0.99 (Repeating) = 1?
- Thread starter Minimike3636
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Seriously, what? It's common practice to make certain machine systems parse it correctly and () has no independent imputation other than "what is inside this is to be dealt with first" or to indicate multiplication. I used to do that all the time. How the hell is a second grader at fault for emphasizing a negative number when they're not supposed to know about parentheses anyway?Dramus said:
Edit: answered while posting.
You would first have to "prove" that repeating numbers exist. If you assert that they do (and you do, with your claim of 1/9 = 0.111r), then you're saying that operations can be performed on them. 9 * 0.111r = 0.999r using conventional maths, although we've already defined that 0.111r = 1/9. 9 * 1/9 = 9/9 = 1. I would consider that a fairly trivial proof of 0.999r = 1, unless you claim that multiplication doesn't work like that on recurring numbers. I can assure you it does, and that no matter how infinitesimally small you go down 0.111r, the end result will always be 0.999r.Dramus said:
We've done this before. I'm going to spoil it for you and say that 0.999... = 1. I'm a mathematics major in college. This is what I do. I'm right. For those of you who understand math, I'll copy paste the proof that I wrote last time...
This causes a lot of people heartache, but actually, there is a simple proof of why 0.9 repeating is equal to 1. Suppose 0.999... is not equal to 1. This tells us that there exists delta such that (**)delta = | 1 - 0.999... | > 0 (i.e. different numbers have some distance between them on the number line). Now, let us also suppose that 0.999... = lim n->infinity of xn, where x1 = 0.9, x2 = 0.99, x3 = 0.999, and so on (so xi = 0 followed by "i" nines).
So, it follows that 1 - x1 = 0.1, 1 - x2 =0.01, and that 1 - xi = "i" zeros followed by a one. This is the same as the series yn = 1 - xn = 10^(-n). By the Archimedean Principle, for every number "A" greater than zero, there is some rational number N such that if n >= N, 1/n 0 such that | 1 - 0.999... | = delta, delta = 0 by contradiction. Q.E.D
sorry for the wall of text
/math nerd
This causes a lot of people heartache, but actually, there is a simple proof of why 0.9 repeating is equal to 1. Suppose 0.999... is not equal to 1. This tells us that there exists delta such that (**)delta = | 1 - 0.999... | > 0 (i.e. different numbers have some distance between them on the number line). Now, let us also suppose that 0.999... = lim n->infinity of xn, where x1 = 0.9, x2 = 0.99, x3 = 0.999, and so on (so xi = 0 followed by "i" nines).
So, it follows that 1 - x1 = 0.1, 1 - x2 =0.01, and that 1 - xi = "i" zeros followed by a one. This is the same as the series yn = 1 - xn = 10^(-n). By the Archimedean Principle, for every number "A" greater than zero, there is some rational number N such that if n >= N, 1/n 0 such that | 1 - 0.999... | = delta, delta = 0 by contradiction. Q.E.D
sorry for the wall of text
/math nerd
In the sense that 0.999... = 1, yes it is the quotient of two integers. Also, a number doesn't need to be the quotient of two integers to be a number... look at the square root of two.Dramus said:
I literally just posted on the other thread and then saw this one, and will impart my wisdom again.
Limits of infinite sequences and series, you cover it in calculus, and your math teacher needs to go to back college before being allowed to teach again.
Limits of infinite sequences and series, you cover it in calculus, and your math teacher needs to go to back college before being allowed to teach again.
The limit of n/(n+1) as n->infinity is, in fact, 1, but it equals .9 repeating. The limit is only the theoretical existence of a number at a particular point, so it will never neccessarily get there. By the time it does reach .9 repeating (as in a truly infinite number of nines), it will equal one for all intents and purposes, but will never actually get there (as infinity cannot be reached).Dramus said:
