Royston Posted July 30, 2008 Posted July 30, 2008 0.9Recurring5 is between 0.9recurring and 1 Recurring means indefinite, so 0.9Recurring5 doesn't actually make sense. If you wish to stick a 5 on the end of a decimal, then it's a finite number. [math]S_n = 0.9[/math] converges to 1, so I suggest you have a look at limits. Apart from that, can we pleeeeeease, not entertain this topic again, it's really very annoying.
Bignose Posted July 30, 2008 Posted July 30, 2008 The original question was about 0.0recurring1. That has a meaning of sorts. No, it doesn't. How can there be an infinite number of zeros... and then a digit after it? It's an infinite number! There is no end! There is no end to put a digit after -- it has no meaning.
DJBruce Posted July 31, 2008 Posted July 31, 2008 If [math].\overline{9}[/math] means [math].9[/math] followed by an infinite number of zeros by definition of infinite there could not be a [math]5[/math] or a [math]1[/math] on that because the series of [math]9's[/math] is never ending.
D'Nalor Posted July 31, 2008 Author Posted July 31, 2008 If [math].\overline{9}[/math] means [math].9[/math] followed by an infinite number of zeros by definition of infinite there could not be a [math]5[/math] or a [math]1[/math] on that because the series of [math]9's[/math] is never ending. [math].\overline{9}[/math] doesn't mean 0.9 followed by an unending string of zeros, it means 0.9 with an unending string of nines. the point of this thread anyway is to enquire into the possiblility of the ridiculously small number [math].\overline{9}[/math]1 . does it or does it not exist, and if it does not exist, should it?
DJBruce Posted July 31, 2008 Posted July 31, 2008 Wow I am an idiot I meant to say that [math].\overline{9}[/math] is [math].9[/math] followed by a infinite series of nines. And no I do not believe that [math].\overline{9}1[/math] exists for the same reason [math].\overline{9}5[/math] does not exist because the nines are never ending there would be no place to put the 1 or the 5.
Country Boy Posted July 31, 2008 Posted July 31, 2008 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) Well, I can't speak for your mental state! 0.9(recurring) MEANS that the string of 9's has no end so how can you have a 5 AFTER it? Since 0.9(recurring) can be interpreted as the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ..., you might try to interpret this, as I did in an earlier post, as the limit of the sequence 0.95, 0.995, 0.9995, 0.99995,... But it is easy to show that the limits of 0.9, 0.99, 0.999, ..., 0.91, 0.991, 0.9991, ..., 0.95, 0.995, 0.9996,... or, indeed, 0.9a, 0.99a, 0.999a, ... where a is any digit, are all the same: 1. And as far as "i" is concerned, far from being "impossible" or even "improbable", it certainly does exist and is used regularly in physics or even engineering.
John Cuthber Posted July 31, 2008 Posted July 31, 2008 "And as far as "i" is concerned, far from being "impossible" or even "improbable", it certainly does exist and is used regularly in physics or even engineering." No its not, it's used in mathematically modeling physics and engineering. Once you convert the model back to reality "i" goes away. If you calculate the displacement of a bridge in a strong wind is something like (5+3i)mm you have got the answer wrong.
Kyrisch Posted July 31, 2008 Posted July 31, 2008 "And as far as "i" is concerned, far from being "impossible" or even "improbable", it certainly does exist and is used regularly in physics or even engineering." No its not, it's used in mathematically modeling physics and engineering. Once you convert the model back to reality "i" goes away. If you calculate the displacement of a bridge in a strong wind is something like (5+3i)mm you have got the answer wrong. While this may be true, [math]i[/math] can be defined as [math]i^2=-1[/math]. [math]0.\bar 95[/math] can't really be defined as anything. And, as I said before, how can you compare it to [math]0.\bar 9[/math]? Is [math]0.\bar 9[/math] really bigger than [math]0.\bar 95[/math]? They should both have the same number of 9's (an infinite amount). Given that, it can be construed that [math]0.\bar 95[/math] is in fact greater. But then, if there is a place to put the 5, then there must be a place to put another 9, in which case it can be construed that [math]0.\bar 95[/math] is in fact smaller than [math]0.\bar 9[/math] by a factor of [math]0.\bar 04[/math]. So, even if the number [math]0.\bar 95[/math] existed, it cannot be evaluated as either greater than or less than [math]0.\bar 9[/math], which, I believe, was the original point of you bringing it up.
D'Nalor Posted August 4, 2008 Author Posted August 4, 2008 Ok, maybe we're looking at this the wrong way. what is [math]0.\overline{0}1[/math] as a fraction. It should be 1 divided by infinity(I haven't figured out how to draw that yet, any help?). that should be right. by the same princible, that would work with any number as the numerator, with that number on the end insteed of 1.
ajb Posted August 4, 2008 Posted August 4, 2008 Infinity is not a number so you can't divide anything by infinity. What you can do is consider the limit as something tends to infinity. So, what you have is something like [math]\lim_{x\rightarrow \infty}\frac{1}{x}= 0[/math]
D'Nalor Posted August 6, 2008 Author Posted August 6, 2008 I Don't think that that really matters. infinity is just a concept, and [math]0.\overline{0}1[/math] is also a concept. infinity doesn't exist because once you try to write it out, you'll never be able to finish, because you'll be able to write in at least one more digit. [math]0.\overline{0}1[/math] is like that, exept you just keep adding 0s in between the decimal point and the 1.
ajb Posted August 6, 2008 Posted August 6, 2008 What do you mean by "infinity does not exist"? All numbers (and other maths things) are "concepts". It is true that infinity is not a real number. By "exist" do you mean that you "cannot assemble an infinite collection of physical objects"? That is you cannot "see" a set of infinite cardinality. Remember your days back when you were first shown how to count, "one apple", "two cars" etc. What you were doing is defining a map from a set of objects to the natural numbers. You call this the "number x". Really this is the cardinality and should not be identified with the number itself. That is the natural numbers "exist as a mathematical concept" independent of being assigned as a cardinality of a set. Don't confuse poor notation with fundamental ideas, [math]\infty[/math], I just wrote the mathematical symbol for infinity and it did start and end on this page! You should adopt the attitude that mathematics "exists" independent of the notation used. I think model theory deals with issues like this.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now