dknguyen

0.999 fixed to 0.999...

ljcox

Correct me if I'm wrong, but I don't see how you can subtract 2 numbers that have an infinite number of decimal places. To do a subtraction, you start at the last digit. eg. 7.89 - 3.45, you start with 9 - 5.

But with an infinite number of decimal places, there is no last digit.

I will be interested to see your convergence of an infinite geometric series.

__________________
Len

Optikon

It is not hard to argue with because it is not mathematically correct.

Your first & last statement say that C = 0.999~ and also that C =1 implying that 1 = 0.999~ which is NOT true.

It is true however that Lim (n->inf) {0.999~} = 1 where n is number of repeating 9's. So your statements as they stand are NOT correct and the convergence of the geometric series is not the issue here technically. The geometric series convergence depends on the existence of its limit. You are showing no limits which is why it is wrong.

rumiam

ok if 1 does not = 0.99999~ then explain to me;
why is it that
1/3 = 0.333~
2/3 = 0.666~
3/3 = 0.999~

3/3= 1
so
1= 0.999~

Prove this to be wrong

ljcox

Your error is in adding the decimal values for 1/3 and 2/3

These have an infinite number of decimal places and so I don't see how you can add them.

__________________
Len

ljcox

I was intrigued by this so I did the maths to prove it. See attachment. What I forgot to write is that the first 3 terms in the last line are the first in the infinite series for e. The terms after 1/n go to 0 as n approaches infinity.

Attached Images

Binomial.png (69.2 KB, 13 views)

__________________
Len

ljcox

I think the answer to this issue is in whether the attached series converges to 1 or to 0.9999~.

It is many years since I studied maths that I don't recall how to do the test for convergence of this series and don't have the time to revise it.

So if any knows how to do it, please enlighten us.

Attached Images

Series.png (53.3 KB, 13 views)

__________________
Len

Sceadwian

I don't believe this post is still going.

__________________

arod

Yes, isnt that the point of the proof? Showing that .999~ = 1...
I don't think you understand. When I type 0.999~, it means that the 9s repeat to infinity.

Instead of me reinvinting the wheel, here is a page showing the convergence of an infinite geometric series proving that 0.999~ = 1

tkbits

How about an alternate proof?

ljcox

Thank you.

__________________
Len

rumiam

As you can see..I just did add them

bloody-orc

Originally Posted by ljcox
Your error is in adding the decimal values for 1/3 and 2/3

These have an infinite number of decimal places and so I don't see how you can add them.

Why can't you add two totally normal numbers as:
There are no infinite numbers i can see in them... just 1,2 and 3...

__________________
Need Help?
Press F1
If that doesn\'t help you, ask me... I might know better.

Glyph

Rewrite 0.99~ as (1-x)

(1) Therefore (1-x) =0.99~

(2) Therefore x = 0.00~1

(3) Assumption (to be proven later) x != 0

(4) Therefore, although x may be infinitely small is still not 0

(5) so if 0.99~ = 1 (as asserted by OP) then substituting into (1) the statement (1-x) =0.99~ becomes (1-x) = 1.

(6) Rearranging gives 1-1 = x

(7) therefore x = 0

(8) VIOLATION of original assumption (3) where X !=0

Proof by "Reductio ad absurdum" that 0.99~ does not equal 1

---------------------------------

Proof that assumption (3) in the above proof is itself correct, where x is not 0

(1) Axiom: 1^infinity = 1

(2) (1+0)^infinity = 1 since (1+0) = 1, therefore from (1) it must be 1

(3) base of natural logaritms "e" is give by (1+1/n)^n as n approaches infinity as given by http://en.wikipedia.org/wiki/E_%28ma...al_constant%29

(4) let x = 1/n

(5) as n approaches infinity, x will approach 0, therefore x = 0.000~1

(6) substitute (4) into (3) to give (1+x)^n = e since x =1/n and from (3) (1+1/n)^n = e

(7) if x = 0 then (1+x)^n = (1+0)^n = 1^n

(8) as n approaches infinity 1^n = 1 from statement (1)

(9) VIOLATION: if x = 0 then from (3) the base of natural logarithms must be e = 1 from (7) and (8) but this violates (3)

Proof that x =0.000~1 is not zero even though it is very small

proof by "reductio ad absurdum"
----------------

Therefore from the above two proofs it is shown that 0.99~ is NOT 1

if it were true then from the above proofs the base of natural logarithms should be 1.... you can check your calculator and spreadsheets to show that it is not 1.

If this proof doesn't satisfy you i honestly don't think any proof will.

Sceadwian

I don't care how much of a math geek you are. If you can't come to the realization that .99~ is not 1 without prompting.... There's just no hope for you as a thinking human being

__________________