Right, it's only a problem because we chose base ten (a rather inconvenient number). If we did math in base twelve, 1/3 in base twelve would simply be 0.4. It doesn't repeat. Simply, then, 1/3 = 0.4, then (0.4 × 3) = (0.4 + 0.4 + 0.4) = 1 in base twelve. No issues, no limits, just clean simple addition. No more simple than how 0.5 + 0.5 = 1 in base ten.
One problem in base twelve is that 1/5 does repeat, being about 0.2497.... repeating. But eh, who needs 5? So what, we have 5 fingers, big whoop, it's not that great of a number. 6 on the other hand, what an amazing number. I wish we had 6 fingers, that'd be great, and we would have evolved to use base twelve, a much better base!
It's a weird concept and it's possible that I'm using it incorrectly, too - but the context at least is correct. :)
Edit: I think I am using it incorrectly, actually, as in reality the difference is infinitesimally small. But the general idea I was trying to get across is that there is no real number between 0.999... and 1. :)
It is possible to define a number system in which there are numbers infinitesimally less than 1, i.e. they are greater than every real number less than 1 (but are not equal to 1). But this has nothing to do with the standard definition of the expression "0.999...," which is defined as the limit of the sequence (0, 0.9, 0.99, 0.999, ...) and hence exactly equal to 1.
They said its the same number though, not basically the same. The idea that as you keep adding 9s to 0.9 you reduce the difference, an infinite amount of 9s yields an infinitely small difference (i.e. no difference) seems sound to me. I think they’re spot on.
Yes, thats what we're saying. No one said it's an infinitesimally small difference as in hyperbolically its there but really small. Like literally, if you start with 0.9 = 1-0.1, 0.99 = 1-0.01, 0.9... n nines ...9 = 1-0.1^n. You'll start to approach one, and the difference with one would be 0.1^n correct? So if you make that difference infinitely small (infinite: to an infinite extent or amount): lim n -> inf of 0.1^n = 0. And therefore 0.999... = lim n -> inf of 1-0.1^n = 1-0 = 1.
I think it's a good way to rationalize, why 0.999... is THE SAME as 1. The more 9s you add, the smaller the difference, at infinite nines, you'll have an infinitely small difference which is the same as no difference at all. It's the literal proof, idk how to make it more clear. I think you're confusing infinitely and infinitesimally which are not at all the same.
Technically you're both right as there are no infinitesimals in the real number system, which is also one of the easiest ways to explain why this is true.
That's what it means, though. For the function y=x, the limit as x approaches 1, y = 1. This is exactly what the comment of 0.99999... = 1 means. The difference is infinitely small. Infinitely small is zero. The difference is zero.
The tricky part is that there is no 0.999...9 because there is no last digit 9. It just keeps going forever.
If you are interested in the proof of why 0.999999999... = 1:
0.9999999... / 10 = 0.09999999...
You can divide the number by 10 by adding a 0 to the first decimal place.
0.9999999... - 0.09999999... = 0.9
because the digit 9 in the second, third, fourth, ... decimal places cancel each other out.
Let's pretend there is a finite way to write 0.9999999..., but we do not know what it is yet. Let's call it x. According to the above calculations x - x/10 = 0.9 must be true. That means 0.9x = 0.9. dividing both sides by 0.9, the answer is x = 1.
The reason you can't abuse this to prove 0=1 as you suggested, is because this proof relies on an infinite number of 9 digits cancelling each other out. The number you mentioned is 0.9999...8. That could be a number with lots of lots of decimal places, but there has to be a last digit 8 eventually, so by definition it is not an infinite amount of 9 digits before. A number with infinite digits and then another digit in the end can not exist, because infinity does not end.
Your way of thinking makes sense but you're interpreting it wrong.
If you can round up and say "0,9_ = 1" , then why can't you round down and repeat until "0 = 1"? The thing is, there's no rounding up, the 0,0...1 that you're adding is infinitely small (inexistent).
It looks a lot less unintuitive if you use fractions:
No, because that would imply that infinity has an end. 0.999… = 1 because there are an infinite number of 9s. There isn’t a last 9, and therefore the decimal is equal to 1. Because there are an infinite number of 9s, you can’t put an 8 or 7 at the end, because there is literally no end. The principle of 0.999… = 1 cannot extend to the point point where 0 = 1 because that’s not infinity works.