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0.999...= 1

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Re: 0.999...= 1

#81

Post by bhl »

ant6n wrote: post #23
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..

Well, here you reduced 1=.9999... to 1/3=0.333... What if I don’t believe that second equation.

What if we recursively defined 1/3? This will allow us to ignore the infinite 0.333... for a second. As an example, let 1/3 = 0.33 + 1/100(1/3).

A definition of a third that most people agree with is that if we multiplied that value by 3, we should get 1. Let's check the right hand side: 3 * (0.33 + 1/100(1/3)) = 0.99 + 1/100 * 1 = 0.99 + 0.01 = 1. Great!

What other expressions for a 1/3 can we come up with? If you agreed with the previous statement, then you must surely also agree that 1/3 = 0.333 + 1/1000(1/3).

Inductively, we should be able to come up with a general formula that 1/3 = bar(3, n) + 1/pow(10, n)(1/3), where bar(3, n) = sum i = 1 to n 3/pow(10, i). We can check that bar(3, 2) = 0.3 + 0.03 = 0.33, and that our first example fits this formula. Intuitively, this formula is giving us a way to represent 1/3 in terms of n decimal places of accuracy and a recursive term.

The question is now, what happens when we run that formula with n to infinity? An infinite level of accuracy! That expression is equal to 0.333... as we have defined.

The right term, 1/pow(10, n)(1/3), goes to 0, so we can discard that. The left hand side, is a geometric series with 1/10 as the power, and a scalar multiple of 3. Using a closed sum formula for that [1], we can see that the left hand side goes towards 1/3. (Apply the formula from Wikipedia, but remember our index starts off at 1, not 0.)

In the end, we have found that 0.333... = n->infty bar(3, n) + 1/pow(10, n)(1/3) = 1/3

[1] https://en.wikipedia.org/wiki/Geometric_series#Sum

Re: 0.999...= 1

#82

Post by tomtomtom1 »

atq2119 wrote: post #64
tomtomtom1 wrote: post #31

Earlier quoted context omitted.

why? 1/3 * 3 could still be equal to one. but 1/3 != 0.33333... that is, 1/3 is not representable in base 10. Which makes way more sense. I wonder if taking 0.9999.. != 1, that is 0.0000...1 exists would allow us to reslove, the fact that some possible events have probability 0?

The crux of the matter is that you have to define what things like "0.3333..." mean in the first place. Any reasonable definition of it as a representation of a real number is going to lead to it being equal to 1/3. If you want to redefine it explicitly as not a real number, you can do that, and maybe even get to some amusing math that way, but you're no longer talking the same language as the rest of the world.

>"but you're no longer talking the same language as the rest of the world"

yes, in the standard real numbers 1 = 0.999.., but people have dealt with numbers like "pi" and "sqrt(2)" before the standard real numbers were defined.

Hence the question, if we define such a system such as 0.333... != 1/3. what are the consequences?

by 0.3333... I mean a countably infinite sequence of 3s.

Re: 0.999...= 1

#83

Post by jhanschoo »

berkeleynerd wrote: post #43

Perhaps the natural discomfort many face when confronted by this challenging formulation instead indicates that a limitation of the real number system has been perceived? I would encourage those who have this reaction to study hyperreal and other alternative systems as mentioned in the article. If this clicks for them they may help lead us in new direction mathematically and advance the state of the art.

This comes from the fact that most people don't learn limits. Once the definition of the decimal representation is understood to be in terms of a limiting process, the meaning becomes clear.

For most purposes hyperreals are too much machinery when learning and manipulating limits would be simpler.

Re: 0.999...= 1

#84

Post by tomtomtom1 »

saagarjha wrote: post #73
tomtomtom1 wrote: post #31

Earlier quoted context omitted.

why? 1/3 * 3 could still be equal to one. but 1/3 != 0.33333... that is, 1/3 is not representable in base 10. Which makes way more sense. I wonder if taking 0.9999.. != 1, that is 0.0000...1 exists would allow us to reslove, the fact that some possible events have probability 0?

> but 1/3 != 0.33333... But the issue is that this is easy to verify experimentally via (in this case infinitely) long division that you can do by hand. So it’s hard to convince people of this.

but the long division algorithm never terminates.

why would it terminate at countable infinity?

Re: 0.999...= 1

#85

Post by vfinn »

jhanschoo wrote: post #79
vfinn wrote: post #62

Sorry for my naivety, but why one couldn't prove by induction that adding 9s never close the gap, or let's say, that by definition the operation is such, that it never closes the gap. If you can always halve the pie, then you can continue eating forever. To me it would be much easier to accept that (1/3)*3 is not 1.

It's because the rigorous mathematical definition is more subtle. It's defined as the smallest real number such that repeating this process (putting more 9's in the end) can't result in a number that's greater than it. So as long as you have a number that's smaller than 1, there's a gap there, and repeating the process of adding 9's will eventually give you a number that lies in that gap.

Yeah, it makes sense if the whole foundation of mathematics is laid out in such way that the above has to be true (kinda circular, but still), and this would mean that what I'm talking about is actually not mathematics but something else :).

Re: 0.999...= 1

#86

Post by ant6n »

notRobot wrote: post #29
larschdk wrote: post #25

Earlier quoted context omitted.

I don't mean to troll you, but if you were doubtful that 0.999... = 1, then you should also be doubtful that 0.333.. = 1/3. Any argument that 0.999... is not quite 1 can also be used to argue that 0.333... is not quite 1/3. I think it's mostly a matter of definition, since mathematicians consider sums of infinite series equal to their limit (if it's finite), i guess for many practical reasons. If you accept this, the…

> if you were doubtful that 0.999... = 1, then you should also be doubtful that 0.333.. = 1/3 I disagree. Any middle school student can calculate 1/3 to be 0.33333... using long division, but there's no immediately obvious way to go from 1 (or 1/1) to 0.9999...

> Any middle school student can calculate 1/3 to be 0.33333... using long division, ...

...the same way That Chuck Norris can count to infinity... twice!

Re: 0.999...= 1

#87

Post by atq2119 »

tomtomtom1 wrote: post #82
atq2119 wrote: post #64

Earlier quoted context omitted.

The crux of the matter is that you have to define what things like "0.3333..." mean in the first place. Any reasonable definition of it as a representation of a real number is going to lead to it being equal to 1/3. If you want to redefine it explicitly as not a real number, you can do that, and maybe even get to some amusing math that way, but you're no longer talking the same language as the rest of the world.

>"but you're no longer talking the same language as the rest of the world" yes, in the standard real numbers 1 = 0.999.., but people have dealt with numbers like "pi" and "sqrt(2)" before the standard real numbers were defined. Hence the question, if we define such a system such as 0.333... != 1/3. what are the consequences? by 0.3333... I mean a countably infinite sequence of 3s.

I think an important distinction is that in those "old days", people were largely working in what we now know to be subsets of real numbers, and the same conclusion applies there.

If you want to go to supersets of real numbers, you may be interested in https://en.wikipedia.org/wiki/Surreal_number

Re: 0.999...= 1

#88

Post by rcxdude »

rini17 wrote: post #8
klodolph wrote: post #4

An interesting consequence of this in proofs. You’ll see various proofs involving real numbers that must account for the fact that 0.999…=1.0. There are, of course, many different ways to construct real numbers, and often it’s very convenient to construct them as infinite sequences of digits after the decimal. For example, this construction makes the diagonalization argument easier. However, you must take care in you…

I never understood the fixation on diagonalization. Why can't ever exist another way for mapping any set to countables?

The point is that diagonalization works whatever map you have come up with: no matter how you construct your list of the reals, you can come up with another real not on the list.

Re: 0.999...= 1

#89

Post by agumonkey »

saagarjha wrote: post #77
agumonkey wrote: post #67

Earlier quoted context omitted.

And I thought mathematicians would reject normal arithmetic operation over the domain of `N...` elements. They're always so ultra rigorous to classify what is or is not, what is defined, the domain .. but then they let an infinite sequence be treated as any finite number.

> but then they let an infinite sequence be treated as any finite number Technically it’s an infinite series (sum of an infinite sequence), which is a finite number in certain cases like this one.

certain cases, so define this special one but frankly it doesn't make my brain happy (but who am I)

Re: 0.999...= 1

#90

Post by MaxBarraclough »

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?

The obvious one seems terrible enough, that division is no longer the inverse of multiplication: (1/3)*3 != 1
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