Simple mathematical problem (?)

[geni]

Don't you see the flaw in your reasoning? This would mean that there's a number larger than infinity, which is impossible by definition

That statement depends on what infinity you are using there are a number of infinities. The one you are using ( the total number of real numbers) is far from being the largest.

 

[Suggestologist]

Also, could you please tell me what 1 - (0.999...) is? Can you please just write out that number in decimal notation for me?

Adam

1-.999999..... would look like: 0.0000000.......000001; where the .....'s represent infinite expanses of digits.

 

[Suggestologist]

What, exactly, do you disagree with?

Surreal numbers can represent 1-epsilon (or iota, depending on the notation used).

Where, exactly, did I say that? [/B]

You didn't. You cleverly omitted it.

 

[Suggestologist]

Don't you see the flaw in your reasoning? This would mean that there's a number larger than infinity, which is impossible by definition.

There are numbers larger than any "infinity" you want to talk about. Surreal numbers even let you do weird things like take the square root of "infinity"+1. Of course, infinity is not a "real" number.

 

[xouper]

Suggestologist: 1-.999999..... would look like: 0.0000000.......000001; where the .....'s represent infinite expanses of digits.

Wrong. The reals are closed under addition and 0.000...1 is not a real number. Try again.

Here's the challenge. Time to put up or shut up. You claim that 0.999... is less than one. Since 0.999... is a rational number, then please show us two positive integers A and B such that

A / B = 0.999...

and also that A does not equal B.

If you cannot do that, then you must concede that 0.999... equals 1.

 

[Cabbage]

Surreal numbers can represent 1-epsilon (or iota, depending on the notation used).

This is where I remind you that originally you were making the claim for the existence of a real number, not a surreal number:

Does there exist a "real" number that is BOTH Less Than and closer to 1 than ANY other real number (which is less than 1)? The surreal number system says that such a number must exist.

You didn't. You cleverly omitted it.

Excellent! Now that that's out of the way, would you care to address my original point? That point being that, in general, we don't give surreal numbers decimal expansions, because surreal numbers aren't the type of number that have decimal expansions (the type with decimal expansions being real numbers).

 

[Zep]

Agreed. That's exactly what I said in my previous reply to you, but in different words. That quote from your book says the sum is S. In this context, the word "is" is synonymous with "equals". Referring back to the geometric series theorem you quoted from your book, if the infinite series is convergent, then it equals a/(1-r). Are we in agreement on this point yet?

I think we are referring to two different things.

First, the formula a/(1-r) refers to the limit value of the convergent series, not the partial sums themselves. It is the value towards which all the partial sums are (desperately) trying to converge. However it is not the value that the partial sums will ever "equal" in the true arithmetical sense.

Second, let me just requote a bit of my reference again:

By the phrase "|L - an| is arbitrarily small for all sufficiently large values of n," we mean that to any positive number E there corresponds an index N such that

|L - an| < E for all n > N

That is, all terms after the Nth lie within the distance E from L.

This, to me, seems to be the mathemetician's way of saying that you can get the series to go as close as you wish to the limit value by taking as many terms as you need to do it. Hmmm....

In that case, I would value your input on the validity of the above expression when the variable E is set to 0 (zero), which would be the case if 0.9~=1. The term on the other side of the inequality becomes...I'm not sure - meaningless? Can you have an absolute value less than zero? Would that then mean that the expression only has validity if E is some positive non-zero number? If that is so, then the absolute expression must be "slightly smaller" in order to remain true? Which says that there IS still "a gap" between the end of the series and unity?

My thinking is getting warped here - what's smaller than the smallest value of all? I think someone pointed out that "iota/2" is sorta silly!

 

[T'ai Chi]

And can you now define tiny for me and give me an example?

Tiny is any e>0, and obviosuly less than .1, since 1-.9 = .1.

An example is 1-.999..., where you tell me the number of 9's you want to consider. If you tell me they 'go on forever', I say 'fine', and therefore the difference .000... goes on forever, and so we never get to the end number 1, and so the difference 1-.999... = 0.

Really, forget the possible rounding problems with .999... I simply haven't seen any good reasons yet to why 1/3 does not equal .333... Doing it long division should reveal that. And then just notice that every digit of .999... is 3 times as large as every digit in .333..., and of course 1/3 * 3 = 3/3 = 1.

 

[Patricio Elicer]

Originally posted by Vorticity
0.999(rec)
= Sum_(n=1)^Infinity 9*(1/10)^n
= 9*( Sum_(n=0)^Infinity (1/10)^n - 1 )
= 9*( 1/(1-(1/10)) - 1 )
= 9*( 10/9 - 1 )
= 9*( 1/9 )
= 1

Mathematicaly speaking, it's correct

[edited to delete a previous erroneous post]

 

[xouper]

Zep: First, the formula a/(1-r) refers to the limit value of the convergent series, not the partial sums themselves.

That's not correct. The expression a/(1-r) is the sum of the infinite series. The geometric series theorem establishes an equality between the series and the formula a/(1-r). That's the whole purpose of that theorem.

The sum is also the limit of the sequence of partial sums. But the infinite series itself is exaclty equal to a/(1-r).

However it is not the value that the partial sums will ever "equal" in the true arithmetical sense.

Correct. The sequence of partial sums converges on the limit. The infinite series, however, is the limit of the sequence of partial sums.

Second, let me just requote a bit of my reference again:This, to me, seems to be the mathemetician's way of saying that you can get the series to go as close as you wish to the limit value by taking as many terms as you need to do it. Hmmm....

You are confusing series with sequence. In that quote, they are referring to the limit of an infinite sequence, not an infinite series. An infinite series does not "get close" to the limit value, it is the limit value (of the sequence of partial sums).

In that case, I would value your input on the validity of the above expression when the variable E is set to 0 (zero), which would be the case if 0.9~=1.

The whole idea of a limit is to avoid that little problem of E=0.


Perhaps a jargon review would be helpful. If not, then skip the following.

A sequence and a series are two different things.

A sequence is simply an ordered list of numbers.

A series is the sum of all the terms in a sequence.

For example, this is an infinite sequence {0, 1, 0, 1, 0, 1, ... }

As you get farther down the list, does the number tend to converge on a limit? In this example, no.

How about this infinite sequence? { 0.9, 0.99, 0.999, 0.9999, ... }

Yes, each successive number in the ordered list gets closer and closer to a unique value, thus this sequence does converge on a limit. And that limit is the infinite series (0.9 + .09 + .009 + .0009 + ... )

To repeat, an infinite series is the sum of all the terms in an infinite sequence.

Let's look again at the following infinite sequence:

{0.9, 0.09, 0.009, 0.0009, ... }

Does it converge on a limit? Yes, each term in the sequence gets closer and closer to zero.

But if we sum all the terms of that sequence, we get the infinite series (0.9 + 0.09 + 0.009 + ... ) = 0.999... = 1

Now, when we talk about the sequence of partial sums, we are talking about the following ordered list of sums:

{
0.9,
0.9 + 0.09,
0.9 + 0.09 + 0.009,
0.9 + 0.09 + 0.009 + 0.0009,
...
}

As we go down the list of items in this sequence, each sum gets closer and closer to 1.

The limit of this sequence is the infinite series (0.9 + 0.09 + 0.009 + ... ) = 0.999... = 1

To summarize, we have talked about three different things here.

1. The infinite sequence T = {0.9, 0.09, 0.009, 0.0009, ... }
   
2. The infinite series S = (0.9 + 0.09 + 0.009 + ... )
   
   and
   
3. The infinite sequence of partial sums R = { 0.9, 0.99, 0.999, 0.9999, ... }

The limit of R is S. The limit of R is equal to 1. And since S is equal to the limit of R, S=1. But S also equals 0.999..., therefore 1 = 0.999...

If you do not agree with this conclusion, then somewhere in all the above jargon review, you are likely misunderstanding something.

 

[xouper]

T'ai Chi: An example is 1-.999..., where you tell me the number of 9's you want to consider. If you tell me they 'go on forever', I say 'fine', and therefore the difference .000... goes on forever, and so we never get to the end number 1, and so the difference 1-.999... = 0.

OK, good, you agree that 1 = 0.999...

We'll overlook that you tried to use 0.000...0001 which is not a real number.

Really, forget the possible rounding problems with .999...

Easy to do since there are no rounding problems here.

0.999... is exactly equal to 1, so there is no rounding to be done. It's an exact equality.

 

[xouper]

Patricio Elicer: Mathematicaly speaking, it's correct

Excellent image, Patricio. Mathematically speaking, it's more rigorous than the other demonstrations in this thread. It shows in symbolic notation one of the standard proofs, which can be found all over the net and in textbooks, and which I posted last July when we had this conversation. Boooeee posted a link to that old thread in this thread, then Vorticity posted his version of the proof, DanishDynamite reposted mine again from the other thread, and LW posted an expanded explanation of it. And now you posted the symbolic version. If someone doesn't get it by now, I don't hold much expectation they ever will.

 

[aerosolben]

There are numbers larger than any "infinity" you want to talk about. Surreal numbers even let you do weird things like take the square root of "infinity"+1. Of course, infinity is not a "real" number.

Alrighty, then. If you are correct, please prove your point by providing an example of a real number < 1 and > 0.9~.

I am not interested in rates of approaching infinity, or alternate number systems, or any infinities not directly relevant to the problem at hand. If infinity+1 is applicable to the problem, please answer my question. Find a real number x, such that 1 > x > 0.9~.

 

[Patricio Elicer]

Thanks Xoup. I had no idea of the previous debates on the matter. I just took Vorticity's idea, and elaborated a little further on it

 

[xouper]

aerosolben: ... please answer my question. Find a real number x, such that 1 > x > 0.9~.

If what he says is true, that 1 does not equal 0.999..., then since both numbers are rational, and since there are an infinite number of reals between any two rational numbers, should be a piece of cake, eh? Or he could simply agree that no such real number x exists. No shame in that.

 

[!Xx+-Rational-+xX!]

If what he says is true, that 1 does not equal 0.999..., then since both numbers are rational, and since there are an infinite number of reals between any two rational numbers, should be a piece of cake, eh? Or he could simply agree that no such real number x exists. No shame in that.

◊◊◊◊ you!@ Xouper can't see posts from travelltroll!

 

[slimshady2357]

1-.999999..... would look like: 0.0000000.......000001; where the .....'s represent infinite expanses of digits.

That is not even a number. It's nonsense! Please try again.

Under the definition of decimal notation, 0.00000....0001 doesn't even make sense.

At best, it's equal to zero, but that's a loose interpretation that is being generous and would have you agreeing with us that 1 = 0.999... anyway!

How can you even suggest that answer when xouper has shown already that 0.0000...0001 is not even a number? And if you had actually read Cabbage's post, you would also realize that it is meaningless under decimal notation.

Traveller

Your post was the first one I've ever reported. Usless.

Adam

 

[whitefork]

Don't they teach math in high school any more? This thread is mind-boggling.

Anyone who seriously believes that .9 recurring is not equal to 1 should read Understanding Infinity by Gardiner. I plug this book all the time. It's clear and gives you a lot of problems to work through.

 

[hal bidlack]

◊◊◊◊ you!@ Xouper can't see posts from travelltroll!

This post has been reported by several people. I responded to Traveller at length elsewhere. Simply put, please reduce your sig block as a matter of courtesy.

This post, as posted, is not a violation of the formal rules. While a tad silly, it can remain.

hal

 

[LW]

Mathematicaly speaking, it's correct

Freeze! This is the typography police! You are under arrest for using italic font to represent a mathematical operator. Everything you say or write may be used against you and you have absolutely no rights at all.

Otherwise, a nice post.