# I know this has been hammered out before, but

1. Apr 29, 2008

### sennyk

[SOLVED] I know this has been hammered out before, but ...

When I was in high school, my math teacher showed us that 0.99999... is equal to 1. He went through the fraction proof and I was literally amazed; however, I've also wondered why 0.99999... can't be considered an irrational number.

My first guess is that there is a least upper bound which is 1. I always thought that irrational numbers never have a least upper bound. But then I thought that the only definition is that an irrational is a number that can't be expressed as a fraction of integers.

Any other thoughts?

2. Apr 29, 2008

### D H

Staff Emeritus
First off, 0.999... must be rational because it repeats. Irrational numbers do not have a repeating decimal. Secondly, this "I always thought that irrational numbers never have a least upper bound" is wrong. Irrational numbers, like rational numbers represent a point on the number line. There is no fuzziness whatsoever when it comes to real numbers. Thirdly, this subject has been treated ad nauseum. I suggest you search "0.999..."

EDIT:
A quick google search for 0.999... limited to this site alone yields 700+ hits; handy link http://www.google.com/search?q=0.999...+site:physicsforums.com".

Last edited by a moderator: Apr 23, 2017
3. Apr 29, 2008

### HallsofIvy

Staff Emeritus
Even if it weren't equal to 1 it wouldn't be irrational. As D.H. said, any repeating decimal is rational.

"least upper bound" of what? sets of numbers or sequences of numbers have least upper bounds. If you are talking about a "least upper bound" of the sequence of "partial decimals", .9, .99, .999, .9999, etc., then, yes, the least upper bound of that sequence is 1. And since the sequence is increasing it converges to that least upper bound-which is the definition of the value of an infinite such a sequence of digits.

But your statement "irrational numbers never have a least upper bound" is completely wrong: Any real number, rational or irrational, can be written as a decimal expansion which has that number as its "least upper bound" and limit.

I'm not sure what you mean by that last. That certainly is one definition and, I think the most commonly used one. But that doesn't mean that you can't the prove other properties that an irrational number must have. One of the most important is that an irrational number, expressed as a decimal, must NOT be "eventually repeating".

A few years ago, on another forum, a person asked, "How do you prove that a rational number can be written as a fraction". I stared at that for some time before I realized that the person asking must have learned "eventually repeats" as the definition of rational number- which is completely valid.

4. Apr 29, 2008

### sennyk

Thanks for the replies. I guess I had forgotten the repeating rule. I'm reading some math wikis and I've seen the proof explaining why 2^(1/2) has to be irrational. It starts by proving that there can't be a rational solution (i.e. written as a fraction). Because of that I wondered if the decimal repeating rule was something that I imagined.

5. Apr 29, 2008

### uman

What's this "fraction proof"?

6. Apr 29, 2008

### D H

Staff Emeritus
Assume $\surd 2$ is rational. This means there exists a pair of relatively prime integers p, q such that p2 = 2q2. Since p is an integer by assumption, the only way p2 can be even is if p itself is even. Writing p=2r, where r is an integer, the relation p2=4r2=2q2 implies that q too is even. Thus both p and q must be even, contradicting the assumption that p and q are relatively prime.

7. Apr 29, 2008

### uman

I know that; I was talking about the "fraction proof" that 0.9... = 1.

8. Apr 30, 2008

### sennyk

1/3 = 0.33...
3/3 = 3 * 1/3 = 3 * 0.33... = 0.99... = 1

9. Apr 30, 2008

### HallsofIvy

Staff Emeritus
Of course, that depends upon first accepting that 1/3= 0.333... which I would consider no more "obvious" than 0.9999...= 1.

The other common "proof":
If x= 0.999...., then 10x= 9.9999.... so 10x- x= 9x= 9 and x= 1
depends upon accepting that those operations, applied to an infinite decimal, are valid. That's true but, again, no more obvious, in my opinion, than 0.999...= 1.

What I would consider a real proof consists of appealing to the definition of "decimal expansion": 0.99999.... is limit of the sequence 0.9, 0.99, 0.999, ... or, equivalently, the sum of the infinite series $\Sum_{n=0}^\infty 0.9(0.1)^n$. That's a geometric series so its sum is
$$\frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1$$

10. Apr 30, 2008

### uman

Mhm Sennyk, that's what I thought was meant by the "fraction proof" and it begs the question as HallsofIvy pointed out. Someone who doesn't believe that $$\sum^\infty_{n=1}\frac{9}{10^n}=1}$$ is equally unlikely to believe that $$\sum^\infty_{n=1}\frac{3}{10^n}=\frac{1}{3}$$ (or at least should be, if their beliefs are consistent)

Last edited: Apr 30, 2008
11. May 1, 2008

### MagikRevolver

In my ignorant opinion, I think this just shows how flawed our knowledge of algebra is. If .999... 1, than I would hope I can answer every math question ever with 0/0 seeing as how 0/0 can be anything, but that is meaningless. So, why isn't .999... meaningless? Just because there is a rule out there that says you can divide by .999..., doesn't really mean you can. Just like there was a theory that you couldn't go faster than the speed of sound, and just like special relativity which will one day be meaningless.

You can never have .999... as a tangible number anyways, so whats the point.

Last edited: May 1, 2008
12. May 1, 2008

### CRGreathouse

I have 0.999... mouths.

13. May 2, 2008

### uart

So by that same logic we can conclude that 0.333... is also meaningless and therefore that 1/3 does not have a decimal representation. Your logic is wrong.