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27 Aug 04
Originally posted by AcolyteI think we can declare 1-0.9999... to be infinitesimal (non zero) and we still have workable mathematicics. See http://mathworld.wolfram.com/NonstandardAnalysis.html
If 0.999... is not 1, then since 1 - 0.999... is not 0, it isn't real, either. The approach from there that causes least damage is to declare 1 - 0.999... to be surreal, and by implication 0.999... is surreal as well. I can see no way in which 0.999... can be real and irrational, unless you accept that the reals are no longer closed under addition, a troubling prospect.
30 Aug 04
Originally posted by THUDandBLUNDERWell... it all depends... :-)
YES or NO?
.
The answer is No. Which ever way you turn it. (except maybe practical calculus, but this doesent seem to be the case in this question)
I think you all get a bit hung up on base 10. In math however the base of the numerical representation lacks importance. Just because
0.9999... is very close to 1 stating it equals one is the same as saying
0.777... equals one if we have octal numbers. leading to
0.777...(oct) = 0.9999... (dec) or
0.1111...(bin) = 0.ffffff... (hex) aso.
this is just plain wrong.
30 Aug 04
Originally posted by chasparosWell, 0.111... (bin) = 1 by the following:
Well... it all depends... :-)
The answer is No. Which ever way you turn it. (except maybe practical calculus, but this doesent seem to be the case in this question)
I think you all get a bit hung up on base 10. In math however the base of the numerical representation lacks importance. Just because
0.9999... is very close to 1 stating it equals one is th ...[text shortened]... (oct) = 0.9999... (dec) or
0.1111...(bin) = 0.ffffff... (hex) aso.
this is just plain wrong.
define S = 0.111... (bin) = 1/2 + 1/4 + 1/8 ...
so 2S = 1 + 1/2 + 1/4 ....
by subtraction, 2S-S = 1
31 Aug 04
Originally posted by SiskinYes. As I said for practical calculus they are *considered* the same.
yes they are the same ..look in any undergraduate textbook on algebra or analysis
In pure numerical theory though this only a way to demonstrate
the imperfections in numeral representation.
If we allow ourselves to consider infinite decimals, we must by the same logic allow us to consider a numeric system with unlimited symbols such that all points on the numeralaxis has one symbol pinpointing it exactly. Would be as hard to learn as writing the entire 0.9.. series :-) but that is beside the point.
Further:
Consider the graphs of the series X(n) where n is number of decimals of 0.9... To be able to reach one the curve has to break the pattern (nine tenths of the remaining distance to 1) and pattern breaking does not appear in math. If you dont change the premise, the patten will be the same.
And finally the inconsistancy:
To allow one infinite entity and disallow another makes
no sense.
0.999.... is 1- 1/(infinity)
31 Aug 04
Originally posted by richjohnsonInteresting...
Well, 0.111... (bin) = 1 by the following:
define S = 0.111... (bin) = 1/2 + 1/4 + 1/8 ...
so 2S = 1 + 1/2 + 1/4 ....
by subtraction, 2S-S = 1
Can you also prove, without using pseudo proof of equality to 1,
that 1/2+1/4+1/8.... = 1/7+1/49+1/343.... (base 7)?
I think you will find it a challange :-)
31 Aug 04
Originally posted by rgoudieSorry for this heavy posting for such a minor thread, but I found this a bit intriguing (spelling :-( )
This assertion is immediately refuted by evaluating the fraction 1/9. This ratio is obviously equal to 0.111...
Multiplying by 2:
2/9 = 2 x 0.111... = 0.222...
Multiplying by 9:
9/9 = 9 x 0.111... = 0.999... = 1.000...
Yes, I realize that this exact argument has been used in an earlier post.
-Ray.
On to my reply:
Why obviously?
By the same logic cant you say its obviously NOT 0.1111...
for the same reason 9*0.111111....=0.999999... not 1
Since the basic rules of vanilla math stipulates
multiplication and division as each others inverse methods
(1/X)*X=1
I would use your exact argument to prove 0.99999 is NOT 1 :-)
It all depends on which side of the fence you start.
31 Aug 04
Originally posted by richjohnsonHow about SxS ?
Well, 0.111... (bin) = 1 by the following:
define S = 0.111... (bin) = 1/2 + 1/4 + 1/8 ...
so 2S = 1 + 1/2 + 1/4 ....
by subtraction, 2S-S = 1
every factor following:
1/2-> 1/4 + 2/8 + 2/16 + 2/32 + 2/64... = 1/2 + 1/8 + 1/16 + 1/32...
1/4-> 1/16 + 2/32 + 2/64 + 2/128... =1/8 + 1/32 + 1/64...
1/8-> 1/64 + 2/128 + 2/256 + 2/512... = 1/32 + 1/128 + 1/256
.
.
.
As you can see this converges to the same series (1/2+1/4...), BUT
it does so at a slower rate. (at any point there is a hole in the series:
(1/2+1/4+1/8)^2 =1/4 + 2/8 +3/16 +1/64=1/2+1/8+1/16+1/64)
this puts the disputed (S*S)=1 further down the series than S=1 would be. This Forces us to concede that one of:
1*1=1 or
S=1
is false.
Make your pick :-)
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31 Aug 04
Originally posted by chasparosFollowing the 0.9999 theme, shouldn't that be:
Interesting...
Can you also prove, without using pseudo proof of equality to 1,
that 1/2+1/4+1/8.... = 1/7+1/49+1/343.... (base 7)?
I think you will find it a challange :-)
1/2 + 1/4 + 1/8 = 6/7 + 6/49 + 6/343... ?
I think it is less of a challenge to prove those equal!
31 Aug 04
I traced another interesting point.
The entire concept of 1-dimensional numeral representation is based on this fact:
f(n) = Base^n is exactly 1 larger than the sum of (Base-1)*f(n) for (0...n-1)
see:
Base=2,n=3
2^3=8
1+2+3=7
Base=10,n=3
10^3=1000
(9*1)+(9*10)+(9*100)
Thus we know that any attempt
to put 1/10+1/100....
to a common denominator will result in
an expression like:
(X-1)/X witch is not 1.. Again by definition.
31 Aug 04
Originally posted by iamatigerYes, ofcourse. Sorry my misstake.
Following the 0.9999 theme, shouldn't that be:
1/2 + 1/4 + 1/8 = 6/7 + 6/49 + 6/343... ?
I think it is less of a challenge to prove those equal!
But I would be intressted in the proof anyway.
I'd really think that disprooving 1/2+1/4... < 6/7 + 6/49...
Constitutes a problem/challange?
Important note:
I am not a mathmatician. Merly intressted in math.
I am not trying to ruffle any feathers here. This thread
contradicted something I belive to be true. And I found that very intriguing. I would really like to hear why I am wrong :-)
31 Aug 04
You should be carefull when dealing with infinite sums 🙂
I can make a sum that comes out on 0, mess around with it so it ends up in 1, then mess it up again to is becomes less then 0, and again to make it more then 1...
As for the 0.9999...thing:
1/9 = 0.111... multiply both sides with 9 and you get:
1 = 0.999...
Another way:
a = 0.999...
10a = 9.999... substracting gives:
9a = 9 wich implies a=1.
If you are familiar with limits (with the limits here i mean limits for n increasing to infinity):
0.999... = lim 1 - 1/(10^n) = 1 - lim 1/(10^n) = 1 - 0 = 1
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31 Aug 04
3 edits
Originally posted by chasparos1/2+1/4+1/8..... = a
Yes, ofcourse. Sorry my misstake.
But I would be intressted in the proof anyway.
I'd really think that disprooving 1/2+1/4... < 6/7 + 6/49...
Constitutes a problem/challange?
Important note:
I am not a mathmatician. Merly intress ...[text shortened]... at very intriguing. I would really like to hear why I am wrong :-)
multiply both sides by 2: 1 + 1/2 + 1/4 + 1/8.... = 2a
substitute in a from the first equation: 1 + a = 2a
take a from both sides: 1 = a
-----------------
6/7 + 6/49 + 6/343.... = b
multiply both sides by 7: 6 + 6/7 + 6/49 + 6/343... = 7b
substitute in the value of b: 6 + b = 7b
take b from both sides: 6 = 6b
divide by 6: 1 = b
---------------------
therefore a = b = 1
more generally with base x:
(x-1)/x + (x-1)/(x^2) + (x-1)/(x^3) = c
multiply by x
(x-1) + (x-1)/x + (x-1)/(x^2) + (x-1)/(x^3) = xc
substitute in the value of c above:
(x-1) + c = xc
take c from both sides
(x-1) = (x-1)c
divide by (x-1)
1 = c
31 Aug 04
Originally posted by TheMaster37Yes. I am not questioning basic algebra or the practicality of using limits in calculus. It works out because of imperfections.
You should be carefull when dealing with infinite sums 🙂
I can make a sum that comes out on 0, mess around with it so it ends up in 1, then mess it up again to is becomes less then 0, and again to make it more then 1...
As for the 0.9999...thing:
1/9 = 0.111... multiply both sides with 9 and you get:
1 = 0.999...
Another way:
a = 0.999 ...[text shortened]... its for n increasing to infinity):
0.999... = lim 1 - 1/(10^n) = 1 - lim 1/(10^n) = 1 - 0 = 1
My first point is that by definition 0.9... is not 1.
Look at 1/9=0.1111...
I could just as easily say that equality is false since
9/9 = 1 not 0.99...... or is this reversal faulty?
My second point is that if you allow yourself one infinity, the infinite
series 0.99..., in one breath and in the next breath uses limits to eliminate that infinitly small differance to proove 1=1. Seems like a waste of time.