What is the largest number less than 1?

[Magnus]

How can you represent the largest # that is less than 1?

 

[PrudensOptimus]

Originally posted by Magnus
How can you represent the largest # that is less than 1?

lim x
x-->1

 

[Magnus]

Can you represent that in decimal form?

 

[one_raven]

Wouldn't .9 (with a line over the 9, sorry, I haven't read the "How to post math functions" thread) be sufficient?

__
.9

 

[Magnus]

The line over it means infinity.

So couldn't you also say .999... ?

 

[Hurkyl]

Among the real numbers (or the rational numbers, or the irrational numbers) there is no largest number less than 1.

Among the integers, 0 is the largest number less than 1.

 

[Magnus]

How can there be no largest # less than 1? that doesn't make any sense.

 

[Hurkyl]

Why would you think there is a largest?

Here's a short proof there isn't a largest number less than 1:

Assume that there is a largest number less than 1. Let's represent this number by \inline{x}.

Now, consider the number \inline{y=(1+x)/2}

First, notice that \inline{y>x}:

<br /> y = \frac{1+x}{2} &gt; \frac{x+x}{2} = x<br />

Now, notice that \inline{y&lt;1}:

<br /> y = \frac{1+x}{2} &lt; \frac{1+1}{2} = 1<br />

So consider carefully what we have proven:

If we assume there is a largest number less than 1, we can find a number less than 1, yet larger than the largest number less than 1.

That is a contradiction; our assumption that there is a largest number less than 1 must be false.

 

[one_raven]

Hurkyl,
I understand your reasoning, and it does make complete sense...

I do have a question, however..
Is there a number greater than .999... and less than 1?
How would you represent that number?

 

[chroot]

Originally posted by one_raven
Hurkyl,
I understand your reasoning, and it does make complete sense...

I do have a question, however..
Is there a number greater than .999... and less than 1?
How would you represent that number?

The number \inline{0.99\overline{9}} is equal to one.

The limit \inline{\lim_{x \rightarrow 1} x} also equals one.

Hurkyl is correct; there is no largest real number less than one. No matter how many nines you put in a row, I can make a number with more. In the limit as the number of nines reaches infinity, the resulting quantity is one.

- Warren

 

[Magnus]

I can't believe that there is no largest # less than 1. There has to be, in theory.

If you can say that .999... = 1 then how can you not in mathamatics represent the largest # that is less than 1?

the .999... = 1 rule only works because you NEVER reach the end of infinity.

I like sound laws and such, and I love math and physics... and at the same time am a hard core programmer at heart. Logic is key to me.

I could never write .999... = 1, I could say 1 = 1 and .999... = .999... but not 1 = .999...

I would say that .999... is the largest # less than 1 because as you go out there in decimal places whatever place your at you can just add a .----1 to that to achieve your value of 1... but yea, you'll never even approach infinity cause it extends forever. Damn you infinity!

 

[krab]

...I love math ...

...I would say that .999... is the largest # less than 1 because...

You may love math, but you don't know what it is if you don't realize that math is based on rigor. If you want to hold on to your opinion about there being a largest number less than 1, then you must find a fault with Hurkyl's proof.

If you don't understand Hurkyl's proof, here's another thing to think about. Consider the mapping

y={1\over 1-x}

and for x, use the real number interval [0->1) (This means, all the real numbers from 0 to 1, but excluding 1.) This interval gets mapped to [1->infinity). You will see that there being no largest real number x less than 1 is analogous to saying there is no largest real number y. (I am assuming you agree that there is no largest real number.)

 

[jcsd]

All the proofs so far have been perfectly adequate, but here's another one:

for a nunber x where:

0 &lt; x &lt; 1

we know that:

0 &lt; \sqrt{x} &lt; 1

and

x &lt; \sqrt{x}

Lets say that there is a largest number between 0 and 1, what is it's sqaure root? if it greater than it's square root, it is greater than 1, if it is equal to it's square root it is equal to 1 and if it's square root is greater than itself then we have generated a number that is larger than the largest number less than 1 so it can't be the largest number less than 1!

 

[HallsofIvy]

I like sound laws and such, and I love math and physics... and at the same time am a hard core programmer at heart. Logic is key to me.

I could never write .999... = 1, I could say 1 = 1 and .999... = .999... but not 1 = .999...

I would say that .999... is the largest # less than 1 because as you go out there in decimal places whatever place your at you can just add a .----1 to that to achieve your value of 1... but yea, you'll never even approach infinity cause it extends forever.

Logic is key? Your last sentence has no logic in it at all. You seem to be confusing logic with handwaving. In particular what is your DEFINITION of .999...?

 

[theEVIL1]

Originally posted by Magnus
The line over it means infinity.

So couldn't you also say .999... ?

Sorry, the largest WHOLE number less than 1 is, by definition, 0.
.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 etc. infinitum is not a whole number

 

[jcsd]

Originally posted by theEVIL1
Sorry, the largest WHOLE number less than 1 is, by definition, 0.
.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 etc. infinitum is not a whole number

0.9... recurring IS a whole number it is equal to 1.

 

[theEVIL1]

Originally posted by jcsd
0.9... recurring IS a whole number it is equal to 1.

um.sure it is...how then can you explan that .999 infinitum will NEVER reach 1?
Must be that new math

 

[jcsd]

Originally posted by theEVIL1
um.sure it is...how then can you explan that .999 infinitum will NEVER reach 1?
Must be that new math

No it's very old maths, most peope are taught the following sometime during their secondary eductaion:

x = 0.99999...

=>

10x = 9.9999... =>

10x -x = 9x = 9 =>

x = 1

 

[Hurkyl]

I can't believe that there is no largest # less than 1. There has to be, in theory.

Why?

If you can say that .999... = 1 then how can you not in mathamatics represent the largest # that is less than 1?

I don't see the connection.

the .999... = 1 rule only works because you NEVER reach the end of infinity.

No, the rule works because it is a logical consequence of the definitions.

I could never write .999... = 1, I could say 1 = 1 and .999... = .999... but not 1 = .999...

Can you write \inline{1/2=2/4}? \inline{1} and \inline{0.\bar{9}} are two different representations of the same number, just like \inline{1/2} and \inline{2/4}.

I would say that .999... is the largest # less than 1 because as you go out there in decimal places whatever place your at you can just add a .----1 to that to achieve your value of 1... but yea, you'll never even approach infinity cause it extends forever. Damn you infinity!

This is where your problem lies; you are imagining \inline{0.\bar{9}} as some sort of process instead of as a number.

While it is certainly true you can get the value \inline{0.\bar{9}} through a process (such as taking the limit of \inline{0.9, 0.99, 0.999, \ldots}), \inline{0.\bar{9}} is a number. It does not change, it does not approach anything; it is simply a number.

 

[HallsofIvy]

...how then can you explan that .999 infinitum will NEVER reach 1?
Must be that new math

No one needs to explain it- it's not true.

.999 "infinitum", by which I presume you mean the infinite sequence of 9s, is, by definition, the infinite series .9+ .09+ ...+ 9(.1)ⁿ+... which can be proven to be exactly equal to 1 (it's a very easy geometric series- you should have learned how to sum those in secondary school).

 

[Organic]

Hi Magnus,

Let us look at the opposite side of this problem.

x = 0

Can you find the smallest number, which is bigger then x?

 

[Magnus]

Excellent point.

I do believe you guys.

The way I see .999... is basically.. a number that extends forever, it starts in the tens, goes to hundereds, thousands, etc. etc.. each time becoming closer and closer to 1. I see it becoming infinitely close to 1 as itself extends infinitely. Its just so hard to picture a # that starts off not as 1 become one just because it has no end.

IE: if you line it up.
1.000
0.999...

1 != 0
. = .
0 != 9
0 != 9
0 != 9

Thats my hangup.

I understand the big picture that infinity has no bounds. It's just mind boggling really.

Kina like, what's outside of the universe?

 

[Integral]

Consider this.

First of all we must work with Real numbers, this is a matter of how the Real number system is defined. So what is a Real number? One important feature of Real numbers is the identity of each digit with an integer. To restate, there is a one to one coorespondenc beteen the digits of a Real Number and the integers. This where the construction which has "an infinite number of zeros followed by a 1" fails the test of a Real number, what integer cooresponds to that one?

In that sense the smallest Real number cannot be written specifically but we can write:
10^{-N} , \in \bold {N}
and claim that in general this is the form taken by the smallest Real number. Of course we cannot actually represtent the smallest Real number as there is no largest Integer.

Ok, here is the whole point of this post.

consder this

.1 + .999... = 1 + .0999...
.01 + .999... = 1 + .00999...

Can you see that if I have added a small number to .999... to get one plus a number consisting of a finite number of zeros followed by an infinite "tail" of 9s.

Now we can do this in general to get

10^{-N} + .999... = 1 + . (N-1 zeros)..999...
I can now write
10^{-N} + .999... &gt; 1 \forall N \in \bold N


So no matter how small of a Real number I add to .999... I get 1 + a bit more.

There is only one number for which it is true, 1. Thus .999... =1

 

[HallsofIvy]

One important feature of Real numbers is the identity of each digit with an integer.

No, that is not an important feature of the real numbers. That is a feature of the symbols used in one specific way of representing the real numbers. One could represent the real numbers in Roman numerals and they would still have the same properties. The properties of the real numbers are independent of how they are represented.

 

[russ_watters]

Originally posted by Magnus
I can't believe that there is no largest # less than 1. There has to be, in theory.

Here's another way to think of it without the mathematical proof (sorry guys). Can you think of a number greater than .9 and less than 1? Sure: .99. How about greater than that and less than one? Sure: .999. How about...

As you can see, you can keep doing that forever. Thats how infinity works. Its about the same as asking if there is any number greater than infinity: Nope.

 

[Integral]

Originally posted by HallsofIvy
No, that is not an important feature of the real numbers. That is a feature of the symbols used in one specific way of representing the real numbers. One could represent the real numbers in Roman numerals and they would still have the same properties. The properties of the real numbers are independent of how they are represented.

It certianly is a key feature, I do not care what number system you use there is only a Countable number of digits in a Real number. That is the key feature, it is indeed indepentent of the representation but the countablility of the digits is essential and that is what I am referring to.

Edit;
I thing I am seeing the point of confusion. Perhaps you thought I was speaking of the actual digits, ie 0,1,2,3,4,5,6,7,8,9. That was not my meaning at all.

The fractional part of every Real number can be represented as

\sum_{n=0}^\infty d_n 10^{-n}

Were the d_n \in \{0,1,2,3,4,5,6,7,8,9\}

The correspondence with the integers I am speaking of is the index n. This is of course a base 10 Real number if you choose to represent the number in a different base the number rasied to a power will change as will the set of basic digits.

I am not sure that Roman numerals dealt well with fractions! Seems it was the Arabic numerals and the place value system which allowed this development.

 

[Organic]

Hi russ_watters,


There are at least two kinds of infinities: actual infitiny, potential infinity.

As mauch as i know, Math language uses only potential infinity.

More about these kinds of infinity you can find here:

http://www.geocities.com/complementarytheory/RiemannsLimits.pdf



Organic

 

[Shahil]

correction on the limit example:

wouldn't it be:

lim x
x --> 1 (x approaches 1 from the negative side!)

 

[Organic]

Hi Shahil,

If you are talking about my pdf on actual and potential infinities, then (-oo,0) is the mirror image of (0,oo), so we need only one of them for the example.

 

[Hurkyl]

One important feature of Real numbers is the identity of each digit with an integer. To restate, there is a one to one coorespondenc beteen the digits of a Real Number and the integers.

This cannot be an important feature of the real numbers because you cannot even write the statement (or even the spirit of the statement) "there is a one to one coorespondenc beteen the digits of a Real Number and the integers" in the basic theory of real numbers.

What HallsofIvy was trying to say is that this is an important feature of the decimal representation (or base-n representation) of the real numbers, not an important feature of real numbers themselves.

There are at least two kinds of infinities: actual infitiny, potential infinity.

As mauch as i know, Math language uses only potential infinity.

It would be great if anyone knew what you meant by actual or potential infinity.

 

[Shahil]

hey organic

actually my correction was on the suggestion prudenceoptimus made!

 

[Organic]

When 1.000... and 0.999... are two representations of the same number then:

1.00... = 0.999...

0.100... = 0.0999...

0.0100... = 0.00999...

0.00100... = 0.000999...

0.000100... = 0.0000999...

0.0000100... = 0.00000999...

Therefore we can write:

0.100... + 0.0100... = 0.0999... + 0.00999...

0.0100... + 0.00100... = 0.00999... + 0.000999...

But this is not true because:

0.1100... not= 0.0999... + 0.00999... = 0.10999...8

0.01100... not= 0.00999... + 0.000999... = 0.010999...8

and so on ...

The unreachable digit is 8 and not 9, therefore 1.000... cannot be represented by 0.999...

 

[jcsd]

by defintion a number represented by an infinite number of decimal places does not have a last decimal place, i.e. it does not terminate.

 

[Organic]

This is not the last digit but the limit digit or the unreachable digit of 0.999...


Therefore 1.000... is not the limit of 0.999...

 

[suyver]

You don't give up, do you? Where did you get the idea for an 'unreachable' digit?

This limit stuff is not a process that is happening. It's not as if nature (or math) is contineously writing nines after your 0.999999... For all intents and purposes, that has already happened.

Look again at jcsd's very elegant proof:

For any nunber x where:

0 &lt; x &lt; 1

we know that:

0 &lt; \sqrt{x} &lt; 1

and

x &lt; \sqrt{x}

Lets say that there is a largest number between 0 and 1, what is it's square root? If it greater than it's square root, it is greater than 1, if it is equal to it's square root it is equal to 1 and if it's square root is greater than itself then we have generated a number that is larger than the largest number less than 1 so it can't be the largest number less than 1!

As long as you can't tell what's wrong with that, maybe you should just accept that you're wrong. Because you are.

By the way: what's the use of your double-posting?

 

[uart]

This is not the last digit but the limit digit or the unreachable digit of 0.999...

Just one question Organic. Is that a potential unreachable digit, an actual unreachable digit or just a betoid unreachable digit ?

 

[Organic]

Hi suyver,

jcsd's very elegant proof is about the non-existence of the largest number smallest than 1

where 1 is the limit of [0.999...,1.000...).

0.99999...
+
0.09999...
=
1.09999...8

and we use the interval notations not between two numbers but among range of different scales, represented by some number, and in this case the number is [1.0999...8) and the infinitely many digits of 9 cannot exist in the above addition if the limit digit 8 does not exist.

 

[suyver]

Hi Organic,

Question: why are you posting this? All the math shown previously in this thread you either didn't understand or ignore, but I am sure that by now even you must realize that nobody believes that you are correct!

Why don't you just give it up and go do something fun. Maybe read a book?

 

[Organic]

Hi suyver,

Please read this:


http://pespmc1.vub.ac.be/INFINITY.html

 

[suyver]

You're responding in a thread on MATH and you're referring to a site on PHILOSOPHY. Don't you realize how absurd that is?

In mathematics and philosopy we find two concepts of infinity: potential infinity, which is the infinity of a process which never stops, and actual infinity which is supposed to be static and completed, so that it can be thought of as an object.

Only the second kind (in this definition) is meaningful in this debate. Again, using this I can prove that
\sum_{i=1}^\infty 9\cdot10^{-i} \; = \; 1
but I guess that you won't believe that either...

 

[Organic]

Math is based on different consistant systems of axioms,which are propositions regarded as self-evidently true without proof.

So the "true" of the axioms is out of the scope of any mathematical research, therefore can be examined only by PHILOSOPHY.

 

[suyver]

O, now I see!
Yes, you must be completely right, Organic.


I give up. Anybody else wants a go?

 

[Organic]

Also you wrote:

but I am sure that by now even you must realize that nobody believes that you are correct!

What is the connection between belief and Math ?

 

[uart]

Thanks for finally posting a link to explain that concept that you've been talking about Organic. So now I know what the concept of "potential" vs "actual" infinity is. I must say however that the disinction is much more a philosophical one then a mathematical one.


As for your arguments of "unreachable digits", such as 1.0999...8 , I can't see how this is any different from the usual old argument that 0.9999' can't be equal to 1 becuae it "clearly" differs by 0.000...1

It does not make any sense to talk about having an infinite number of zeros followed by a one, just the same as it doesn't make any sense to talk about an infinite number of nines followed by an eight.

If you want to set it up as a proper limit then that's fine, but the result you will get is the same as everyone has already proven.

0.999...8 = Lim as n->infinity 9 * (10^(-1) + 10^(-2) + ... 10^-n) + Lim as n->infinity 8*10^(-n-1) = 1 + 0

 

[Organic]

Hi uart,

You write:

0.999...8 = Lim as n->infinity 9 * (10^(-1) + 10^(-2) + ... 10^-n) + Lim as n->infinity 8*10^(-n-1) = 1 + 0

If you write 8*10^(-n-1) then you don't understand my argument, which is based on the idea
of the open interval http://mathworld.wolfram.com/Interval.html .

Instead of using it between two different numbers, i use it on one number, represented by base 10 (we can use any other base value instead).

Through this point of view i clime that:

0.99999...
+
0.09999...
=
1.09999...8

and we use the interval notations not between two numbers but among range of different scales, represented by some number, and in this case the number is [1.0999...8) and the infinitely many digits of 9 cannot exist in the result of the above addition if the limit of digit 8 does not exist.

 

[HallsofIvy]

If you write 8*10^(-n-1) then you don't understand my argument, which is based on the idea
of the open interval http://mathworld.wolfram.com/Interval.html .

Instead of using it between two different numbers, i use it on one number, represented by base 10 (we can use any other base value instead).

Then you are using it incorrectly. There is no open interval consisting of one number.

You are still using the phrase "among range of different scales" without defining "different scales". As long as you do not define your terms no one will understand what you are saying.

 

[Organic]

HallsofIvy

The Indian-Arabic number system, based on some base > 1 and powered by
0 to -n or n, is actually a fractal with -n or n finite levels or
-aleph0 | alaph0 infinite levels, where each level has a different scale, depends on base^power.

Any infinite fraction is some unique sequence of digits along these scales, and there is no mathematical law that does not allow me to use the open interval idea on this range of digits, existing in these infinite levels of scales.

Therefore [1.0999...8) is a legal notation, which is the result of

[0.99999...9)
+
[0.09999...9)
=
[1.09999...8)

 

[NateTG]

Originally posted by Magnus
How can you represent the largest # that is less than 1?

Under convetional ordering, in the reals or the rationals, this number does not exist.

Let's say that there is a real number x with that property.
Then we have x < (1-x)/2 + x <1, which contradicts the desired property of x.

However, if you choose a different ordering on the real numbers then there can be a number x such that x is the smallest number less than one.

 

[Hurkyl]

you don't understand my argument

How can he understand it if you don't understand it?

 

[Integral]

What HallsofIvy was trying to say is that this is an important feature of the decimal representation (or base-n representation) of the real numbers, not an important feature of real numbers themselves.

The countability of digits is certainly is a fundamental feature of the real numbers. I do not care how you represent it. The sum

\sum_{i=0}^\infty d_i b^{-i}

is the general representation of the fractional part of a real number, b is the base the d_i is a selection from a set of digits. For example
d_i \in \{0,1,2,3,4,5,6,7,8,9\} if b =10
or

d_i \in \{0,1\} if b =2
The one thing that is consistent across all representations is that the summation index i is countable. Thus any representation of a Real number can have only a countable number of d_i associated with it. Am I still not clear enough?