weirder.earth repost: 0.999...=1

[packbat]

(This explanation assumes you are confidently capable of long division. We can add an appendix about that if one is needed.)

If you want to divide one by three, you have a problem: either you can't because 1 < 3, or you can't because the long division never ends - you just keep getting more 3s. So, we make a convention: when we have a repeating part that never ends, we just indicate what repeats and let that stand for what we would get if we could write infinity decimal places. And the nice thing is that this works - you can do addition, subtraction, multiplication, and division the same way you did before, you just have to figure out what the result and its new recurring decimal will look like.

But a weird thing happens sometimes. If you multiply one-third by three, you get one. But if you multiply 0.333... by 3, all those threes become nines and you have 0.999.... So either our nice new strategy just broke ... or we have to declare that 0.999... equals 1. But in math, you can't just declare it, you have to show that it works to do it that way.

So, this is important. All of elementary school mathematics is riding on this. Can we prove 0.999... equals 1? We know it should - a third times three is one - but can we prove it?

Here's two arguments, and I think you can make both hold up in court.

First: can 0.999... be anything else?

Well, if two numbers are different, there should be numbers between them. But if you subtract anything from 1, you get a number less than 0.999... - whatever you subtract is more than nothing, and therefore you can put enough 9s on the end of 0.999... to reach it. Same with addition: whatever you add will make a carry and leave part of the tail of 9s intact.

And hey, let's explicitly take the average!

1+0.999...=1.999...
1.999.../2=... well, doing the long division, 19 is more than 2*9=18, so you get 0.9 and .1999... left over ... and then 0.99 and .01999... left over...

...it's just 0.999... again! The average is equal to what you had before!

Second proof: let's do a little algebra.

x=0.999...

10x=9.something...

*squints*

...the next number is always gonna be a 9, right? it's 9.9something, 9.99something, 9.999something ... it's all the same repeating decimal, and we know how to notate repeating decimals.

10x=9.999...

Now, let's subtract x from both sides.

10x-x=9.999...-x

The left side is easy - that's 9x:

9x=9.999...-x

The right side ... well, we know x=0.999..., so let's substitute:

9x=9.999...-0.999...

This is ... weird, but it looks like it should be simple, right? You just go one digit at a time.

first decimal place: 9-9=0
9.999...-0.999...=9.0something

second: 9-9=0
9.999...-0.999...=9.00something

third: 9-9=0
9.999...-0.999...=9.000something

Wait, this is all zeros! We can use our notation!

9.999...-0.999...=9.000...

And adding zero does nothing, we know that, so we can add infinite zeroes!

9x=9

x=1

0.999...=1

---

Like, you can kinda see the process here, right?

We had a thing we couldn't do - write out an infinite decimal - so we came up with a substitute we could do - recurring-decimal notation. It worked without struggle in ordinary cases, but we found a weirdness that made us concerned. Then we asked, "Can we resolve this weirdness?"

In order for our recurring decimals to work at all, a third has to be 0.333..., and that means 3*0.333... has to be 1, even though going digit by digit we get 0.999.... But, if we can prove that 0.999... is 1, that they're the same, then it works and our recurring decimals are saved.

So we start fucking around to see if we can prove the thing we want to be true is true. And if we can find arguments that hold up, then we can proceed.

Something we did very deliberately here that we wanted to highlight: we didn't actually define "0.333..." as "a decimal point followed by an infinite number of threes". We defined it as "what you would get if you could keep appending threes forever".

The reason we did it this way is because you can't write an infinite number of threes, but you can see when your process for creating a decimal is gonna loop - is gonna keep giving you the same numbers over and over.

One of the regular ways that, for example, the multiply-it-by-ten proof fails to convince people is that they go, "it's not 9.999..., it's 9.999...0". But that ellipsis isn't indicating actual 9s in our notation, it's indicating hypothetical 9s. There is no after-the-ellipsis because we never get that far writing 9s, and we refuse to try.

And that's why we were careful to prove that 10*0.999... was 9.999..., also. Because that step isn't obvious to everyone.