0.999...= 1 • Hacker News Board

Re: 0.999...= 1

#171

Post by danharaj » Tue, Apr 28, 2020, 2:50 PM UTC

rrmm wrote: post #48

traderjane wrote: post #38

Professor N.J. Wildberger is probably among the most well known "ultrafinitist" on YouTube. https://www.youtube.com/watch?v=WabHm1QWVCA I mention him because I would think he sympathizes with those who have concern over the meaning of this kind of notation.

Wildberger is great. His lectures that he teaches at UNSW (i think) are interesting, and he usually keeps a clear dividing line between std math and his own predilections. It threads the line between being a kook and legitimate published mathematician very finely. I actually have some sympathies with his contention that real numbers (limit points of infinite series) are somehow a different animal than rational number…

If I give you two representatives of real numbers, say turing machines that write out on their tape the binary digits of those real numbers, in general you will not be able to order them.

Re: 0.999...= 1

#172

Post by dahart » Tue, Apr 28, 2020, 2:50 PM UTC

CogitoCogito wrote: post #165

dahart wrote: post #157

Earlier quoted context omitted.

> 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 t…

> 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 t…

> why can’t you have an infinite amount of 9s and then just put a 7 after that?

You can. The proof still works if you do that.

What you’re refusing to accept here is the definition of infinity.

Re: 0.999...= 1

#173

Post by marcosdumay » Tue, Apr 28, 2020, 2:50 PM UTC

tomtomtom1 wrote: post #68

knzhou wrote: post #44

Personally 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…

if we say that infinitesimals exist. that 1/3 != 0.33.. and 1 != 0.9999... and the probability of possible events is never 0. what are the properties that we would lose?

I have never seen a number system with infinitesimals where the addition wasn't updated to ignore smaller classes if they come with larger ones.

That is, for any number system I've seen, 1 = 1 + dx, and infinity = infinity + 100.

Re: 0.999...= 1

#174

Post by johnhattan » Tue, Apr 28, 2020, 2:51 PM UTC

ginko wrote: post #9

I remember being doubtful when being presented with this in middle school, but after being shown this as fractions makes it obvious: 1/3 = 0.333.. 3 * 1/3 = 3 * 0.333.. 3/3 = 0.999.. 1 = 0.999..

I remember a conversation I had with my daughter in the car when she was starting out with algebra...

Me: Is 9.999... the same as 10, or is it just really close to 10?

Kid: Really close. It never gets all the way there.

Me: Well then how close? What do you get when you subtract 9.999... from 10?

Kid: (pause) An infinite number of zeroes. . .and then a one. . .wait, you can't do that.

Me: Right. You just have an infinite number of zeroes. Which is zero.

Kid: (pause) Oh, that's mind-blowing.

Re: 0.999...= 1

#175

Post by tomtomtom1 » Tue, Apr 28, 2020, 2:52 PM UTC

mFixman wrote: post #97

tomtomtom1 wrote: post #68

Earlier quoted context omitted.

if we say that infinitesimals exist. that 1/3 != 0.33.. and 1 != 0.9999... and the probability of possible events is never 0. what are the properties that we would lose?

If we say that infinitesimals exist, it still happens that 1 = 0.999…. It just happens that 0.999 ≠ 1 - 𝛚. 0.999… = 1 is a property of the way we write some rational numbers, not of the number system itself.

wouldn't 0.999.. be equal to 1 - 10^w since it's only a countably finite series of nines.

Re: 0.999...= 1

#176

Post by gspr » Tue, Apr 28, 2020, 2:52 PM UTC

amelius wrote: post #149

OskarS wrote: post #134

Earlier 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…

> which is indeed false, you can also interpret the expression as: > lim[eps->0] ((1-eps) Assuming you don't mean some special notion of limit, I would guess that by `(1-eps) If so, `f` does indeed have a limit from above that is `false` and a limit from below that is `true`. Where do you wanna go from here?

Edit: Corrected stupid wrong assertion about limit from below, d'oh.

Re: 0.999...= 1

#177

Post by hilbertseries » Tue, Apr 28, 2020, 2:53 PM UTC

bognition wrote: post #168

ping_pong wrote: post #160

My 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…

No .666666 is not equal to .6700000 0.666... is equal to 0.666...7

Not sure if you’re joking, but 0.666...7 is not a real number. Can you define it?

Re: 0.999...= 1

#178

Post by ColinWright » Tue, Apr 28, 2020, 2:53 PM UTC

CogitoCogito wrote: post #124

empath75 wrote: post #101

Earlier quoted context omitted.

It's not a rational number. Real numbers are defined as an equivalence class such that if the differences of two infinite sequences of rationals tend toward zero, then they are equal. The difference between 0.999... and 1.000... clearly tends towards zero as it heads of to infinity, and so they are equal. If you want to argue that it doesn't then you have to come up with some other definition for numbers which have a…

> It's not a rational number. > Real numbers are defined as an equivalence class such that if the differences of two infinite sequences of rationals tend toward zero, then they are equal. The difference between 0.999... and 1.000... clearly tends towards zero as it heads of to infinity, and so they are equal. > If you want to argue that it doesn't then you have to come up with some other definition for numbers which…

I think you have mis-stated yourself. Either that, or I don't understand what you're saying.

Trying to rearrange it and remove as many negatives as possible, I started with your statement:

> I don't see how a proof involving the standard arithmetic operations found within the rational numbers, but not including any concepts of limits, completeness, etc. is invalid.

I think what you mean is that any proof that does not use the concepts of limits and completeness is going to be invalid.

That seems clear to me, the reason being that one needs to define what one means by the sequence of symbols "0.9999...".

You can say "It's infinitely many 9s stretching off to the right", but that doesn't tell me what it means.

People seem to think it does, but when I dig deeper, they usually don't have any sense of what it means. And therein lies the problem (as I see it). People blithely write the glyphs, but don't have a concrete interpretation.

Re: 0.999...= 1

#179

Post by danharaj » Tue, Apr 28, 2020, 2:54 PM UTC

amelius wrote: post #163

OskarS wrote: post #159

Earlier quoted context omitted.

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.

I think you should look more closely at what I did with the order of operations, and the fact that "<" now is part of the expression acted over by the limit.

It is not reasonable to interpret the notation that way. For one, variable binding becomes incoherent.

Re: 0.999...= 1

#180

Post by tomtomtom1 » Tue, Apr 28, 2020, 2:54 PM UTC

johnhattan wrote: post #174

ginko wrote: post #9

I remember being doubtful when being presented with this in middle school, but after being shown this as fractions makes it obvious: 1/3 = 0.333.. 3 * 1/3 = 3 * 0.333.. 3/3 = 0.999.. 1 = 0.999..

I remember a conversation I had with my daughter in the car when she was starting out with algebra... Me: Is 9.999... the same as 10, or is it just really close to 10? Kid: Really close. It never gets all the way there. Me: Well then how close? What do you get when you subtract 9.999... from 10? Kid: (pause) An infinite number of zeroes. . .and then a one. . .wait, you can't do that. Me: Right. You just have an infin…

> An infinite number of zeroes. . .and then a one. . .wait, you can't do that.

why not? why can't an infinitely small number exist?