It is 1/9 and that is a frational fraction.
You are right, but what fraction is .999999.... ?
It is 1/9 and that is a frational fraction.
You are right, but what fraction is .999999.... ?
Chinglu, you are mistaken -- I do not need to concatenate an infinite number of 9's to form 0.999... because by assumption the digits in 0.999... are always 9 and already exist at all positions right of the decimal point.
Because of the axiom of completeness for the real numbers, the following holds true for the non-negative real numbers.
$${a_0.a_1a_2a_3\dots}_{\tiny 10} \quad = \quad \sum_{k=0}^{\infty} \frac{a_k}{10^k} \quad = \quad \textrm{sup} \; \left{ x : \; \exists n \in \mathbb{N} \; \wedge \; x = \sum_{k=0}{n} \frac{a_k}{10^k} \right} \; = \; \lim_{n\to\infty} \sum_{k=0}^{n} \frac{a_k}{10^k} \; \in \; \mathbb{R}$$
Therefore $$0.999... = \quad \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} = 1$$.
This follows from continuity and monotonicity of the exponential because $$\forall k\in \mathbb{Z} \; \frac{1}{10^{k+1}} \lt \frac{1}{10^k} $$ and so if $$0 \lt 0.999... \lt 1$$ then $$n \quad = \quad 1 - \left\lfloor \log_{\tiny 10} \left( 1 - 0.999... \right) \right\rfloor \quad \in \mathbb{N}$$ and therefore $$1 - \frac{1}{10^{n-3}} \lt 0.999... \lt 1 - \frac{1}{10^n} \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} \lt 0.999...$$ which is a contradiction that proves $$0.999... \not\lt 1$$.
do you find chinglu's ravings to be abusive?How can Chinglu's ravings be acceptable but Tach's not? I truly don't understand the standards here.
tryHere is one of may very simple proofs, this is from wiki:
$$\frac{1}{9} = 0.111...$$
$$9 \times \frac{1}{9} = 9 \times 0.111...$$
$$ 1 = 0.999...$$
It isn't really that hard to understand.:shrug:
That is 1.
Where is the .999....?
How can Chinglu's ravings be acceptable but Tach's not? I truly don't understand the standards here.
do you find chinglu's ravings to be abusive?
try
1/9 = 1(0.111...)
9 x 1/9 = 9 x 1(0.111...)
1= 1(0.999...)
1=1
using quantity for quantity...
Mix the context from quantity to value and you will always be able to fudge a result.
I think my proof that 1 = 1 is better than your proof that 0.999... = 1
the proofs mixing context from wiki are flawed... sorry ...and you math guys should be ashamed of yourselves for falling for it. IMO.. [chuckle]
OK, can you prove the irrational .9999.. is a rational number being 1? That is the issue.
no ... not at all.. look againYou do realize that when you wrote:
1= 1(0.999...)
1=1
That this means that 1 x 0.999... = 1 or 0.999...= 1
So you apparently now agree with the mathematicians. [chortle]
Here is one of may very simple proofs, this is from wiki:
$$\frac{1}{9} = 0.111...$$
$$9 \times \frac{1}{9} = 9 \times 0.111...$$
$$ 1 = 0.999...$$
It isn't really that hard to understand.:shrug:
Please give us the' big dummies guide', to how the finite, 1.0 is equal to the inferred infinite 0.999....
They are pulling the mathematical illusion over their own eyes at same time the attempt to pull it over ours. imho.
This issue is really a no brainer, yet they appear to want to present them selves as being beyond the simple person and simple logic and simple common ergo the send us to a link of complex mathematics, and words most of humanity has never seen, and will never hear of, and say to us, see, did you read that. that is the proof.
Or they may post a half page of complex algebraic symbols and other who knows what symbols, and again say, see,did you read that, that was the proof.
Uh, yeah sure dude, when you come back to Earth with something for 80% of humanity to actually even begin to grasp then you can cry proof because you certainly have not proved anything to me.
origin.."You have been given about 4 or 5 proofs that show 0.999... = 1."
Like I said many proofs have been given - your arm waving gibberish is not a proof.
Just because the proof is beyond your ability that does not make it wrong. It just means it is beyond your ability.
Lucky for us science and math is not dictated by what you can understand, if that were true I fear we would be back in the dark ages.
Sure it does. All the $$a_k$$ terms are non-negative because they are digits, right?I looked at your supremum operator and it does not fit in your equation.
As I did where? Also, why not?You can't compare sequences based on n and then all of a sudden compare them to infinite sequences as you did.
Where am I using the pinching theorem? I'm saying that IF $$0.999... \lt 1$$ then there exists an n where $$1 - \frac{1}{10^{n-3}} \lt 0.999... \lt 1 - \frac{1}{10^n}$$ and this contradicts $$0.999... \; = \; \textrm{sup} \, \left{ 0, 1 - \frac{1}{10}, 1 - \frac{1}{10^2}, 1 - \frac{1}{10^3}, \dots \right}$$ which says that for all n, $$1 - \frac{1}{10^n} \lt 1 - \frac{1}{10^{n+1}} \leq 0.999...$$ -- there is no pinching, just contradiction in assuming $$0.999... \lt 1$$.The Pinching Theorem requires a comparison to n for all sides in the sequence to assert the conclusion. You violated this requirement.
All sources on modern mathematics agree $$0.999... = 1$$ so it appears you are trying to invoke an authority you yourself ignore.Your proof has no basis in modern mathematics.
You have not given a proof and certainly not one that I nor 90% of humanity can understand, nor repreduce.
The equality 0.999... = 1 has long been accepted by mathematicians and is part of general mathematical education.