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.
Actually 0.99… is the same as 1. They both represent the same number
https://en.m.wikipedia.org/wiki/0.999…
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.
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That’s simply not true, as I demonstrated in my example.
There was also a veritasium video about this.
It was interesting.
His videos always are
If .99…9=1, then 0.999…8=0.999…9, 0.99…7=0.999…8, and so forth to where 0=1?
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.
Wonderful explanation. It got the point across.
That is the best way to describe this problem I’ve ever heard, this is beautiful
This is the kind of stuff I love to read about. Very cool
Maybe a stupid question, but can you even divide a number with infinite decimals?
I know you can find ratios of other infinitely repeating numbers by dividing them by 9,99,999, etc., divide those, and then write it as a decimal.
For example 0.17171717…/3
(17/99)/3 = 17/(99*3) = 17/297
but with 9 that would just be… one? 9/9=1
That in itself sounds like a basis for a proof but idk
Yes that’s essentially the proof I learned in high school. 9/9=1. I believe there’s multiple ways to go about it.
0.999…8 does not equal 0.999…9 so no
If the “…” means ‘repeats without end’ here, then saying “there’s an 8 after” or “the final 9” is a contradiction as there is no such end to get to.
There are cases where “…” is a finite sequence, such as “1, 2, … 99, 100”. But this is not one of them.
I’m aware, I was trying to use the same notation that he was so it might be easier for him to understand
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:
1/3 = 0.3_
0.3_ * 3 = 0.9_
0.9_ = 3/3 = 1
Huh… Where did you get “0.999… = 0.999…8” from? There’s a huge difference here.
Read carefully. I wrote a finite number(0.999…9)
Think carefully
What does 0.99…8 represent to you exactly
If it’s an infinite amount of 9s then it can’t end in an 8 because there’s an infinite amount of 9s by definition so it’s not a real number
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.
If you really wants to understand the concept , you need to learn about limits
There is no .99…8.
The … implies continuing to infinity, but even if it didn’t, the “8” would be the end, so not an infinitely repeating decimal.