James R liked this post
But there is no last digit, if there was the sequence would be finite instead of infinite. So can you see how that means there is a contradiction?In any of your posts, that's where.
Consider, 0.999... = 0.9 + 0.09 + 0.009 + 0.0009 + 0.00009 + ...... upto infinite terms.
Can you tell me, which term in the RHS above is not having the last digit as 9?
Hansda, consider $$0.999... = \sum_{k\geq 1} \frac{9}{10^k}$$
If you agree that the left side is a number, then the right side must also be just a number -- that's what the equals sign means.
So what is the number? I'll pretend I don't know so I'll call it $$S$$.
$$0.999... = S = \sum_{k\geq 1} \frac{9}{10^k}$$
That didn't do much but first it saves some typing and it lets me write the family related quantities $$S_n = \sum_{k=1}^{n} \frac{9}{10^k}$$.
Now every n specifies how many terms we use before we stop the sum. $$S_1 = \frac{9}{10}, \; S_2 = \frac{9}{10} + \frac{9}{100}, \; S_3 = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000}, \; \dots$$
And for every n greater than or equal to 1, we can prove $$S_{n+1} = S_n + \frac{9}{10^{n+1}}$$ and $$\frac{9}{10^{n+1}} \gt 0$$ so $$S_{n+1} \gt S_n$$.
So presumably $$S_1 \lt S_2 \lt S_3 \lt \dots \lt S_{n+5}\lt S_{n+6}\lt S_{n+7} \lt \dots \lt S$$ and indeed this is provable from the rules of inequalities.
So any member of the family of $$S_n$$ is less than $$S$$, in other words each member is a lower bound on what the value of $$S$$ could be. But none is the greatest possible lower bound on what $$S$$ could be.
$$S_n$$ gets ugly to compute the hard way -- imagine the pain if n = 1000. And yet most of all counting numbers are actually larger than 1000. So I want to prove that:
$$S_n = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}}$$ for all counting numbers, n.
First let me prove that $$S_1 = \frac{9}{10} \times \frac{1 - 10^{-1}}{1 - 10^{-1}}$$. The proof writes itself: $$S_1 = \frac{9}{10} = \frac{9}{10} \times 1 = \frac{9}{10} \times \frac{1 - 10^{-1}}{1 - 10^{-1}}$$.
Next, I want to prove that if for a particular counting number k, that $$S_k = \frac{9}{10} \times \frac{1 - 10^{-k}}{1 - 10^{-1}}$$ is true, that $$S_{k+1} = \frac{9}{10} \times \frac{1 - 10^{-(k+1)}}{1 - 10^{-1}}$$ follows as a result. So from the definition of the family, I know that $$S_{k+1} = S_k + \frac{9}{10^{k+1}}$$ and from assumption: $$S_{k+1} = \frac{9}{10} \times \frac{1 - 10^{-k}}{1 - 10^{-1}} \, + \frac{9}{10^{k+1}}$$, So we have $$S_{k+1} = \frac{9}{10} \times \frac{1 - 10^{-k}}{1 - 10^{-1}} \, + \; \frac{9}{10} \times \frac{1}{10^k} = \frac{9}{10} \times \left( \frac{1 - 10^{-k}}{1 - 10^{-1}} + 10^{-k} \right) = \frac{9}{10} \times \left( \frac{1 - 10^{-k}}{1 - 10^{-1}} + \frac{10^{-k} - 10^{-k-1}}{1 - 10^{-1}} \right) = \frac{9}{10} \times \frac{1 - 10^{-k} + 10^{-k} - 10^{-(k+1)}}{1 - 10^{-1}} = \frac{9}{10} \times \frac{1 - 10^{-(k+1)}}{1 - 10^{-1}}$$ which is what I wanted to prove.
Using the principle of finite induction, I now know that $$S_n = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}}$$ is true for all counting numbers, n.
But because $$\frac{9}{10} = 1 - 10^{-1}$$ this also proves that $$S_n = 1 - 10^{-n}$$ for all counting numbers, $$n$$.
Thus our lower bounds on what $$S$$ can be looks like this:
$$1 - 10^{-1} \, \lt \, 1 - 10^{-2} \, \lt \, 1 - 10^{-3} \, \lt \dots \lt \,1 - 10^{-(n+5)} \, \lt \, 1 - 10^{-(n+6)} \, \lt \, 1 - 10^{-(n+7)} \, \lt \dots \lt S$$
Now I want to prove that if $$X$$ is less than 1 then there are members of the family $$S_n$$ which are larger than it.
Say $$X \lt 1$$. Then $$0 \lt 1 - X$$. Then $$1 - X$$ is a
positive number. Then there is a real number $$\epsilon$$ such that $$e^{-\epsilon} = 1 - X$$ because $$f(x) = e^x$$ is a continuous function takes on all positive real values. Also $$f(x \ln 10) = e^{x \ln 10} = 10^x$$. So $$X = 1 - e^{-\epsilon} = 1 - 10^{- \frac{\epsilon}{\ln 10}} $$. So the question of if there are members of the family $$S_n$$ which are greater than $$X$$ reduces to the question of asking if there are counting numbers which are larger than the real number $$- \frac{\epsilon}{\ln 10} = - \frac{\ln ( 1 - X) }{\ln 10}$$ and the answer is yes, most of them.
So every number less than 1 is less than some (or all!) members of the family $$S_n$$ and thus less than $$S$$.
Thus 1 is greatest lower bound on what S could be, because all numbers less than 1 fail to qualify.
One of the biggest principles of the real numbers is that if a non-empty set of numbers has a upper (or lower!) bound then it has a least upper bound (or a greatest lower bound). By this, we see that the set of positive rational numbers whose square is less than 2 has greatest lower bound of 0 and a least upper bound of √2. 0 is not positive and √2 is not rational, but they are the minimal bounds of that particular set of positive rationals. That's why $$S$$ which is the least upper bound of the family of $$S_n$$ need not be a member of the family and can be 1.
Likewise $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} (1 - 10^{-n}) = 1 - \lim_{n \to \infty} 10^{-n} = 1$$ provides a concise line of reasoning for what S is.
Claims that $$1 - S$$ are some sort of positive infinitesimal have been asserted but not reasoned from first principles. For one, there are no rules for manipulating infinitesimals in conjunction with the real numbers, so there are no first principles under discussion.
Secondly, positive infinitesimals make no sense within the context of other numbers we have experience with. Let's assume that near 0 there is the collection of all positive infinitesimals that differ qualitatively from finite rational and real numbers, $$P$$. 0 is the greatest lower bound of $$P$$ and let $$p$$ be the least upper bound of $$P$$. So $$p$$ is either finite or infinitesimal. But because $$p \gt 0$$ then $$p + p > p$$, both $$2 p$$ and $$10 p$$ must therefore be finite and positive while $$p/2$$ must be infinitesimal. But then there is always some counting number $$n$$ such that $$p \leq 10^{-n} \leq 10 p$$. Then $$p \gt 10^{-(n+2)}$$ so there must be some members of $$P$$ that are between $$10^{-(n+2)}$$ and $$p$$. But that contradicts the assumption that infinitesimals are qualitatively different than finite numbers. Something must be wrong in the assumptions being made, but the easiest assumption to deny is that there are any positive infinitesimals. And indeed there aren't in the real numbers.
I was just logging out and checking for typos when I caught your prompt reply. Just quickly then.
