Recently I stumbled across a, to me, rather strange idea. I was messing around with the proof of 0.999...=1, when I figured that what 0.999... means is that those are all nines. That way I came upon a weird idea. Say a=0.999..., then a=1−x. Yet, how do we define what x is? It should say x=0, but my theory did not. If 0.999... is just an endless sequence of nines, then why can't we say x just is an endless sequence of zeroes, ending with a 1, like 110∞?
If we take the equation n=0.9
Thanks in advance.
Sjoerd Dorrestijn.
EDT: I prefer 110∞ to use as an indication of 0.000...01, even though 10∞=∞ in some way, I just seem to find this more clear.
EDT2: Just to be clear, I read a proof that 0.999...9=1 because it'd be 1−0.000...0=1−0=1. What I tried to prove here is that it is not equal to 1, because otherwise maths would collapse. My question was whether I am right or wrong. Since the original statement uses infinity (an infinite amount of nines) I think it is a must to use infinity as well. So, the question is if either the original statement is false, or if I made a mistake somewhere.
Answer
The (proven true) statement that 0.999…=1 means that limn→∞0.999…9⏟n=1. That in turn is equivalent to limn→∞(1−0.999…9⏟n)=0. Now note that 1−0.999…9⏟n=110n, so we have limn→∞110n=0.
However, although the above reasoning works, it is easier and more straightforward to conclude that limn→∞110n=0 directly from the definition of the limit, without ever considering 0.999….
Finally, it is advisable to avoid the notation 110∞ altogether, as it is not defined given the usual definition of its constituent symbols.
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