# Why doesn't .9999~ = 1?

x-ray vision…in what way does that web page purport that .999… doesn’t equal one? The entire page is devoted to showing that it does equal one! Just like your first cite.

Anyway, your more recent cite contains my favourite argument as to why 0.999…=1, namely:

I have yet to see anyone identify a number between 0.9999… and 1, nor do I expect anyone to do so. Not that this convinces anyone, of course.

We can always bring out the formula for those not bothering with the other threads. We use the basic method for turning a repeating decimal into a fraction.

let x = 0.9999…
then 10x = 9.99999…

Now we subtract
10x - x = 9.99999… - 0.99999…

Solving both sides
9x = 9.00000

Reducing
x = 9/9 = 1

Therefore 0.9999… = 1

No calculus involved.

RC: If you take calculus as the study of limits, then strictly you are using calculus as soon as you write
0.999…
and also when you multiply it by 10 to obtain
9.999…

Again, how is this irrelevant? If a number isn’t actually zero (and never will be), then it isn’t zero.

You stated that …
“following the basic rule of math that states that any number divided by itself equals unity, or 1.”
Zero divided by zero is not 1 nor is it zero for that matter. I believe mathematicians call the answer either “indeterminate” or “undefined”.
And Achernar, the proper phrase for “…” is ellipsis and not ellipses.

Orbifold, I meant to write does, not doesn’t. I need some sleep!

The limit of the sequence:

{ 0.9, 0.99, 0.999, 0.9999, … }

is 1. This is really the nomen ludi.

You realize that “limit” isn’t just a term mathematicians throw around, right? It has a strictly-defined meaning, and it’s not ambiguous. It may not be the most intuitive meaning to you, which could be causing the confusion. For instance, it is not correct to think of the limit as the last term in an infinite sequence. The limit does not have to be found in the sequence at all. Anywhere.

This entire discussion hangs on a pretty thin thread – the limitations of the decimal number system.

Q.E.D. and Jabba, adding the decimal values for 1/3 and 2/3 makes no more sense than multiplying the decimal value of 1/3 by 3. Why strain yourself? 1/3 * 3/1 = 3/3 = 1. (Duh.) 1/3 + 2/3 = 3/3 = 1. (Duh again.)

Nicely done, RealityChuck!!!

Why not? It’s perfectly valid. Since .9999… - 1 saying .33333… x 3 = .99999… = 1 or .33333… + .66666… = .99999… = 1 are both entirely correct.

That’s logical?

Perfectly. mathematically, it is asking you to solve for 1 - .9999… = x. What would you have the value of x be?

None of the terms of the sequence is 0, but the limit of the sequence is. Sometimes the limit of a sequence is one of the terms. For example,
1, 1, 1, …
certainly tends to 1 and so does
1, 1/2, 1, 2/3, 1, 3/4, 1, …
But it is by no means necessary that this be so.

Perhaps a look at the formal definition of a limit will help. For a sequence of real numbers, we say that a[sub]n[/sub] tends to the limit l if:
Given any real [symbol]e[/symbol]>0, we can choose a natural number N such that |a[sub]n[/sub] - l| < [symbol]e[/symbol] whenever n>N.
In other words, if you give me any positive margin of error, I guarantee that all the terms of the sequence from some point on will approximate the limit number l to within that margin of error.

Let us apply this to the two sequences we have been discussing. I have claimed that
1, 1/2, 1/3, 1/4, … converges to 0
and that
0.9, 0.99, 0.999, 0.9999, … converges to 1.
As an example take [symbol]e[/symbol], the magin of error, to be 0.001
For the first sequence, any term after the 1000th is less than 0.001 and so is within 0.001 of the proposed limit.
For the second, every term after the 3rd is within 0.001 of 1.

If instead we take [symbol]e[/symbol] = 0.000001, then we just have to go further along the two sequences: the 1,000,000th term will do for the first sequence; the 6th term for the second.

In general, whatever positive [symbol]e[/symbol] you choose, I can guarantee that all terms from some point on will be within [symbol]e[/symbol] of my claimed limit. That is what we mean by convergence, and in particular by the value of an infinite sum.

.000~1 I’ll accept that. And since .0000…1 = 1/infinity = 0, you’re correct.

Now now, I’m with you Q.E.D., but performing arithmetic on infinity isn’t going to strengthen our cause.

Q.E.D., what does 2 - .9999… = x. What would you have x be?

1

Scratch the “what does”.

1 - .9 = 0.1

1- .99 = 0.01

1 - .999 = 0.001

I can go on forever and never get 0, however, 1 -1 =0.

My conclusion is 1 cannot be the same as .999…