If x equals 0.9 repeater, and ten x equals 9.9 repeater, 9x equals nine and therefore x=1 = how did x change?

Answer 1

in the limit x = 1 - 0.1^n and when n goes to infinite x goes to 1

Answer 2

x does not change.

x = 0.999... (1)
10x = 9.999... (2)

The operation (2)-(1) is undefined and not possible. Our arithmetic deals only with finite number representations. So the result that x=1 is untrue.

I am unsure what your level of mathematics is, but the main argument suporting this erroneous belief is that if one cannot find a number between 0.999... and 1, then these must be the same number. As it so happens, there are infinitely many numbers between 0.999... and 1.

Here are some examples:

Let p = 100.

The number represented by:

99/p + 99/(p^2) + 99/(p^3) + ....

is between 0.999 and 1.

Check:
9/10 < 99/100 < 1
9/10+9/100 < 99/100+99/10000 < 1

You have an infinite number of choices for p.
Thus there are infinitely many numbers between 0.999... and 1.

Others will argue that the limit of all these series (which in fact represent numbers) is 1. This is a no-brainer. Duh?! When numbers are compared, their partial sums are compared, not their limits. As an example, try to compare any irrational number - does the series representing the irrational number have a limit we can compute? No! It has a limit but it cannot be computed. Some common examples are: pi, e, sqrt(2), and so on.

You can find many of these lame arguments in the Arguments section at Wikipedia unless they have for some unknown reason deleted the link.
You can make up your own mind after you study the material. It is quite detailed and contains a lot of different viewpoints.

Let's see where Mesperanto goes wrong:
"He says x=0.9999999... has the same number of 9's in 10x=9.999999..... no matter how big infinity is at the point of the calculation."

I wonder how he arrives at this conclusion? In 9.999... we have shifted the nines to the left so how do we know there are the same number of nines after the period? We don't. Now notice when Mesperanto performs arithmetic with finitely represented numbers, he arrives at the correct value, i.e. x = 0.999... It's when he thinks that he can compute using numbers to infinity that he arrives at the wrong conclusion, i.e. x = 1.

Now mathematician made a bold claim that we do in fact compare limits. Yet he failed miserably to provide a limit for any one of the examples I provided, i.e. pi, e, sqrt(2). That's right mathematician, these are numbers represented by series whose limits are not computable. So yes, we have always compared 'partial sums', never the limits. In fact when we compare limits, we are not comparing the numbers. Wikipedia is dead wrong and 0.999... is definitely less than 1. Nothing will ever change this fact.

Sorry, just one more response: mathematician claims that "...*if* we can show that x_n and y_n differ by more than 2*10^-n for some n, then we do know that x and y are different. But this is not the case for 1 and .999...."

Well, if this is how we determine that real numbers are different, we have serious problems because the reals are then not continuous. However, we know they are, so this argument is invalid, i.e. the fact that the difference of the partial sums of 0.999... and 1 tends to zero does not mean the numbers are the same. The fundamental problem lies with real analysis. d(x,y) = 0 iff x=y - this property is not only untrue but it makes no sense whatsoever because a point has no extent. That numbers can be expressed as Cauchy sequences relies on this property.

BTW: There is no such thing as an infinitesimal. It does not exist, not theoretically and not in practice. Be careful what you read on Wikipedia!!

I am not going to refute any other argument. Read the arguments page - you might learn something!

Answer 3

Tom is completely incorrect here.
It is simply true that 1=.99999.....

We do not compare numbers through partial sums, we *do* compute via limits. In this case, surprisingly enough, Wikipedia is correct.

They way to see if two numbers expressed as limits are equal is to see if the limit of the differences of their partial sums goes to zero. In this case, the difference between 1 and .999..9 (a partial sum) is of the form 10^-n for some n. As n goes to infinity, this goes to zero, so the numbers are equal. Again, Tom is completely wrong here. The flat out definition of an infinite decimal expression is the limit of the partial sums. So sqrt(2) is the limit of the partial sums of its decimal expansion, just like every other number.

Now, if we have some number x, and let x_n be the n-th partial sum in the decimal expansion, we do have that x and x_n differ by an amount less than or equal to 10^-n. If some other number y has partial sums y_n and *if* we can show that x_n and y_n differ by more than 2*10^-n for some n, then we do know that x and y are different. But this is not the case for 1 and .999....

Answer 4

If x=0.9999999...., then there must be the same number of 9's in 10x=9.999999..... no matter how big infinity is at the point of the calculation.

So, if we start off with 2 decimal places we get:
x=0.99
10x=9.9
Subtracting we get 9x=8.91 therefore x=0.99

If we start off with any finite number of 9's we get the same result:
x=0.999999
10x=9.99999
Subtracting we get 9x=8.999991 therefore x=0.999999

This will happen no matter where we stop the recurring 9's. The paradox arises because that's not what you actually did when you caused it to recur to infinity. You cheated on the multiplication by 10 and added in another 9 to end of the second multiplication, so we actually had this:
x=0.999
10x=9.999 (which is really 10x+0.009)
So, subtracting we got 9x=9 therefore x=1 exactly because we borrowed the final 9 which wasn't really there (even at infinity).

The two methods differ, so you actually got two different results as a result! It's a very clever slight of hand illusion!

Answer 5

X = 0∙9999999999....
10X = 9∙999999999....
9X = 9(0∙9999999999...) = 8∙999999999....
So X = 9X/9 = 8∙999999999.../ 9 = 0∙999999999...

X changed because your numbers were rounded off.

Answer 6

It means that 0.999... = 1

Answer 7

0.(9) doesn't exist!
Good joke!