Does 0.99r = 1 more than 1/3 * 3?

Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaah... the horror came true. We're doomed!

My dual-PSU PC is so powerfull that the neighbourhood dims when I turn it on

they are the same because we can argue that 1/3=.33333r so 3*.333r= .9999999r which equals .999999r. Thus they are the same- 1 in fact.

Hilbert space is a big place.

You really know how to f u c k up a thread, don't you?

<b><font color=blue>~ <A HREF="http://forums.btvillarin.com/index.php?act=ST&f=41&t=324&s=58e94ba84a16bedfebbf0f416d5bac48" target="_new">Nice sig 81.</A> ~<font color=blue></b>

I pride myself on my ablity to do so

In battle Israel uses F-16's and big ass tanks, Palestine uses small children strapped to bombs. Which do you support?

what? am I not allowed to respond to the question?

In battle Israel uses F-16's and big ass tanks, Palestine uses small children strapped to bombs. Which do you support?

lol we can't kill him tho, you know what the dusty book says about that! But if we hire a hitman...

In battle Israel uses F-16's and big ass tanks, Palestine uses small children strapped to bombs. Which do you support?

No, this is the same question as the one over a year ago about "does .999 repeating =1". NO. If you understood infinity, you would understand why. There is an infinately small distance between .9999 repeating and 1. Regardless of how small it is, the difference is still there.

It's not 1 more. It's not .1 more. It's 1 to the power of -infinity more.

<font color=blue>Watts mean squat if you don't have quality!</font color=blue>

None of your proofs make sense if you really understand infinity!

.9+.1=1
.99+.01=1
.999+.001=1
.9999+.0001=1

Notice in each of these cases, the 1 goes out THE EXACT SAME NUMBER OF DECIMAL PLACES AS THE 9. Let's try that with subtraction:

1-.1=.9
1-.01=.99
1-.001=.999

Hmm, we see the same thing happening, I wonder why that is?

Could it be that no matter how many decimal places the 9 repeats, if you add a 1 to the number on the end, it comes out to 1.0? Hmm.

.999r, or repeating. Hmm, that word, repeating...means it repeats forever. Sound a lot like infinity. Let's see, forever vs. infinity. Yep.

Say you made a string of number 9's, in 8 font, from here to Pluto. Now say you made the same number of 0's, except that the last 0 was a 1. You add those two number, you get 1.0.

To say that .1^(-infinity) can't exist is to say that negative infinity can't exist. But for any number to exist, it's inverse must exist. Therefore, if -infinity doesn't exist, infinity doesn't exist. And if infinity doesn't exist, your 9's can't repeat forever, because forever=infinity. Doh!

So I've proven that if you take repeat a 9 any number of places from the decimal point, and add a one to that same number of decimal places, you get 1.0. Now, take that to infinity. .99r + .1^(-infinity) = 1.00r

<font color=blue>Watts mean squat if you don't have quality!</font color=blue>

Now I'm starting to recall why I loved the original so much.

<b><font color=blue>~ <A HREF="http://forums.btvillarin.com/index.php?act=ST&f=41&t=324&s=58e94ba84a16bedfebbf0f416d5bac48" target="_new">Nice sig 81.</A> ~<font color=blue></b>

I understand infinity perfectly well thank you. Infinity is different from the complex numbers (although we really only have to deal with the reals here) as it is not a point on the number line.

The proofs are perfectly fine. I have proven that adding any infinitessimal amount to 0.9r will take you away from 1. There is only one logical conclusion one can draw from that.

First of all your .1(-infinity) is so wrong it's not even funny. .1^(-1) = 10, .1^(-10) = 10000000000... If you increase the exponent to infinity, you will get an infinitely large number. However, I'll just assume that you didn't bother to figure out exactly what you were writing, and that you really meant .1^infinity

Your logic regarding the existence of .1^(infinity) is also very flawed. Infinity exists, negative infinity exists, .1^x exists, lim x-> inf .1^x is zero, but .1^(infinity) does not exist.

As well, later on in your "proof" ou state that for any number to exist, it's inverse must exist. Do you even know what an inverse is? The inverse of x is NOT -x. Let y = inverse of x. y is any number such that xy = 1.

I guess 0 must not exist now huh?

You can't just do regular arithmetic with infinity as though it were some real number. It just doesn't work. Answer me this:

(-1)^1 = -1 = negative
(-1)^2 = 1 = positive
(-1)^3 = -1 = negative
(-1)^infinity = ? and is it positive or negative?

You can't just use infinity in normal math. It must be dealt with very carefully.

<A HREF="http://mathforum.org/dr.math/faq/faq.large.numbers.html" target="_new">Here</A> is a little info on infinity, one where they are a little more rigorous than: "forever=infinity"

Take your string of 9's to Pluto. That is quite easily not an infinite string. Why did you stop at Pluto? Because you needed a place to tack on that 1. You NEED the string to be finite in order to say where the 1 is.

Let n represent the index of the position of any value after the decimal point in either of our two strings. Let's also define the [] function to extract the value from the index.

ie: [4] of 3.14159 = 5

I can, for any n tell you that [n] = [n+1] = [n+k] = [n+k+1] = 9 for any k in the 0.9r string.

I can also tell you that [n] = [n+1] = [n+k] = [n+k+1] = 0 for any k in the 0.0r string.

It can also be proven that you can describe the decimal expansion for any rational number q in this way.

71/99 = 0.7171717171... So [2k] = 1 and [2k-1] = 7

Now, let's assume your arithmetic with infinity is valid.
{
.1^(infinity) = (1/10)^infinity = 1^infinity/10^infinity
therefore, .1^(infinity) can be written as q = x/y and is therefore rational (again, all of this is using your logic that x^infinity is a real number)
Further, because it is rational, you must be able to write it as I have just written 0.9r, 0.0r, and 0.71717171...

You cannot.

The [n] = [n+1] part is fine for this "number", but let's try it for k. You would say that [k] = 1, and we're done, but what about [k+1]?
}

You can't have a 1 at the end of an infinite string. As soon as you tack that 1 on there, the string becomes finite.

Most schools require arithmetic as a prerequisite to real analysis...

Some day I'll be rich and famous for inventing a device that allows you to stab people in the face over the internet.

LOL, every number has an end...those 9's that repeat forever, stop at a point called forever, and if you add a one to that point, what happens?

The fact that this point called forever doesn't exist in reality is...pointless.

<font color=blue>Watts mean squat if you don't have quality!</font color=blue>

They can't stop at a point called "forever" or infinity because infinity is NOT a point. You really don't understand the meaning of infinite if you use it in the same context as stop, unless you add some kind of indication that it doesn't stop.

And actually, no number has an end. All numbers are infinite decimals. However, just for ease of writing, we don't write them that way. 11 = ...000011.00000...

You're just spouting drivel. Reply with something mathematically sound, preferably with some degree of formality, or do the internet a favour and quit wasting bandwidth.

Some day I'll be rich and famous for inventing a device that allows you to stab people in the face over the internet.

Sorry about the harshness, I forgot I was probably talking to a 15 year old.

<font color=blue>Watts mean squat if you don't have quality!</font color=blue>

BTW, a proof is "this is true, that's true, therefore this is always true". In order to prove my point, I used check points. As in, .9 and .1. .0009 and .0001. Etc. The last place I stopped was at pluto. The "point" at the end of infinity is imaginary, kind of like your logic.

<font color=blue>Watts mean squat if you don't have quality!</font color=blue>

Again, due to your lack of clarity, I'm going to assume you mean 0.000...0001 and not ...0000.1 as you actually said.

First of all, the number 0.000...0001 does not exist. It's as simple as that. As I've stated numerous times already, you can't have in infinite string that doesn't end that ends. How about this? If you can say that your notation is valid, I'll say this:

0.000...0001 + 0.999...9999...9999... > 1

You have an infinite amount of zeros ending with a 1. I have an infinite string of nines ending with an infinite string of nines, ending with an infinite string of nines...

Of course, none of that actually makes any sense to anyone who understands mathematics beyond grade school.

Some day I'll be rich and famous for inventing a device that allows you to stab people in the face over the internet.

yes, well, the difference in theory is based on the fact that I allow you to use a reference point while you don't.

<font color=blue>Watts mean squat if you don't have quality!</font color=blue>

The main problem with you is you're seeing 0.9r as a progression towards something. It's not moving. 0.9r doesn't get closer to 1. It doesn't move, you don't add nines, your computer doesn't add them to a string. 0.9r represents a 0 with a decimal point followed by an infinite number of nines, not a string of nines that you increase to infinity.

You are still thinking of it as a sum like S = sum n = 1..k (9/10^n) and then letting k grow. This is NOT what we are talking about.

Your computer explanation is flawed. A computer has a set speed and you are talking about a finite time. There is no was for the computer to spit out an infinite string in a finite time. Therefore, it can't be used to describe 0.9r.

We are by definition talking about a decimal point followed by an infinte string of nines. The nines are described to be already infinite, not of length k as k approaches infinity. Why don't we just define the length as "forever"?

This is also in error. We are NOT talking about a progression with time. If you were to say that You had a frog who could jump 9/10 the distance to the pond with each jump and then let it go, of course it would never each the pond. This is definitely not what we are talking about though.

Again, you can't use a finite example to show what happens with an infinte string. Words like "when" are completely out of place here as nothing is moving. The bullet will never have an infinite distance behind it whereas our string is already infinite.

0.9r is a number, not a progression. How can a number approach another number? What is the limit of 2 as 2 approaches 3? If you want to deal with length of strings, take 2.0, 2.00, 2.000, ... , 2.000... Talking about the limit of 0.9r is the same as talking about the limit of 2.0r

Number theory states that the density of the number line is infinite. ie, you can always find an infinite number of points between two unique points. No two points exist such that there is only one point in between them. If there are no points between two numbers, they are the same. That is the case with 0.9r and 1. There are no numbers in between 0.9r and 1, and they are therefore the same.

Some day I'll be rich and famous for inventing a device that allows you to stab people in the face over the internet.

The real number line contains all numbers.
It goes from negative infinity to positive infinity.

The answer to this question is that

1. The limit of .9r as it goes to infinity is 1
2. .9r dne 1

<font color=red>
<A HREF="http://kevan.org/brain.cgi?dhlucke" target="_new">The French are being described as cheese-eating surrender monkeys.</A></font color=red>

THANK YOU! That's exactly what my Calc teacher would have said. Bless her soul, I understood the concepts but forgot the math.

<font color=blue>Watts mean squat if you don't have quality!</font color=blue>

Maybe... or maybe you say that because you don't want to beleive that he doesn't agree with you anymore.

My dual-PSU PC is so powerfull that the neighbourhood dims when I turn it on

Either way, it doesn't make the two any less equal...

Some day I'll be rich and famous for inventing a device that allows you to stab people in the face over the internet.

Heh, hot damn! I just graduated to Ancient Poster...

Some day I'll be rich and famous for inventing a device that allows you to stab people in the face over the internet.

Maybe it doesn't sound like 1 to you, but it is 1.

Some day I'll be rich and famous for inventing a device that allows you to stab people in the face over the internet.