In calculus there are often values that approach ever closer to,
but never reach some point. Yes. Would this not be the
same, so 0.9999... approaches, but never reaches 1.0?
No. 1/3 is
exactly 0.3333... and three times that is exactly 3/3 or 0.9999..., and
3/3 is 1. I think it's more obvious if you do it with
ninths:
1/9 = 0.1111...
1/9 * 9 = 0.1111... * 9
9/9 =
0.9999...
1 = 0.9999...
Another proof:
(1) 0.9999... =
x
(2) 9.9999... = 10x (mutiply both sides by 10)
(3) 9.9999... -
0.9999... = 10x - x (same on both sides, by (1).)
(4) 9.9999... - 0.9999...
= 9 (both have the same number of digits: infinity)
(5) 10x - x = 9x (simple
algebra.)
(6) 9 = 9x (substituting from (4) and (5) into (3).)
(7) 1 = x
(divide both sides by 9.)
(8) 1 = 0.9999.... (substituting from
(1).)
Now, of course, the part that may give you pause is (4). But it's
correct. Remember, infinity doesn't end. If you add or subtract one from it, you
still have infinity. So, if you shift 0.9999... one place to the left
(multiplying by 10), you still have an infinite number of nines following the
decimal. Not infinity-minus-one, because there is no such thing--it's just
infinity. So, when you do the subtraction, all the nines after the
decimal go away, because there's exactly the same number of them.
You
can find this and several other proofs on YouTube, here.
--- Do not
meddle in the affairs of Wizards, for it makes them soggy and hard to light. [ Reply to This | Parent | # ]
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