Is 0.999... Truly Equal to 1 in the Realm of Infinity?

[ram2048]

suppose that .999~ this number has (infinity(d)-1) of the digit nine by your definitions, then you can agree that .4999~ has (infinity(d)-2) of the digit nine right?

k then by your reasonning .999~ - .4999~ has to have 1 digit 9 right?

but in fact it has 0 digit nine since .999~ - .4999~ = .500~

no you had it right. look here. .999~ and .0999~ exhibit a has 1 more 9 than exhibit b.

the 9 in that place is destroyed by the 4 when you do the subtraction of .999~ - .499~

that's why there's no 9 left over. there is no contradiction, visually or otherwise.

 

[hello3719]

exactly, you were proving the problem visually, look again at your proof.

 

[ram2048]

not understanding you.

please point it out for me

 

[Hurkyl]

9.999~ consider this to have "default infinity" number of digits 9

Then what you mean by 9.999~ is not what mathematicians mean by 9.999~, unless you can somehow show that having a "default infinity" number of digits means exactly the same thing as saying that there is exactly one digit corresponding to each positive integer.

999~ this number has (infinity(d)-1) which is demonstrably true when you subtract .999~ from 9.999~ you are left with exactly 1 digit 9.

Where is that 1 digit 9?


And I forget if I asked this:
How does your system of decimals tell the difference between 9 + 0.999~ and 10 * 0.999~, seeing how they're both 9.999~?

both. it is both not logical and provably untrue as shown above.

It can't be both; nonsensical statements are just that, nonsensical, and cannot be logically manipulated. In particular, you can't prove them false. This is more evidence that you are not talking about logic.

because in base 10 calculations, multiplying by 10 creates a process which shifts ALL the digits. after defining a default infinity you can see this creates a distinguishable "gap" in comparison to the standard transient infinity.

Then, again, you aren't talking about the standard decimals. (Assuming "transient infinity" means the standard mathematical way of doing things)

This is called a "strawman argument". Your "decimals" are something different than the decimals of mathematics. While your statements may then be true about what you mean by "decimal", you are merely attacking a strawman; there are no grounds to suggest any of your statements also apply to the mathematical meaning of "decimal".

the 0 is at the "end" of an infinite number of 9's.

How can there be an end if every digit has another digit to its right?

And here's another stickler. If there's an "end" then what happens to the last 9 in 0.999~ when you divide by 10?

x = .666~
10x = 6.666~
10x - x = 6.666~ - .666~
9x=6
x=2/3

according to calculus. 10x - x = 6 BUT 9x = 5.999~4

This has nothing to do with calculus; it's arithmetic. But 9x is 5.999~ = 6, and 5.999~4 is something that has no meaning in standard mathematics.

no you had it right. look here. .999~ and .0999~ exhibit a has 1 more 9 than exhibit b.

If you consider .999~ to have "one more" 9 than .0999~, then why don't you consider 9.999~ to have "one more" 9 than .999~?

What about this proof?
x = .999~
x/10 = .0999~
x - x/10 = .999~ - .0999~ = .9
(9/10) x = .9
9x = 9
x = 1

 

[ram2048]

What about this proof?
x = .999~
x/10 = .0999~
x - x/10 = .999~ - .0999~ = .9
(9/10) x = .9
9x = 9
x = 1

assume .999~ in line 1 to be default infinity number of 9's
in line 2 assume x/10 would shift all decimals right one digit and create a 0, yet maintain the same "default infinity" number of digits 9
.999~ - .0999~ =.8999~1

9/10 = .9
9/10 x .999~ = .8999~1

 

[ram2048]

How does your system of decimals tell the difference between 9 + 0.999~ and 10 * 0.999~, seeing how they're both 9.999~?

they wouldn't be the same.

in case 1, 9.000000000~ even is added to .999~ . in case 2, 10 instances of .999~ are added.

if we were to set .999~(d) as default infinity number of 9's (from now on i'll use this notation) adding 9 would not cause the shift in digits, 10 x .999(d) would. such that 9+.999(d) - (10 x .999~(d)) = .000~9

 

[hello3719]

remember this?

9.999~ consider this to have "default infinity" number of digits 9
.999~ this number has (infinity(d)-1) which is demonstrably true when you subtract .999~ from 9.999~ you are left with exactly 1 digit 9.

you SAW that there is 1 digit 9 left, this is visual and doesn't mean that "infinity(d)" - ("infinity(d)"-1) = 1

assume .999~ in line 1 to be default infinity number of 9's
in line 2 assume x/10 would shift all decimals right one digit and create a 0, yet maintain the same "default infinity" number of digits 9
.999~ - .0999~ =.8999~1

9/10 = .9
9/10 x .999~ = .8999~1

k so you just admitted that .999~ and .0999~ have the same "default infinity" number of digits 9. You are now giving your infinity the same property as our infinity.So "infinity(d)" - ("infinity(d)"-1) isn't anymore equal to 1.

 

[ram2048]

i don't think you're understanding that i can take any "infinity" and define that as my "default infinity" and work calculations from there.

default infinity isn't the same for every calculation, it MUST be designated to be used.

all it does is create a base to work from, it doesn't matter what you use as a base, the logical transformations from there stem outward and resolve to the same conclusion.

the notation system does need work i admit that, but there are no contradictions in the system, visually or otherwise.

 

[hello3719]

So in fact it is useless since you can't prove anything usefull about your system.

 

[Hurkyl]

i know nothing of continuums.

not going to even go there

Ok, I rememebred some easier to describe problems that have similar trappings:

Suppose I have a circle. Point A is inside the circle and point B is outside the circle. Can you prove the line segment joining A and B intersects the circle?

Prove that x^2 = 2 has a solution. (In other words, prove that √2 exists)

 

[Hurkyl]

Back to your "default infinity":

You suggest that ∞(d+1) = ∞(d) + 1.
Is this true in general? That ∞(a + b) = ∞(a) + b?
If so, note that we could define ω := ∞(0), then ∞(d) = ω + d.

Do any of the following make sense? If so, what are they?
∞(1/2)
2 * ∞(0)
(1/2) * ∞(0)
π * ∞(0)
∞(0)^2
∞(0)^π
√∞(0)
2^∞(0)
sin ∞(0)
∞(∞(0))

If this question makes sense, is &infin;(0)^2 < 2^&infin;(0)? How would you prove your answer?

To what default infinity does this sequence approach:
0, 2, 2, 4, 4, 6, 6, 8, 8, 10, 10, 12, ...
and what about this sequence:
1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, ...
Which of the two infinities is bigger? Or are they the same?

What do you have to say about this sequence:
0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...
and how does it compare to
1, 2, 3, 4, 5, 6, ...
or to
1/2, 1, 3/2, 2, 5/2, 3, ...?


Can I define a number x to have a 9 in every finite position, but a 0 in every infinite position? If so, how does x compare to 10x - 9? If not, why not?

Can I define x to have 9 in every position and do the same thing? That means not only does it have a 9 in every finite position, but also in every infinite position. (So, for instance, there is no d so that the &infin;(d+1)-th position doesn't have a 9 in it)
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[ram2048]

Ok, I rememebred some easier to describe problems that have similar trappings:

Suppose I have a circle. Point A is inside the circle and point B is outside the circle. Can you prove the line segment joining A and B intersects the circle?

i don't remember how to plot a circle, but I'm assuming this "proof" would involve me computing an exact value of π

Prove that x^2 = 2 has a solution. (In other words, prove that √2 exists)

a square with side length 1 has a corner to opposing corner diagonal or exactly √2 length using pythagorean. I'm not sure what you want me to do about it though, visually and logically they both exist, but can they be rationally measured?

i'll work on that but don't expect results anytime soon ;D

 

[Hurkyl]

a square with side length 1 has a corner to opposing corner diagonal or exactly &radic;2 length using pythagorean. I'm not sure what you want me to do about it though, visually and logically they both exist, but can they be rationally measured?

Blarg, right, square roots aren't a problem; the problem I meant is showing x^3=2 has a solution!


Actually, there is some issue related to square roots, I'll see if I can remember what it is.

 

[Hurkyl]

Given a point O, and another point R:

Any point X is said to be:
Inside the circle if X = O.
Inside the circle if the segment OX is shorter than the segment OR.
On the circle if the segment OX is the same size as the segment OR.
Outside the circle if the segment OX is longer than the segment OR.


And since you bring up pi, here's another problem which is much trickier than the above one:

Given a point P, does there exist a point Q such that the line segment PQ has length equal to the circumference of a circle of radius of length 1?

 

[Hurkyl]

Oh, I almost forgot to ask this again: how does your decimal system represent irrational numbers, like &radic;2 or &pi;?

 

[Hurkyl]

Blarg, you got me with the red herring.

You've presented no reason whatsoever for one to suppose that your number system has anything to do with ordinary Euclidean geometry. (which is part of the reason why I was asking these particular questions!)

Thus, I am still interested in seeing you prove that your decimal system contains a number whose square is 2. (Since the proof from Euclidean geometry does not suffice)


Also, I really should rephrase my question about circle continuity:

In your system, I'm defining the "ram plane" to be ordered pairs of "ram decimals"; so each point has an x and y coordinate that is one of your decimal numbers.

I define the square distance between the two points (x, y) and (u, v) to be (x - u)^2 + (y - v)^2

So then, a "ram circle" would be the defined as I did above: Given a point O and another point R in the ram plane, for any point X I say:

Let x be the square distance between O and X.
Let r be the square distance between O and R.
Then, if x < r, X is said to be inside the circle OR
Then, if x = r, X is said to be on the circle OR
Then, if x > r, X is said to be outside the circle OR

 

[ram2048]

i didn't MEAN to red herring you, as all it did was make more problems for me ;D

too much at once. just pick ONE

and my notational system for decimals only represents irrational decimals created from fractional values to create rational ones.

i could possibly devise a way to represent pi as an accurate decimal as well, but i don't intend to bite more than i can chew.

if by some miracle people accept THIS then we can work on new stuff

 

[ram2048]

by the way, what's the accepted calculation for obtaining digits of pi ?

 

[Zurtex]

by the way, what's the accepted calculation for obtaining digits of pi ?

Well ram2048 you have totally failed to answer most of the questions and any of the questions with answers that make sense because most of them do show you clearly wrong. There are many ways of calculating Pi, here is one:

\pi = \sum_{k=0}^{\infty} \frac{{-1}^k}{2k + 1}

 

[matt grime]

i know nothing of continuums.

not going to even go there.

me: <looks up definition of continuum> if a continuum is blah blah then you could blah blah and so blah blah
you: no you can't
zurtex: WRONG hahaha you are WRONG SEE? YOU SUCK HAHAHA
me: <sad>

how about no?

so, you're going to spout on at length about what 'our' mathematical system is, and isn't, yet you aren't going to learn anything about it that might help in your understanding?

 

[ram2048]

don't start, Grime

i'm taking this a step at a time, i don't need to be stretched in directions i don't have to unnecessarily.

and like I've said before, with my current knowledge i have already spotted flaws in the system and come up with solutions for them that work. i'll get to continuums when time and resources permit.

Well ram2048 you have totally failed to answer most of the questions and any of the questions with answers that make sense because most of them do show you clearly wrong.

hmm i am clearly wrong when i can do calculations that come up with the EXACT correct answer and you are RIGHT when you come up with approximations.

yes that makes a lot of sense Zurtex.

 

[Zurtex]

hmm i am clearly wrong when i can do calculations that come up with the EXACT correct answer and you are RIGHT when you come up with approximations.

yes that makes a lot of sense Zurtex.

You do realize "a lot" isn't a real word?

As you say sums to infinity are wrong please give us how you would define \pi or \sqrt{2}. Also please define this so called default infinity. Also please could you show us if multiplying 10 by 0.\overline{9} "creates" a 0, which if you'd listened you'd know was absurd, where this 0 actually happens considering all the 9s go on forever.

These are just 3 things you can't do, your system isn't more accurate it is just plain wrong. I have tried to show you this as have many people, but you apparently haven't taken in a single point and just carry on spewing out rubbish.

 

[Zurtex]

So you see all infinities are different and there is some infinity which follows this:

\frac{1}{\infty} = 0.\overline{0}1

As all numbers of the form 0.000...0001 can be written of the form:

\frac{1}{10^n}

Where n is a natural number. Such that n is how many digits the significant number 1 is below the unit digit. So taking your previous unique infinity can you define for us:

\log_{10} \infty

So we know how far down these infinitesimal numbers are. Once again if you think about this you will realize how absurd this whole thing is.

 

[matt grime]

yes, ram, you've managed to sort out all the problems of the real number system without bothering to find out what the real numbers are, that's exactly what you've done. have you looked up any of these other things you've been told about where people have actually done these things properly? robinsonian analysis? infinitessimals, conways arithmetic of infinite cardinals?

for instance, associate a number n with a set of cardinality n, then addtion of n+m is the cardinality of the union of the sets, thus if we take infinity as aleph-0, for the sake of argument, then aleph-0+aleph-0= aleph-0=aleph-0 + 1.

of course w and w+1 are distinct in the ordinals (w is the first infinite ordinal).

but we wouldn't expect you to have checked that.

so go on, infinity is a 'real number' is it? which one?

 

[Zurtex]

If multiplying by 10 "creates" a 0 then this will surely apply so that it "creates" a 0 in 10\pi. Also multiplying your default infinity by \pi will give us a number such that it will only have 0s as digits below the unit digit "after the decimal point". So please tell us in the multiplication:

\infty \pi[/itex]<br /> <br /> What is the number in the unit digit \pi? Or another way of putting it is &quot;what is the last digit&quot; in \pi?

 

[hello3719]

with my current knowledge i have already spotted flaws in the system and come up with solutions for them that work.

Can you show us and explain us these "flaws" in our system?
Show me 1 contradiction in our system and not something you don't like because you think it is illogical without even backing it out.

 

[ram2048]

okay this is just stupid.

you: show me where we're wrong
me: <here> see?
you: you've proven nothing show me where it's wrong
me: wtf <here>
you: you know nothing, show me where we're wrong

i'm not showing you any more until you can come up with a disproof for my theory that consists of more than "that's not how we do things, you must be wrong" or "there is no end, you must be wrong"

my numbers are the EXACT same ones in subtraction as the ones gotten from multiplication. yours are NOT "blah blah 8.999~ IS 9..." if it was 9 then it would have calculated out to BE 9.

 

[Zurtex]

ram2048 that is total rubbish. We have shown you over and over again why it is what it is and shown you over and over again why your so called disproofs are wrong. But you just keep posting over the same thing over and over again.

 

[ram2048]

that consists of more than "that's not how we do things, you must be wrong" or "there is no end, you must be wrong"

no, not really

 

[Zurtex]

no, not really

Then at least reply to the last few posts once again showing how unbelievably wrong you.

Edit: Why have you quoted yourself AGAIN? And disagreed with yourself.