for some large q. How do you know that there is such a q? This cannot be proven using Dirischlet's theorem.

 

And what do you mean by "this is not true" as a response to number 3? Are you saying that I read it wrong, or that you wrote it wrong?

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Thanks for all that. It seems to me that an understanding of "the infinite" is fraught with all sorts of problems and difficulties (especially language). You first need to understand such concepts as "a real value", "a variable", and "regression" and so on. The concept of "taking a limit" was explained to me, I recall, in a similar way to the explanation I used above. That the hypotenuse's slope (sorry should have specified that), approaches the slope of the curve at some point. This point doesn't have to be "reached", the triangle just has to be small enough and it is easy to "extend" this idea of "small enough" to "small as possible", or "infinitisemally small", except you don't have to get that far, just accept that it is logically possible. There are semantic difficulties when using "ordinary" language that require a more formal grammar, which is all that number and set theory I need to get my head around. My earlier posts on all this have factual errors, I see.

 

In Hilbert's hotel, are the guests all natural numbers, or have I jumped to an incorrect conclusion? What if they are all infinite series with non-finite results (values)? Or infinities?

 

The guests are basically placeholders, while the rooms are the natural numbers. The guests represent the idea of a set isomorphism.

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Pioneer - do you have any mathematical or physical reason, any reason at all, to believe that? I'm sorry, but if you cannot back up what you're thinking, then no one is going to believe you.

 

However, there is reason to believe you are wrong - the only currently known reason (that I know) for things to get out of a black hole is Hawking radiation. Nothing can get out of a black hole - because that is the definition of a black hole. Light cannot escape, by definition. If light cannot escape - that is, if light does not have enough energy to escape - then nothing can have enough energy. Things can, however, fall in and eliminate mass - like photons/etc.

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Fred the hypotenuse is a length. The slope is a y-value divided by an x-value. These are different units, so it's impossible for the two to be the same. Are you saying that the slope will be approached?

There is no such thing as an infinitely small triangle. A triangle always has finite size, by the definition of a triangle. There really isn't an infinitesimal quantity - just a limit to 0. Nothing more, nothing less. All infinities are is really saying "If you get really close" or "if you get really, really large...", not a limit to an actual number.

and of all the topics which have been on, so many others are the ones I see contain wisdom and knowledge. This very topic renders me so emotional that I could run out of the database memory. I simply can not comment a single thing without embarking upon a sojourn of consensus ridicularum.

 

i even see people's replies which I always am interested in seeing what they have to say but this is just too much. enjoy the posts till it becomes degenerate and the inductions and deductions mentioned from the OP's post will become redundant.

 

 

 

Sorry, just have to, cos i started reading it all instead of browsing but from bottom up i think.

 

I mean, the point of infinity... one specific point? lol , yes lol. I mean if you are speaking from a metaphysical point , then this becomes interesting. If you mention it as a valid mathematical definition, then there is just no sense in it.

 

 

(not the part about funtion behaviour, but how you worded 'the' point AT infinity).

Actually, there is such a thing as a point "at infinity" in some senses of the word. Projective geometry is the same as affine geometry, except that all parallel lines meet at a point, known as the point at infinity. Now, you are right in that there are more than one - there is a point for every direction - but there still is one. There even is a line at infinity for the plane, and a plane at infinity for projective space. This geometry is very useful because it is so much more symmetric than affine that its conclusions are much nicer.

 

In other words infinity actually has an infinite number of values, but is also actually none of them (in actuality).

 

How is it none of them?

 

Isn't this a problem when canceling or reducing infinite terms (either side of an equation or ratio)? Unless mathematical or algebraic infinity behaves "normally"?

 

You usually don't cancel or reduce infinite terms directly. Usually, as said before, it's done by reducing terms before the sequence (if it is a sequence) approaches infinity.

For example: the limit as x goes to infinity of x + 1 divided by x is 1, even though both numerator and denominator go to infinity.

 

So how do you say something like "the infinite set (of values)"? How do you describe the "value" of infinity. Or isn't this possible (or logical)? Would it be better to say "infinity can be any indefinite value"? Or that "it is not denumerable"? I'm trying to avoid using terminology from set or number theory, and see if sticking to "ordinary" language runs into lots of problems, or what.

Could you expound a little? If a real number is fixed, how does it "assume" an infinite value? Although I studied calculus to 3rd year, the concept of approaching a limit wasn't covered in any great depth (perhaps because, once you understand how to use it, you don't need to keep thinking about what it really is, sort of).

 

There are a lot of values that infinity, when you try to use it as a quantity, can be. It can be the number of integers - which also happens to be the number of rational numbers, etc. It can be the number of real numbers. It can be just about anything. However, this infinity is different from the infinity used in calculus - this is a set-theory infinity.

A real number doesn't assume an infinite value. A real variable can be increased past any bound. If it does so, then it "approaches" infinity. However, this can only be done with a variable - not an actual real number.

 

The idea of a limit is the following: Let's consider the function f(x), and how it behaves as it increases. If L is a number such that for any epsilon there exists an M such that for any x > M, |f(x) - L| is less than epsilon, then L is considered to be the limit of f(x) as x goes to infinity. This differs from the usual limit because usually there is a range of values that is finite in length, as opposed to an entire ray.

=Uncool-

Quad Plane: a plane made of two triangles (wich is know as plane).

A plane is not made of two triangles; do you mean a surface which is made of two triangles? A plane is always flat.

 

You can move a point on a plane to any position and you ever get a plane, let say a planar surface. However on a quad plane you move a point and you get a 3D surface, so the task would be move that point on X or Y or Z to get back a planar surface.

Gah...all the misunderstandings in here.

Guys, unless you are trying to work on the extended real line, infinity isn't a number. Usually, most people work on the normal real line, which does not include an infinity - only a limit to it. Even with limits to it, there is a lot of stuff you can't do.

Really, infinity is a placeholder for saying "It becomes too large for you to measure. It keeps going up, and up, and up..."

=Uncool-

THE HYPOTHETICAL PROOF OF FERMAT'S LAST THEOREM,

based on the following two lemma-theorems

 

Lemma 1:

For any odd number [math]a[/math] there is an infinite set of such prime numbers [math]q>a[/math], that the numbers [math]a[/math] and [math]q-1 [=m][/math] are mutually prime.

This is easy to prove using Dirischlet's theorem.

Lemma 2:

All last digits in the numbers [math]a^t[/math] in the base [math]q[/math], where [math]t=1, 2, … q-1[/math], are different, i.e., compose the complete set of positive digits in the base [math]q[/math].

Are we assuming that a is a generator of q? Because in general this is not true.

[Now there is no time to recall the proofs of these lemmas - this can be put off to the future. But it is worthwhile to look comments of “shwedka” and “tolstopuz” on the page http://lib.mexmat.ru/forum/viewtopic.php?t=8525&start=150 (in Russian).]

 

Essence of the obtained contradiction of the FLT: in the base [math]q[/math] the number [math]a^n+b^n -c^n=1[/math], but not [math]0[/math].

 

Thus, let us assume that

(1°) [math]a^n+b^n = c^n[/math], where odd [math]n >2[/math] and [math]a[/math] or [math]c[/math] also odd.

 

Proof VTF

(2°) for the odd number [math]3na^n[/math] (or for [math]nc^n[/math]) let us take the prime number [math]q>3na^n>c^n[/math], which satisfies the condition of lemma 1.

Which is it? do you want q greater than both?

Let us compile linear diophantus equation with the mutually prime parameters [math]n[/math] and [math]m=q-1[/math]:

(3°) [math]n(x+tm)-m(y+tn)=1[/math] with the general solution [math](x+tm; y+tn)[/math].

Since m and n are relatively prime, sure.

(4°) Since according to lemma 2 (since one-digit number [math]a[/math] and number [math]m[/math] mutually prime), all one-digit ends in [math]t-1[/math] numbers [math]a^t[/math] - consequently, and the numbers [math]a^x+tm[/math] - are different, there are such values [math]x+tm=r[/math] and [math]x+tm=s[/math] that the one-digit ends in the numbers [math]a^r[/math] and [math]a^s[/math] are equal to [math]b[/math] and [math]c[/math].

What? Could you please try to write this part again? I can't understand what you are saying at all.

But according to Little Fermat's theorem, one-digit ends in the numbers [math]a^n[/math], [math]a^{rn}[/math] and [math]a^{sn}[/math] in the base [math]q[/math] are equal to [math]1[/math], and then the one-digit end of the number

Are you saying that because you think a^n = 1 mod q?

(5°) [math]a^n+a^{rn}-a^{sn}[/math], i.e., [math]a^n+b^n-c^n=1[/math], but not [math]a^n+b^n-c^n=0[/math].

 

FLT is proven.

Why do a^{rn} and a^{sn} substitute for b^n and c^n?

=Uncool-

As to the first question: The circumference would still be exactly y. If it is y, then it will always be y. The thing about pi is that pi itself cannot be exactly enumerated as a rational number. However, that does not mean that other things that might relate to pi cannot be - for example, pi*(1/pi) = 1 can be. The problem with your example is that you forgot about the diameter - the diameter is what cannot be measured in your example.

 

No, we cannot form pi from constructions. This deals a little in Galois theory, but the only "constructible numbers" are those whose "minimal polynomial" is of a degree that is a power of 2. pi is a transcendental number - that is, it is not a number that solves a rational polynomial a_0 + a_1x + . . . + a_n*x^n, and therefore pi cannot be constructed.

Just remember - a circumference may be exactly calculable in terms of decimals. Just not a circumference and a diameter at the same time.

=Uncool-

Hrm...just a question on the math of it.

It seems to me that if you have a uniform sphere, then gravitational attraction will always be towards the cener. Think about it this way: Say the radius of the sphere is R. Then if you integrate the forces over each part of the ball, try integrating it by spheres around the object (not the sphere). For each spherical surface around the object until the spheres hit the wall, there is no force. However, once you hit the surface, the force drops off in one direction - so the total force for those spheres is towards the center. Therefore, since the force in each sphere is towards the center, the total force must be towards the center.

 

Even for a hollow sphere, the total force shouldn't be 0...

=Uncool-

Umm...all I can say to what you have is...What?

What are you trying to say? The math in your paper seems to be completely unrelated to your final conclusion, and your conclusion itself makes no sense at all. I can't even read the end to your paper.

=Uncool-

Then a corresponding example would be:

A = 28, B = 29, n = 3.

A is equivalent to n mod 9, so A^3 will be the same mod 9, as will B^3. Then A+B will also be equivalent, and they will have a common factor of 3.

Also, with n = 5:

A = 127, B = 128. The common factor will be 3.

=Uncool-

You said you had a lemma where (A + B, (A^n + B^n)/(A + B)) = 1, where (a, b) means gcd(a, b). But if A = 1, B = 2, and n = 3, then you have a problem. I am sure I could find several more problems than that.

=Uncool-

I'm still trying to find what you mean by no common factors.

Does that mean that every pair of numbers is relatively prime? or that no number besides 1 divides all 6?

 

Also, you haven't looked at the error I found.

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The way I did it:

[hide]M must be 1, as it is an extra place out. Then M + D + E = D, so M + E is 0, so E is 9. Then, A must be 8, as that is the only one that will give the 1 for M. D + A + E + carry(A + N + V) = V + 20. A + N + V + 1 = E + 10*carry, so 9 + N + V = 9 + 10*carry, so N + V = 10, and the carry is 1. 9 + 8 + 5 + 1 = V + 20, so V = 3, and N = 7.

O = 0

M = 1

V = 3

D = 5

G = 6

N = 7

A = 8

E = 9

MOVED = 10395.

[/hide]

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