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The illogical thing is that you can only get a bunch of 9's growing by arithemetic, you claim you can get them all. Moreover, when I question the ending 9 in that "all", you go back to the statement that you are still growing the 9's.
Induction only gives you finite numbers increasing, how can you get them all, Ricky?
Put it in other words, if you have got them all, you have already passed the induction phase.
To post 9
You have to first distinguish what your infinite means, Ricky.
We are in a circle now. You keep imagining induction guarantees you all the natural numbers, but it only gives you a growing yet FINITE amont of numbers at every given STEP. If you imagine you have got them all, you have to have gone through infinite steps, which means you have included infinity in your natuaral number set.
Step1 {1}
Step2 {1,2}
Step3 {1,2,3}
....
Tell me Ricky, how many steps you have gone through so far and how large your ending natuaral number is at your latest step?
I remembered a book called
1,2,3 and then infinity
Ha, such casual argument should belongs to poets only.
you just proved
Δ(Cosx)/Δx=-Sinx+o(Δx)
where o(Δx)/Δx->0 thus neglective
So you have proved
d(Cosx)/dx=-Sinx
According to Leibniz's definition
This can be called as
d(Cosx)=-Sinx*dx - differential form
I tried to apply for a trader's position in a European arbitrage hedge fund, but failed to meet their standard of arithmetric speed!
Just irrelevant. Where is the guy who posted this topic anyway?!!!
Howabout
Q-(Q-)
mrphysics, i used the similar method to locate the last 9 in 0.999...
you see, the new 0.9999... has exactly one more 9 in it so we wanna find where it is by compare and contrast
0.999...
0.9999...
and through a proof I posted in previous thread the one 9 previous 0.999... lacks has no suceeding 9 after it or it would be ...09
in other words, it has a last 9
Start from here,
and think it this way,
try to sum up 9's from the last 9 backwards, you will get 0, or infinitesimal, which doesn't matter. And no matter how many or much you have summed up, you will never be able to come up with an even small number larger than 0.
Is it a proof
0.9+0.09+0.009+...+last9=0?
This is my paradox
I know Ricky you would argue the cardinality (only a quible name for amount when it comes to infinity) of the digits of 0.999... is N0. But I know how this Hebrew alpha 0 is defined by Cantor, and I have a similar disproof for so called mapping integers to evens.
BTW, the "proof" of ∞=∞+1 is fake, or two proofs, one is mapping, the other is stating larger than finite.
for the former one
0.999...->0.0999... every 9 has already done one'o'one mapping
and +0.9 strictly adds in one more 9 to the latter
the only reason now to equate the 0.999... and the new 0.9999...
is by their similarity or unexaughstive counting, by simply looking at
0.9 0.99 0.999 ...
0.9 0.99 0.999 ...
and say since first second and third 9's are the same, and the examine process can go on beyond our human's ability to count all over
the 9's in the two are the same amount.
such finite "examination" is plausible because this method itself is defected since it is not done at all. (Have it examined "all" digits for either 0.999...?) You cannot force us to testify someone innocent by stressing s/he was not seen in the crime scene. And another piece of evidence might be sufficient to nail down the case.
The only exhaustive method is by wholistic mapping
when you do the division by 10 to 0.999...
every digit is already assumed moving rightwards by one digit
(or if you try to move the 9's one by one, you never can finish the division in your life
and you should not claim you can do the division in your life if you are rigorous),
I want to mention here not a single one more 9 is created or less 9 is diminished in this division
otherwise it is not defined as "every" or "all" (however it can be relaxed if Ricky you want)
then add in one more 9
yes just the one more 9 to compose 0.9+0.999.../10
and now we have already created ∞+1>∞
and then you can use my proof to just locate where the one more 9 is
yes the one more 9 is 0.9
but taking it at a different angle, suppose the 9's representing the same 9*10^-r play the same role and be treated the same in both 0.999...'s and from previous knowledge we know there is one 9's difference, one not and one has, the former can be filled with 0.
And this pair of 0 and 9 will not have a succeeding 9 'n' 9 pair to compose the paradox
...09
...99
Logically it follows this pair is on the right end. A 9 has no 9's on its right is clearly the right end of a bunch of 9's
Simple,huh? And I did not use countive mapping to try to conclude not enough evidence for the last 9. Additional evidence has already been provided to pin it down.
appendix
"I want to mention here not a single one more 9 is created or less 9 is diminished in this division
otherwise it is not defined as "every" or "all" (however it can be relaxed if Ricky you want)"
Now I relax it, but could anyone tell me which 9 can be diminished? The 9 at the right end?
Well Ricky, I don't expect everyone can accept my new idea. The guy who discovered square root of 2 was thrown to the sea by his teacher. Although I argue he was deceived by seeing is believing, the reason he was rejected was not he was wrong, but people are simply reluctant to accept new ideas.
I will go on with my inquiry. Today I discovered this counting method help solve another important problem. But I guess I would like to leave it to some journal to decide whether I am right or wrong.
For Dude, you have a good curiosity, this is nice. And the next post is for you. Hope you find it make sense.
1 No
2 No
The reason why? Right-end paradox I used in 0.999... post.
The reason why I ignored your post is simple, you continuously misunderstand potential infinity and realized infinity, which I have tried to distinct in the original post.
You think you have got infinite amount of digitals afterwards, but in fact you are getting only more and more yet finite amount of digitals. If you think you have got them whole, without bother to state the infiniteth digit and instead pretend you have only digits mapped to all natural numbers, you are ignoring the 0.999... post.
let me see
0.111... is on 2^No
and
1/10 or
1/1001=0.00011 0011 0011 recurring
is somewhere between
2^(No-1) and 2^No
hmm...i think that what zeno was referring to is that all change can be reduced to .0000...on to infinity...1. since the limit of this is zero, we can never move, hence the paradox. the problem with this is that all change cant in fact be reduced to .0000...1. which of course, leads us to the nature of the continuity of change. this leads us on to continuous functions on the complex plane, i think...or maybe im just a freshman who knows nothing.
Yes you get it.
1 It is not empirical, that means no evidence in our real world consolidates its existence
2 It is paradoxical. This denies its soundness in the world of concepts. If you accept two numbers, each is larger than another, I have nothing to say. But if not, please note the inconsistency of the concept of infinite decimals.
Well how about pi itself unresolvable logic inconsistency? Do you still accept it? Or at least will you state it is proven?
That's nice. I remember we have a similar proof in Calculus for multi-variable function's Taylor expansion. It adds in lamda as well. Such method is called "adding parameter", which shares the same delicacy as adding a line to solve geometry problems.
Fixed Cost:
(1000+200+2000)/25days = 128 per day
Variable Cost:
10+15*3+50=105 per hour
Variable Income: (under full operation)
1500 per hour
So even you operate one hour a day, you still have a lot to earn.
I think a more realistic assumption is the occupancy level, that is how empty your bus is can break you down.
There is no decision to be made when your bus is full, there is decision to be made when it is half empty, and you need to decide whether to run it or not. (When is pure empty you lose money)
Strange. (x,y)'s shall form a circle with radius r, origin at (0,0).
So you just charge $1500 per hour? That's your steady income?
I think your variable cost is linked either to operation hours or how many buses you have. Making this distinction would help clarify your case.
Are you arguing that such a ratio cannot exist?
Yes I do dude. Seeing is not believing. Please read my post in "This is cool" and you will find my argument that the existence of a circle in not testified or verified by science discoveries so far. And read my recent disproof that infinite decimals has its consistency problem.
Ricky, just imagine we have already count over 2[sup]N[/sup] of the binaries between 0 and 1, j in the similiar way that you imagined you count over all the infinite digits of 0.111... . My set is well ordered and also infinite.
In fact, it has 2[sup]No[/sup] of elements as Cohen proved in Cantor's Continuous Assumption(Theorem), where No means the infinite amount of all natural numbers (primary infinity). He did state all the binaries between 0 and 1 represents all real numbers between 0 and 1, and his idea is to switch on and off each digits to show there should be 2^the length of the digits (primary infinity) of combinations. Just suppose you have infinite bulbs in a line and how many combinations you would have.
What I did, is just hold all the digits after the first off at 0, and gradually each by each, allowing them to be turned on as 1, and by this way I am sure to order all the 2[sup]N[/sup] elements, just as Cantor did to order all the N*N elements of fractions (rationals).
Here I shall stress the symbol "N" or "No" here stands for Hebrew letter "alpha", it doesn't mean any natural number you think of, it represents the primary infinity you would tag at
1,2,3, ...->∞
or the amount of all natural numbers. (They call it cardinality, same thing)
Thank you Ricky. This is mainly because I haven't group all the ideas into a system. Now I just have inspirations on this on that and I am still reading new chapters. I happened to read Cohen's proof on Cantor's Continuous Assumption, which is the amount(cardinality) of all real numbers is 2 to the exponential of the amount of natural numbers. And he did used forcing, the binary digits. And from there I start speculation that although it is long, it is possible to order the 2[sup]N[/sup] of all real numbers between 0 and 1.
Thank you JaneFairFax for your clarification and wording. Although I don't agree with your opinion.
It's a proof that pi is not a rational number or a fraction of two integers. However, it is not a proof that pi exists. You have to add in existential assumption to believe it exists.
This is also a reply to luca.
That doesn't mean that they aren't there, though
How do you find they are there? Prove it.