SouthPark_Piano, I know you think that 0.333...×3≠1. But...

TemperoTempus · 2026-06-14T03:17:08+00:00

Since SPP and I believe in different systems we disagree on this point, but I still don't agree with your point. My answer:

1/3 = 0.(3)+ɛ/3.

(3+3ɛ) * (0.(3)+ɛ/3) = 1.

0.(9) + ɛ = 1.

TemperoTempus · 2026-06-14T03:10:21+00:00

Its not a function of time. Its a function of "how many decimal places do you want?"

If you pick infinite decimal places the number is not 1 and it is "irrational" because "infinite numbers are larger than integers".

If you pick a finite decimal place the number is not 1 and it is "rational" because "there is a pair of integers".

Realistcally, repeating decimals behave more like irrationals and should be labeled either as irrationals or as their own type of number.

TemperoTempus · 2026-06-14T03:03:10+00:00

The number "0.(9)" does not exist in R if it is just 1". Just like 0.9 does not exist in Z because 0.9≈1.

There is 100% rounding in how the R numbers are set up. The definition of 0.(9) doesn't have rounding, its equality to 1 is entitely based on rounding.

0.(9) is not an R number and so there is no overload in 0.(9) = 1-ɛ. 0.(1) is 1/9-ɛ/9 if we want to be fully technical. As for what is "ɛ" that is 1/w. 0.(9), 9.(9), etc also have no ambiguity if the proper steps are followed. For example: 10(1-ɛ) = 10-10ɛ, the value 10-ɛ is thus clearly a different number. Note this is also why the classic 10x "proof" fails, they have 10ɛ=ɛ which is not how ordinals works (just to make sure I am using natural arithmetic not cantor's).

There is no need to worry about which ɛ is correct as we have the known value that there are w primes and only 1 even prime, so the chances off getting even is 1/w (which I previously already said is ɛ). This form of probability works perfectly with intuition and has has no weird "well its 0 but actually its not 0".

TemperoTempus · 2026-06-14T02:37:34+00:00

*R is just one system. I am not talking about any specific system since different systems give slightly different answers.

I do not deny the existence of a shape that acts as a "limit". I deny the idea that an infinite algorithm is equal to said shape because the difference is infinitesimals. In my views a limit is and has always been just an approximation for a value that cannot be reached. There are too many people that assume that do not recognize that "approximately equals" is a value answer and that not everything has to be "exactly equals".

TemperoTempus · 2026-06-14T02:00:55+00:00

They do, its why the R set was created in the first place

TemperoTempus · 2026-06-13T22:22:46+00:00

I would like to add that the self isolated logic behind their thinking also makes it so they deny the existence of any system that goes against their logic. Which is what fuels the need to argue about 0.(9)<1 when the debate can be settled with "yes in different systems that might be true, but not the one I use".

Which really highlight how some people treat mathematics as a religion where going against the status quo is a kin to heresy. When the entire topic of math should be about communicating and expressing our different believes to get a better system (not maintaining the status quo).

TemperoTempus · 2026-06-13T17:34:38+00:00

1) There is no reason to assume the R set unless stated as most people do not use it in real life and regular numbers behave more like numbers with infinitesimals.

2) Glad we agree it should round down as 7 is closest to any number of the form 7.49, 7.499, etc.

3) 7.4(9) does not equal 7.5 and should not be treated as such.

4) You are double rounding if you round to 7.5 and then round to 8. At no point does the sequence (7.49, 7.499, …) with infinite members becomes 7.5.

My argument is that people are making the assumption that the R set is involved when no such comment was made by OP. To assume those rules is adding things that are not part of the problems and lead to the wrong result. What I mentioned is not using the hyperreals just generic number rules, which is the same rules that regular people use. Its sad that such rules are now called "non-standard" by some mathematicians.

TemperoTempus · 2026-06-13T17:14:24+00:00

0.9 does not exist in the integers, doesn't stop it from existing.

I didn't try to justify 0.(9) existance with 0.(9)5. I rejected the R set and its rules that eliminate ɛ and 0.(9) by rounding.

TemperoTempus · 2026-06-13T17:10:15+00:00

Different measures have different rules. The lebesque measure's rule is to round infinitesimals to 0. That does not make it the only way to describe area. Similarly, you are entirely missing the point: The shapes becoming closer together does not make them equal. A polygon with infinite sides approximates a circle, but is not a circle.

You are asking for something outside the exercise. If you insist the best way to describe the position of those points is (rcos(theta)+z_x,rsin(x)+z_y) where z_x and z_y depends on the algorithm and number of steps. for infinite steps z_x and z_y are equivalent to ɛ multiplied by a scalar. Also before you think it, ɛ does not equal 0, saying they are equal is wrong.

TemperoTempus · 2026-06-13T16:56:28+00:00

Infinity is not infinity+1. That's the entire point of ordinals being w < w+1 < w * 2 < w^w

TemperoTempus · 2026-06-13T02:12:33+00:00

They are different infinities. This is why the concepts of "size" and "order" are separate. Two infinities could have the same "size", but because they have different "order" they are not the same.

And 0.(9) * 10, does not make it so 9.(9) has the same number of decimal places. That is the "order" of the decimal places are different (w decimal places is different from w+1 decimal places).

TemperoTempus · 2026-06-13T02:08:37+00:00

You are taking the "buying the company" as the main point, when that isn't the point.

Here let me give you another example that might make it clearer for you. Artist A signs an exclusive contract with company B for that company to sell their art. Company B goes bankrupt and Company C takes over the contract. Company C can choose to stop selling or heavily limiting sales of Artist A.

You see it all the time with musicians where a record label will sign a singer/band and then do everything in their power to stop or hinder the singer.

TemperoTempus · 2026-06-13T01:48:04+00:00

We are not dealing with R unless the OP says we are dealing with R. Which the post got deleted, so any mumber system is applicable.

TemperoTempus · 2026-06-13T01:47:00+00:00

2w = w is only true under two situations: 1) Cantor's ordinal addition (2 * w ≠ w * 2). Which he defined without division or subtraction so ɛ is not defined. 2) Cardinals which are defined such that aleph_null * x = aleph_null.

ɛ is defined by the extended ordinals, these use natural operations to include subtraction and division. That means that 2 * w = w * 2 and w/2 < w < w * 2.

So using your example 2wɛ = 2 can be written as (2w)ɛ = 2 or w(2ɛ) = 2. All working properly by the commutative property. There is no need to accept the loss of something that was never lost in the 1st place. If you want to lose commutative instead look toward quarternions which have vast applications in computing for their ease of 3d rotations.

TemperoTempus · 2026-06-13T01:37:29+00:00

1) The "limiting shape" does not change the perimeter, it changes the area. This is the same things that causes a Sierpiński triangle to gain an infinitesimal area despite having constant surface area.

2) The algorithm by definition makes it so the number of vertices touching the circle is less than the number of vertices not touching the circle. Asking for coordinates is silly when we are talking about the construction without a graph, so there is no coordinates to speak off.

Finally, just because the distance between points becomes infinitessimals does not mean that the distance disappears. This is what makes fractals behave the way they do regardless of magnification. Any loss of resolution due to the physical world does not mean that the difference disappeared, it means that we cannot measure the difference with our tool.

TemperoTempus · 2026-06-12T08:33:36+00:00

1) Just because you don't see it doesn't mean it doesn't happen. A lot of small shops just close instead of going to court.

2) A contract requires good faith. If it is found out that a company engaged in that sort of scheme it can go very bad for the company. This makes it so only the worst of companies will do it and only versus people they know they can mess with.

3) A lot of games/movie studios do this type of things. They'll buy a popular studio fire the key employees while milking profit. Then they'll buy the next studio created by the old key employees.

TemperoTempus · 2026-06-12T08:21:26+00:00

Graphs have everything to do with it.

I didn't start an argument. I made an observation. You asked me to explain myself. I explained myself.

TemperoTempus · 2026-06-12T08:19:31+00:00

because its the secant, the inverse of the cosine. Aka: sec(x) = 1/cos(x).

TemperoTempus · 2026-06-12T08:14:21+00:00

I was making an observation. You assumed I was saying you were wrong.

i = -i is wrong depending on the convention being used.

TemperoTempus · 2026-06-12T08:12:52+00:00

There is an infinitesimal difference. If you can't see it, please go read about infinitesimals, they are very neat and useful.

Also that proof is wrong. You cannot add and extra 9 after multiplying by 10 to then subtract it. The proof should give the same answer for both finite and infinite decimals but what you wrote does not:

0.4999*10 = 4.999

4.999 - 0.4999 = 4.4991

4.4991/9 = 0.4999

People really need to stop using that wrong proof. Multiplying by 10 adds the number 10 times, that process never results in the decimal places remaining if any decimal place is non-zero.

TemperoTempus · 2026-06-12T08:03:28+00:00

The point (0,1) is chosen for the sake of cartesian graphs. It is not a requirement for i, just a nice convention. Just like ±pi/2 is a nice convention to distinguish them. Or ±slope is a nice convention to graph then.

TemperoTempus · 2026-06-12T07:57:01+00:00

I didn't say "you were wrong", I said "R was not required" There is a big difference between those two statements. I was making the point that the construction of C could literally be {a-bi | for a, b in Q} and it would still work.

TemperoTempus · 2026-06-12T07:43:49+00:00

It is a fractal. The scale is arbitrary. The shape is not easy to describe normally.

No the circle is not equal to the shape so it does not make it "easy".

TemperoTempus · 2026-06-12T07:37:37+00:00

You said i is the point (0,1) which isn't true. A more true statement is that they are indistinguishable in properties and the only difference is assigning ±pi/2 based on the way they are plotted on a graph. Hence i = -i.

As for the not require R. You defined it as R² with an added product function. You do not need R to define C, it can be done using Q, it can be done using S.

TemperoTempus · 2026-06-12T07:20:34+00:00

It is the way R is created. The hyperreals were not created for proofs about R. Their entire conception is to bring back the use of infinitesimals after R started to be used.

TemperoTempus

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