My Math Forum Square roots of numbers < 1

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 March 29th, 2014, 10:04 AM #11 Senior Member   Joined: Nov 2013 Posts: 160 Thanks: 7 I already mentioned a book by Rudy Rucker: Infinity And The Mind. He is a mathematician and writes on his book page 79 that the number K with decimal expansion 0.99999... is the same as 1, his argument is 10K=9.99999.... - K=0.99999.... ------------------------ 9K =9 K =1 He continues "maybe this argument is misleading. What if there is some number call it 1 - 1/ω, that is greater than any finite string of 0.9.....9 of nines, yet less than 1? If K were equal to 1-1/ω, there is infinitesimal quantity that does not get canceled out, and the above argument does not work": 10K=10 - 10/ω - K= 1 - 1/ω ---------------------- 9K = 9 - 9/ω K = 1 - 1/ω Last edited by TwoTwo; March 29th, 2014 at 10:18 AM.
 March 29th, 2014, 08:54 PM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra You are creating a sequence where $a_{n+1}= sqrt{a_n} ,\, 0 < a_0=< 1=$ If the sequence approaches a limit, L, we have \begin{align*} L &= \sqrt{L} \\ L^2 - L &= 0 \\ L\left(L-1\right) &= 0 \end{align*} So L = 0 or L = 1 $a_{n+1}^2 - L^2= a_n - L$ So $0 < a_n < 1 \;\Rightarrow\; 0 < a_{n+1} < 1$ Thus the sequence is bounded above. $a_{n+1}^2 - a_n^2= a_n - a_n^2 = a_n \left(1 - a_n \right)$ So $0 < a_n < 1 \;\Rightarrow\; a_{n+1}^2 - a_n^2 > 0$ Thus the sequence is increasing. Since the sequence is both bounded above and increasing, it must converge to a limit. And we showed above that the limit must be 1. However, since the sequence is increasing, we can say that there is no N such that $a_N= 1$ Last edited by v8archie; March 29th, 2014 at 08:59 PM.
 March 30th, 2014, 06:44 AM #13 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 In general, the successive values are approximate, so when the calculator internally obtains, say, 0.999999999896, and cannot display that many digits, it may well display 1 instead. Eventually, even the internally calculated value may become exactly 1.

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