Is 0.999... Really Equal to 1? Exploring the Mathematical Proofs

[Tom McCurdy]

It seems like the same principles used here are the ones used to solve zeno's paradox... I posted some of zeno's paradoxes in another part in the math forum... correct me if I am wrong.

 

[chroot]

Tom,

Yep, that's correct. A proper understanding of the sum of infinite series solves this problem and Zeno's paradoxes the same way. An infinite series can have a finite sum.

- Warren

 

[Rogerio]

Well, 0.999... is just a representation.
It means infinite 9's after the decimal point. It's not a growing number - all the 9's are already there! It's not required a high level of abstraction to understand it (or is it?!), and children commonly accept it.

So, anyone who can understand that
1/3 = 0.333... ,
should understand that
3 * 1/3 = 0.999... , too.

And then, they should understand that
1 = 3 * 1/3 = 0.999...

But, incredibly, some people have serious difficulties in understanding these equalities. Personally, I believe if an adult refuses to accept the first equality, then even all the arguments in the world are not enough to go beyond.

 

[Integral]

I think that in general the people who have trouble with these concepts cannot see infinity as anything but a large finite number. Until they can be convinced that infinity is NOT simply a large finite number there is no argument that they will be able to comprehend.

 

[musky_ox]

Okay let me answer this dumb thread for you guys. 0.9 repeating does NOT =1. The problem comes in the base 10 math system. 1/3 is not exactly 0.3 repeating. 1/3 cannot be accurately represented by the base 10 system, base 12 would be much better than base 12 since we could represent 1/3, and 1/4.

I can understand that infinity means that it has forever 9s. It has no limit to the number of 9s after the decimal point. However, what some thick headed idiots don't realize is that 1 is a limit.

 

[Integral]

You have just called everyone with a reasonable Math degree a thick headed idiot. I take that personally and have issued you a warning for insulting behavior.

Please take a math course beyond high school calculus.

Edit: Meanwhile, study the proof starting on page 2 of the link I posted in Post #5 of this thread. It does not rely on taking any limit, but does rely on the definition equality on the real number line.

 

[musky_ox]

Okay 1 = 5, 3 = 7. Whatever you want to think. However, if 0.9 repeating = 1 then they wouldn't have named them differently in the first place. They are 2 numbers that could be represented as 2 lines, one line being just a 'point' longer than the other.

 

[Hurkyl]

However, if 0.9 repeating = 1 then they wouldn't have named them differently in the first place.

Ah, I see. So $\frac{6}{4}$, $\frac{3}{2}$, $1\frac{1}{2}$, and 1.5 are all different numbers! I can't believe I thought they were all the same.

They are 2 numbers that could be represented as 2 lines, one line being just a 'point' longer than the other.

How can a number be represented as a line? Oh, I see, you meant line segment. Well, as I'm sure you know, between any two distinct points there is another point. If both have their left endpoint at the origin, what lies between the right endpoint of the line segment representing 0.999... and the line segment representing 1?

 

[musky_ox]

Consider this inequality.

$$1 - .1^n < x < 1+ .1^n$$

It seems clear that there is only a single number for which this is true for ALL values of n>0, x= 1 . This is a simple statement that any number added to x results in something greater then 1, or any number subtracted from x results in something less then one. I doubt that you will find many people who will argue with the truth of the statement, x=1.

Now, in the link I posted above, I show that using simple arithmetic, involving only valid rational numbers, one can construct this inequality.

$$1- .1^n < .999... < 1+.1^n$$

notice that is the exact inequality as above, thus we have x=.999... If the original statement is correct that the inequality can only be satisfied by 1 you must be lead to the conclusion that
1=.999...

I don't get what you are saying here. It seems to me that while 1- .1^n = .999... maybe be true, this doesn't make sense... 1- .1^n < .999... < 1+.1^n. You are saying that 1cm - (and infinitely small amount of space aka a point) is greater than 0.999...cm, which is the same thing as far as i can tell.

 

[musky_ox]

Ah, I see. So $\frac{6}{4}$, $\frac{3}{2}$, $1\frac{1}{2}$, and 1.5 are all different numbers! I can't believe I thought they were all the same.




How can a number be represented as a line? Oh, I see, you meant line segment. Well, as I'm sure you know, between any two distinct points there is another point. If both have their left endpoint at the origin, what lies between the right endpoint of the line segment representing 0.999... and the line segment representing 1?

Alright, maybe i have a bad example there, if space is quantisized. Anyways, since mathematics is abstract, i don't see why there can't be an infinitely small amount. Wouldnt this be 1/infinity? If you had an infinite amount of 1/infinities you would get 1, but if you have (infinity - 1) amount of 1/infinities you would get 0.9... so there is a difference theoretically.

 

[Integral]

I don't get what you are saying here. It seems to me that while 1- .1^n = .999... maybe be true, this doesn't make sense... 1- .1^n < .999... < 1+.1^n. You are saying that 1cm - (and infinitely small amount of space aka a point) is greater than 0.999...cm, which is the same thing as far as i can tell.

Simply pick an n, any n in the integers and do the arithmetic. The whole point is that for ANY n you choose (you must pick an n) the relationship holds.

Please show me an integer for which the bolded statement holds? Remember that infinity is not a valid integer, or is it a valid real number. I am not talking about anything physical, we are discussing math and not physics here.

 

[musky_ox]

Simply pick an n, any n in the integers and do the arithmetic. The whole point is that for ANY n you choose (you must pick an n) the relationship holds.

Please show me an integer for which the bolded statement holds? Remember that infinity is not a valid integer, or is it a valid real number. I am not talking about anything physical, we are discussing math and not physics here.

Well you say that n cannot be infinity, but we are dealing with infinity and infinitly small values here so i don't see how you can have this stipulation on the equation...

 

[Hurkyl]

Anyways, since mathematics is abstract, i don't see why there can't be an infinitely small amount.

Just because it's abstract doesn't mean you can do whatever you please. Mathematical objects and ideas usually have very precise definitions.The definition of the real numbers (and thus the decimal numbers) permits no infinities or infinitessimal numbers.

If you so desired, you could start talking about some different abstract idea in which you do have infinities and infinitessimal numbers. You may or may not be able to show that 0.999... is different than 1 in one of these different abstract systems. But the point is that you're speaking about something other than the real numbers.

Incidentally, in one of the more useful alternate systems, the hyperreal numbers, there are two analogues of 0.999...: you could have a hyperdecimal number that has some specified transfinite number of 9's (followed by zeroes). Such numbers would then be infinitessimally close to 1, but not equal.

However, the direct analogue of 0.999... is the one with a 9 in every position, and it is equal to 1, just as in the reals.

 

[musky_ox]

So basically you are saying that mathematics (or whatever system of it we are talking about) is incapable of handling infinities, and is thus not capable of representing our universe? Then what is the use of everyone studying it, if there are other systems that can better account for everything in life? In higher level mathematics/physics do they switch to different systems so we can handle the infinite?

 

[musky_ox]

In base 5, 1/4 = (0.111...), and 4·(0.111...) = (0.444...). In every base, dividing 1 by the largest digit of the system gives 0.111... In hex, 1/(F) = (0.111...), and (F)·(0.111...) = (0.FFF...).

0.999... = 1
(0.444...)5 = 1
(0.FFF...)16 = 1

Its just a flaw that exists in any base system, because you can never acount perfectly for all the fractions. All you are proving is that the mathematical system rounds 1/3 to 0.33...

 

[Integral]

I am not sure how you arrive at that conclusion? Why cannot we handle the infintiies correctly? Perhaps it is you who cannot deal with the infinities correctly?

 

[musky_ox]

Well i can see that 0.99... is infinitely close to 1. It is just the way that our base system rounds up 0.33... to be 1/3. Obviously whatever mathematics we are talking about cannot handle infinity if it has to precisely say that infinity doesn't have a place in it. We know that infinity has a place in abstract ideas and in the universe.

 

[Hurkyl]

So basically you are saying that mathematics (or whatever system of it we are talking about) is incapable of handling infinities

I'm saying this: zero is the only infinitely small real number, and no real number is infinitely large.

Well i can see that 0.99... is infinitely close to 1.

Well, you see wrong. Or, more precisely, your intuition is not in agreement with the definitions.

Its just a flaw that exists in any base system, because you can never acount perfectly for all the fractions.

No, this is a flaw in your version of the decimals. The very purpose of the decimals is to exactly represent any real number (and thus all fractions). The properties of the decimals were carefully chosen so they fulfill this purpose.

You state that 0.333... is not an exact representation of 1/3, and similarly for other decimals, but you will always get exactly the right answer if you replace your fractions with decimals, compute, then convert back to fractions. (this includes using 0.999... = 1, and similar equalities)

We know that infinity has a place in abstract ideas and in the universe.

And that place is not as an element of the real numbers.

 

[Rogerio]

It is just the way that our base system rounds up 0.33... to be 1/3. Obviously whatever mathematics we are talking about cannot handle infinity if it has to precisely say that infinity doesn't have a place in it.

Commonly, anyone who knows divide by hand, knows that 1/3 is 0.33... EXACTLY!
The three points at the end (...) means infinite 3's , and there is no rounding here.
It is not the same result we get with a calculator.

 

[Locrian]

So basically you are saying that mathematics (or whatever system of it we are talking about) is incapable of handling infinities, and is thus not capable of representing our universe? Then what is the use of everyone studying it, if there are other systems that can better account for everything in life? In higher level mathematics/physics do they switch to different systems so we can handle the infinite?

There are tools to deal with infinities. For instance, limits. You probably learned about them in high school.

The original poster, however, clearly did not.

 

[musky_ox]

Yea sure i learned about limits. However, in limits you are saying that the limit to something is for example 0 as you approach x=infinity, not that it actually equals 0. Anyways, limits has nothing to do with it. Here is what I am saying.

"In base 5, 1/4 = (0.111...), and 4·(0.111...) = (0.444...). In every base, dividing 1 by the largest digit of the system gives 0.111... In hex, 1/(F) = (0.111...), and (F)·(0.111...) = (0.FFF...).

0.999... = 1
(0.444...)5 = 1
(0.FFF...)16 = 1

Its just a flaw that exists in any base system, because you can never acount perfectly for all the fractions." You can only pick on this example as long as you use the decimal system.

 

[chroot]

You cannot find a number between 0.999... and 1, and therefore, by the definition of the real numbers, 0.999... and 1 are the same number. It's just that easy.

- Warren

 

[Hurkyl]

you can never acount perfectly for all the fractions.

I challenge this. Would you care to demonstrate an actual error that arises from using decimals instead of fractions, even if you accept 0.999... = 1?

 

[musky_ox]

I can think of a number between them...

0.99...9 + 0.00...1 = 1
0.99...9 + nothing = 0.99...9

 

[musky_ox]

I challenge this. Would you care to demonstrate an actual error that arises from using decimals instead of fractions, even if you accept 0.999... = 1?

The error is in the tread name:

"0.99... = 1"

In base 12, 1/3 is a terminating number... so you don't get the small rounding error when you use it for calculations.

 

[Hurkyl]

I can think of a number between them...

0.99...9 + 0.00...1 = 1
0.99...9 + nothing = 0.99...9

Warren didn't ask for a number between 0.{terminating sequence of 9's} and 1. He asked for a number between 0.{9 in every allowable position} and 1.

(And, incidentally, you didn't produce a number between 0.99...9 and 1)

In base 12, 1/3 is a terminating number... so you don't get the small rounding error when you use it for calculations.

This doesn't address my challenge. Maybe if I restate it, it will be more clear.


I am asking you to produce an arithmetic calculation involving only fractions.
It must have the property that, if I convert the fractions to decimals, do all the arithmetic according to decimal arithmetic, then convert back to a fraction (including the use of 0.999... = 1), I get the wrong answer.

 

[musky_ox]

Warren didn't ask for a number between 0.{terminating sequence of 9's} and 1. He asked for a number between 0.{9 in every allowable position} and 1.

(And, incidentally, you didn't produce a number between 0.99...9 and 1)

Okay try this then. Using your logic:

0.99...9 = 0.99...8
0.99...8 = 0.99...7

Now, i can think of a number between 0.99...9 and 0.99...7, which is 0.99...8.

In base 12, 1/3 is a terminating number... so you don't get the small rounding error when you use it for calculations.

This doesn't address my challenge.

I am challenging you will this. I have just shown you the answer to your "challenge." 0.99... = 1 is the error that it causes.

 

[Hurkyl]

It's easy to find a number between two different terminating decimals of the same length: simply append a 5 to the lesser of the two.

0.999... is not a terminating decimal. It has a 9 in every allowed position. There is no place left to put a 5.

 

[chroot]

Okay try this then. Using your logic:

0.99...9 = 0.99...8
0.99...8 = 0.99...7

But these numbers do not (and cannot) exist. There is no such thing as a number with an infinite number of nines, followed by an eight. There is no position to put an eight in "after" an infinite number of nines, because an infinite number of nines never ends.

- Warren

 

[musky_ox]

They don't exist? Okay. So an infinitly large number exists, but an infinitly small number doesnt? Theoretically, what is 1/infinity? Its limit is 0, but we know that it will never reach 0, doesn't this make it 0.00...1?

Its like my example with the guy counting. He is counting out 0 forever after his decimal point, always with the conscious idea that he is going to say 1 as his last digit, even though he will never make it.

Also, please address my posts about using base 12 to eliminate the 1/3 glitch.

 

[BoulderHead]

I can think of a number between them...

0.99...9 + 0.00...1 = 1
0.99...9 + nothing = 0.99...9

Let me guess...somewhere, wayyyyy out there at the end of infinity there's room for just one more digit ?

 

[Hurkyl]

Okay. So an infinitly large number exists, but an infinitly small number doesnt?

No, no infinitely large (real) number exists, and no infinitely small nonzero (real) number exists.

 

[musky_ox]

How do you get from 1.00... to 1.10...? Do you not have to first pass through 1+infinitesimal to get there?

 

[musky_ox]

No, no infinitely large (real) number exists, and no infinitely small nonzero (real) number exists.

So how many 9s are after the decimal point in 0.99...? There is no real number for them, so the number can't exist?

1 2 3 4 5 6 ... infinity
2 3 4 5 6 7 ... infinity + 1 = ?

Why can't we add an 8 after the 9s? Would there still be an infinite number of decimals after the decimal point?

 

[Hurkyl]

So how many 9s are after the decimal point in 0.99...?

I don't know^*. All I know is that there is a 9 in every legal position.

Why can't we add an 8 after the 9s?

There is a 9 in every legal position in the decimal 0.999...; in particular, there is no "after". Any decimal number (with a 0 to the left of the decimal point) with an 8 in it is smaller than 0.999..., because 0.999... has a 9 where that number has an 8.



*: actually, I do know, but this question is irrelevant to the discussion at hand. A different number system, called cardinal numbers, is used for counting the "size" of things. This number system is an extension of the natural numbers, and has rather poor arithmetic. For example, x + x = x for nearly every cardinal number.