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Well here's the question OP. Prove to me it's not approximately 1.


I'll assume for now that 0.999... makes sense at all. If it does, then

0.[9 repeated n times] < 0.999... ≤ 1

for any finite integer n. Now

1 - 0.[9 repeated n times] = 1 / 10^n

so the distance between the number and 1 must satisfy

0 ≤ 1 - 0.999... < 1 / 10^n

for any n. For any positive integer n we have 10^n > n so we also have

0 ≤ 1 - 0.999... < 1 / n

for any positive integer n no matter how large. In other words, 1 - 0.999... is infinitesimal. If it is also not zero, then we can divide by it, obtaining

1 / (1 - 0.999...) > n

for any n no matter how large, meaning 1 / (1 - 0.999...) is infinite.

Now whether infinite and infinitesimal quantities exist in real life we don't know. But we define the real numbers not to include infinite or infinitesimal quantities. If something has an infinite length, we can't measure its length with a real number, and if two lengths differ by only infinitesimal amount, then we use the same real number to describe both lengths. So it's possible that real numbers only approximately describe reality, but we have no way of measuring an infinitesimal difference, so we wouldn't know.

The idea that infinite real numbers don't exist is called the Archimedean property, named after Archimedes who used it a lot in his proto-calculus derivations of things like the volume of a sphere. It's a fundamental property that we assume about the real numbers (although some textbooks derive it from other postulates). Formally it's expressed as the statement that for any real number x, there's an integer n greater than x. The opposite of the Archimedean property would be a real number x that's greater than every integer n (if it were lesser then n would be greater than it, and if it were equal, n+1 would be greater). Such a number is called an infinite number. There are systems of numbers that include infinite numbers, but the real numbers don't. A consequence of there being no infinite numbers in the reals is that there are no nonzero infinitesimals in the reals because their reciprocals would be infinite, as we saw for 1 - 0.999... .

Since 0.999... is supposed to identify a real number and 1 - 0.999... can't be a nonzero infinitesimal if it's a real number, the only option is for 1 - 0.999... = 0, which means 0.999... = 1.


poll's back


Looks like "Bully nerds" got cut off.


nerds screwing us over again


Typically two numbers are considered to be different only if there is some other number between them. Since there is no number between 1 and 0.9999… they are not different numbers.


File:1623226401408.png (16.99 KB,87x87)

whats the smallest number bigger than PI?


pi + 1


It's 4......


In the set composed of all integers and pi, it’s 4


Typically Two real numbers are considered to be different only if there is some other rational number between them. Since there is no rational number between 1 and 0.9999… they are not different numbers.


I chose to use “typically” in case someone wanted to be obtuse and say that their numbers are discrete and not continuous. Like only the naturals are numbers. I’m not sure you can use the same reasoning with a discrete set.


Well, there's more to the changes than that. Saying that between every two rational numbers there's another rational number or that between every two real numbers there's another real number are both true. But it's not clear to me how you'd use them to prove 0.999... = 1. For the first, you would need to prove that 0.999... is a rational, and for the second, you would need to prove there is no real number between 0.999... and 1. Those are both true, but I don't think they would be obvious to someone who hasn't proved 0.999... = 1 yet.

For the version I suggested, we need to prove that there is no rational number between 0.999... and 1. We can do so as follows: Suppose p/q is such a rational number. Then we must have 1 - 1/10^k < p/q < 1 for every natural number k. Without loss of generality we can take q to be positive. Subtracting 1 from our inequalities and multiplying by -q, we obtain q/10^k > q - p > 0. Since q - p is an integer, this makes q/10^k ≥ 1. Let k = q. We then have q ≥ 10^q, contradicting the fact that 10^q > q. So there is no such rational number.


File:a26.png (411.3 KB,680x680)

the joke has gone on too long now, mathemeticians
it's time this ends


Are the nerds being bullied the two in the OP image? Because I'd definitely like to do some bullying to them.


pi + 1 - 0.999...


0.333... = 1/3
0.666... = 2/3

Then clearly
0.999... = 3/3 not 1


File:79884942_p0.png (312.6 KB,674x859)

"....It was fun to solve
mathematical problems, but in a deeper sense mathematics was boring
and empty because for me it had no purpose. If I had worked on applied
mathematics I would have contributed to the development of the
technological society that I hated, so I worked only on pure mathematics.
But pure mathematics was only a game. I did not understand then, and I
still do not understand, why mathematicians are content to fritter away their whole lives in a mere game. I myself was completely dissatisfied with such a life. "


Does lim(0.999...^n) = lim(1^n)?


0.999 repeating (I'm assuming that's what the post is referring to) is at a point where it has to be 1.

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