# I Non unicity of decimal expansion and extremes of intervals

1. Feb 27, 2016

### DaTario

Hi All,

The famous proof of the theorem: $1 = 0.9999999...$ seems to point to a statement more or less like this:
"There is no uniqueness in decimal expansions of real numbers, specially if one wishes to compare numbers (and their decimal expansions) extremely close of one another."

Is this is correct, shouldn't this imply that it is kind of useless to assign to intervals adjectives as closed or opened?

It seems that one of the extremes of an interval may be proved to be equal to any one of its two immediate neighbors.

Best wishes,

DaTario

2. Feb 28, 2016

### Samy_A

Can you give an example of "one of its two immediate neighbors" for some real number?
Take 1 as an example. What are the "immediate neighbors of 1?

3. Feb 28, 2016

### Staff: Mentor

This isn't a theorem -- it's just a statement about two representations of a single number.

1 and 0.999... are not merely extremely close -- they are the same number
No. For example, the half-open, half-closed interval [0, 1) contains .9, .99, .999, and other numbers very close to 1, but does not contain 1.
No that's not true at all. The interval of my example has a least upper bound of 1, but 1 is not a member of this interval. Unlike integers, real numbers do not have immediate neighbors. Between any two real numbers, there are an infinite number of other real numbers.

4. Feb 28, 2016

### HallsofIvy

Your basic mistake is in thinking that there is such a thing as an "immediate neighbor" for a number.

5. Feb 28, 2016

### jbriggs444

Yet another take on the matter...
There is uniqueness in the decimal expansions of real numbers in the sense that almost all real numbers have unique decimal expansions. The ones that do not are those whose decimal expansions end in an infinite string of zeros or nines. For instance, 1/3 has a unique decimal expansion.

There are only countably many real numbers whose expansions are not unique. There are uncountably many real numbers with unique expansions.

Last edited: Feb 28, 2016
6. Feb 28, 2016

### DaTario

Thank you all,

I would like to express my gratitude for the answers.

Is there a real problem with calling the problem of proving the statement $1= 0.9999999....$ a theorem?

I guess I understand the problems with the concept of immediate neighbors in real numbers. But this famous problem evokes an apparent identity between decimal expansions (or numbers !?) that are good candidates to be immediate neighbors. I (although smiling here now) would not expect any other number to live between 1 and 0.9999999.....

Because it involves the "almost all" clause, do you think we should treat this uniqueness as partial?

Even so, it seems to me that it may be a sufficient reason to treat the uniqueness as partial.

As a final remark in this post, I would say that if we are adopting an identification between two near but different points on the real axis, the fact that whether these numbers are expanded with nines or not seems to be of less importance. If I imagine an infinite line of standing dominoes and call this line 0.99999....., being each nine one domino standing, I would not accept an identification of this system with an infinite line of fallen dominoes, which I would call 1.000000......, being each zero one domino fallen (and the 1 would be the unique domino lying on a perfect horizontal).

I am sorry, dear coleagues, if I am pushing too far in thinking in a non rigorous manner. This is a very interesting and misterious subject.

Best wishes,

DaTario

7. Feb 28, 2016

### Staff: Mentor

Why would you think it should be a theorem? If it should, then I guess we would need similar "theorems" to show that 2 = 1.999..., 3 = 2.999..., .5 = .4999..., and on and on.
Good, because there aren't any.
If something is unique, there is only one of it. An attribute either is unique or isn't -- it can't be partially unique, any more than someone can be "partially pregnant."
Your analogy is inapt. There is no difference between .999... and 1.000... None.
If you subtract .999... from 1.000... you are apparently thinking that there is a 1 digit out there somewhere.
Can you tell me its exact position?

8. Feb 28, 2016

### WWGD

But the rule whereby two Real numbers are considered equivalent to each other is not necessarily the same rule you would use to consider arrangements of dominos to be equivalent to each other.

9. Feb 28, 2016

### Staff: Mentor

The endless repetition of nines is related to our commonly used decimal system. The same situation exists in other number systems. For example, in binary (base-2), .111...2 = 12. The number on the left is $\frac 1 2 + \frac 1 4 + \frac 1 8 + \dots$, an infinite sum that can be shown to converge to 1.

Similarly, in base-3, 1.222...3 = 2.000...3. In any given number system, if the fractional part contains an endless string of whatever the highest digit is in that number system, that number has another representation that has an endless string of 0 digits.

10. Feb 28, 2016

### DaTario

Sorry, I was not aware that the number of theorems above the inverted pyramid of an axiomatic structure should be small.

Theorem may be defined as "a formula, proposition, or statement in mathematics or logic deduced or to be deduced from other formulas or propositions"
(http://www.merriam-webster.com/dictionary/theorem)
So, I will not take your saying, in this specific point, as a serious contribution, sorry.

I understand this professorial tone of yours, but from what jbriggs444 has said the question of this uniqueness was disputable in some sense.

I guess we both believe in that famous demonstration of equality.

11. Feb 28, 2016

### Staff: Mentor

You may take this serious. Your claim to call the equality that simply arises from the fact that our descriptions with decimals are not perfect a theorem is not serious.

No, it is not.

12. Feb 28, 2016

### jbriggs444

Do all real numbers have unique decimal representations? Certainly not. Do some real numbers have unique decimal representations? Assuredly so.

The point that I think Mark44 is trying to make is that there is not a one to one correspondence between real numbers and decimal expansions. If you look at one decimal expansion and note that there is another decimal expansion that is the next one, that does not demonstrate that given one real number you can find a next one. Nor does it serve to somehow make the real numbers fuzzy so that open and closed intervals become meaningless.

13. Feb 28, 2016

### DaTario

So, Mark44, if we are allowed to use different and "well chosen" basis for different sets of numbers in the real axis in such a way that, after several sessions of demonstrations, perhaps we could demonstrate that each point in the real axis may be seen as a number having in its decimal expansion an infinite number of whatever the highest digit of the system used in that turn. This creates a system of infinite number (Aleph one) of demonstrations for each point in the real axis. In these demonstrations, each point would accept two decimal expansions.

My point (which I am suffering here to defend ) is not only to put this notion in check, but to connect it with our language in mathematics that uses the terms open and closed to intervals.

Best wishes.

14. Feb 28, 2016

### DaTario

If you demand students to prove that between 1 and 2 there is no integer then it is a theorem. Don´t you agree?

15. Feb 28, 2016

### jbriggs444

There only countably many possible bases and in each one, only countably many real numbers with dual expansions in that base. That is not Aleph one. That is Aleph null.

In any case, the number of representations that exist for a real number has nothing at all to do with whether or not that real number is distinct from every other real number , whether it has adjacent neighbors or whether it can be the endpoint of an open interval.

16. Feb 28, 2016

### Staff: Mentor

No, it is by construction. You can write 1 in decimal digits in two ways. So what? It only means that decimal digits aren't a perfect way to write them. That's all. You cannot write π at all in decimal digits. Does it mean it doesn't exist?

17. Feb 28, 2016

### DaTario

Sorry, I guess I haven´t made myself clear, I was also referring to choose $\sqrt{2}$ for example as unity.

18. Feb 28, 2016

### DaTario

I am not addressing anything like this. I am not arguing that a number, expressed in some manner, doesn´t exist.
There are representations that, used with some numbers, provide an infinite string of digits or provide a much less compressible package of information in that basis.

19. Feb 28, 2016

### DaTario

Where did you find that if some statement admit proof by construction it cannot be called a theorem?

20. Feb 28, 2016

### Staff: Mentor

Ok, let's pretend, that "There is no integer between 1 and 2." is a theorem.
Please start and define the term "integer" as well as "1" and "2". That is needed before we can think of a proof.