What if you have 0.9̅4? Can we say 0.9̅5 > 0.9̅4 > 0.9̅3? More on what happens if you allow this: https://mathwithbaddrawings.com/2013/08/13/the-kaufman-decim...
0.999...= 1
151–160 of 647 posts
Re: 0.999...= 1
#152"In other words, "0.999..." and "1" represent the same number." - Ok, I'm done with this world.
Re: 0.999...= 1
#153A formally rigorous proof of this (in Metamath) is here:
http://us.metamath.org/mpeuni/0.999....html
Unlike typical math proofs, which hint at the underlying steps, every step in this proof only uses precisely an axiom or previously-proven theorem, and you can click on the step to see it. The same is true for all the other theorems. In the end it only depends on predicate logic and ZFC set theory. All the proofs have been verified by 5 different verifiers, written by 5 different people in 5 different programming languages.
You can't make people believe, but you can provide very strong evidence.
Re: 0.999...= 1
#154Earlier quoted context omitted.
Another secondary school 'proof' x = 0.9999..... 10x = 9.9999..... (10x -x) = 9x = (9.9999.... - 0.9999....) = 9 x = 9/9 = 1
> 10x -x subtracting infinities is dangerous, you can achieve any result from it https://www.youtube.com/watch?v=-EtHF5ND3_s
You're not subtracting anything infinite. Whatever you think of 0.99999... (and the correct thing to think is that it equals 1), I hope we can both agree that it is at least finite! I mean, we can agree that it's less than 2 and more than 0, right?
That subtraction is just as valid as saying 0.333... + 0.333... = 0.666..., or that 1/3 + 1/3 = 2/3.
Re: 0.999...= 1
#155Earlier quoted context omitted.
Both are false. 0.99999... is not less than 1. It is the same as 1.
Yes, because someone defined it that way. It is because the "limit" in 0.999... = lim[eps->0] 1-eps is implicit and defined as being applied before anything else. But you might as well define that implicit limit as applying over the entire expression. UPDATE: So instead of interpreting the expression as: (lim[eps->0] 1-eps) which is indeed false, you can also interpret the expression as: lim[eps->0] ((1-eps) which is…
> But you might as well define that implicit limit as applying over the entire expression.
What is this supposed to even mean?
Re: 0.999...= 1
#156Earlier quoted context omitted.
(STATEMENT OF PERSONAL IGNORANCE [SOPI]: Anyone who actually understands this stuff please correct my mistakes below. Thanks.) In the real numbers, which are not always simple or intuitive, 0.99... = 1. That's true and I seem to understand the proof. But the real numbers aren't the only system that might be sitting behind "0.99..." and "1" when I write those symbols down and talk intuitively to people in my family. T…
Those number systems do exist, but I'm not sure it's right to say they work just as well for everyday purposes. They work only as long as you use them in a way that reduces to treating them as real numbers, either never computing an infintesimal in the first place or calculating 23 + 6ε and saying "oh that's basically just 23".
Sure. And it's true, 0.99... is equal to 1.
All I'm saying is [SOPI below] it's all a little more technical than the junior high school proof. For example if 23+6\epsilon = 23, then how do I define 23 + 6\epsilon - 23? I can choose different approaches here, but "zero" is going to be pretty inconvenient when I go to do an integral.
[SOPI] Statement of Personal Ignorance. I don't quite know what I'm talking about. If you do know, please step in and help correct me.
Re: 0.999...= 1
#157Personally I've always thought "proofs" using "arithmetic" are right, but kind of stated backwards. The point is that in elementary school arithmetic, you define addition, multiplication, subtraction, division, decimals, and equality, but you never define "...". Until you've defined "...", it's just a meaningless sequence of marks on paper. You can't prove anything about it using arithmetic, or otherwise. What the "a…
> Personally I've always thought "proofs" using "arithmetic" are right, but kind of stated backwards. I've never considered them right at all. By saying something like 0.9... x 10 = 9.9... and then saying that 9.9... - 0.9... = 9 you're basically just a priori defining 0.9... to be 1. In other words you're basically just defining 0.9... as a symbol to be some number x which has the property that 10x - x = 9. So you'r…
> you’re basically just a priori defining 0.9... to be 1.
I think the point is not defining 0.9... to be 1, the point is that “...” means an infinite number of 9s. If you shift the decimal point by 1, then nothing changes, there are still an infinite number of 9s. If you shift the decimal point by 5 places, there are still an infinite number of 9s to the right. And here is the logical (induction) step: if you shift the decimal point by an infinite number of places, then there are still an infinite number of 9s to the right. This works for any repeating fraction, in groups of more than 1 repeating digit.
> I’ve never considered them right at all.
Do you mean you disagree with the result, or that you agree with the result but don’t believe the proof is really a proof?
Re: 0.999...= 1
#158> ...infinitely many 9s... How about we prove that an infinite number of 9s is impossible? Assume that we have a finite number of 9s. Add a 9. The result is not infinite. Add another 9. The result is still not infinite. We can repeat this process for an infinite amount of time and still not have an infinite number of nines. Any process that can not be completed in a finite amount of time can not complete and can not…
> How about we prove that an infinite number of 9s is impossible?
First you need to define what you even mean by this statement. The rest of what you wrote makes no sense either.
Re: 0.999...= 1
#159Earlier quoted context omitted.
Both are false. 0.99999... is not less than 1. It is the same as 1.
Yes, because someone defined it that way. It is because the "limit" in 0.999... = lim[eps->0] 1-eps is implicit and defined as being applied before anything else. But you might as well define that implicit limit as applying over the entire expression. UPDATE: So instead of interpreting the expression as: (lim[eps->0] 1-eps) which is indeed false, you can also interpret the expression as: lim[eps->0] ((1-eps) which is…
It's still not true. No no sense is it true. It is true that 0.999... = lim[eps->0] (1 - eps), but it is ALSO true that lim[eps->0] (1 - eps) = 1. That's how limits work.
If you accept both of those two (which you should, because they're correct), then since equality is transitive, 0.999... = 1. Therefore it is not less than 1, it is equal to 1.
Re: 0.999...= 1
#160My 5 year old stumped me with this, and I had to look it up. He asked me why 1/3 + 1/3 + 1/3 = 1, since it's equal to 0.333... + 0.333... + 0.333... which is 0.999... How can that possibly equal 1.000...? And is 0.66... equal to 0.67000...?
I didn't have a good enough answer for him, so I had to look it up and found this page. I tried to explain it to him but since I'm a terrible teacher and he's only 5, it was hard for me to convince him. Luckily he has many years before it matters!