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.
It’s so dumb and it makes perfect sense at the same time. There is an infinitely small difference between the two numbers so it’s the same number.
There is no difference, not even an infinitesimally small one. 1 and 0.999… represent the exact same number.
They only look different because 1/3 out of 1 can’t be represented well in a decimal counting system.
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!
I mean, there is no perfect base. But the 1/3=0.333… thing is to be understood as a representation of that 1 split three ways
Well, technically “infinitesimally small” means zero sooooooooo
Edit: this is wrong
An infinitesimal is a non-zero number that is closer to zero than any real number. An infinitesimal is what would have to be between 0.999… and 1.
You are correct and I am wrong, I always assumed it to mean the same thing as a limit going to infinity that goes to 0
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. :)
I think you did use it right tho. It is a infinitesimal difference between 0.999 and 1.
“Infinitesimal” means immeasurably or incalculably small, or taking on values arbitrarily close to but greater than zero.
The difference between 0.999… and 1 is 0.
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.
Wait what
I always thought infinitesimal was one of those fake words, like gazillion or something
It sounds like it should be, but it’s actually a real (or, non-real, I suppose, in mathematical terms) thing! :)
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.
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.
deleted by creator
That’s simply not true, as I demonstrated in my example.