What is the largest number less than 1?

[Organic]

Hi russ_watters,

Please show me what have you found in my work, which is not self-consistent.

Thank you.


Organic

 

[russ_watters]

Originally posted by Organic
Hi russ_watters,

Please show me what have you found in my work, which is not self-consistent.

Thank you.


Organic

1.000...1 is not self consistent. We've been over this before though - there can't be an infinite amount of zeros before the 1 by the definition of infinity.

 

[jcsd]

Organic is the number 0.999...9 rational or irrational?

 

[BigRedDot]

Let us assume, for the sake of contradiction, that 0.999... and 1 are distinct real numbers. Then, since the rational numbers are dense in the reals, it follows that there must exist a rational number q such that 0.999... < q < 1.

Organic, would you please demostrate to us explicity a rational number which lies between 0.999... and 1?



I should also mention that formal developments of the real numbers do not rely on decimal expansions, per se. Indeed, Strichartz points out why such a development is often avoided: "However, it has two techincal drawbacks. The first is that the decimal expansion is not unique: 0.999... and 1.000 are the same number." One favorite method to define the real numbers is in terms of equivalence classes of Cauchy sequences. That is, real numbers are idetified with sequences of rational numbers that do not converge to rational numbers (yet satisfy the Cauchy criterion), but more than that, more than one sequence is identified with the same real number, since many sequences can belong to the same equivalence class. This is fine however, since members of equivalence classes are, well, equivalent.

In a certain sense, your position is tenable. People have certainly investigated non-standard analysis, and in particular Abraham Robinson founded a logically satisfactory basis for the real numbers using "bonafide" infintesimals. However, this kind of non-standard analysis is actually harder to justify (it requries us to assume more) and gains us nothing in what we can prove. I'm also pretty sure you didn't have any of this in mind however.

Now,I happen to believe that mathematics is by and large, if not entirely, a human creation, whose sole justification is pragmatic sanction. So the whole question is somewhat meaningleess. You are free to believe whatever you want about things you call numbers. The only important questions are: is your system useful and is it not obviously inconsistent. I don't know about your real number system, but the real number system where 0.999... does equal 1.000... (the one that is almost universally used and that all the analysis textbooks describe) has answered both those questions in the affirmative long, long ago.

 

[Lonewolf]

I can't believe this thread is still going...

 

[Hurkyl]

Even in the hyperreals, there is no largest number less than 1.

 

[R12]

Is "bumping" (posting on threads that haven't been posted on in a while) allowed here? Didn't find it in the rules.

Okay, so in theory, .999... = 1. This could mean that .888... = 1 as well. After all, it is only .111... different.

But wait, .111... supposedly equals 1, so .888... would be 1 (.999..., to clarify.) minus 1 (.111...) But wouldn't that be zero?

Also, he just asked how to represent the greatest number less than 1, not if .999... was a real number. So it seams as if

lim x
x-->1

was the answer, as PrudensOptimus said.

Forgive me if I'm wrong. Trying to wrap my head around this is hard, considering I am only 16.

 

[gb7nash]

Okay, so in theory, .999... = 1. This could mean that .888... = 1 as well.

Not quite. There's a multitude of reasons, but one reason .888... is not equal to 1 is that there exists a number between .888... and 1 that is not equal to .888... or 1.

This is a property of the real numbers that you'll learn in real analysis:

If two real numbers are distinct, there are an infinite number of real numbers between them. The contrapositive* of this statement is that if there are not infinite number of real numbers between two different numbers, then they are, in fact, the same number.

For .999... and 1, there's no number that's between them, so they're the same. There's a ton of other reasons too why this is true but this is one of them.

 

[Phrak]

Is "bumping" (posting on threads that haven't been posted on in a while) allowed here? Didn't find it in the rules.

Not that I know of, and it's a very good thread.

Okay, so in theory, .999... = 1. This could mean that .888... = 1 as well.

No. Within the system of numbers within the discussion of the year 2003, where .999...=1, then 0.888...=0.889.

 

[HallsofIvy]

Is "bumping" (posting on threads that haven't been posted on in a while) allowed here? Didn't find it in the rules.

Okay, so in theory, .999... = 1. This could mean that .888... = 1 as well. After all, it is only .111... different.

So? 0.1111... is not 0. 0.8888... is, by definition of the "decimal representation of the real numbers" the infinite sum
$$\frac{8}{10}+ \frac{8}{100}+ \frac{8}{1000}+ \cdot\cdot\cdot$$
which is the "geometric series"
$$\sum_{n=1}^\infty \frac{8}{10^n}= \sum_{n=1}^\infty 8(0.1^n}$$

There is a simple formula for the sum of a geometric series:
$$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$$

Here, the sum starts at n= 1 rather than n= 0 but that is easy to fix- with n= 0 $8(.1^n)= 8(.1^0)= 8$ so we only have to subtract off the missing first term, "8".
$$\sum_{n=1}^\infty 8(0.1^n)= \frac{8}{1- 0.1}- 8= \frac{8}{.9}- 8$$
$$= \frac{80}{9}- 8= \frac{80}{9}- \frac{72}{9}= \frac{8}{9}$$
which is NOT equal to 1.

But doing exactly the same thing with 0.9999... rather than 0.88888... gives
$$\sum_{n=1}^\infty 9(0.1)^n= \frac{9}{.9}- 9= \frac{90}{9}- 9= 10- 9= 1$$

But wait, .111... supposedly equals 1, so .888... would be 1 (.999..., to clarify.) minus 1 (.111...) But wouldn't that be zero?

No, you have simply misunderstood everything that was said here.

Also, he just asked how to represent the greatest number less than 1, not if .999... was a real number. So it seams as if

lim x
x-->1

was the answer, as PrudensOptimus said.

Forgive me if I'm wrong. Trying to wrap my head around this is hard, considering I am only 16.

As anyone who has taken basic Calculus or precalculus knows, $\lim_{x\to 1} x= 1$ so it is NOT "less than 1".

And, of course, 0.99999... is a real number- any number written in decimal notation like that is a real number.

 

[Dragonfall]

The largest number less than 1 is $$\frac{9,007,199,254,740,991}{9,007,199,254,740,992}$$. This is a fact.

 

[TylerH]

$$1 > \frac{9,007,199,254,740,992}{9,007,199,254,740,993 } > \frac{9,007,199,254,740,991}{9,007,199,254,740,992 }$$

 

[HallsofIvy]

I suspect that DragonFall meant that as a joke!

 

[TylerH]

I would hope so! :P I was just going along. I should have added a sarcastic "This is a fact." at the end to make it clearer, I guess.

 

[Dragonfall]

No computer scientists here? Tough crowd, tough crowd.

Seriously though, if I'm not mistaken, the number above should be the largest number less than one expressible in double precision. $$1-2^{-53}$$

Or one minus "machine epsilon".

 

[HallsofIvy]

I was under the impression that "machine epsilon" could vary from one processor to another.

 

[TylerH]

Dragonfall's going by the IEEE standard for a double precision floating point. So, even though there are other machine epsilons, the one he used is the standard.

 

[Stocko]

A simple proof that 0.999... = 1 (for ... read "recurring". I'm new here and haven't figured out how to do symbols yet).

1/9 = 0.111...

Therefore 9 x 1/9 = 0.999...

But 9 x 1/9 = 1

So 0.999... = 1.

My small, modest and very late contribution to this remarkably long lived thread.

 

[camilus]

This question is rather simple, you just have to be more precise.

Analyze "What is the largest number less than 1?" in terms of set theory, you can't go wrong. Greatest number less than 1 in:

1. N? none because 1 is the smallest element in the set N.
2. Z? 0
3. R? none because the set of Reals is uncountable.

 

[n.karthick]

This question is rather simple, you just have to be more precise.

Analyze "What is the largest number less than 1?" in terms of set theory, you can't go wrong. Greatest number less than 1 in:

1. N? none because 1 is the smallest element in the set N.
2. Z? 0
3. R? none because the set of Reals is uncountable.

What about in the set of rational numbers Q which is countable

 

[Nebuchadnezza]

Sorry for not reading through all of the pages...

But can't you simply prove that $$0.999...$$ is equal to $$1$$ in this matter?

$$0.999...$$ can be expressed as a geometric series. As such

$$\frac{9}{{10}} + \frac{9}{{{{10}^2}}} + \frac{9}{{{{10}^3}}} + ... + \frac{9}{{{{10}^n}}}$$

An infinite geometric series is an infinite series whose successive terms have a common ratio.

The formula for the sum of a infinite geometric series is as follows: $$\frac{{{a_1}}}{{1 - k}}$$ where $$k$$, is the ratio between the terms and $$a_1$$ is the first number in the sequence. Plugging in everything we have. We now get.

$$= 0.999...$$

$$= \frac{9}{{10}} + \frac{9}{{{{10}^2}}} + \frac{9}{{{{10}^3}}}+...+\frac{9}{{{{10}^n}}}$$

$$k = \frac{{{a_n}}}{{{a_{n - 1}}}} = \frac{{\frac{9}{{{{10}^n}}}}}{{\frac{9}{{{{10}^{n - 1}}}}}} = \frac{9}{{{{10}^n}}}:\frac{9}{{{{10}^{n - 1}}}} = \frac{9}{{{{10}^n}}} \cdot \frac{{{{10}^{n - 1}}}}{9} = \frac{{{{10}^{n - 1}}}}{{{{10}^n}}} = {10^{n - 1}} \cdot {10^{ - n}} = {10^{\left( {n - 1} \right) - n}} = {10^{ - 1}} = \frac{1}{{10}}$$

$${a_1} = \frac{9}{{10}}$$

$$S = \frac{{{a_1}}}{{1 - k}} = \frac{{\frac{9}{{10}}}}{{1 - \frac{1}{{10}}}} = \frac{9}{{10}}:\frac{9}{{10}} = \frac{9}{{10}} \cdot \frac{{10}}{9} = 1$$

 

[camilus]

What about in the set of rational numbers Q which is countable

$$\mathbb{Q}$$ is a densely ordered set, which means that for any x and y such that x < y, the exists a z in $$\mathbb{Q}$$ where x < z < y. So there in no greatest number less than 1 because whatever number x you give me when y=1, I can always find a z.

 

[olivermsun]

$$\mathbb{Q}$$ is a densely ordered set, which means that for any x and y such that x < y, the exists a z in $$\mathbb{Q}$$ where x < z < y. So there in no greatest number less than 1 because whatever number x you give me when y=1, I can always find a z.

Just to connect this to the discussion above, also notice that every number in the series 0.9, 0.99, 0.999, 0.9999, ... is a rational number, so there are your arbitrarily close rationals right there.

 

[Thinker8921]

Could someone please clear a doubt I have with infinity being taken as a value.
I would like to start from the beginning to be clear. The sum of a geometric series is given by:
S=(rB-A)/r-1
Where S=sum of the series, r= ratio between terms, B=last term, A=first term.
Now B=A(r^N-1). So,
S=[{rA(r^N-1)}-A]/r-1
i.e, S=[{(r^N)-1}A]/r-1
Now in the case of the infinite geometric series 0.99… ,
N= infinity A= 0.9, r= 0.1
Then there is as the 1st term 0.9, 2nd term is 0.09, 3rd is 0.009 etc. Then the sum of this series will be 0.99...
Also in the numerator is 1-(r^N). This is instead of (r^N)-1. And in the denominator it is 1-r instead of r-1. This is because r is less than 1.
So finally there is:
S=[{1-(r^q)}A]/1-r
Where q represents infinity.
Now since there is infinity as the power of r, this is seen to become:
S=A/1-r
But the problem is it correct to take infinity as a value and approximate like this in the case of 0.99… ?
I thought infinity never had a definite value, that it was instead just an ever-rising concept. Could this 'problem' make the 0.99… equal 1?
How can 0=1/q, so q*0=1 be explained then? (q= infinity)
*I don't know how to write the equations with symbols yet. So sorry for the messy work.

 

[olivermsun]

The infinite sum (if it converges) is defined as the limit of the partial sums
$$\sum_{n=1}^\infty{a_n} = \lim_{N\rightarrow\infty} \sum_{n=1}^N{a_n}.$$

Thus in your expression you take the limit where the exponent goes to infinity and call that the value of the infinite series. This is different from stating that the exponent is actually infinity.

 

[Thinker8921]

Thanks for clearing that up. I got it now.

 

[dalcde]

If you use hyperreal numbers, it might make sense that the largest number is (1-epsilon), where epsilon is an infinitesimal. However, according to my understanding, there isn't a one unified infinitesimal. An infinitesimal can be smaller than another in some weird sense.

 

[Hurkyl]

If you use hyperreal numbers, it might make sense that the largest number is (1-epsilon), where epsilon is an infinitesimal. However, according to my understanding, there isn't a one unified infinitesimal. An infinitesimal can be smaller than another in some weird sense.

Hyperreals satisfy the same theorems (interpreted internally) that the reals do -- there is no largest hyperreal smaller than 1.

e.g. (1 - epsilon) < (1 - epsilon / 2) < 1