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.
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.
No, it’s not “so close so as to basically be the same number”. It is the same number.
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.
No, there is no difference. Infitesimal or otherwise. They are the same number, able to be shown mathematically in a number of ways.
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.
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.
Infinity small is infinity small. Not zero
That’s simply not true, as I demonstrated in my example.