Phil Scott
Member
Seriously, steve_bank, this is basic undergraduate level real analysis. If you're not familiar with this stuff, there's no shame in it. But you shouldn't be trying to educate anyone else about real numbers if so.No. Numbers can have infinite decimals and still be computable. If there is a program that can compute the decimal to arbitrary precision, then the number is said to be computable.
And to correct an earlier post of yours: 0.999... is a real number. It denotes the limit of a particular geometric series, and that limit is 1.
How does 0.999... resolve to 1 as a limit?
Formally: any infinite decimal x.x0x1x2x3... literally denotes the limit of the series
\(x + \frac{x_0}{10} + \frac{x_1}{100} + \frac{x_2}{1000} + \frac{x_3}{10000} + \cdots\).
The limit of a series, in general, is the limit, should it exist, of the sequence of initial prefix sums of the series. So for decimals, that means that
x.x0x1x2x3...
is the limit of the sequence
\(x, x + \frac{x_0}{10}, x + \frac{x_0}{10} + \frac{x_1}{100}, x + \frac{x_0}{10} + \frac{x_1}{100} + \frac{x_2}{1000}, x + \frac{x_0}{10} + \frac{x_1}{100} + \frac{x_2}{1000} + \frac{x_3}{10000} + \cdots\)
The limit of a sequence, in general, is defined as that number, should it exist, that the sequence eventually becomes arbitrarily close to. Formally, given a sequence { xn }, the sequence converges to some limit l if and only if, given an arbitrary rational error threshold ε > 0, there exists a point N such that, for all n > N, |xn - l| < ε.
In other words, no matter what error threshold you give me, I can name a point in the sequence after which I stay within that threshold of the limit.
This limit is guaranteed to exist in the general case of infinite decimals due to the least upper bound property, which must itself be either taken as an axiom, or else established by a Construction of the real numbers.
However, for the trivial case of 0.999..., we have a very simple geometric series, the limit being defined as the limit of
0
0.9
0.99
0.999
...
and that limit is easy to establish. I say it is 1. Because if you give me a rational error threshold ε = p/q with p,q > 0, I can take the (q+1)th term in the sequence. The difference between 1 and this term is 1/10^q < 1/q < p/q. That difference decreases for all later terms, so from the (q+1)th term, we stay within the error threshold, satisfying the defining property of the limit.
Hence, 0.999... = 1.
I assure you that none of these are tricks, nor any of it esoteric calculus. These are the formal definitions. They are precisely what mathematicians have in mind when they talk about real numbers and limits, and what they require to meet their demands for rigor. They are not simply running with their vague intuitions gathered from playing with calculators.
This stuff may not apply when you're doing number crunching with a computer, and it might not apply in pharmacy, but it's what the actual mathematicians have been talking about for the last century and a half when making and reasoning with these distinctions.