# RAA Proof that .999'=1

1. Mar 10, 2005

### Pyrovus

Hi everyone! Apologies to those annoyed by my starting one of those .999'=1 threads :)

This is a little reductio ad absurdum proof that .999'=1 that I came up with, and I'm just wondering what everyone thinks with regards to the soundness of it.

Assume .999' < 1
Now, for any 2 real numbers, a and b, such that a < b, a number (a+b)/2 exists such that a < (a+b)/2 < b
So, letting a=.999' and b=1:
(a+b)/2 = (.999' + 1)/2
= .999'/2 + 1/2
= .4999' + .5
= .999'
And, since a < (a+b)/2:
.999' < .999'
So, .999' is less than itself!
Adding an arbitary constant to both sides:
.999' + c < .999' + c
Hence, no number is equal to itself.

2. Mar 10, 2005

### HallsofIvy

Staff Emeritus
Not bad. The weaknesses (not errors) are in the use of ".999'/2= .4999' " and
".4999'+ .5= .9999' " those require calculations on an infinite number of digits that, while valid, are exactly what people, who deny that 0.999'= 1, would object to.

3. Mar 10, 2005

### Icebreaker

0.999... / 2 will result in a decimal such that the LAST digit is always 5. i.e, 0.9/2 = 0.45, 0.99/2 = 0.495, 0.999/2 = 0.4995, for any FINITE number of decimals. I don't know what happens if 0.9 periodic.

4. Mar 10, 2005

### Zurtex

Think of it as 0.495 as: 4/10 + 4/100 + 5/1000. So in general we have that the final term in the decimal representation is like this:

$$5 \cdot 10^{-n}$$

Rewriting:

$$5\frac{1}{10^n} \leq \frac{1}{n} \, \, \, \forall n(>2) \in \mathbb{N}$$

And therefore from the Archimedean property we get that the final term of the sequence being:

$$\lim_{n\rightarrow\infty} 5\cdot 10^{-n} = 0$$

5. Mar 10, 2005

### Icebreaker

I guess if you wanna prove 0.999...=1 completely, you will have to include a second part where you assume 0.999...>1 and, using the same method, prove the absurdities; therefore 0.999... must equal to 1. Kind of like the squeeze theorem.

6. Mar 10, 2005

### Integral

Staff Emeritus
The proof I favor uses the nested interval theorem (your squeeze theorem?).

It is easy to show that both :
$$1 - 10^{-n }< .999... < 1+ 10^{-n}$$
and
$$1 - 10^{-n }< 1 < 1+ 10^{-n}$$

hold for all n>0

Apply the Nested interval theorem and you are done.

7. Mar 10, 2005

### dextercioby

I see these threads regarding recurring decimal as purely useless.
$$0.(9)\equiv 1$$ follows from the definition of the recurring simple decimal
$$0.(a)=:\frac{a}{9}$$ by plugging a=9...

No sqeeze of any theorem,jus a plain simple DEFINITION...

Daniel.

8. Mar 10, 2005

### Integral

Staff Emeritus
Spoken like a true Physicist.

9. Mar 10, 2005

### NateTG

Ah, but can you show that
$$0.\bar{9}$$
is a real number in the first place?

10. Mar 10, 2005

### Icebreaker

Mmm... if you can show that $$0.\bar{9} = 1$$, then $$0.\bar{9}$$ is a real number.

11. Mar 10, 2005

### dextercioby

Divide 4/9...Is the result 0.(4) a real number...?Then inverse the notation into the definition:

$$\frac{4}{9} =:^{not} 0.(4)$$...

Then evidently

$$0.(4)=:\frac{4}{9}$$...

IIRC,R^{*} is closed under the division...

Daniel.

12. Mar 11, 2005

### matt grime

Actually, taking Daniel's approach isn't just "the physicists". The real numbers aren't after all decimals, but a set with certain properties. If we offer the decimals with the equivalance of the nines to a string with zeroes then all we are doing is producing a model of the real numbers. So another dismissal of people's objections is simply if 0.9... and 1 aren't equal it simply isn't a model of the axioms, as it fails the archimidean principle.

13. Mar 11, 2005

### HallsofIvy

Staff Emeritus
Yes, by using a DEFINITION of "real number" which is the best way to prove 0.9999... = 1 in the first place. (The crucial point that people who claim 0.999... is NOT 1 alway miss is that you have to DEFINE the real numbers before you can sensably talk about them.)

For this purpose, the simplest definition of "real number" to use is:

Consider the set of all non-decreasing sequences of rational numbers with upper bounds. We say that two such sequences {an}, {bn} are equivalent if and only if the sequence {an- bn} converges to 0. The "real numbers" are the equivalence classes of rational numbers corresponding to that equivalence relation and we can define addition and multiplication in obvious way. There is an obvious 1 to 1 function from the rational numbers to these equivalence classes (the rational number r corresponds to the class containing the sequence r, r, r,...) so that we may regard the rational numbers as being a subset of the real numbers.

Now: it is clear that the sequence 0.9, 0.99, 0.999, ... is a non-decreasing sequence and has 1 as an upper bound. Therefore, it is in one of these equivalence classes and corresponds to a real number.

Further more, since for {a_n}= 1, 1, 1, ... and {bn}= 0.9, 0.99, 0.999... {an- bn}= 0.1, 0.01, 0.001, ... converges to 0, 0.999... and 1 belong to the same equivalence class and so are the same real number: 0.999... = 1.0.

Last edited: Mar 11, 2005
14. Mar 14, 2005

### nnnnnnnn

or could you use the binomial thm for a more general case?