matt grime

What counter argument? Cantor's theorem is precisely that there is not a 1-1 correspondence between the set of binary strings indexed by N and N.

Be careful abusing terms as in 'nxn where n is infinite'

Check, it please, it was just a guess on my part.

But if you've got something in the alg. closure of the small field satisfying an irred poly with coefficients somehow involving the a's, then you've gotten swapping roles over a poly that a satsifies over the small field, contradicting the fact that a is transcendental.

Just consider the minimal polys. write out the definitions, if something has a poly in F[a] with certain properties then it has one in F too since 'a' is transcendental.

and it's hard exactly because you need to memorize lots of definitions; but this makes the theorems harder to state and thus easier to prove.

Apart from relating y=x, what else can x be substituted?

Absolutely any other letter, or any other reasonable function of x (ie don't pick one that is not differentiable).

Now, what was the point of all this?

You don't give a formula in post three apart from one that is trivially always true

[math]\int_a^bf(x)dx=\int_a^bf(y)dy[/math]

that is gotten by subsituting y=x in to the integral.

Is it a Geometric rv? You are after all *composing* shuffles. A proof that it is indeed geometric would not go astray (nor should it be hard to provide, if it exists, I think; it is a group action here after all).

Look, people, decimal representations are just representations of Real numbers. Stop thinking of physical things, it is unnecessary and apparently misleading. Whether or not there are 'an infinite number of nines' from measurements is irrelevant, this is maths, it is an abstraction, not real life. You're deducing things about apples from looking at sheep, never mind oranges.

Why is 0.9..... equal to 1? Because these are decimal representations of elements of the field of real numbers, and the former is the limit of a cauchy sequence that is equivalent to 1 in the completion of the rationals.

For the layman who didn't understand that: if we were not to make these identifications then these would not represent real numbers: you would have two distinct elements satisfying |x-y|<1/n for all n in the natural numbers, and that is not possible because of the restrictions we place upon the real numbers.

Just to really hammer home the point. If these were base 11 representations of real numbers they would not be equal.

This is about the decimal representations of Real numbers, as I keep saying, so why is no one even bothering to think what decimal, representation, or Real means?

What are the limits they go from x=a to x=b, so if you set y=-x, then why are the limits y=a to y=b the same after this subsitution? You've just asserted that a=x=-y=-a and b=-b similarly.

Certainly the formula in post 3 is true, but it is substituting y=x.

Doing y=-x does not effect that change of integral, when f(x)=x^2, the substitution also changes the dx to a -dy, doesn't it?

Moral of story: don't mix and match, changing x to -y must be done at all points the x appears.

The fact that

[math]\int_a^bx^2dx=\int_a^b(-x)^2dx[/math]

does not happen because of a substitution, it is because x^2=(-x)^2

I'm sorry, but that makes no sense.

To start with 0.999999 is not 1. You're missing the recurring thing.

What has relativity got to do with structure of the real number system and decimals as representations as real numbers? Instead of trying to figure out relativity why don't you find out what the real numbers really are? It'd do you good, perhaps, to see the structure of them.

Well, let's see if it is reasonable, suppose that b_n = n, and a_n=an+c for some choice of a and c, then the limit of the difference converges to a, and a_n/b_n also converges to a, so it seems perfectly reasonable to me. Note neither a_n or b_n converge.

Now suppose that b_n is some sequence converging to 1, say, and a_n=kb_n for some k, then the difference quotient converges to k, as does the quotient, and a_n and b_n converge, so it seems really quite reasonable, doesn't it?

Can we delete that bloody powerpoint slide that is big, a waste of badnwidth and off topic?

I have a hunch that there's something up with the question. If [math]

\lim_{n \to \infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=G[/math]

 

then [math]

\lim_{n \to \infty}(\frac{a_{n+1}}{b_{n+1}-b_n}+\frac{-a_n}{b_{n+1}-b_n})=G[/math]

 

so [math]\lim_{n \to \infty}(\frac{a_{n+1}}{-b_n}+\frac{-a_n}{-b_n})=G\lim_{n\to\infty}b_{n+1}[/math]

 

and[math]\lim_{n \to \infty}\frac{-a_n}{-b_n}=\lim_{n \to \infty}\frac{a_n}{b_n}=(G\lim_{n\to\infty}b_{n+1})-\frac{a_{n+1}}{-b_n}[/math]

 

see what I mean?

no, not at all. the conclusion after the word 'so' is absolutely not valid;

sorry but i prefer to talk about the maths and not about the means to convey it via the internet.

 

i guess i should work it out by my own' date=' as allways.[/quote']

 

 

if you do not express yourself accurately and concisely then you cannot claim to be 'talking about the maths'. if you want help and want to ask in a written forum like this you have the responsibility to make sure that what you write is unambiguous and correct; it is not up to the audience to attempt to interpret if, by a_n+1, you mean

 

[math]a_{n+1}[/math]

or

 

[math]a_n +1[/math]

 

if you want to communicate clearly take the effort to either learn the latex or typeset clearly using braces. the latter of these two options costs you zero time to do and greatly eases the time of those who voluntarily choose to help you.

 

dont' try and pretend it is up to others to figure out what you mean.

what i have typed is:

lim a_n+1-a_n/b_n+1-b_n=G

yes that is what you typed now, not what you first wrote, and the tree's interpretation of either string of characters is strictly correct, it does equate to 1/1, because you have not braced off the subscripts. Sure, the correct meaning is clear if you stare at it ifor a while but you shouldn't have to do that.

As far as I'm aware the only decent WYSIWYG for maths is OO-Math which only seems to be supported by OO. So I figure the best thing would just have been to post it and use the Latex feature.

Well, modulo the fact that there is no such thing as a decent WYSIWYG editor for latex at all, what about winedit for the windows based, lyx, and the many others out there.

If you want to make latex documents use emacs or (g)vi(m). or just a plain text editor.

however, i don't see what this WYSIWYG debate or otherwise has to do with anything. I wasn't even talking about usability but about the quality and universality of the end result. word documents are overblown and render maths horribly. And the OP didn't even type anything that couldn't have been done in plain text here as can be seen by the fact that I literally cut and pasted the text into the thread window.

Oh look, a word document. You should avoid using word to do maths since it is rubbish at it (and most other things). I work in Linux, Unix, and OS X, I might be able to decode that propietary encoded document on one of them. Try pdf, or ps, or just type them out here; it isn't very hard to learn latex to the required standard for this forum.

Here, let me do it for you using cut and paste.

Decoded in oowriter:

I have four questions about limits, or rather trying to see if my approach/solution is correct.

1) the are given two series: {a_n}, {b_n} in both of them n=1 till infinity, when {b_n} increases b_n+1>b_n up to infinity, and lim (an+1-an)/(bn+1-bn)=G as n approaches infinity.

Prove that lim (an/bn) as n approaches infinity exist and that it equals G.

 

2) define: a1=1 a2=2, an+2=sqrt(anan+1), calculate lim an as n approaches inf.

 

3)let a1=0 and for n>=1; a_2n+1=0.5+a_2n; a_2n=(a_2n-1)/2, calculate lim sup a_n, and lim inf a_n.

 

4)calculate lim (1-1/(n^2-4))^(3n^2+5) as n approaches infinity.

 

My apparoches/solutions:

 

1)every lim will apparoach infinity:

lim (a_n+1-a_n)=G(lim(b_n+1-b_n))

lim a_n=lim a_n+1-G(lim(b_n+1-b_n))=a-G(lim(b_n+1-b_n)=g

lim b_n=G(lim(b_n+1)=f

if an->g and b_n->f and b_n doesn’t equal 0 then for all n natural (a_n/b_n)->(g/f) therefore there exist a limit.

 

Lim (a_n/b_n)= Lim |a_n/b_n|=|(a-Glim(b_n+1-b_n))/Glim(b_n+1)|=

=a/(Glim(b_n+1))-(lim(b_n+1-b_n)/lim(b_n+1)= a/(Glim(b_n+1))-1+lim(b_n/b_n+1)=lim(a/(G(b_n+1))-1 as you can see, not equal G, so where did I blow it?

 

2)here I know it equal the famous sqrt(2*sqrt(2*sqrt2….)

but how do I prove it.

 

3)limsup a_n=1; lim inf a_n=0, am I right?

4)I got e^3.

 

you didn't even appear to use any maths typesetting, so why the cumbersome file format?

i'll put my indented comments in another copy:

1) the are given two series: {a_n}, {b_n} in both of them n=1 till infinity

a series is a sum of terms, these are presumably sequences. that the index goes over n in N is automatic

when {b_n} increases b_n+1>b_n up to infinity, and lim (an+1-an)/(bn+1-bn)=G as n approaches infinity.

Prove that lim (an/bn) as n approaches infinity exist and that it equals G.

let me latex up the next one click to see the code

2) define:

[math]a_1=1 a_2=2, a_{n+2}=\sqrt{a_na_{n+1}}[/math]

at least I think that is how it ought to read. let me retype the next one as well so it is easier to read, let me know if i get it right.

, calculate lim an as n approaches inf.

 

3)let a_1=0 and for n>=1; a_{2n+1}=0.5+a_{2n}; a_{2n}=(a_{2n-1})/2, calculate lim sup a_n, and lim inf a_n.

 

4)calculate lim (1-1/(n^2-4))^(3n^2+5) as n approaches infinity.

 

My apparoches/solutions:

 

1)every lim will apparoach infinity:

lim (a_{n+1}-a_n)=G(lim(b_{n+1}-b_{n}))

no, you can't do that, those limits might not exist (ie they may be infinite ans so you can't treat them as numbers.

lim a_n=lim a_n+1-G(lim(b_n+1-b_n))=a-G(lim(b_n+1-b_n)=g

lim b_n=G(lim(b_n+1)=f

if an->g and b_n->f and b_n doesn’t equal 0 then for all n natural (a_n/b_n)->(g/f) therefore there exist a limit.

 

Lim (a_n/b_n)= Lim |a_n/b_n|=|(a-Glim(b_n+1-b_n))/Glim(b_n+1)|=

=a/(Glim(b_n+1))-(lim(b_n+1-b_n)/lim(b_n+1)= a/(Glim(b_n+1))-1+lim(b_n/b_n+1)=lim(a/(G(b_n+1))-1 as you can see, not equal G, so where did I blow it?

 

2)here I know it equal the famous sqrt(2*sqrt(2*sqrt2….)

but how do I prove it.

 

3)limsup a_n=1; lim inf a_n=0, am I right?

4)I got e^3.

the last one sounds plausible, though it ought to be -3 I think.

for 2 show it is bounded and increasing, or decreasing and hence result.

How can you argue against a valid proof. This is maths, not an episode of CSI. It is a consequence of definitions, like all maths. Whilst certain conjectures in maths do have evidential support (ie statements like: For all *something* we believe the following.... do lead to evidential support, ie many cases where it is verified) but this is not one of them, it is a straightfoward consequence of the definitions, you do not need to rely on 'evidence' at all. The 'multiplying 1/3 by 3 argument' is not evidence for this result since no one who cites it ever proves that decimal representations are a model for the real numbers, since such a proof would require you to declare 0.9... and 1 to be the same representations. It is certainly a persuasive argument for why we need to declare them to be the same (otherwise the system is inconsisitent: i can do the same thing two supposedly equivalent ways but get different answers), but the reason they are is because of the properties (that we declare them to have) of the real numbers. In short a proof exists, it is easy to state, possibly hard to understand. But lacj of rigour shouldn't be tolerated like this if it lets these myths continue.

Moral of the Story: numbers are not decimal expansions, decimal expansions are representations of numbers.

y is a dummy variable, so is x, that is why you can do substitutions.

It isn't necessarily a question of competency, more that this is such an infamous and pointless debating point on this and other math forums that it's best to nip it in the bud.

If you don't ask questions you won't learn. And sometimes it is better to be rash and find out quickly. Unfortunately you picked 'the bad question' to ask.

One reason why this question keeps coming up is that you aren't taught what the Real numbers really are until university, and only if you do maths (it is a delicate question), yet you have to use them at school.

Snail, JustStuit, stop it. There is no grey area, there are no competing thoeries. 0.99... and 1 are different representations of the same number, just like 1/2, 2/4, 3/6, and 1234/2468 are all different representations of the same number.

The Real Numbers, the place where this argument takes place, are a complete metric space, 0.99... is the limit of the sum

S(n)=9/10+9/100+...+9/10^n

and lim S(n) as n tends to infinity is 1. That is easy to prove since the distance between S(n) and 1 is 1/10^n and this converges to zero. Since limits are unique this shows that these two symbols represent the same number.

Well ordered means any non-empty subset has a least element doesn't it? The reals are totally ordered not well ordered. Or am I misremembering this one.

But, you're correct, any field automorphism of R preserves the ordering, for the reason you give. And yes, this tells you there is only one automorhpism. Sorry I was no help earlier.

The substitution y=-x gives the same answer. Just try it. It changes the integral to

[math]\int_{-a}^{-b}-y^2 dy[/math]

Why do you think you 'must keep the limits constant' (which doesn't really make sense).

Theorem: let F be any field, then any map F->is either zero or invertible. Proof: the kernel of a homomorphism is an *INSERT WORD HERE*, and fields only have TWO OF THESE THINGS.

Can't really give more hints than that without outright telling you the answer.

No way is this an epsilon type question.

If a_n a nd b_n converge then showing the limits are as required can be done by taking limits in the defining equations.

Secondly, just remember that any increasing sequence bounded above or decreasing sequence bounded below converges.

Eh? Point out where I said that infinity is best used as a verb? I corrected you once on that. I said not to use it at all, as is perfectly possible.

So you're not going to explain why it is 'the result of an interative process' then?

Your premise about what maths 'ought to be' is fine, for you I suppose. The rest of the world doesn't seem to agree, and I see no compelling arguments to accept your ideas.

So how about posting in speculations if you're not going to respect what this forum is (mathematics).

And what about other considerations? does 1/3 now have a decimal expansion? it can't in your system. limits do not exist, this means there are no numbers like pi, or e, or even the square root of two (all non-rational real numbers are limits of rational ones, that is how they are defined, so you can't have them).