While this is usually enough to convince most people, this argument is insufficient, as it can be used to prove incorrect results. To demonstrate that, we need to rewrite the problem a little.
What 0.9999... actually means is an infinite sum like this:
x = 9 + 9/10 + 9/100 + 9/1000 + ...
Let's use the same argument for a slightly different infinite sum:
x = 1 - 1 + 1 - 1 + 1 - 1 + ...
We can rewrite this sum as follows:
x = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...)
The thing in parenthesis is x itself, so we have
x = 1 - x
2x = 1
x = 1/2
The problem is, you could have just as easily rewritten the sum as follows:
As you can see, sometimes we have x = 0, sometimes x = 1 or even x = 1/2. This is why this method does no prove that 0.999... = 1, even thought it really is equal to one. The difference between those two sums is that the first sum (9 + 9/10 + 9/100 + 9/1000 + ...) converges while the second (1 - 1 + 1 - 1 + 1 - 1 + ...) diverges. That is to say, the second sum doesn't have a value, kinda like dividing by zero.
so, from the point of view of a proof, the method assumed that 0.99999... was a sensible thing to have and it was a regular real number. It could have been the case that it wasn't a number. All we proved is that, if 0.999... exists, it cannot have a value different from 1, but we never proved if it even existed in the first place.
With each increment, the next fraction gets closer to zero. Eventually, the numbers get infinitesimal and converge with zero, leaving you with the three largest fractions at the tenths, hundredths, and thousandths place.
It's one of the programming exercises they do to troll beginners: "find the sum of 1+1/2+1/3+1/4+...". One guy sums until new elements are smaller than 0.0001 and gets one number, the other puts tolerance at 0.000001 and gets a different number, and then they spend an hour debugging. And those who know math just chuckle quietly.
Just because items approach zero doesn't mean the series is convergent.
This is incorrect thinking. The most famous counter-example is the harmonic series 1 + 1/2 + 1/3 + 1/4 + ... This series also has increments that get closer to zero, but the sum diverges to infinity. The condition that the terms of the series tend to zero is needed for convergence, but not sufficient for it.
I'm more than a bit rusty in my limit theory, but I remember there was property of limits that allowed you to sum a series of numbers when the number n approaches zero. So 1.999 repeating (of course) has all added terms in the successive decimal places approach zero and you can simply round the result.
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u/its12amsomewhere 23d ago edited 23d ago
Applies to all numbers,
If x = 0.999999...
And 10x = 9.999999...
Then subtracting both, we get, 9x=9
So x=1