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large decimals


D'Nalor

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Is it at all possible to have the number 0.0(reoccuring)1? or for that matter any other number at the end? my idea is that the number is 0.0 with an uneding string of 0s on the end, but if it did have an end, it would end in 1.

 

Is this possible?

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Is it at all possible to have the number 0.0(reoccuring)1? or for that matter any other number at the end? my idea is that the number is 0.0 with an uneding string of 0s on the end, but if it did have an end, it would end in 1.

 

Is this possible?

 

If zero's are recurring, at which point would you be able to attach a zero at the end? ;)

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Is it at all possible to have the number 0.0(reoccuring)1? or for that matter any other number at the end? my idea is that the number is 0.0 with an uneding string of 0s on the end, but if it did have an end, it would end in 1.

 

Is this possible?

 

"but if it did have an end"? And you just said "unending string of 0s"? Surely you see the logical falacy! :)

 

The nearest I can come to matching that is to consider the sequence of finite decimals: 0.1, 0.01, 0.001, 0.0001, 0.00001, etc. Its limit could be considered to be "an unending string of 0s with a 1 on the end".

 

Of course, it is easy to see that the limit is simply 0. In fact, the limit of the sequence 0.a, 0.0a, 0.00a, 0.000a, 0.0000a, etc. is 0 no matter what digit 0 is. It really is just an unending string of 0s!

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I wonder if we are heading towards infinitesimals? These are not real numbers, but "numbers infinitesimally close to zero". We use them all the time in calculus, differential geometry, deformation theory... the list goes on....

 

You really have to deal with them quite formally as "external parameters". For example, [math]dx[/math] is really a formal object.

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0.0recurring1 may be a bit meaningless, but it's the difference between 0.9recurring and 1

Since 0.9recurring is equal to 1, 0.0recurring1 is equal to zero.

 

0.9recurring is not equal to 1, it is just a tiny bit smaller than 1, but the difference is so small it is just considered to be 1.

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0.9recurring is not equal to 1, it is just a tiny bit smaller than 1, but the difference is so small it is just considered to be 1.

 

That's actually not the case... [math]0.\overline{9}[/math] is actually equal to 1. There are plenty of proofs for this.

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0.9recurring is not equal to 1, it is just a tiny bit smaller than 1, but the difference is so small it is just considered to be 1.

 

0.99999999... is, by definition, the limit of the sequence 0.9, 0.99, 0.999, 0.999, etc. That is a geometric sequence and it is easy to show that its limit (and so the value of 0.9999... ) is exactly 0.11/(1- 0.9)= 0.1/0.1= 1.

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".9recurring is not equal to 1, it is just a tiny bit smaller than 1, but the difference is so small it is just considered to be 1."

Er, no. The difference is exactly 0.0recurring1

However as the difference is also zero then it follows that 0.0recurring1 is also zero.

 

If you don't like limits of series try this

X= 0.9...

Multiply both sides by 10

10X= 9.9...

Subtract X (ie 0.9...) from both sides

10X-X =9

Simplify

9X=9

leaving you with

X=1

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0.9recurring is not equal to 1, it is just a tiny bit smaller than 1, but the difference is so small it is just considered to be 1.

 

Wonderful, we're going to start this again.

 

Please answer this question which cuts right to the heart of the matter. If 0.99999... (infinitely repeating 9's) is NOT equal to 1, what is the number that comes between them? If there is no number that comes between them, then they are equal. If they are unequal, then there must be a number between them. So... since you say they aren't equal, what is the number between them?

 

Same thing with your 0.0000 (infinitely repeating 0's)1 number. First of all, there really isn't any such number. If there are infinitely repeating 0's, how can there be a 1 on the "end"? It can't, there are infinitely many 0's. But, even if it were possible? What number comes between 0.00000 ... etc. and 0? Again, there isn't any such number, so they are equal.

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Same thing with your 0.0000 (infinitely repeating 0's)1 number. First of all, there really isn't any such number. If there are infinitely repeating 0's, how can there be a 1 on the "end"? It can't, there are infinitely many 0's. But, even if it were possible? What number comes between 0.00000 ... etc. and 0? Again, there isn't any such number, so they are equal.

 

 

Yes, it I such more like what I said. In a rough sense you need to "add" such "numbers" to the real line. This is the theory of infinitesimals.

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"Please answer this question which cuts right to the heart of the matter. If 0.99999... (infinitely repeating 9's) is NOT equal to 1, what is the number that comes between them? "

OK.

Within the bizarre context of this thread the number 0.9recurring5 exists and meets that criterion.

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0.(9) is greater than 0.99999, which is in turn less than 1. (If I have read your reply right)

 

The basic premise is that you can't find a real number that is closer to 1 than 0.(9). In fact, you can show that 1 and 0.(9) are actually just different representations of the same number. You can even see a proof of this on wiki.

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"(If I have read your reply right)"

No, it seems you have not.

Have a look at what this whole thread is about.

It's about the weird world of numbers that have an infinite number of, for example, 9s after the deimal point then another number.

 

"In fact, you can show that 1 and 0.(9) are actually just different representations of the same number. You can even see a proof of this on wiki."

Nevermind wiki, I posted a proof of that earlier in this thread (did you read the thread btw?).

However if you accept that "zero point nine recurring with a five on the end " has a meaning, then it falls between 0.9recurring and 1

If you don't accept that the term has a meaning (and anywhere other than this thread I would go allong with that) then the thread is meaningless anyway.

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If 0.9(recurring)5 is meaningless, then why do I understand what it means? I know that although the acctual number is impossible(or improbable), isn't the number i (the square root of -1) in physics also an impossible(or improbable) number?

 

(by the way, Kyrisch, nice paradox)

Edited by D'Nalor
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If 0.9(recurring)5 is meaningless, then why do I understand what it means?

 

Because you don't? The number itself is a contradiction. You have an endless repetition of 9's, and then at the end, a 5. But there is no end, so there can't be a 5 at the end.

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The original question was about 0.0recurring1. That has a meaning of sorts. It's the difference between 0.9recurring and 1

It's the limit (though not, I think, in the strict mathematical sense) of

1-.9=0.1

1-.99=0.01

1-.999=0.001

and so on to

1-.9recurring= "0.0recurring with a 1 tagged on the end".

From that point of view it's a perfectly reasonable concept. It happens to be another way of writing zero.

 

Once you accept that you can label some of these weird (and very near useless)figments of imagination as having certain properties you can look at those properties.

For example, in the same way that 0.95 is between 0.9 and 1 you can see that

0.9Recurring5 is between 0.9recurring and 1

 

It's not physically meaningful, but (as has been pointed out) nor is i. That doesn't stop mathematicians playing with it.

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