>Of course then most students will insist that 0.999⋯≠1 which shows how ineffective this method is to understanding what the real numbers are.
Okay, you got me. Why is 0.999... != 1 incorrect?
Okay, you got me. Why is 0.999... != 1 incorrect?
Stuff you write down aren't numbers, they're representations or names of numbers. There is a number named "1", there is a number named "2". "2" is not the number itself. You can't infer the two-ness of "2" by staring at the symbols.
It turns out that our way of naming numbers isn't perfect. Each number has multiple names. "2" and "2.0" are both names for the same number, as is "2.00", "2.000", and "2.0000". It turns out that you can name "2" a different way, as "1.9999999999999999999999999999999999999999999999999999...", where there are an infinite number of nines. The names may be different, but the numbers they name are the same.
If you want to talk about this further and to understand what's going on, I'm happy to try to work with you. My email is in my profile.
If the fractional part is not finite, we need to compute an infinite series
n
0.999… = lim Σ 9/10^k = lim ( 1 - 1/10^n ) = 1 - lim 1/10^n = 1 - 0 = 1
^ n→∞ k=1 ^ n→∞ ^ n→∞ ^ ^
| ^ | ^ | | |
definition | induction | algebraic basic now we're
| | limit theorem proof getting silly
0.9 = 1 - 0.1
0.99 = 1 - 0.01
… = …
∞
Σ a⋅r^k = a / (1 - r)
k=0
Another way to look at this is via the constructive definition of the reals as equivalence classes of Cauchy sequences of rational numbers:
The sequences (1, 1, 1, …) and (0.9, 0.99, 0.999, …) are just two of an infinite number of Cauchy sequences which make up the real number 1.
You can see that 1 = 9/9 = 9 * 1/9 = 9 * 0.111... = 0.999...
More intuitively, what is the difference (as in subtraction) between 0.999... and 1? I think you'll find that it doesn't make sense for it to be anything other than zero. And if the difference between two numbers is zero then surely they must be equal.
x = .99999... 10x = 9.99999... 10x - x = 9.999999... - .9999999... 9x = 9 x = 1
10x = 9.99999....
10x - x = 9.9999... - 0.99999....
9x = 9
x = 1
therefore, 0.999... = 1
That last value is, in fact, just zero, only written in a less-condensed form than usual. It is always the case in the real numbers that the only time a - b = 0 is when a = b, so that must mean 1.000... = 0.999... in the real numbers.
More verbosely: At every point in the decimal expansion of the result of the subtraction, the result must be 0 because both 1.000... and 0.999... go on forever. This means that the result must be smaller than any positive real number, which means, because the real numbers contain no non-zero infinitesimal values, it must be zero, which means 1.000... must equal 0.999... because a - b = 0 if and only if a = b.
There are other sets of numbers where this is not the case, such as the p-adic numbers for some p.
Wikipedia has a whole article devoted to this: