Question about .9999

[Psyguy22]

I accept that .999...=1, but what is the importance in that definition? What does it help us accomplish?

 

[haruspex]

It's not so much a definition as a consequence of other definitions. If it were not the case we'd be in something of a mess, like 1 = 3*(1/3) + 3*(.3333...) = .999... not being 1.

 

[Psyguy22]

So is that all its there for? Just so we can have those fractions make complete sense?

 

[Vorde]

Nothing 'is there for -> this'. People developed fractions and decimals and someone noted that by the rules they had created, .99999... = 1.

 

[Edgardo]

I accept that .999...=1, but what is the importance in that definition? What does it help us accomplish?

0.999... = 1 is not a definition. It is a result as mentioned by haruspex.

0.999... is defined as $\sum_{k=1}^{\infty}9/10^k$, i.e. it is the limit of the sequence $s_n = \sum_{k=1}^{n}9/10^k$.

s₁ = 0.9
s₂ = 0.99
s₃ = 0.999
...
One can show that this sequence converges to 1, i.e. if I give you a small number $\epsilon$, then you could find an index m such that |s_m - 1|< $\epsilon$.
For instance, I give you $\epsilon$=0.0001. Can you find an m?

Intuitively this means that $s_n$ moves arbitrarily close to 1.

 

[Hurkyl]

We use decimal notation because it is convenient notation for real numbers. If you follow through how decimal numerals correspond to real numbers, you find that the numerals 1.000... and 0.999... both correspond to the same real number.

 

[Psyguy22]

0.999... = 1 is not a definition. It is a result as mentioned by haruspex.

0.999... is defined as $\sum_{k=1}^{\infty}9/10^k$, i.e. it is the limit of the sequence $s_n = \sum_{k=1}^{n}9/10^k$.

s₁ = 0.9
s₂ = 0.99
s₃ = 0.999
...
One can show that this sequence converges to 1, i.e. if I give you a small number $\epsilon$, then you could find an index m such that |s_m - 1|< $\epsilon$.
For instance, I give you $\epsilon$=0.0001. Can you find an m?

Intuitively this means that $s_n$ moves arbitrarily close to 1.

Ok. I understand how to prove that .999..=1, but my question is what is the importance behind it?

 

[micromass]

Ok. I understand how to prove that .999..=1, but my question is what is the importance behind it?

It's simply true. Why should it have importance?? What is the importance of 1+1=2? Or what is the importance that a cat has (usually) 4 legs??

 

[Psyguy22]

It's simply true. Why should it have importance?? What is the importance of 1+1=2? Or what is the importance that a cat has (usually) 4 legs??

So its nothing more than that? Just that its true?

 

[micromass]

So its nothing more than that? Just that its true?

I'm confused. What more do you expect?? What would you think is a good importance??

 

[Psyguy22]

I thought that it would resemble some kind of importance like e^(pi*i)=-1

 

[micromass]

I thought that it would resemble some kind of importance like e^(pi*i)=-1

How exactly is that important in the first place??

 

[lavinia]

I accept that .999...=1, but what is the importance in that definition? What does it help us accomplish?

The sequence of decimals .9 .99 .999 ... is a Cauchy sequence. That means that even though the sequence is infinite,the numbers in it cluster together and the further out the sequence the more closely clustered they get. To say that the infinite sequence equals one, says that it actually converges to the number 1. A prioi one might imagine that such an infinite sequence might end up nowhere, But in fact any infinite decimal sequence will converge to some number. This means that the numbers have no holes, that there is nothing missing in them. That to me is the importance of that expression.

 

[0xDEADBEEF]

It's actually a bit more complicated. 0.999... is 1 because of our definition of real numbers. You can use for example the definition following Dedekind called Dedekind cuts. Using this definition a real number is something that divides all the rational numbers (fractions) into those that are larger and those that are smaller. As you will see, the limit of .999... will define the same cut as the number 1 therefore they are the same. If you find another number that will cut the rational numbers in this place it will also be 1.

 

[homeomorphic]

The importance of it might be that you have to be careful with decimal expansions to handle exceptional cases like that. I can't think of a very good elementary example, but here is what I can come up with.

Maybe you would like to prove that the cardinality of the power set of the natural numbers is the same as that of the real numbers. The power set of a set is just the set consisting of all its subsets. What you would like to do is define a function from the power set of the natural numbers to the interval of real numbers from 0 to 1. The way you would LIKE to do this is to identify the power set with the set of all the functions from the set to the set consisting of 0 and 1. 0 means that number is not included in the set. 1 means it is included in the set. So, those functions are the same thing as subsets. Now, to map to the interval, you would LIKE to do is just "think in binary" and say that those functions, in turn, just look like binary decimal expansions of numbers in that interval. But, you have to be careful because there's some duplication. So, you can get inspiration for the proof this way, but actually, it gets slightly more complicated because of this issue of decimal expansions not quite being unique, since 0.1111... in binary is the same as 1.000000...

So, you can get this general idea to work, but it's not as simple as you might think at first glance.

So, this illustrates that if you want to get everything right logically, it is important to realize that 0.9999 equals 1.