1
   

0.99999.....=1

 
 
Kolyo
 
  1  
Reply Tue 5 Nov, 2013 10:13 pm
@OmSigDAVID,
OmSigDAVID wrote:

What r u doing addressing me
or even reading my posts, if u have me on Ignore???? That 's cheating.


I'm sure you've been at a2k long enuf to have heard that question asked and answered a million times by now:

(1) ignore doesn't make it impossible to see a person's posts
(2) ignore isn't permanent
(3) ignore doesn't stop you from seeing a person when surfing the site in offline mode
OmSigDAVID
 
  1  
Reply Tue 5 Nov, 2013 10:19 pm
@Kolyo,
Kolyo wrote:
infinitely many 9's after the decimal WILL get you to 1.
Infinite .9s will get u all the way to ALMOST 1. (Beyond that point is fakery and your wishful thinking.)
Kolyo
 
  1  
Reply Tue 5 Nov, 2013 10:59 pm
@OmSigDAVID,
Everything you said could apply equally to the case of 0.333... and 1/3.

0.3 Not Equal 1/3
0.33 Not Equal 1/3
No matter how many 3's you add after the decimal point you will not get to 1/3.

And yet most people in math are fairly comfortable with the idea that 0.333... = 1/3.
OmSigDAVID
 
  1  
Reply Wed 6 Nov, 2013 12:19 am
@Kolyo,
Kolyo wrote:
Everything you said could apply equally to the case of 0.333... and 1/3.

0.3 Not Equal 1/3
0.33 Not Equal 1/3
No matter how many 3's you add after the decimal point you will not get to 1/3.
Agreed. This thread (falsely) alleges that "O.99999.....= 1".
I rose in DENIAL of liberal math (i.e., in denial of a hoax).
I am content with that; I ratify my position of negation.




Kolyo wrote:
And yet most people in math are fairly comfortable with the idea that 0.333... = 1/3.
Let THEM defend their errant, liberal comfort. I am not responsible for their errors.
For that matter, thay elected obama (with no help from me).
Its not MY fault.





David
0 Replies
 
OmSigDAVID
 
  1  
Reply Wed 6 Nov, 2013 01:04 am
@Kolyo,
OmSigDAVID wrote:
What r u doing addressing me
or even reading my posts, if u have me on Ignore???? That 's cheating.
Kolyo wrote:


I'm sure you've been at a2k long enuf to have heard that question
asked and answered a million times by now:

(1) ignore doesn't make it impossible to see a person's posts
It does, UNLESS u peek,
thereby nullifying n voiding the whole point of Ignore.


Kolyo wrote:

(2) ignore isn't permanent
The hell, its not. In the last 1O years,
I 've put about 2O people on Ignore; its almost always been a 1-way trip.
I 'm a fairly patient fellow; I 've only dumped the very most perniciously toxic
abusers into the dungeon.
Seldom have I let any out again.
In truth, there r a couple who have proven their worthiness to be free.
One sick mind that I can think of, after about a year of freedom
was grossly noxious to someone else. I stuffed him back into Pandora's Box,
where he 's remained for the last 2 years or so.

U were a bit ruff with someone when I first saw your post,
so I addressed u in the same tenor b4 u detonated; your choice.
I love the filosofy of Patience and Prudence, to wit:
"Got along without u b4 I met u, gonna get along without u now."


Kolyo wrote:
(3) ignore doesn't stop you from seeing a person when surfing the site in offline mode
I never thought of that.
If u put him on Ignore, then Y wud u WANT to do that ??





David
Logicus
 
  1  
Reply Wed 6 Nov, 2013 08:04 am
@OmSigDAVID,
I still don't see many reasons supporting your claim.
DrewDad
 
  1  
Reply Wed 6 Nov, 2013 08:44 am
@OmSigDAVID,
David, based on what you are saying there is no e or pi, because both of them are irrational numbers.

.999... is simply a special case of an irrational number.

Just like 3.14159..yadayada = pi, we have .99999999... = 1

I'm sorry that math and the rest of the universe is incompatible with your conservative views. I'm sorry that your world view does not encompass anything more complex than flint hatchets, but the rest of the universe is not so constrained.

I suspect you'll have similar difficulty with the infamous double-slit experiment.

Fortunately, your ignorance doesn't affect reality; have fun in your little fantasy of how math works, but I won't be joining you.
raprap
 
  1  
Reply Wed 6 Nov, 2013 09:00 am
@DrewDad,
0.99999999.... is not irrational. It's a rational number. An irrational number cannot be expressed as a ratio of two integers

0.999999...=1/1

so by definition its rational.

to OmSigDAVID perhaps math is too crisp for most attending attorneys---fortunately that wasn't true for Fermat.

Rap
DrewDad
 
  1  
Reply Wed 6 Nov, 2013 11:27 am
@raprap,
Oops.
0 Replies
 
OmSigDAVID
 
  1  
Reply Wed 6 Nov, 2013 12:09 pm
@Logicus,
Logicus wrote:
I still don't see many reasons supporting your claim.
U don t have to put people on Ignore, if u dont want to.
No one is forcing u.
OmSigDAVID
 
  1  
Reply Wed 6 Nov, 2013 12:21 pm
@DrewDad,
DrewDad wrote:

David, based on what you are saying there is no e or pi, because both of them are irrational numbers.

.999... is simply a special case of an irrational number.

Just like 3.14159..yadayada = pi, we have .99999999... = 1
In the name of logic, u exhort inconsistency.
DEFINITIONALLY, .9999 etc. is something less than 1. It is almost, but not quite, 1.
In the same breath, u deny that it is one and u affirm that it is. U rape logic.



DrewDad wrote:
I'm sorry that math and the rest of the universe is incompatible with your conservative views. I'm sorry that your world view does not encompass anything more complex than flint hatchets, but the rest of the universe is not so constrained.
I don t recommend all that sorrow,
but I respect your sovereign right to choose your own emotions.





DrewDad wrote:
I suspect you'll have similar difficulty with the infamous double-slit experiment.

Fortunately, your ignorance doesn't affect reality; have fun in your little fantasy of how math works,
but I won't be joining you.
Good bye.
DrewDad
 
  1  
Reply Wed 6 Nov, 2013 12:57 pm
@OmSigDAVID,
OmSigDAVID wrote:
DEFINITIONALLY, .9999 etc. is something less than 1.

Only in your bullet-addled mind.
Kolyo
 
  1  
Reply Wed 6 Nov, 2013 01:09 pm
@OmSigDAVID,
OmSigDAVID wrote:

DEFINITIONALLY, .9999 etc. is something less than 1. It is almost, but not quite,


Great! Feel free to share your definition of what 0.999... is with us.

I've given a definition already. I think the sum 0.9 + 0.09 + 0.009 + ... is how we should formally define 0.999... in order to get at the heart of what 0.999... represents.
What's your definition?
0 Replies
 
Tryagain
 
  2  
Reply Wed 6 Nov, 2013 01:53 pm
Dayvid, I always had you pegged as one of them liberal types who would never shoot a child for stealing a crust of bread without shouting a warning first!

But as I consider any New Yorker my brother and seaing as how you are under sustained fire from a pack of a>kademmicks; not that I’m sure they are all Irish - so I will supply subdiffusion in support of your premise thus allowing you time for a tactical withdrawal…

Whilst waiting for Democrats to get out their crayons, consider:

.999...= x.
x can NOT BE EQUAL TO ANYHING ELSE BUT .999

Do you know why?
Because I just define x to be equal to .999.... So adding any other number would break the constraint I placed it. That is the only possible solution.

X represents a number in which it's value may not be known but it doesn't have to be equal to any number in the number spectrum. It is not a paramater. We are aloud to make x be define by making it equate to something

.999.. (multiplied by ten) = 9.999.....
x (multiplied by ten) = 10x

I think we can agree with x being equal to ten x and that is not a problem. So for the one above the argument often is that a 0 will be added to the end and shift everything to the decimal place.

But let's not forget that we know there is an infinite amount of 9's. So let's say we add a 0 at the end of an infinite amount of nines. That would make sense but let's not forget it is placed after the infinite amount of 9's.

So .999(after infinite amount of 9's) 0 = 10x
Ok now let's get to the interesting part
9.999 (after infinate amount of 9's ) 0 = 10x
- .999 (infinte amount of 9's) = X

Let's agree now that we can add a infinite amount of zeros after the infinite amount of 9's which is always assumed that we can add as many zeros afterwards in the decimal place and it does not affect the number .0010 = .001

So that means this can be rediscribed as
9.999 (after infinate amount of 9's) 0 = 10x
- .999 (after infinate amount of 9's) 0 = x _______________________________________________
9 = 9x
Now because of this we can do a simple subtraction to get this result. And of course from here.
1 = x


Ipso facto quid pro quo!

Enjoy your freedom; just don’t go to Florida... that’s America's wang!
OmSigDAVID
 
  1  
Reply Wed 6 Nov, 2013 04:51 pm
@Tryagain,
Tryagain wrote:
Dayvid, I always had you pegged as one of them liberal types
who would never shoot a child for stealing a crust of bread
without shouting a warning first!
Thay can have the crusts
after the service staff severs them from the bread.




Tryagain wrote:
But as I consider any New Yorker my brother and seaing as how you are under sustained fire from a pack of a>kademmicks; not that I’m sure they are all Irish - so I will supply subdiffusion in support of your premise thus allowing you time for a tactical withdrawal…

Whilst waiting for Democrats to get out their crayons, consider:

.999...= x.
x can NOT BE EQUAL TO ANYHING ELSE BUT .999

Do you know why?
I do. If thay add more 9s,
then that 'll be a DIFFERENT NUMBER, which is non-X.




Tryagain wrote:
Because I just define x to be equal to .999.... So adding any other
number would break the constraint I placed it. That is the only possible solution.
No doubt.




Tryagain wrote:
X represents a number in which it's value may not be known
but it doesn't have to be equal to any number in the number spectrum.
It is not a paramater. We are aloud to make x be define by making it equate to something

.999.. (multiplied by ten) = 9.999.....
x (multiplied by ten) = 10x

I think we can agree with x being equal to ten x and that is not a problem.
So for the one above the argument often is that a 0 will be added to the end and shift
everything to the decimal place.

But let's not forget that we know there is an infinite amount of 9's.
So let's say we add a 0 at the end of an infinite amount of nines.
That would make sense but let's not forget it is placed after the infinite amount of 9's.

So .999(after infinite amount of 9's) 0 = 10x
Ok now let's get to the interesting part
9.999 (after infinate amount of 9's ) 0 = 10x
- .999 (infinte amount of 9's) = X

Let's agree now that we can add a infinite amount of zeros after the infinite amount of 9's which is always assumed that we can add as many zeros afterwards in the decimal place and it does not affect the number .0010 = .001

So that means this can be rediscribed as
9.999 (after infinate amount of 9's) 0 = 10x
- .999 (after infinate amount of 9's) 0 = x _______________________________________________
9 = 9x
Now because of this we can do a simple subtraction to get this result. And of course from here.
1 = x


Ipso facto quid pro quo!

Enjoy your freedom; just don’t go to Florida... that’s America's wang!
He was a Law Professor I knew; he became a citizen. He came from China.

Thanx for your observations





David
0 Replies
 
Logicus
 
  1  
Reply Wed 6 Nov, 2013 05:52 pm
@OmSigDAVID,
Ignore. Your reasons are subjectively-based, and not really explaining the point, but, if you feel that's how you think 0.999... = 1, okay. I wasn't sure of it either.
Kolyo
 
  1  
Reply Thu 7 Nov, 2013 07:22 am
@Logicus,
Ultimately, according the laws of real analysis, I don't think there is any way to define 0.999... to be a real number and have it be less than 1.
However, there may be hope for David's view in the realm of hyperreal numbers.

A hyperreal number is an infinite sequence of real numbers.

Standard real numbers, when considered in their hyperreal representation, are just constant sequences:

1 = (1, 1, 1, 1, ...)
0.9 = (0.9, 0.9, 0.9, 0.9, 0.9, ...)

I'm wondering whether the natural representation for 0.999... as a hyperreal would be:
0.999... = (0.9, 0.99, 0.999, ...).

This would give 0.999... two desirable properties:
(i) 0.999... < 1
(ii) For all real x < 1, x < 0.999...

I believe that David has echoed the views of many from this site who have not posted in this thread. It is intuitive that 0.999... < 1. When the laws of mathematics (its definitions and axioms) give us a result that disagrees with most people's intuition, one answer is to find a new set of axioms and definitions.
OmSigDAVID
 
  1  
Reply Thu 7 Nov, 2013 09:00 am
@DrewDad,
DrewDad wrote:

OmSigDAVID wrote:
DEFINITIONALLY, .9999 etc. is something less than 1.

Only in your bullet-addled mind.
Meaning no dis-respect,
I 'd not wanna hire a man with your filosofy
to be my accountant.
OmSigDAVID
 
  1  
Reply Thu 7 Nov, 2013 09:08 am
@Kolyo,
A chosen axiom shud be accurate.





David
0 Replies
 
DrewDad
 
  1  
Reply Thu 7 Nov, 2013 10:21 am
@OmSigDAVID,
OmSigDAVID wrote:
accountant.

Ah. You're discussing arithmetic. We're discussing math.
 

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