Does 0.9999999... = 1?

[SoyLeche]

Hey, Ryan.

I've been trying to work through a more rigorous proof, but my skills at proving things mathematically are pretty rough (I've always been more of an application guy).

Here's what I want to prove (it's been seen here before):

x < y if and only if there exists a w such that x < w < y

Those are strict < signs, not <= signs, just to make sure there is no confusion. I have no doubt that this statement is true, but can it be proven?

Applying that to this situation we have:

.9999... < 1 if and only if there exists a w such that .9999... < w < 1

A little manipulation of the inequality leads to 0 < w - .9999... < 1 - .9999...

If Halcyon is to be believed, 1 - .9999... = .0000... which is greater than 1, so we have (set z = w - .9999...): 0 < z < .0000...

The next step, once the first proof is in place is to prove that w and z cannot exist. The proof for either is sufficient.

Can you help me with this? I asked my wife, who took some higher math classes than I did, and did a bit more on the theoretical side than me, but she said if I'd have asked her 3 years ago she probably could have done it. Now she's too rusty

 

[SoyLeche]

Hey, Ryan.

I've been trying to work through a more rigorous proof, but my skills at proving things mathematically are pretty rough (I've always been more of an application guy).

Here's what I want to prove (it's been seen here before):

x < y if and only if there exists a w such that x < w < y

Those are strict < signs, not <= signs, just to make sure there is no confusion. I have no doubt that this statement is true, but can it be proven?

Applying that to this situation we have:

.9999... < 1 if and only if there exists a w such that .9999... < w < 1

A little manipulation of the inequality leads to 0 < w - .9999... < 1 - .9999...

If Halcyon is to be believed, 1 - .9999... = .0000... which is greater than 1, so we have (set z = w - .9999...): 0 < z < .0000...

The next step, once the first proof is in place is to prove that w and z cannot exist. The proof for either is sufficient.

Can you help me with this? I asked my wife, who took some higher math classes than I did, and did a bit more on the theoretical side than me, but she said if I'd have asked her 3 years ago she probably could have done it. Now she's too rusty

We may need to apply this to show that .9999... is not less than 1, and then for fun apply it the other way:

0.9999... > 1 iff there exists a w such that 0.9999... > w > 1

although I doubt that anyone will be arguing that 0.9999... > 1. After this we'll know that 0.9999... is not less than 1 or greater than 1, so it must be equal to 1.

 

[Ryan2065]

x and y are both positive numbers and x is greater than y.

We have x-y=w w >= 0
If w = 0 then
x-y=0
x=y

If w > 0 then
x-y=w
x=w+y
y=x-w

Since x=w+y and w is not 0, w > y
Since y=x-w and w is not 0, w<x

So, if x and y are not equal there exists some nonzero real number w that satisfies the expression x>w>y

 

[Halcyon]

So, if x and y are not equal there exists some nonzero real number w that satisfies the expression x>w>y

Sure, it's called infinitum.

 

[SoyLeche]

Sure, it's called infinitum.

Let's move on then.

0 < infinitum < 0.000...

In order for 0 < infinitum to be valid, there must exist a w such that 0 < w < infinitum, right? If such a w doesn't exist, then 0 = infinitum, and 0 = 0.0000...

 

[Halcyon]

Let's move on then.

0 < infinitum < 0.000...

In order for 0 < infinitum to be valid, there must exist a w such that 0 < w < infinitum, right?

Sure, that's called infinitum too. Once you have an infinite decimel, you don't need a filler between it and the next number because it never reaches the next number, it is always just short.

 

[SoyLeche]

Sure, that's called infinitum too. Once you have an infinite decimel, you don't need a filler between it and the next number because it never reaches the next number, it is always just short.

If it is "just short" then there must be something in between. If there is nothing in between, it isn't "just short", it is equal.

 

[SoyLeche]

Sure, that's called infinitum too. Once you have an infinite decimel, you don't need a filler between it and the next number because it never reaches the next number, it is always just short.

There you are with that "reaches" again too. 0.9999... is not continually adding 9s. All the 9s are already there. It isn't moving anywhere or reaching anything. It has arrived.

 

[Halcyon]

There you are with that "reaches" again too. 0.9999... is not continually adding 9s. All the 9s are already there. It isn't moving anywhere or reaching anything. It has arrived.

I know, kinda (in a way there are 9's being added all the time, if there weren't then it would be complete and so not infinite, it just happens that the 9's have always been, will always be added and have always been added, while at the same time not being added) what i'm saying is that it will never reach 1, because it never stops - yet it never becomes whole either, by its very nature it is infinitely short of 1.

 

[SoyLeche]

I know, kinda (in a way there are 9's being added all the time, if there weren't then it would be complete and so not infinite, it just happens that the 9's have always been, will always be added and have always been added, while at the same time not being added) what i'm saying is that it will never reach 1, because it never stops - yet it never becomes whole either, by its very nature it is infinitely short of 1.

Because it never stops it does equal 1. If it were to ever stop it would be "just short" of 1

 

[Halcyon]

Because it never stops it does equal 1. If it were to ever stop it would be "just short" of 1

Nah, it would have to stop to become 1, as 1 is a whole.

Imagine it like two infinite parallel lines. you've got

1.0000^ (infinite zeroes, no 1 on the "end")

alongside

0.9999^

They never converge, they just continue forever side by side.

 

[SoyLeche]

Nah, it would have to stop to become 1, as 1 is a whole.

Imagine it like two infinite parallel lines. you've got

1.0000^ (infinite zeroes, no 1 on the "end")

alongside

0.9999^

They never converge, they just continue forever side by side.

For any number of digits behind your decimal < infinity you are absolutely right. We aren't talking about that though.

 

[Halcyon]

For any number of digits behind your decimal < infinity you are absolutely right. We aren't talking about that though.

What are we talking about then?

 

[SoyLeche]

What are we talking about then?

an infinite number of digits

We aren't talking about lim(n -> infinity)sum[(from i = 0 to n).9*(.1^i)]

We are talking about sum[(from i = 0 to infinity).9*(.1^i)]

Also, your visual is off. They would not be parallel lines. They would be lines that continually get closer together.

note: please excuse the notation - it isn't easy to do this here.

 

[Mercy Not Sacrifice]

I would stay away from trying to use a calculator to prove something rather than concepts. Most calculators and computer programs will only calculate up to the 14th decimal place or less. They will also round the last digit, so if you enter 7/9 you will actually get .7777777778, or something similar depending on your calculator.

That's why I called it a demonstration, not a proof.

 

[Ryan2065]

Sure, it's called infinitum.

Thats not a real number =p (hense me saying "real number")

Sure, that's called infinitum too. Once you have an infinite decimel, you don't need a filler between it and the next number because it never reaches the next number, it is always just short.

But the thing is, that isn't math... There are no two real numbers not equal with no points between them. I just gave the proof of that (well for positive numbers because I didn't want to think too hard for negatives =P )

Then concept you cannot seem to grasp is that we are not adding 9's to the end of a number... they are there. This changes the example from "Imagine we are adding an infinite number of 9s to the end of the number" to "Imagine there are an infinite number of 9s at the end of the number".

You are right in what you say though Hal. If we are adding an infinite amount of 9's to the end of .99 then the number is not 9 and never will be 1 (without limits of course =p). If we already have an infinite amount of 9s at the end of .99 (which is what we are discussing) then .99 is 1.

 

[Yoco]

Really this all kind of depends on how one looks at the concept of "infinity", as we are essentially taking .9^ out to an infinite decimal. The funny thing about this is that the only way for .9^ to not eventually be indistinguishable from 1 is to view infinity as being at some point a finite number. As everyone can agree that infinity is never finite, it follows that at on any scale, on any given number line, .9^ becomes indistinguishable from 1.
It also follows that thinking of .9^ as being "infinitely close" to 1 but not being 1 is thinking that at some point infinity is a finite value.

So Yes, .9^ is "infinitely close" to 1. However as infinity has no finite value, there is no way of distinguishing between .9^ and 1. This makes them equal on every level that can be quantified.

 

[SoyLeche]

Really this all kind of depends on how one looks at the concept of "infinity", as we are essentially taking .9^ out to an infinite decimal. The funny thing about this is that the only way for .9^ to not eventually be indistinguishable from 1 is to view infinity as being at some point a finite number. As everyone can agree that infinity is never finite, it follows that at on any scale, on any given number line, .9^ becomes indistinguishable from 1.
It also follows that thinking of .9^ as being "infinitely close" to 1 but not being 1 is thinking that at some point infinity is a finite value.

So Yes, .9^ is "infinitely close" to 1. However as infinity has no finite value, there is no way of distinguishing between .9^ and 1. This makes them equal on every level that can be quantified.

You're right. If you think of infinity as "a really large and increasing number" - you get exactly what you are explaining. That implys that infinity is a changing value though. It isn't. It's just infinity. Infinity is not getting any bigger.

If you have a very large and increasing number of 9s following the decimal, that number "aproaches" one. It can approach something because it is moving.

If you have an infinite number of 9s following the decimal that number is equal to 1.

1/3 in decimal form is a decimal followed by an infinite number of 3s. A decimal followed by a very large and increasing number of 3s approaches 1/3. There is a difference.

It might be more enlightening to think of this infinite sequence of 3s a different way: Instead of trying to think about writing all the 3s, think about trying to look at them. At any point you get to there are still an infinite number of 3s in front of you, not because someone is at the end putting them on, but because they are already there.

 

[SoyLeche]

Infinity isn't bigger today than it was yesterday, no matter how much that line of thinking helps you conceptualize it.

 

[SoyLeche]

I just had a brainstorm of sorts (while I should be figureing out how to model these redundant reserves, but...)

I would try and do the long division here, but that just isn't going to work in this medium. But...

If you were to do long division on 1/3, what do you get? You will start writting down 0.33333... and continue to add 3s till you are sick of it (that usually happens for me at about 4 of them). You may try and tell me that because you can't write all of the 3s that are required 0.3333... doesn't actually equal 1/3, it approaches 1/3.

What I realized is, let's do long division on 250/5. The first step in the long division will get you a 5 in the tens place. The next step will get you a 0 in the ones place.

While we were still on the first step was 250/5 aproaching 50 or was it still equal to 50 and we just hadn't gotten there yet?