# Is this statement valid?

1. Jul 13, 2013

### iScience

|1-0.$\bar{9}$|=dx?

2. Jul 13, 2013

### WannabeNewton

No. First of all $0.\bar{9} = 1$. Secondly, as you were told in the other thread, $dx$ is not a real number.

3. Jul 13, 2013

### micromass

No. We have that $1=0.999...$. So $|1-0.999...|=0$. On the other hand, $dx$ is not a real number.

4. Jul 13, 2013

### Infrared

See https://www.physicsforums.com/showthread.php?t=507002 [Broken] for a good explanation of why $0.999...=1$

Last edited by a moderator: May 6, 2017
5. Jul 14, 2013

### iScience

how does 0.999...=1?... .....how?.....

just to be sure, im not rounding..

that's like saying i have a piece wise function.. f(x)={2x , 0<x<5}
{3 , x=5 }
{2x , 5<x<∞}

the statement 4.999999....=5 is NOT even saying as x approaches 5... no. This is like saying at 4.999999...... wer're already AT 5.
how does this make sense??

and if 4.99999.... = 5 , then this must also be true 5.00000000.......00001 = 5.
okay so.. now we have three "fives" on the number line... let's work our way outward shall we

4.999...$\equiv$5 so now.. 4.999......9998 is also 5.. and 4.999.....9997 is also 5...
5-$\Sigma$dxi=5 no matter what the value of i is.. even as it approaches ∞ this is still five.
and now we have an infinite number of fives on the number line. the implications do not make sense to me.
if i am wrong with my analysis, i assume i am wrong from the point when i start to work my way outward after the point where 4.999.... is said to be equal to five..

Last edited: Jul 14, 2013
6. Jul 14, 2013

### pwsnafu

See the FAQ HS-Scientist linked to in post #4.

Also Wikipedia article.

7. Jul 14, 2013

### micromass

That is actually the second part of the FAQ, so be sure to read the first part as well: https://www.physicsforums.com/showthread.php?t=507001 [Broken]

Last edited by a moderator: May 6, 2017
8. Jul 14, 2013

### WannabeNewton

You really have to stop treating $dx$ like it's some kind of number. It's fine for hand wavy arguments in subpar physics books but not for what you're interested in here.

9. Jul 14, 2013

### micromass

If you want to be rigorous about it, then for any function $f:\mathbb{R}\rightarrow \mathbb{R}$, you can define

$$df:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}: (p,h)\rightarrow f^\prime(p) h$$

In particular, we have that

$$dx:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}: (p,h)\rightarrow h$$

The main reason for this definition is to make the following true:

$$df = \frac{df}{dx} dx$$

If you want to do physics, then infinitesimals come in very handy. Those infinitesimals are made rigorous by the hyperreal and surreal number system. But they are not denoted as $dx$.

In short: you need to stop treating $dx$ as a number, or even worse: as a real number. It isn't, and it's very confusing.

10. Jul 15, 2013

### mathsciguy

About 0.999...=1, is representing it as a geometric infinite series a valid proof?

11. Jul 15, 2013

### HallsofIvy

Yes, it is probably the simplest strictly valid proof.

0.9999...= .9+ .09+ .009+ .0009+ ...= .9(1)+ .9(.1)+ .9(.01)+ .9(.001)+ ...= .9(1)+ .9(.1)+ .9(.1^2)+ .9(.1^3)+ ...

That is a greometric series of the form $\sum ar^n$ with a= .9, r= .1. A geometric series with |r|< 1 (which is the case here: |r|= .1< 1) converges to a/(1- r).

Here, a/(1- r)= .9/(1- .1)= .9/.9= 1.

12. Jul 16, 2013

### iScience

guys what level of math are you all in? ie what's the highest math courses you've taken? i just finished diff eq and can't wait to get to where you guys are at.. things are so confusing for me right now :'(

13. Jul 16, 2013

### WannabeNewton

Don't worry, it will all come in due time. Most of the people answering you are doing their PhDs/masters or already have their PhDs so don't fret, you'll learn it all in sooner or later.

14. Jul 16, 2013

### HallsofIvy

Yes, at PhD level, but the "geometric series" proof that 0.999...= 1 is one that I learned in PreCalculus.

15. Jul 17, 2013

### Mandelbroth

I think that $|1-0.\bar{9}|=0$, which admittedly IS close to 0 if you want to think about it that way, seeing as the distance between the two is 0.

For now, it's best to treat $dx$ and $df$ as variable holders for notation. Later, in differential geometry, you'll be introduced to the concept of the "exterior derivative," which extends the idea of the differential to a meaningful, mathematically rigorous context. For now, though, just treat it as notation.

Whereas you and micro seem to do better than most of us without a degree. :tongue:

It's not about your level of education in school. It's about how much you study by yourself. You can get a lot further by self study than you can by a formal college education, in my opinion.