28 Nov 04
Originally posted by Fat mans revengeNo it doesn't..it is a direct consequence of the least upper bound axiom of standard analysis, there have been several non-algebraic proofs on this thread already
Well, after reading the original post, and a few there after it. I decided to explain my reasonings here.
This whole problem derives from the assumption that 1/3 =0.333333...
05 Dec 04
Originally posted by TestriderYeeeeeeees, but you see, if there are an infinite amount of zeroes you would never get to the 9 how long you continued, so it would be
I don't think you can simply multiply it with 10 unless it would be somethin like this:
0.99999... = x
multiply by 10
9.99999... = 10x
subtract one equation from the other
9.000...9 = 9x
with an infinite amount of zero's
9.0000.... = 9x
So, there cannot be a 9, also because 9.0000....9 is per definition finite (you end somewhere, in this case at the 9), when of course it shouldn't be.
06 Dec 04
1 edit
Originally posted by pidermanIn that case you simply can't multiply .999... by 10 😉 (and get an outcome/solution/whatever word i should use in en.)
Yeeeeeeees, but you see, if there are an infinite amount of zeroes you would never get to the 9 how long you continued, so it would be
9.0000.... = 9x
So, there cannot be a 9, also because 9.0000....9 is per definition finite (you end somewhere, in this case at the 9), when of course it shouldn't be.
06 Dec 04
But my solution at the bottom of the 7th list of posts works too, right? Money could theoretically equal .99999... of another form of currency on a certain day, at least I think so... the exact value of $$$ changes daily due to the economy, at least in America, I think... Now I'm just confusing myself... 😕
07 Dec 04
Originally posted by Testridersubtract one equation from the other
I don't think you can simply multiply it with 10 unless it would be somethin like this:
0.99999... = x
multiply by 10
9.99999... = 10x
subtract one equation from the other
9.000...9 = 9x
with an infinite amount of zero's
9.000...9 = 9x
Where did you get that last 9? All 9's after the decimal point subtract away.
07 Dec 04
Originally posted by TestriderWhy not? Since we calculate in decimal notation the multiplication by 10 is simply moving the decimal point one place to the right.
In that case you simply can't multiply .999... by 10 😉 (and get an outcome/solution/whatever word i should use in en.)
So 10*0.999999... = 09.999999... = 9.9999999...
07 Dec 04
Originally posted by pidermanPiderman, I think the problem is that people are cautious when we divide/multiply. Many a trick involves incorrect use of those operations. Or maybe they are just hesitant in multiplying infinite numbers (just as you have to be careful with calculating infinite sums).
Why not? Since we calculate in decimal notation the multiplication by 10 is simply moving the decimal point one place to the right.
So 10*0.999999... = 09.999999... = 9.9999999...
For Testrider I'll post a previous proof given by me, wich involves nothing but addition and rules of Standard Analasis:
The difference 1 - 0.999... is smaller then any positive number you can name. The only non-negative number wich has that property (being smaller then any positive number) is 0 (*). Having a difference 0 means that the two numbers are equal.
(*) PROOF: Let's say there is a second number with that property (non-negative and smaller then all positive numbers), say O. Then because of the ordening on R we have either O<0, O>0 or O=0.
CASE 1: O<0. X<0 means that X is a negative number, so O is a negative number, contradiction with the fact that O is a non-negative number.
CASE 2: O>0. Make a number A = O/2. Then 0<A<O, wich is a contradiction with the fact that O is smaller then any positive number.
The only possibility left is CASE 3; 0=O wich must be true because we must have case 1,2 or 3.
QED.
07 Dec 04
Originally posted by opsoccergurl11Equally, ''by algebra'...
well, by algebra,
if .999999=x, then
9.999999=10x
and 9=9x after you subtract the two equations
when you divide by 9, you get 1=x
so, i just stated x=.99999999 and x=1 by just modifying the original equation
if .333333 = y then
3.333333 = 10y
and 3 = 3y therefore .33333 = 1
Clearly .33333 = 1 = .99999
but .33333 x 3 = .99999
hence 3 = 1
Do all numbers = 1?
08 Dec 04
Originally posted by KINGSIthat was the most horrible logic i ever saw, you suck at math
Equally, ''by algebra'...
if .333333 = y then
3.333333 = 10y
and 3 = 3y therefore .33333 = 1
Clearly .33333 = 1 = .99999
but .33333 x 3 = .99999
hence 3 = 1
Do all numbers = 1?
.333333333333 = y (ok, fine)
then 3.33333333333 = 10y (still fine, okay)
and 3 = 3y WTF ARE YOU TALKING ABOUT, YOU JUST SAID y=.333...???
so 3y= 1!!!!! WHAT ARE YOU SMOKING????
Clearly .333333 = 1 ACTUALLY THEY ARENT EQUAL AT ALL!!!!
and yes .999999 repeating does equal 1
08 Dec 04
1 edit
Originally posted by Testriderthe whole is that there is no smallest possible number because any number can always be divided by two
If you can't devide the smallest possible number by 2 (wich seems logical) that must mean you cant multiply the number that is the closest to 1 by 10
EDIT: except for zero! which means the two numbers are equal!