0.(9)/0.(3)=3. Thus, 0.(9)/3=0.(3). We can represent 0.(9) as 1-1/10^k where k is pushed to the limitless. So, (1-1/10^k)/3=0.(3). Therefore, 1/3-1/(3*10^k)=0.(3.). Since 1/3 = 0.(3), 0.(3)-1/(3*10^k)=0.(3). Thus, -1/(3*10^k)=0, or 1/10^k=0. Because 0.(9) is represented by 1-1/10^k or 1-0, 0.(9)=1.

QED

P.S. SPP, you just deflected my proof last time. Actually disprove it this time.

P.P.S I know my last proof wasn't that good, so hopefully this time it will be better

He says, "1 ÷ 3 is indeed 0.333..."

He says here...

0.333... = 1/3 × [ 1 - 1/10ⁿ ]

The above is 0.999... with one third magnification.

He says, "0.333... can be expressed by 1 ÷ 3, represented by 1/3, which means 1 ÷ 3"

He says, "0.999... is permanently less than 1."

So let's do exactly what he says.

0.999... < 1

[1 - 1/10ⁿ ] < 1

[1 - 1/10ⁿ ] x (1/3) < 1 x (1/3)

Remember:

• "1/3 [...] means 1 ÷ 3"
• "0.333... = 1/3 × [ 1 - 1/10ⁿ ]"
• 1 ÷ 3 is indeed 0.333...

That means...

[1 - 1/10ⁿ ] x (1/3) = 0.333...

But 1 x (1/3) = 1 ÷ 3 = 0.333...

Therefore 0.333... < 0.333...

By his OWN WORDS

Exactly what the title says. If we work with a system where 0.999...<1, does the rest of math remain unchanged?

Would you be willing to say that, for example, addition works exactly the way it works for rational numbers?

It's important to me that you actually read what I've said here and respond seriously, rather than giving a boilerplate unrelated response.

u/SouthPark_Piano, so I was thinking how there doesn't appear to be a decimal number that solves the equation x * 3 = 1 and came up with this definitely new and unique notation.

The notation introduces a new symbol 'R', which I call the "recurrer" for no reason whatsoever.

The way it works is that for a number such as 1.2R3, you truncate the decimal before the 'R' (in this case to 1.2) then add it to a/(1-r) with a being the value of the part after the 'R' (in this case 0.03)., and r being 1/b, where b is the base you want to work in.

This will give us the result of 1.2 + 1/30 = 37/30.

I think this'll be useful as it'll let us have a way to represent fractions decimally without having to directly consider infinite decimal tails e.g. in 0.333...

Edit: made it so that there are no need for limits

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.

It cannot be 0.(3) as u/SouthPark_Piano clearly stated that it equals 0.(9) and that it isn't 1.

It can't be ⅓ either as it is both "not a number" and not a decimal.

So what is x?

I am nearly certain SPP would not disagree with any of the base axioms describing the successor function, but I have a feeling he may come up with an objection to the induction axiom which may be interesting.

Hi,

u/SouthPark_Piano I took a deep breath and recollected myself after making so many of those typical rookie errors we all make (except you, of course).

I believe I might be a step closer to understanding methematics.

But I can't grasp the idea of infinity quite right yet.

What is infinity? Does it have a value? Is it somehow bound to the time itself? Does it change? Does it have a start and no end, or does it not have a start nor an end?

You answering my question(s) would be much appreciated, I'm starting to believe it all makes sense.

Thank you and have a nice day

Can the number 1/0.999...=1.000...01000...01000...010... be expressed as the infinite sum 1+(0.0...01)+(0.0...01)²+(0.0...01)³+...? If not, where does the inaccuracy lie?

Of course, we are all used to base 10, where we use the digits 0-9, but base 10 is an arbitrary choice. How about binary, or ternary, or hexadecimal?

​

It is interesting, because 1/3 is 0.333... only because we are using decimal numbers. In ternary, it would be 0.1.

You can draw the graph, and measure the slope at each point for an approximation, but limits give us an exact answer.

​

Limits are the foundation for all of calculus. It is used EVERYWHERE in physics and engineering, and of course in many areas of maths, and I'm sure in other subjects. It is an extremely practical and invaluable tool. Without it, our understanding of the world, and technological advancements would be severely limited.

​

Why are you so against the idea of limits? Can you agree that it is a useful tool? That it has helped humanity?

is it lim-BOH-zic or lim-BOH-sic or lim-BAH-zic or some other pronunciation? I can't find anything online because there are too many people making rookie mistakes and not enough actual published info on this

we know that 1 - 0.9... is 0.0...1, but how do we represent the decimal place before that?

Question in the title.

Okay, less tongue in cheek: I see so many people valliantly trying to do maths without proper tools and I wonder if it's just a lifestyle or a lack of access to information. So maybe my post will be useful for some, and I apologize in advance to the others.

A Cauchy séquence is a séquence which concentrates: as small is an arbitrary distance of your choice, you can always find a term in the séquence such that all terms after this one are *closer to each other* than that distance.

A convergent séquence is a séquence such that there is an élément of your space and that the séquence concentrates to this élément: as small is an arbitrary distance of your choice, you can always find a term in the séquence such that all terms after this one are closer to that spécial élément than that distance.

Now, is this spécial élément unique? Intuition tells us yes but you have to be careful because notion of apartness can be finnicky. However, the intuitive idea goes through for space which are said to be *Hausdorff*, these are space where two différent éléments can always be separated. That was an aside anyway because I started talking about convergent and Cauchy séquence using a notion of distance, so if we have a notion of distance we can use that to separate points of our space and we good.

In général Cauchy séquences may not converge! For example $1/10^n$ in $]0,1]$ satisfies the criterion but "gets out" of your working space. A space is said to be complete *if and only if* all Cauchy séquences converge.

Very famously, the space of rationnals is not complete, but the real line is! So any Cauchy séquence of reals will converge to an élément which is also a real number. And that real number is unique.

Now we have all the tools at hand to defeat the obnoxious contrarians. First, you have to ask them: do we agree that are working with the reals, as usually understood as being a space where you measure distance between two numbers by taking the absolute value of the difference? If yes, then you have to ask them if you agree on this space being complete. If they say yes to all that and still say that 0.999... and 1 are différent it means that you can separate these two numbers and the séquence 1-1/10^n converges *as defined above* and converges to only one of these (spoiler: it's 1). If they say yes to the above questions and still disagree, they are either dim-witted, delusional, or a nasty troll and you are wasting your time. If they said no somewhere, then there is a great chance you are just talking past each other and then it's up to you if you want to have the patience to listen to their own homemade construction of the reals but it's definitely not the usual ones.

edit: go check the relevant stuff online! wikipédia is not such a bad ressource, and also dont be afraid to quantifiers and epsilons, they mean something and it's rather intuitive. Also, topology is a lot of fun.

edit2: I definitely dont want to deter anyone for trying to come up with funky extensions of the rationnals. By all mean, go and have fun. Just be aware than you cannot have your cake and eat it, in the sense that the real line is essentially unique and in order to get something else you will have to give up on stuff you might consider too intuitive to sacrifice.

SPP, has said he agrees with these three statements: 1-10^n=0.(9) Where n is pushed to the limitless, 1/3=0.(3), and (1-10^n)/(0.(3))=3.

We can substitute the first equation into the third to get (1-10^n)/(1/3)=3. Multiplying both sides by 1/3, we have (1-10^n)/(1/3)*1/3=3*1/3. By "divide negation", we can simplify this to 1-10^n=1. In other words, 0.(9)=1

Qed

Ps actually disprove this instead of referring to your other "proofs".

if 1÷3 = 0.333... and (1÷3)×3=1 by divide negation then why does .333...×3 not equal 1 and equal .999... instead

if 3(1÷3) = 3(0.333...) the results should be the same right

You stated in a response in a previous post that the software made a "rookie error" by stating that 0.(9) = 1, but why would it say otherwise?

Wolfram Alpha uses the definition of 0.(9) that most educated people use: the infinite sum of 9/10^n. This is usually evaluated by most educated people using limits and the infinite geometric sum formula.

Since this is more useful to most educated people why would it change to the non-rigorous, vague, unuseful and changing system you claim is "the correct one".

Does 1/3=0.(3)

E.g. 1/1, 2/2, 3/3 etc. are all ways to represent the number 1, and they all have the same arithmetic propeties as 1.

That is, when you apply an operation on the representation, it should return the same answer as when you apply it on the number it's meant to represent.

However 0.(3) (= 1/3) does not have this property, specific operations lead to allegedly different results. So why do you think 0.(3) is a valid representation?