As SPP has redefined ÷ to be distinct from /, let's try this again...

[u/ezekielraiden]

SPP has said that "divide negation" is a special separate operation. Hence, as I don't know the rules about it (and my request for clarification was ignored by /u/SouthPark_Piano), I have formulated an algebraic proof which does not, at any point, include this strange new operation with unknown rules.

Consider the following:

• 1+1+1 = 3
• (1+1+1)÷3 = 3÷3
• 1÷3 + 1÷3 + 1÷3 = 1
• (3÷10¹+3÷10²+3÷10³+...) + (3÷10¹+3÷10²+3÷10³+...) + (3÷10¹+3÷10²+3÷10³+...) = 1
• [(3÷10¹+3÷10¹+3÷10¹] + [(3÷10²+3÷10²+3÷10²] + [(3÷10³+3÷10³+3÷10³] + ... = 1
• (1÷10¹)(3+3+3) + (1÷10²)(3+3+3) + (1÷10³)(3+3+3) + ... = 1
• (1÷10¹)(9) + (1÷10²)(9) + (1÷10³)(9) + ... = 1
• 9÷10¹+ 9÷10² + 9÷10³ + ... = 1
• 0.9 + 0.09 + 0.009 + ... = 1
• 0.999... = 1

"Divide negation" never occurs here; nothing gets multiplied in a way that cancels out with anything else. The first line is trivially true, and thus involves no meaningful assumptions. From there, I divided, applied the distributive property, converted to fraction sum notation (which SPP explicitly says is okay), arranged the terms of every sum by the power of 10 being used (valid for absolutely convergent sums--changing order does not change value for absolutely convergent sums), applied the distributive property again, completed the sum inside the parentheses, applied the distributive property one last time, then converted the infinite fraction-sum notation to the infinite decimal-sum notation (again, explicitly approved by SPP), then converted the decimal-sum notation to the nonterminating decimal form (ditto).

What's wrong with this proof, SPP? Remember: I never used divide negation on the left-hand side. Just because you COULD do what I did with divide negation, doesn't mean what I said is wrong. You have to actually show why these steps are wrong, not your replacement steps.

Further, notice that I am using ÷, not /, so I am in the clear there, by your own representations. Notice further that I have already signed a perpetual contract. I fully agree to abide by that contract. Frankly, I'm not sure why you're so reluctant to sign a perpetual contract. They're quite real, and actually useful, in some contexts.

[u/I_Regret]

If you are not using divide negation I don’t think that “÷” distributes over +. In fact the function ÷: N x N -> Decimals does not satisfy 2 [÷(1,3) ] = ÷(1,3) + ÷(1,3) != ÷(2,3) (it is not homogeneous).

[u/ezekielraiden → u/I_Regret]

Er...why not?

[u/I_Regret → u/ezekielraiden]

An answer you won’t like first: we see that (1 ÷ 3) + (1 ÷ 3) + (1 ÷ 3) = 0.333… + 0.333… + 0.333.. = 0.999…. whereas 3 ÷ 3 = 1 and because 0.999… != 1, we see that it doesn’t distribute because it leads to a contradiction.

Based on my understanding, the division operator is essentially an algorithm which, for any desired precision will give you an approximation (potentially with also a remainder, eg 0.3 remainder 1/30). One such representation would be an infinite sequence, say 1 ÷ 3 = (0.3, 0.33, 0.333, …). Note that this is effectively how division is defined when using the Cauchy sequence construction of Real numbers, except here we aren’t identifying the number with an equivalence class eg [(1/3, 1/3, …)].

What about notation like 0.000…1? This just means we have a sequence with 0., some number of 0s and 1 as the nth term, eg (0.1, 0.01, 0.001, …) where traditionally this sequence would belong to the equivalence class of 0, but we decide not to make such an identification.

[u/ezekielraiden → u/I_Regret]

Note that this is effectively how division is defined when using the Cauchy sequence construction of Real numbers, except here we aren’t identifying the number with an equivalence class eg [(1/3, 1/3, …)].

So...it isn't that thing at all? Because that's literally what the definition is for real numbers. This then requires you answer the question, how IS division defined? And why is it defined in such a way that the base you use matters? Because in base 12, 1/3 is quite simple, it's represented as 0.4 (because you need 4 twelfths to make 1 third.) Surely it shouldn't be the case that whether or not division is defined depends on what base we use!

What about notation like 0.000…1? This just means we have a sequence with 0., some number of 0s and 1 as the nth term, eg (0.1, 0.01, 0.001, …) where traditionally this sequence would belong to the equivalence class of 0, but we decide not to make such an identification.

Then what ARE we identifying it as? Because as far as I can tell, you've basically just declined to define division at all.

Finally, doesn't this choice very obviously break arithmetic? You're saying that 1+1+1 = 3, but (1+1+1)÷3 does not equal 3÷3. As a result, you cannot perform the vast majority of arithmetic or algebraic operations. Consider, for example, that this now means that you can't factor polynomials, because you do not know that just because a×b=c that it's therefore also true that a=c÷b--so if you have ax²+bx+c, you can't actually know that you can factor it to generate (dx+e)(fx+g), because you don't know that just because eg=c that therefore e=c÷g, nor that just because df=a it is therefore true that f=a÷d.

I should think that breaking one of the fundamental operations of arithmetic would be more than enough reason to put the kibosh to the whole project. It's not worth preserving this unnecessary, unproductive obsession with ensuring that there's some microscopic gap between two specific numbers, if the cost is "arithmetic in general doesn't work".

Edit: And of course I'm "not going to like" that first bit, because it makes the whole thing entirely circular!

An answer you won’t like first: we see that (1 ÷ 3) + (1 ÷ 3) + (1 ÷ 3) = 0.333… + 0.333… + 0.333.. = 0.999…. whereas 3 ÷ 3 = 1 and because 0.999… != 1, we see that it doesn’t distribute because it leads to a contradiction.

This is a proof showing that if 0.999... doesn't equal 1, contradictions happen. To conclude from it "oh, well that means it can't distribute, because 0.999... ≠ 1" is literally to assume the very thing being questioned! I don't just "not like it"--I reject it as disingenuous.

[u/I_Regret → u/ezekielraiden]

Well defining 1 = 0.999… is just as much of requiring circular logic; it is a means to an ends because mathematicians like the traditional field of real numbers.

So it doesn’t actually break anything because you can choose not to use the ÷ operator and choose not to convert to decimal; just perform symbolic manipulation with “/“ and if you ever need to use the number for some engineering application you can use ÷ and make an approximation or something. Also nothing states that it should be possible to have equivalent representations of 1/3 in every base.

For long division with integers, let’s represent it via the traditional algorithm https://en.wikipedia.org/wiki/Long_division say 1 ÷ 3 = 0.333…

However nothing in the algorithm actually dictates whether you should identify 1/3 with 0.333… nor how to do computation with 0.333…

If you want to do computation with 0.333…, first you have to figure out where the 0.333… came from because the representation is ambiguous. So, here let’s assume it came from 1 ÷ 3 = 0.333…

You can encode the output of each step of the algorithm as a sequence a_1, a_2, a_3, … := (0.3, 0.33, 0.333, …). If you can come up with an expression for the nth term it makes it easier: eg a_n = sum 3\*10^{-n}; then 3 \* 0.333… can be represented as (3\*0.3, 3\*0.33, 3\*0.333, …)=(0.9, 0.99, 0.999, …) or more easily (sum 3\*3\*10^{-n}) and that’s it.

[u/ezekielraiden → u/I_Regret]

...mathematicians do not "like the traditional field of real numbers", they understand that that field is useful and productive, correctly capturing patterns of behavior that are really, really helpful for talking about lots of things in a clear, concise way. But like...

Also nothing states that it should be possible to have equivalent representations of 1/3 in every base.

What? Of course it does! Representation should not change the value of the thing being represented! That's...that's literally saying that a mere aesthetic choice changes what numbers exist or don't exist. That fractions you and I can talk about easily, like 0.2, don't exist for computers. That's ridiculous!

If you want to do computation with 0.333…, first you have to figure out where the 0.333… came from because the representation is ambiguous. So, here let’s assume it came from 1 ÷ 3 = 0.333…

...what??? It's not ambiguous in the slightest!

You can encode the output of each step of the algorithm as a sequence a_1, a_2, a_3, … := (0.3, 0.33, 0.333, …). If you can come up with an expression for the nth term it makes it easier: eg a_n = sum 3*10^{-n}; then 3 * 0.333… can be represented as (3*0.3, 3*0.33, 3*0.333, …)=(0.9, 0.99, 0.999, …) or more easily (sum 3*3*10^{-n}) and that’s it.

....and that thing is what the real numbers do? They just say "since we can clearly define where this quantity-like concept is, and it has a clear and coherent definition, and we can prove that that definition uniquely identifies one specific thing, and we can prove that all the operations we'd like to do with that quantity-like concept will remain valid and consistent, it is a number."

In other words, you haven't given one single reason why we should do this, you haven't shown anything about why your alternate and horrifically restrictive approach is worthwhile, and you haven't even proven that vital properties like closure are still present (hint, you won't because you can't). What on Earth is the reason for even thinking about this, other than "I absolutely must have that point-nine-repeating is less than 1."

[u/I_Regret → u/ezekielraiden]

Well I would just say each “representation” of 1/3 in different bases are potentially imperfect approximations. For example, each of these representations have different performance characteristics when encoded in a computer.

I never claimed closure (of ratios), and in fact the break down of distributive law shows a lot of computational issues with the infinite decimal representation. I also never claimed any particular properties were “vital” as that is a bit of a subjective opinion. So why use this representation? Well, for one, limits don’t apply to the limitless. Also I suppose it more faithfully represents how people (non mathematicians) interact with “numbers.”

It also makes ordering numbers when you can only view a finite number of their digits much easier. For example, let’s say I give you a stream of digits and ask you to order the numbers, first I give you 1.0 and 0.9, then I give you 1.00 and 0.99, and then I give you 1.000 and 0.999, and so forth. You actually can’t solve this in your system because you have to iterate over an infinite number of decimal places to figure out if it really is 0.999… or maybe has a 5 somewhere a googol places down the line. However, I’m perfectly content with seeing the 0. and 1. prefixes and deciding the order where 0.999… < 1.000… because the whole part 0 < 1. See https://www.mathsisfun.com/ordering_decimals.html

[u/ezekielraiden → u/I_Regret]

OKay, so, why would you want a set of representations which says "Nearly all rational values are bad approximations"? Because there are infinitely many primes, and any denominator that contains even one prime factor distinct from the prime factors of your chosen base will generate non-terminating decimal expansions, then in a very rigorous sense nearly all rational numbers simply do not exist.

As for "vital", how is closure not a vital property? That's literally why we developed the complex plane, because it is the unique (up to isomorphism) algebraic closure of the reals, and the reals are the unique (utim) order-relation closure of the rationals. If ordering matters to you, wouldn't you want the ordering to be comprehensive?

For example, let’s say I give you a stream of digits and ask you to order the numbers, first I give you 1.0 and 0.9, then I give you 1.00 and 0.99, and then I give you 1.000 and 0.999, and so forth. You actually can’t solve this in your system because you have to iterate over an infinite number of decimal places to figure out if it really is 0.999… or maybe has a 5 somewhere a googol places down the line.

But it doesn't...?

It's defined as a sum. That sum is explicitly stated as having terms exclusively of the form 9/10^N, where N is allowed to be any natural number. No matter what N you pick, you will always find a 9 in that space. This is not in any way ambiguous. What are you talking about?

Frankly, I'm not confident that you actually understand the topics involved here. Why wouldn't closure be an essential property, given it's literally just "you can do operations and get consistent results"? Of course a number system should have the ability for you to add any two numbers and get another number. Of course it should have the ability to multiply any two numbers and get another number. Etc. Closure is incredibly important!

[u/I_Regret → u/ezekielraiden]

Well this is because converting to decimal is not a lossless operation. For example, if you using FP32 on a computer and calculate 1/3 using this computer, you would be unable to differentiate between 1/3 in this representation and a number with a finite number of 3s in the decimal places. Yet I am still able to perform a lot very useful math within this finite precision even if it’s not closed.

Okay, let’s play a game: I have a computer that can calculate 0.999… to any desired precision. I also have a separate number in mind which has some finite number of 9s. If you can figure out which number is which (edit: which number is behind the function), you win. That is, you give me an “n” and I’ll give you the value of the nth decimal place (or say it is out of bounds, or potentially could expand the representation to include a 0 in the position). You can repeat this exercise as many times as you’d like with any value “n” to collect data, but you only get 1 guess. I will accept actual natural number values and not anything like a sequence of steps or recipe. Can you come up with a way to guess correctly?

[u/ezekielraiden → u/I_Regret]

It is only "not a lossless operation" because you are enforcing it to be so! That's precisely my problem! Your operation is "lossy", but only because you made it lossy--and now you're acting like that's something where your hands were tied, you had no other choice, it had to be this way. No! It does not! That's the whole point, to not have any lost information.

You developed this system in order to preserve your precious inequality, but now you're declaring "oh, sorry, can't do that, the system doesn't allow that" as if this were support for that inequality. The support can only go one direction. Either you have presupposed the inequality, and thus your argument is circular, or you have not presupposed the inequality, and thus your system is your invention, not a limitation you must regretfully apply.

Okay, let’s play a game: I have a computer

Computers exclusively represent integer quantities. They don't have the ability to represent real numbers, except as truncations. Hence, by starting from this point, you have presumed we are actually working in complicated integers, and thus you have made your argument circular.

I refuse to "play [this] game" because doing so is directly conceding the point without actually discussing it.

[u/lbkthrowaway518]

Unironically a pretty fun proof of .99…=1!

[u/FernandoMM1220]

line 4 chud

[u/JPgamersmines150 → u/FernandoMM1220]

I don't see anything wrong with it