0.9999....(recurring) = 1?

mikau · 2007-06-08 14:48:22

its as good a proof as I could ask for, Identity. Though its pretty much the same as the standard proof only it features summation notation. However, Anthony R. Brown insists that infinity + 1 is greater than infinity, so its not really worth arguing with him.

George,Y · 2007-06-08 15:38:48

mikau wrote:

However, Anthony R. Brown insists that infinity + 1 is greater than infinity, so its not really worth arguing with him.

Then what do you think of infinity, mikau? A docile, flexible amount rather than a static one?
Sorry but I've refuted this kinda of infinite 9's many posts before.

Last edited by George,Y (2007-06-08 15:39:10)

mikau · 2007-06-08 16:10:26

you've refuted what kind of infinite 9's? what do you mean?

I'm saying that infinity + 1 is essentially a meaningless or redundant expression. Am I not correct?

What I think of infinity doesn't matter. How do mathematicians define infinity at present?

Ricky · 2007-06-08 16:35:02

George has yet to rigorously define any type of infinity, so any refutation is meaningless.

As for your statement, mikau, it does depend of the type of infinity you are talking about, as there are many different branches of mathematics. However, typically the statement that ∞ + 1 = ∞ comes out of analysis when evaluating the limits of functions. Also, set theory reaffirms this since there exists a bijection between A and A U {x}, where x is some element not in A. The only type of math I know of where this doesn't hold is in the surreals (and possibly hyperreals, not sure on that one) where the numbers are expanded using set theory to include all different types of infinites. Of course, this is not standard mathematics in any shape or form, although I'm told it is useful in game theory.

But in general, it's ok to say that ∞ + 1 = ∞.

mikau · 2007-06-08 16:41:39

interesting.

You see, I think what this all comes down to is this. I've shown in my proof that 0.999... = 1 -1/10^infinity ( I don't think there are any arguments there). This is equal to 1 only if 1/10^infinity is zero. We know that in the finite case that 1/10^n wher n > 1 can never actually equal zero, but is it true in the infinite case?

Ricky · 2007-06-08 16:44:49

Don't confuse limit with algebraic expression. You cant' just "substitute in infinity". That isn't allowed. And it sends your mind off in the wrong direction thinking of 0.999... as approaching something rather than being a single number.

Instead, 0.999... = limit of ...

A limit is a single number. Does that make sense?

mikau · 2007-06-08 16:47:59

sure! But I think the contrary arguments are saying that its not equal to the limit, its just equal to itself. x_x

Ricky · 2007-06-08 16:58:47

We define the limit of the sequence to be the real number we wish to represent. So for example:

3.1
3.14
3.141
3.1415
...

The limit of this sequence we call "pi". Similarly, for 0.999..., the sequence is:

0.9
0.99
0.999
0.9999
...

And the limit of this is 1.

As I have said before, you don't have to accept such a definition. But we use definitions to ensure that we are all talking about the same thing. If you don't accept this definition, then you are talking about something different than most mathematicians are, and so any talk about 0.999... then becomes meaningless.

George,Y · 2007-06-08 17:11:28

Okay, I think you two guys have an understanding problem and a memory problem.

Not refuted?

Post 513

I have clearly discussed the Only 3 ways to get a 0.999... with Infinite Digits.

So actually I have done defining the infinite amount implicitly.

On the term of infinity, my defination is always the same: The infinite amount is the amount larger than any amount that can be expressed as a clear number. This has been shown in my apple pile example-Ricky you should have known this if your memory is not that poor.

George,Y · 2007-06-08 17:16:08

That's why I have gotten so tired of this kinda of forum debate:

Because you have to redebate over and over again.

I am in the middle of being tired in debates and being regarded as a fool by someone who hasn't reviewed history yet.

Hey guys!
Can you help me out of this??

George,Y · 2007-06-08 17:25:46

having refuted flexible infinity as well

"sure! But I think the contrary arguments are saying that its not equal to the limit, its just equal to itself. x_x"
Great, you have got there mikau!! That's the point!! Defining somewhere a series approaches as the series itself or some mixed other series is the whole trick of the defination of Reals. That's what I called the "approaching pretending to be being" problem long ago.

Where are you going to?
-I am walking on the road.

George,Y · 2007-06-08 17:35:26

10*0.999... is also a good false proof discussed in the first or the second page of this thread.

Last edited by George,Y (2007-06-08 21:46:31)

mikau · 2007-06-08 17:42:43

I see what you are saying, George y, but I think we're just bogged down in definitions here. O_O

If you can't define it as a limit, we have to substitute for infinity, as Ricky said. So we're left with the expression 1 - 1/10^infinity. So are you saying 1/10^infinity has magnitude?

George,Y · 2007-06-08 17:54:27

No, I say real infinity has no reason to exist, as most mathematicians agree (they actually avoid equating it in standard maths). In that way 0.999... can only have finite 9's.

However if you say something just happens without a reason like quantic physics. You would like to use 1/10^inf=0 when you need to say 0.999...=1 and 1/10^inf≠0 when 0.999...infinite structure need to be preserved- It can be 0 and non0 alternatively, undefinately, by chance, chaotically. Then I admit this kind of reasoning. But I keep the counter-case: non0 in 0.999...=1 case, 0 in structure occasion. This case has the same reason to exist, however. And then 0.999...=1 cannot be solid.

George,Y · 2007-06-08 18:11:08

0.999... is very self delusive because it hides real infinity inside elipses and stops investigation on the final digit.

Because mathematicians hate to have x=∞ and don't admit ∞ as a real, I perceive they already know infinitething is no good.

However they keep using ∞ in infinite digits and in variables "going" ∞ so as to get "accurate results" as 0.999...=1 case. And that's cheating in a way.

Reference celebrity:

Father Berkeley,
a guy who debated whether dt=0 or not in the spontaneous speed (v=ds/dt) of accerlerating linear movement. He debated against Newton. The argument was that dt is a ghost, can be 0 and non 0 at the same time. (v=ds/dt dt≠0; ds/dt=[k(t+dt)[sup]2[/sup] -kt[sup]2[/sup]]/dt=2kt+k[sup]2[/sup]dt dt=0) The same ghost arguement applies to 1-0.999... as well- really nothing new.

Last edited by George,Y (2007-06-08 18:23:05)

George,Y · 2007-06-08 18:26:52

Compared to older models of integers, rationals, or finite digits, the one of infinite digits is really logically weak. If you truely want to use it mikau, you have to tolerate contradictions.

Last edited by George,Y (2007-06-08 18:28:07)

luca-deltodesco · 2007-06-08 18:45:53

if you say 0.999... has finite 9's, then you're not saying 0.999... thats like saying:

space is infinite but has an end, i.e. space is finite.

it doesnt stop invesitigation of the final digit, because there isnt one

Last edited by luca-deltodesco (2007-06-08 18:46:41)

mikau · 2007-06-08 18:46:57

so could we resolve this by agreeing that 0.999... = 1 - a ghost?

I'd be content with that!

Last edited by mikau (2007-06-08 18:54:59)

George,Y · 2007-06-08 21:51:33

Luca-deltodesco, the real divergency lies in whether the space and time is continuous. If it fails to be so, infinitesimal or accuracy aquired by infinite digits(infiniteth digit) has no use and is apparently false.

Hey guys, where are you now?
It's supposed to be early morning in US

You got up at 6:30??

luca-deltodesco · 2007-06-09 00:05:01

yes actually, i usually get up at that time

George,Y · 2007-06-09 00:44:43

Oh my!!
I've watched sun rise only 3 or 4 times.

Anthony.R.Brown · 2007-06-09 01:37:49

---------------------------------------------------------------------------------------------------------------
THE REASON INFINITE/RECURRING 0.9 IS CALCULATED WRONG By Anthony.R.Brown 09/06/07.
---------------------------------------------------------------------------------------------------------------

The reason Infinite/Recurring 0.9 is Calculated wrong! By everyone that thinks Infinite/Recurring 0.9 = 1 is because of one important thing they are all doing wrong!

And that is by applying multiple Calculations to the 0.9 in various different ways!

Whether it may be the Classic rounding up method ( 90 / 90 ) = 1 which is the same as getting rid of any Number with a decimal point and making the numbers equal 1 ( 123.99 / 123.99 ) = 1 or ( 0.123 / 0.123 ) = 1

And any other multiple calculations! (1 / 0.9 ) x 0.9 = (1 / (9 / 10 )) x ( 9 / 10 ) = ( 10 / 9 ) x ( 9 / 10 ) = 90 / 90 = 1 "Here again (1 / 0.9 ) x 0.9 = 0.999... can be done again! but the whole Example is wrong!

---------------------------------------------------------------------------------------------------------------
For Infinite/Recurring 0.9 you can only Calculate the 0.9 as one single Calculation!
---------------------------------------------------------------------------------------------------------------
As soon as you do any other Calculations! you are changing the Nature of the problem!

( 1 / 0.9 ) this is the Value that Calculates the Infinite/Recurring! x ( 0. 9 ) now being done!

My example above is 0.9 having one Calculation done to it! which is the Answer!!
This is the same ( 1.111... ) x ( 0.9 ) = 0.999...

So there we have it! the Problem has been found! .................................................................

A.R.B

Last edited by Anthony.R.Brown (2007-06-09 01:38:41)

luca-deltodesco · 2007-06-09 01:40:23

wait wait, rounding up method? your saying that 90/90 doesn't equal 1? lol.

Last edited by luca-deltodesco (2007-06-09 01:40:38)

Anthony.R.Brown · 2007-06-09 01:50:47

Quote:" wait wait, rounding up method? your saying that 90/90 doesn't equal 1? lol. "

A.R.B

Of course it does! read the whole Post! " ONLY ONE CALCULATION ALLOWED TO BE DONE TO 0.9 " as in Answering the Question to Calculate Infinite/Recurring 0.9

Ricky · 2007-06-09 02:49:55

Compared to older models of integers, rationals, or finite digits, the one of infinite digits is really logically weak. If you truely want to use it mikau, you have to tolerate contradictions.

You have never shown a contradiction, the only thing you have shown is "I don't like the standard definition of real numbers."

Math Is Fun Forum

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