# What is the largest number less than 1?

1. Nov 26, 2003

### Magnus

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

2. Nov 26, 2003

### PrudensOptimus

lim x
x-->1

3. Nov 26, 2003

### Magnus

Can you represent that in decimal form?

4. Nov 26, 2003

### 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

5. Nov 26, 2003

### Magnus

The line over it means infinity.

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

6. Nov 26, 2003

### Hurkyl

Staff Emeritus
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.

7. Nov 26, 2003

### Magnus

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

8. Nov 26, 2003

### Hurkyl

Staff Emeritus
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 $$\inline{y>x}$$:

$$y = \frac{1+x}{2} > \frac{x+x}{2} = x$$

Now, notice that $$\inline{y<1}$$:

$$y = \frac{1+x}{2} < \frac{1+1}{2} = 1$$

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.

9. Nov 26, 2003

### 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?

10. Nov 26, 2003

### chroot

Staff Emeritus
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

11. Nov 27, 2003

### 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!

12. Nov 27, 2003

### krab

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.)

13. Nov 27, 2003

### jcsd

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

for a nunber x where:

$$0 < x < 1$$

we know that:

$$0 < \sqrt{x} < 1$$

and

$$x < \sqrt{x}$$

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!

14. Nov 27, 2003

### HallsofIvy

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...?

Last edited by a moderator: Apr 21, 2011
15. Nov 27, 2003

### theEVIL1

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

16. Nov 27, 2003

### jcsd

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

17. Nov 27, 2003

### theEVIL1

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

18. Nov 27, 2003

### jcsd

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

19. Nov 27, 2003

### Hurkyl

Staff Emeritus
Why?

I don't see the connection.

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

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 $$\inline{2/4}$$.

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.

20. Nov 27, 2003

### HallsofIvy

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).

21. Dec 1, 2003

### 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?

22. Dec 1, 2003

### 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?

23. Dec 1, 2003

### Integral

Staff Emeritus
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 infinte "tail" of 9s.

Now we can do this in general to get

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

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

Last edited: Dec 1, 2003
24. Dec 1, 2003

### 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.

25. Dec 1, 2003

### Staff: Mentor

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.