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1=.999999...?

Oh yeah oops, I read 0,999... Instead of 0,999
My bad
The infinite sum of that sequence is still 1, though
 
Oh yeah oops, I read 0,999... Instead of 0,999
My bad
The infinite sum of that sequence is still 1, though

No it is not, hence the thread. The limit of the sum is 1, not the sum itself.

An infinite sequence of non-repeating decimals is never going to contain a repeating decimal, though.

You keep using that word. I do not think it means what you think it means.
Inigo Montoya

I think I understand what you meant to say. A non-repeating string of decimals can represent a non-algebraic number. Why not a repeating string? A repeating string will always approach a rational number, but that is not sufficient reason.

J
 
Why is this thread still alive?
 
I think I understand what you meant to say.

I'm just saying that this:

0.999... is an element of the sequence (0.9, 0.99, 0.999, ...)

Is not true.

The sequence you have enclosed in brackets in the quote above only includes non-repeating decimals, such as 0.99. It does not include decimals that go on into infinity, such as 0.999...
 
The sequence you have enclosed in brackets in the quote above only includes non-repeating decimals, such as 0.99. It does not include decimals that go on into infinity, such as 0.999...

Why?

As a practical matter I understand. However, it is a practical, not a theoretical distinction. There is no theoretical reason why the sequence cannot have countably infinite terms. We construct infinite sets all the time. However, using the limit of the sequence makes the situation more tractable.

J
 
:hatsoff: Thank you, it's good to be missed.

it is if you take all n's to infinity
If you take the limit of the sum as n approaches infinity, then you do get one, but then you are talking about the limit.

Why?

As a practical matter I understand. However, it is a practical, not a theoretical distinction. There is no theoretical reason why the sequence cannot have countably infinite terms. We construct infinite sets all the time. However, using the limit of the sequence makes the situation more tractable.

J
The sequence does have countably infinite terms, but it has that regardless of if is 0.9999... is a member. Most sequences however map to the whole numbers, so (0.9,0.99,0.999 ... 0.999...) would be a strange sequence because infinity is not a whole number. 0.99... would have to be the infinitieth term to be in that sequence. It would also have the unique distinction of not having a predecessor -- all whole numbers have a predecessor, but infinity does not. So just because all other terms in (0.9,0.99,0.999 ... 0.999...) are less then one, it does not necessarily follow that 0.999.... is less than one. Because 0.999... does not have an immediate predecessor, you cannot use induction to reason about it. You don't have an infinity - 1 member in the series. (unless you wanna say that infinity is it's own predecessor, but that doesn't help induction any)
 
If you take the limit of the sum as n approaches infinity, then you do get one, but then you are talking about the limit.

This is redundant. The limit of the infinite sum is the limit. What does it add to the conversation?

The sequence does have countably infinite terms, but it has that regardless of if is 0.9999... is a member. Most sequences however map to the whole numbers, so (0.9,0.99,0.999 ... 0.999...) would be a strange sequence because infinity is not a whole number. 0.99... would have to be the infinitieth term to be in that sequence. It would also have the unique distinction of not having a predecessor -- all whole numbers have a predecessor, but infinity does not. So just because all other terms in (0.9,0.99,0.999 ... 0.999...) are less then one, it does not necessarily follow that 0.999.... is less than one. Because 0.999... does not have an immediate predecessor, you cannot use induction to reason about it. You don't have an infinity - 1 member in the series. (unless you wanna say that infinity is it's own predecessor, but that doesn't help induction any)

You used the key word "map". A mapping is a function. You assign elements of one set to elements of of another set, which means you are no longer dealing with the first set. Specifically, you enlarge an open segment by adding a end point. One common theme of the thread is that using the limit is a transform, mapping an uncountable set onto a countable one.

No one has raised induction before. Interesting. IIRC you do not need an infinite term to use induction, merely an arbitrarily large one.

J
 
This is redundant. The limit of the infinite sum is the limit. What does it add to the conversation?
I was agreeing with you and responding to Lohrenswald's remark that apparently suggested that an infinite sum equals something that's distinct from its limit at infinity.

You used the key word "map". A mapping is a function. You assign elements of one set to elements of of another set, which means you are no longer dealing with the first set. Specifically, you enlarge an open segment by adding a end point. One common theme of the thread is that using the limit is a transform, mapping an uncountable set onto a countable one.

No one has raised induction before. Interesting. IIRC you do not need an infinite term to use induction, merely an arbitrarily large one.

J
Yes, a sequence can be thought of as a mapping of whole numbers to a set. In this case an infinite set. I'm not saying that you can't have a sequence that also has a defined value at infinity. It's not a traditional sequence at that point, but that's ok with me.

The problem is, how do you prove that every member of the sequence is less than 1? Normally with infinite sequences you'd prove a property of the whole set it by showing that it applies to at least one element (like the first), and then that it applies to the n+1 element. This is induction. So you would show that 0.9 is < 1. and that 0 point n nines is less than one, using the fact that 0 point n minus one nines is less than one. That would formally show that every element of the sequence is strictly less than one. Except it doesn't prove that the infinitith element is less than one.

So even if you say that 0.99... is a member of the sequence (0.9, 0.99, 0.999, ...) you cannot draw conclusions about 0.999... based on the other members of the "sequence".

Yes, you can also use induction on finite sequences, but that's neither here nor there.
 
I was agreeing with you and responding to Lohrenswald's remark that apparently suggested that an infinite sum equals something that's distinct from its limit at infinity.

No I meant that it in this case it is 1
 
0.999... is an element of the sequence (0.9, 0.99, 0.999, ...).

I think this the root of the problem. 0.99... is not in that sequence. The sequence does not contain any number with infinite number of nines. Every number in the sequence is finite.

0.99.. is just shorthand for a zero followed by an infinite number of nines.

if you agree that lim (0.9+0.09+0.009+...)=1, then it follows that 0.99.. is also 1.
Since every term in the sequence (0.9, 0.99, 0.999, ....) is smaller then 0.99.. we know that lim(0.9, 0.99, 0.999, ....) is less or equal to 0.99..
so we get 0.99 >= lim(0.9, 0.99, 0.999, ....) = 1. i.e. 0.99.. >= 1.
That is the weirdest attempted proof yet. Don't say "just". If you dig in the thread there is some confusion on that point. One thing is clear, 0.999 < 1. Your statement, that 0.99... >= lim(0.9, 0.99, 0.999, ....) is always false. I am unsure how you would think it was true.

J

I agree that 0.999 < 1. But 0.99... >= lim(0.9, 0.99, 0.999, ....) is true.

It is easy to see that 0.99... is larger then every element in (0.9, 0.99, 0.999, ....).
0.99... minus element n is 0.(n zeroes)(infinite nines). Since 0.99... is larger then every element in the sequence the limit of the sequence can't be larger then 0.99... .
 

Even if that set has an infinite number of elements in it, you've defined it to only include non-repeating decimals such as 0.99. (by listing the first three elements the way you did - they're all non-repeating decimals)

So basically what you've got an infinite set with only non-repeating decimals in it, so it can't possibly include 0.999...

Basically what Folket is saying:

I think this the root of the problem. 0.99... is not in that sequence. The sequence does not contain any number with infinite number of nines. Every number in the sequence is finite.
 
Even if that set has an infinite number of elements in it, you've defined it to only include non-repeating decimals such as 0.99. (by listing the first three elements the way you did - they're all non-repeating decimals)

So basically what you've got an infinite set with only non-repeating decimals in it, so it can't possibly include 0.999...

Basically what Folket is saying:

There is a jump in the thread here. Which definition are you referring to?

However, I think you are mistaken. Repeating decimals typically go with the non-repeating decimals, unless the repeat is all zeroes.

J
 
There is a jump in the thread here. Which definition are you referring to? J

This one:

0.999... is an element of the sequence (0.9, 0.99, 0.999, ...)

Your sequence does not include 0.999... , it only includes non-repeating decimals, while 0.999... is a repeating decimal, so it's not a part of that sequence.
 
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