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, then 0.999... = 1. If you don't, then 0.999... can't be assigned a value (but converges to 1), which may be the intuitive understanding of infinite series for some.

> 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...

I can just do it backwards - is 1/3 equal to 0.33333...?

1 / 3 = 0.33333... 3 * 0.33333... = 0.99999... and my child brain "knows" that 1 != 0.99999...

In my child brain this proves that 1 / 3 is not equal to 0.33333..., it's just an approximation.

So I agree with larschdk, those problems are equivalent and one can't be used to prove the other ...

> 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!

I smart middle-schooler is absolutely capable of understanding that dividing 1/3 results in an infinitely repeating sequence of 0.33333... Even without understanding the concept of infinity, they will quickly realize that there's no reason to believe the problem will stop adding a 3 to the end of the result with each iteration.

> I smart middle-schooler is absolutely capable of understanding that dividing 1/3 results in an infinitely repeating sequence of 0.33333..

And how will I smart middle-schooler know that the result of running the long division algorithm is exactly 1/3, rather than some approximation.

Recursively?

> Any argument that 0.999... is not quite 1 can also be used to argue that 0.333... is not quite 1/3.

Yes, but there aren’t good arguments for either of them, and that’s the point. The difference is that you have probably already learned how to divide 1 by 3 and have thus convinced yourself that 1/3 does indeed equal 0.333 repeating. It’s not so simple to come to the conclusion that 0.999 repeating equals 1 from simple long division that you would encounter in grade school.

See my other reply along the same lines. I was thinking about adding this caveat to my original message as well, but I think understanding that decimal numbers with infinite digits exist and that 0.999.. = 1 are separate things. The second being less intuitive.

I agree. The argument ignores the rule that infinity is a point which can never be reached; it can only be approached. So repeating 9s infinitely many times will still not reach 1, it will only approach it.

Another secondary school 'proof'

  x = 0.9999.....
  10x = 9.9999.....
  (10x -x) = 9x = (9.9999.... - 0.9999....) = 9
  x = 9/9 = 1

or,

      x = 0.9999...
    10x = 9.999...
    10x = 9 + 0.999...
    10x = 9 + x
     9x = 9
      x = 1

Presented slightly more clearly

https://en.wikipedia.org/wiki/0.999...#Algebraic_arguments

Yeah that’s way more complicated than it needs to be and I’m tempted to replace that whole section with:

x = 0.9999...

2x = 1.9999...

2x - x = 1

x = 1

It's much more intuitive that 100.9999...=9.9999... than that 20.9999...=1.9999...

You've met the formatting limitations. You mean:

"It's much more intuitive that 10 * 0.9999... = 9.9999... than that 2 * 0.9999...= 1.9999..."

How do you jump from

x = 0.999...

to

2x = 1.999..

Picture an infinite ticker tape in your mind.

Write

   0.9
 + 0.9
 —————-
   1.8

Now keep extending the 9’s. You have to carry a one, so fix the carry and then add more nines.

Where people keep getting tripped up is thinking you can stop when you get tired, or die, or when the universe ends in heat death. You don’t get to stop. You never get to stop. It’s nines all the way down.

It’s less intuitive, but still relatively straightforward from performing the “long” multiplication. You do need to convince people that the “8 at the end“ is irrelevant, however.

You need to convince people there is no 8 at the end because there is no end.

If you see an 8 it’s because you didn’t carry the 1. Keep going.

Another way to think about it is that you have n digits, and I’m using the n and n + 1 digit at the same time. But since n goes to infinity, +1 hardly matters.

The difficult part is to convince that (9.9999.... - 0.9999....) = 9. Someone else mentioned about Zeno's paradox involving Achilles and the tortoise. Since you multiplied x by 10 to get 9.9999...., you know that it's always going to be 1 decimal place slower in pace to reach convergence compared to 0.9999.... What you might get instead is (9.9999.... - 0.9999....) = 8.9999....

I like this one better:

    0.9        = 1 - 0.1
    0.99       = 1 - 0.01
    0.999      = 1 - 0.001
    0.9999     = 1 - 0.0001
    0.99999... = 1 - 0.00000... with a 1 at the end of the infinite series of 0

>[...] at the end of the infinite series of [...]

...uhh

"Eternity feels a bit too long ... especially near the end."

> 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.

Well, if I could choose I wouldn't personally accept 1/3 = 0.333... . But rather, I'd say it equals a limit:

    1/3 = lim(N -> oo) 0.3{N}     (3 is N times repeated)

Especially I would distinguish between infinitely many threes, and N threes, where N goes to infinity. In the first case, you would still be missing an infinitisimal amount, in the latter case you have the usual situation and the sequence has the least upper bound of 1/3.

When you are calculating a limit, you can never just plug in the value for N (say if N is in the denominator and the limit goes to 0). Why should you be able to do this when N is infinity?

At least this is my personal justification why I find non-standard reals interesting. They also justify the nice calculation method where you can cancel out 'dx'es from fractions.

But you absolutely can evaluate that limit as N goes to infinity and correctly conclude that 1/3 does equal 0.333 repeating.

Repeating decimals may be introduced in a mathematics education long before any other infinite series or the methods used to tame them, such ass limits.

I see the introduction of a conceptually incomplete notion as one of the big hurdle of school-level math: without limits, children are asked to just adopt the axiom that there is such a thing as a concept of infinity with any form of practical usefulness, and this riles.

It's only with limits and proper formalism that I was reconciled with maths that frankly were just tending towards approaching an equality with bullshit.

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.

> 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?

It can, and infinitesimals are a part of so-called nonstandard analysis, but you cannot write an infinitesimal using decimal notation. "0.999…1" is simply meaningless, a contradiction. If you have a "…" it means there's no place where you could put a "last" digit. If "0.999…1" doesn't feel impossible enough, then what would "0.999…9" mean?

People in this thread seem to think that “9 repeating” is not an infinity of nines but is instead “write or think of nines until you get bored and then write something else”

Because if it has 1 at the end, then this will mark it’s end, thus making it finitely small.

I personally like using fractions of 9.

  1/9 = 0.111...
  2/9 = 0.222...
  3/9 = 0.333...
  ...
  8/9 = 0.888...
  9/9 = 0.999...

What's neat is that this trick works for any repeating decimal, with any number of digits in the repeating part. For instance:

  123/999 = 0.123123123...
  999/999 = 0.999999999...

Multiply or divide by powers of 10 as necessary to shift the decimal point, and add the non-repeating part.

Once you accept this mapping, it's trivial to treat 0.999... as 9/9 (or 99/99, or 999/999, etc). Which can be simplified to 1.

Nice to see a few different proofs/intuitions here. Not being a big fan of symbol manipulation, I always felt partial to the proof/intuition that you couldn't find another number between the two :-)

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

As in the 0.333... will stop at some point? That would still mean that 3 time 0.333... with a LOT of 3s end up being being 1.

I also figure it's a bit more intuitive for pupils to just try out calculating the decimal representation of 1/3 and seeing that it'll just keep going forever.

more like 1/3 != 0.3333 ....

as in 1/3 does not have a decimal representation. you can only approximate it but never reach it.

This also happens to be the test for whether your calculator is any good.

My high-quality scientific calculator makes this a bit uninteresting:

  1/3            ⅓
  Ans * 3        1