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I read the following in a wikipedia article about 0.9999...:

A third derivation was invented by a seventh-grader who was doubtful over her teacher's limiting argument that $0.999... = 1$ but was inspired to take the multiply-by-$10$ proof above in the opposite direction: if $x = ...999$ then $10x = ...990$, so $10x = x − 9$, hence $x = −1$ again."

If $x=0.999.......9$, then how can $10x=......90$? It must be $9.9999.....9$.

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    $\begingroup$ You have mistaken ...999 for .999.... They are different. The second is a decimal, all nines. The first is an attempt to make an integer with infinitely many digits, all nines. $\endgroup$ – Gerry Myerson Apr 12 '15 at 7:27
  • $\begingroup$ Basically there is no gap between the number 0.99999...... and -1 and 1. You cannot find a number between 0.99999...... and 1 and also between 0.99999..... and -1 $\endgroup$ – user210387 Apr 12 '15 at 7:27
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    $\begingroup$ @SayanChattopadhyay "and also between 0.99999..... and -1" - how about $0$? $\endgroup$ – user207868 Apr 12 '15 at 7:28
  • $\begingroup$ @SayanChattopadhyay Is it possible by any way to find a rational or irrational number between 0.99999.... and 1 $\endgroup$ – Shm Mukhj Apr 12 '15 at 7:38
  • $\begingroup$ You cannot find any number @ShmMukhj this was the first question i encountered in my 6th grade class $\endgroup$ – user210387 Apr 12 '15 at 7:39
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What you have read concerns analysis of functions of $p$-adic numbers, not real ones. Why? Because previous paragraph states that "In the 10-adic numbers, the analogues of decimal expansions run to the left.".

If you think about real numbers as limits of series (here $n$ is an integer): $$x = \sum_{k=n}^\infty \frac{a_k}{10^k} \in \mathbb R, \quad 0 \le a_k < 10$$ then $p$-adic numbers are simply the same series "running to the left". In usual topology (with euclidean metric) it's not convergent, but with $p$-adic metric, it is. $$x = \sum_{k=n}^\infty a_k p^k \in \mathbb Q_p, \quad 0 \le a_k < p.$$

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  • $\begingroup$ is the calculation given here is for finding value of 9 recurring? $\endgroup$ – Shm Mukhj Apr 12 '15 at 7:40
  • $\begingroup$ @ShmMukhj Basically, $\dots 999$ isn't a number in the traditional sense. It's not an integer or a real number, since those can't have an infinite number of digits on the left. But $.999\dots$ is a real number, defined as $\sum_{n=1}^\infty \frac{9}{10^n}$, which equals $1$. $\endgroup$ – Kitegi Apr 12 '15 at 7:50

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