If ...999 was to equal 1 then the .9's would End? But as we Know Reach the End of the Universe ? & Find a Wall ? ...

Then There must be Something behind the Wall !! Example = .........999999999 [1]Wall] .........999999999

]]>I should note though, I don't quite understand the start of the universe, end of the universe thing, mind explaining it?

]]>Another way of looking at & Solving the Problem is 0.999... = (The Start of the Universe) and 1 = (The End of the Universe ? )

]]>anonimnystefy wrote:

No, 3 is allowed, 0.(9) isn't. There cannot be recurring 9's after the decimal point.

So I know more specifically about what your talking about, I'd like to see your response to...

bob bundy wrote:

Why not?

Stefy, supposing what you say is true, then what is the point of this argument in the first place? Why are there hyper real numbers even? You completely invalidate ... everything this controversy has to do with.

Here's a case where this could come in. 1/3 + 2/3 = 1. In decimal form, this can be argued to be 0.¯9 (**not** saying this is 1/3 and 2/3s tho). The issue directly has to do with decimal, because it is in base 10. This is impossible to accurately represent this in decimals in base 10 because the closest you can get is 0.¯3 and 0.¯6. Just simply looking at 0.¯3, and 0.¯6, what does 0.¯3 + 0.¯6 equal (**only** look at it like that is now, do **not** assume this is 1/3 and 2/3s, as I'm trying to represent how people get to this)? Even though it is repeated infinitely, it is irrelevant, because in the "end," it still never equals 1. Now to clear up any confusion, I am **not** saying this is **equal** to 1/3 + 2/3s, otherwise it would be 1. I'm only trying to represent (if you are able to look at it from a different mindset and **not** assume its 1/3 and 2/3s), a way to get to 0.¯9. In other words, 0.¯3 and 0.¯6 are not a fraction of 1, they are just numbers on their own. Now, I still don't know exactly what you are talking about, so I still would like to see what you say to to, "why not?"

Maelwys wrote:

SMBoy wrote:A = ...999

B = ...001

A + B = 1

If A was to Equal 1 on it's own then... A + B = Would Equal 1.1

First off "A + B = Would Equal 1.1" is blatantly false. A + B = 1. ...001, by your notation.

Secondly, that assumes that there is a "last digit". In your notation, the 9's clearly stop. There is a "last nine" in A, and in the same spot in B there is a corresponding 1. The problem is that this contradicts with the very definition of "infinity". If there are an infinite number of 9s, there can be no "last nine". That means that there is no corresponding location in B to put the 1, so B just becomes an infinite string of 0s. And 0.000.... (with an infinite number of 0s) = 0. So if B is 0, and A + B = 1, then A = 1.

I think what SMBoy is saying Maelwys is that if you add that to 0.9, or 0.99, or 0.999, regardless how many 9s there are, it will equal 1. But if you add the same thing to 1, you will get 1.1, or 1.01, or 1.001, etc. More showing how the 2 answers are different and not equal in that sense.

]]>A = ...999

B = ...001

A + B = 1

If A was to Equal 1 on it's own then... A + B = Would Equal 1.1

First off "A + B = Would Equal 1.1" is blatantly false. A + B = 1. ...001, by your notation.

Secondly, that assumes that there is a "last digit". In your notation, the 9's clearly stop. There is a "last nine" in A, and in the same spot in B there is a corresponding 1. The problem is that this contradicts with the very definition of "infinity". If there are an infinite number of 9s, there can be no "last nine". That means that there is no corresponding location in B to put the 1, so B just becomes an infinite string of 0s. And 0.000.... (with an infinite number of 0s) = 0. So if B is 0, and A + B = 1, then A = 1.

]]>B = ...001

A + B = 1

If A was to Equal 1 on it's own then... A + B = Would Equal 1.1

]]>But the a Two differences remain constantly the same!

Is .1 the same as .00001 or .0000000001 or ...

Those differences are not the same. They seem to be vanishing.

]]>Bob

]]>I didn't mean it in that sense. Such a number isn't allowed in mathematics.

What number is that? 3 ?

I think you'll find it is used in quite a few areas of mathematics.

And so is 0.999999999 recurring.

To save me some time, I refer you back to my post # 999.

Bob

]]>So the question becomes "Why is .1111... equal to 1/9?"

Simple solution: Choose base 9 instead of base 10. Then 1/9 = .1 Now we don't have to

deal with an infinite repeating decimal. Of course, now representing 1/10 in base 9 is an

infinite repeating decimal. So to eliminate all these problems just scrap the "decimal systems"

and go back to using only the fractions.

The basic problem (in base 10) is that any unit fraction 1/N where N has prime factors other

than 2 or 5 cannot be written as a finite sum of fractions whose denominators only have

powers of 10.

Suppose we have an N with prime factors not twos or fives. Then the algorithm for dividing

1 by N would never have a remainder of zero since products of 3, 7, 11, 13, 17, 19, ... never

end in zero. So at every stage of the division we have non-zero remainders. Hence the

"decimal representation" of 1/N would have to be "infinite."

Any time the base b and N are relatively prime (this doesn't cover all cases for N's that

cause an "infinite repeating b-esimal") the division algorithm always has a remainder at

each stage. If we stop at any stage, then we have to add in the remainder to get the

original number exactly.

In applications all we ever use in "real world" problems are approximations to "some number

of decimal places." Take pi for instance. We use so many decimal places to approximate

pi in applications.

BUT it sure is CONVENIENT to "use" infinite decimals in mathematical expositions.

]]>

I didn't mean it in that sense. Such a number isn't allowed in mathematics.

]]>But the number 0.9999... itself doesn't exist...

How many times must I say this? No numbers exist. They are all just elements in an imaginary set that mathematicians have invented.

You cannot have a 3 => it doesn't exist.

But it's a jolly useful concept (especially for describing how many apples I have today) so let's go on using it.

Clearly this argument is going to continue for ever.

So:

assign the number 0.9 to post #1

assign the number 0.99 to post #2

.........

assign the number 0.999......9999999 (n 9s) to post #n

.........

and so on ad infinitum.

Bob

]]>.99 + .01 = 1

.999 + .001 = 1

.9999 + .0001 = 1

.99999 + .00001 = 1

But the a Two differences remain constantly the same! just another 9 and 0 ...further down the line does not make them nearer to being equal to 1

]]>