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The problem with this argument is it assumes you can get to infinity 9s by adding them one-by-one. It's like adding sides to a regular polygon to get to a circ
by iopq 10y ago
The problem with this argument is it assumes you can get to infinity 9s by adding them one-by-one.
It's like adding sides to a regular polygon to get to a circle. No matter how many sides you add, you still have a polygon. You can only say that a circle is a polygon with infinity sides, but you'll never get there adding them one by one.
- kazinator 10y agoThe argument is that you cannot get to infinity by adding one by one. (If the argument assumed otherwise, it would only be temporarily, for the sake of setting up a reductio ad absurdum.)
- iopq 10y agoYou can't use induction to get to an infinite amount of nines, so induction doesn't work in this case because you never get there. The number of nines is not in the set of natural numbers, yet you're trying to use induction by covering every natural number.
- kazinator 10y agoYou cannot simultaneously believe that "you never get there" (i.e. an infinite number of nines is not constructible) and believe that the syntax 0.999... denotes an actual infinite string of 9's which has a straightforward value (that happens to be precisely 1). That simply isn't the basis for how 0.999... is regarded as 1. Rather, 1 is the supremum ("least upper bound", LUB) of the countably infinite set { 0.9, 0.99, 0.999, ... } as a subset of the reals. We define that 0.9... denotes that supremum: i.e. that the ellipses suffixed to 0.9 denote the expansion of 0.9 into the set { 0.9, 0.99, 0.999, ... } followed by determining the LUB of subset among the reals. No "infinite string of 9's whose length isn't a natural number" nonsense is involved.