Does .99~ = 1

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Does it? proof please?
 

Hurkyl

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Yes.

Try dividing both sides by 9, or computing the sum of 0.9 + 0.09 + 0.009 + ...
 
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i think ive been here before...
 

HallsofIvy

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0.999... is DEFINED AS the sum of the geometric series [itex]\Sigma_{n= 0}^\infinity 0.9(.1)^n[/itex]. It is fairly easy to show that, as long as |r|< 1, the sum [itex]\Sigma_{n=0}^\infinity ar^n = \frac{a}{1-r} [/itex].
[itex]\Sigma_{n= 0}^\infinity 0.9(.1)^n= \frac{0.9}{1-0.1}= \frac{0.9}{0.9}= 1[/itex]
 
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0.999... is DEFINED AS the sum of the geometric series...
How infinitely many elements can have a sum?

All we can say is that they are approaching the sum, but never reaching the sum.

Shortly speaking, this is the all idea of being infinitely many ... .

Standard Math breaking its own rules by saying that .999... = 1
 

matt grime

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No, organic once more you need to learn what is going on. No FINITE portion of the sum (truncation after n additions) is 1, but that is not the same as the limit, is it? For the, what, tenth time, the real numbers are defined to be (amongst other equivalent definitions) the completion of Cauchy sequences of rationals wrt the Euclidean metric. Thus by definition 0.9999... is 1. There is more to maths than that which can be constructed in finite time. It's just a shame that cranks don't know this.
 
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Cauchy sequences of rationals wrt hte Euclidean metric are "raping" the infinite to be finite, no more no less.
 
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Zurtex

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Organic said:
How infinitely many elements can have a sum?

All we can say is that they are approaching the sum, but never reaching the sum.

Shortly speaking, this is the all idea of being infinitely many ... .

Standard Math breaking its own rules by saying that .999... = 1
Sigh, no it is the limit of n approaching infinity. It never actually reaches infinity just as you can never write 0.999... out in full. This is barely above high school mathematics and really simple to understand if you read it and tried to learn something.
 

pig

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Organic said:
Standard Math breaking its own rules by saying that .999... = 1
which rules exactly?

if you do not believe that 0.999... = 1, it should be very easy for you to prove it. find a real number between them.
 
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if you do not believe that 0.999... = 1, it should be very easy for you to prove it. find a real number between them.
Demonstrate a zero gap between 0.999... and 1
 

matt grime

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We have done, again and again and again.
 
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No, you did not, for example:

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

proof:

If a is not b then |a-b|>0.

Let d be the difference.

Let e = d/2 then |a-b|=d or |a-b|=d/2 are both > 0, hence d > 0
and also |a-b|/2 > 0.

therefore non-zero/2 > 0.
 

matt grime

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Can we make this your last post in this thread? Here we go:

"If a and b are distinct real numbers then for any e > |a-b| = d > 0."

There is no conclusion at the end of that sentence, so we don't know what yo'ure trying to prove.

"Let e = d/2"

this isn't relevant as in the statement you say for all e>|a-b|=d and since d=>0 we cannot have e=d/2


"|a-b|=d/2"

no you defined d to be |a-b| so if d=d/2 then d=0, which is a contradiction to your hypothesis that a and b are distinct.

"hence d > 0"

but that was true as it was the modulus of |a-b|, which by hypothesis is not zero, so there was no need to "prove" this fact

"therefore non-zero/2 > 0"

that is false: -1 is non-zero.

Can you take this back to your other thread now?
 

pig

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Organic said:
Demonstrate a zero gap between 0.999... and 1
don't you agree that for every non-zero real x, it is correct that:

|1 - 0.999...| < |x|

?

since by substracting 0.999... from 1 we get infinitely many 0s after the decimal point, the result is smaller than any number which has a non-zero digit.

unless you consider things which make no sense like "0.000...0001" real numbers.

and that leaves zero as the only possible solution in R.
 
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If you use the open interval idea on a notation of a single number you get:

[.000..., x) where x is any value > 0.

Therefore |[.000..., x)|=1 and no infinitely many levels of zero contents can change the cardinality of [.000..., x) to 0.
 

chroot

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pig: Welcome to physicsforums. There's really not much sense in arguing with Organic, since he simply chooses to use different definitions of terms than everyone else.

The issue of whether or not 0.999... = 1 is of course closed, and has always been closed. They certainly are equal.

- Warren
 
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Sure Chroot, it is closed exactly like Lord kelvin once said about Physics, and then Plank came ...
 

ahrkron

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pig said:
don't you agree that for every non-zero real x, it is correct that: |1 - 0.999...| < |x| ?

since by substracting 0.999... from 1 we get infinitely many 0s after the decimal point, the result is smaller than any number which has a non-zero digit.

unless you consider things which make no sense like "0.000...0001" real numbers.

and that leaves zero as the only possible solution in R.
Good post pig. As Chroot mentioned already, there is no controversy about this among professional mathematicians. Organic has admitted many times that the meaning he gives to many terms is not standard (although sometimes he forgets about it in these discussions).
 

ahrkron

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Organic said:
Sure Chroot, it is closed exactly like Lord kelvin once said about Physics, and then Plank came ...
No.

In physics, there is always the possibility of new and improved measurements showing that our models are not quite right once we "zoom in" deep enough.

With math, the story is quite different. As long as definitions are kept the same, the invention of new concepts do not show old ones wrong.

The case of 0.9999... is, for some reason, a favorite for internet forums, but it poses no controversy or problem in math. It is settled, and it does not require any especially hard math.
 

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