Does .999~ equal 1?

1. Jun 8, 2003

Alex

I was just wondering

2. Jun 8, 2003

Integral

Staff Emeritus
Yes, if you mean .999...

http://home.attbi.com/~rossgr1/proof2.PDF [Broken]

Last edited by a moderator: May 1, 2017
3. Jun 8, 2003

Kalimaa23

My views are slightly different here.

If you consider 0.999999... you must specify what metric space you are working in. If you have a complete one (likeR ), 0.9999... is simply a number in R. Although it is arbitrarily close to one, it does not equal one.

I guess it really depends on your definition of equality. If you take it as x-y<E => x=y, but I think my analysis professor would shoot me if I were to write it up on the blackboard next week.

4. Jun 8, 2003

Integral

Staff Emeritus
Dimitri,
Remember that .999... represents an infinite number of 9s not an increasing number but an infinite number. Of course since nothing else was specified I am speaking of the real number system.

Can you specify the number between .999... and 1?

Did you make an effort to look at my proof?

5. Jun 8, 2003

Kalimaa23

I did, and it relied on the statement that if you can approach something by less then any epsilon, it equals it.
But any analysist will tell you that this is only indeed true for a limit, and not for any quantity.

For example, by that reasoning an real number equals a rational number, because every epsilon interval has infinitely many rational numbers in it.

Anyone have a third opinion?

6. Jun 8, 2003

KLscilevothma

not a rigor presentation, but quite interesting

1/9 = 0.111111......
2/9 = 0.222222......
.
.
.
8/9 = 0.888888......
9/9 = 0.999999......= 1

7. Jun 8, 2003

Hurkyl

Staff Emeritus
Let x = 0.99999...
Integral proved:

[uni]&epsilon; > 0: 1 - &epsilon; < x < 1 + &epsilon;

The rest of the proof is by contradiction. Suppose x [x=] 1.

From the axioms of an ordered ring and the fact (1/2) exists in R, one can prove:

x < (1/2)(1 + x) < 1

Now, if we let &epsilon; = 1 - (1/2)(1 + x) > 0
Then we have x < 1 - &epsilon;. This contradicts Integral's proof that [uni]&epsilon; > 0: 1 - &epsilon; < x

Therefore, our assumption that x [x=] 1 was incorrect, and thus x = 1.

This method of proof predates limits, and is fully rigorous. It predates limits, and was essentially known to the ancient Greeks as the Method of Exhaustion.

The theorem explicitly stated is:

If x and y are real numbers, and:
[uni]&epsilon; > 0: y - &epsilon; < x < y + &epsilon;
Then we may conclude x = y.

and this is something that actually gets used in analysis; sometimes it is much easier to prove the hypotheses of the above theorem than to set up and prove the appropriate limit.

Incidentally, IMHO, everything is much more clear if we appeal to the actual formula for mapping decimals to reals:

0.9999.... = &Sigmai=1..&infin; 9 * 10-i

And we know the sum of this infinite geometric series to be:
= (9 * 10-1) / (1 - 10-1) = 1

8. Jun 8, 2003

jammieg

To me 10,000 equals 1 is I'm just considering the penny differences in price of buying 2 houses, I mean 9 and 10 are not the same number but if it's practical to simply use two 10's then same enough, incidentally this is probably why I'm not very good with math.

Last edited by a moderator: Jun 8, 2003
9. Jun 8, 2003

HallsofIvy

Dimitri wrote:
Yes, my opinion is that you had better go have a long talk with your analysis teacher.

If it is true that |x-y|< epsilon for ALL epsilon, then x= y.

Yes, it is true that, given any real x and positive epsilon, there exist an infinite number of rationals, y, such that |x-y|< epsilon.
However, if they exist a rational, y, such that |x-y|< epsilon for ALL positive epsilon, then x is rational and, in fact, x= y.

The DEFINITION of the "decimal expansion" of a real numbers, say
a.d1d2d3... is that it is the infinite sum a+ d1/10+ d2/100+ d3/1000+... which is, in turn, the limit of the sequence a, a+ d1/10,
a+ d1/10+ d2/100, ...

In particular, the DEFINITION of a decimal expansion such as
0.9999.... is 0+ 9/10+ 9/100+ 9/1000+ ... which is a geometric series and easily summable to 1.

0.999... = 1 in the standard definition of the real numbers.

10. Jun 8, 2003

Alex

I see the proofs and they make sense, but it just seems odd that an infinite number can be equal to a rational number like that. Oh well looks like its just something I'll have to get used to. Thanks guys.

11. Jun 8, 2003

elephant

I think that it depends on the way you look at the expressions.

0.999... = 1 - 1 / infinity

12. Jun 9, 2003

Kalimaa23

Integral, you are off course completely correct. I'm sorry, but I was confusing the matter with some other proof.
What I stated about a limit is true in general. But off course [rr] is a complete metric space, so off course the limit of 0.999... exist and the sum of the aforementioned series is 1.

I blame all that chemistry they make me take in the first year, it has a way of messing with your brain

Discussion closed I guess , 0,999.... DOES equal one.

13. Jun 9, 2003

jonnylane

as a physicsist, 0.999..... does equal 1 to all intents and purposes.

Some of my lecturers used to use some really bizarre approximations.

"look, that 4.3*10^-5 is about 10^-5"

the list goes on

approximations make life worth living

14. Jun 9, 2003

HallsofIvy

To Dimitri Terryn: Yeah, chemistry will do that to you- especially if you inhale!

To jonnylane: As a mathematician, 0.999... IS 1 there is no "to all intents and purposes" about it, it's not an approximation.

15. Jun 9, 2003

jonnylane

I understand.

I was providing a physical view on things.

Physicists make clumsy mathematicians and vice versa

16. Jun 9, 2003

HallsofIvy

What??? Mathematicians make clumsy physicists? How dare you!

Ooops, I didn't mean to knock that cyclotron over!

17. Jun 10, 2003

Dissident Dan

You should realize that infinite, repeating numbers such as 0.9999 are only representations of division in number set in a give base (such as base-10 that we are all so familiar with), so it should not be such a surprise that it ends up being equal to one.

For example, how do you get 0.333......? Divide 1 by 3. You keep continueing with the digits because the same remainder keeps coming up. However, in the base-3 number system, one third is just 0.1

18. Jun 10, 2003

jonnylane

oops, there goes that abacus.

19. Jun 10, 2003

plus

Of course Newton was a mathematician, but he was also a physicist.

20. Jun 10, 2003

Kalimaa23

I just came back from my chemistry exam.

I think that mathematicians can be good physicists and vice-versa, if only there is proper motivation and interest.

My university still has the combined first year mathematics/physics, and I'm very happy they keep this up, even though there is pressure to make the physics program less "abstract"

I will say one thing though, physicists make clumsy chemists

-Dimi