*ruins the foundations of math*

Is there something wrong with that? That’s exactly why it’s an axiom.

>implying "meaning" exists

New axiom: you're a faggot. Can't refute this, because it's an axiom. Consistent too.

Why are you such a brainlet faggot that needs "purpose" for something rather than just accepting a logical objective truth is true regardless of it's engineering "purpose".
Scientists and engineers are fucking brainlets.

[math]\sum_{n=1}^{\infty} \frac{x}{(x+1)^n} < \sum_{n=1}^{\infty} \frac{(x+1)}{(x+2)^n}[/math] for every partial sum.
There are truly an infinite amount of unique numbers that are read as "0.999..." and they all have relative greater than or less than values compared to each other.

[math]0.\bar{9} < 0.\bar{9} < 0.\bar{9} < 0.\bar{9} < ... < 1[/math]

I mean, if you believe infinity is a valid concept and it's usage in mathematics doesn't need to be corrected, you'll just have to live with the fact that [math]0.\bar{9} \neq 0.\bar{9}[/math], as prescribed by infinite property.

I have no idea what your point is.

>implying

*saves mathematics*

...

Is this your first time attempting an argument? Why is this so hard for you?

It would make sense for [math]\infty \neq \infty[/math] since [math]\infty \stackrel{+}{-} n = \infty[/math]. Mathlets just never gotta around to firing enough simultaneous neurons to piece together whether they wanted to treat infinity like a number or treat it by its own arbitrary rules. Get off the fence brainlets.

Define number

Because the existence of transfinite numbers and set of cardinality bigger than of the natural numbers surely don't look like "logical objetive truth" for me. Rather, they almost look like a logical falsehood. So i wanted to know if they at least serve a practical purpose.

Is this your first time responding ti bait? Why is this so hard for you?

>I'm fine with the idea of a set of all natural numbers...
That's literally what the axiom in pic says.

[math] \sum_{n=1}^{\infty} \frac{x}{(x+1)^n} < \sum_{n=1}^{\infty} \frac{(x+1)}{(x+2)^n} [/math]
But that's incorrect. It's should be [math] \leq [/math] , not [math] < [/math]

>x = 0.999....
>10x doesn't equal 9.999... because there is a zero percent chance i've referenced the same exact 0.999..., given there are infinitely many of them with non-equivalent values
>y can equal 9.999... though
>y-x = 9.999...a - 0.999...b
>y-x = 9
> x = 0.999...
> y = 9.999...
?????????????? How domes i prove 0.999 = 1 den?

Yes, but when you use other axioms with it (like the power set axiom), it result in things that I'm not fine with.

For every partial sum, it is never equal. For every partial sum, it is always less than. It is not incorrect. The incorrect statement is to say they are equal in some loose misinterpretation of final summation when there are literally infinite proofs claiming otherwise.

something which can be used in arithmetic.
for example, n+1 = n+1

infinity fails this outright. infinity+1 = infinity.

3/3=1
1/3 X 3 =1
1/3=0.333...
0.333... X 3 = 1 = 0.999...

How does infinity fail the test? inf+1 = inf+1.

>infinity fails this outright. infinity+1 = infinity.
What's the issue?

What's wrong with the power set axiom?

>

(You)
it means infinity is not a number and is no more valid a concept to math than the word "many". How much is many?

Ironically the only use for infinity in math is proving infinity doesn't belong in math using the very same methods mathematicians use infinity with.
it completely cancels itself out as a concept and has amounted to little more than a wafting fart.

There are proofs that the real numbers don't have a bijection with the naturals, I don't understand why you are so angry at this.

By itself? Nothing.
But when you use it with the infinity axiom you create a infinite set that is "bigger" than another infinite set. And that you can't even define all it's members properly with logic because of how "big" it is.
It seems counter-intuitive and unecessary to me.

My problem is exactly with this concept of "real number". I don't see why you would need a set or even the concept of "real numbers". Computable numbers, or definable numbers, plus some rougue elements perhaps would be enough for anything I can imagine and you still would have a cardinality small enough to have a bijection to the set of natural numbers.

Yea because there is a difference between a countably infinite and uncountably infinite cardinality. Why are you so bent out of shape?

My problem is that I see no practical use for uncountable sets and I don't see them as intuitive/obvious enough for me to accept they "exist" or "shoul exist". They seem counter-intuitive for me and I don't see why we would even want them.

It doesn't matter that you have a problem with the concept or if you can't see a use for it or if they're "un-intuitive" (which they aren't you just need to think about them more), there are proofs that the real numbers exist. If you seriously want to deny the existence of e, √2, π or literally the uncountable infinite amount of real numbers just because they make you angry that math isn't discrete I don't know what to tell you. This universe and it's physical laws are NOT what math is, math is a platonic thing that exists outside of space and its laws are a superset of the laws of physics.
Just by accepting the existence of the naturals and we can build all the other sets from them. How can you deny √2? If we accept the naturals, what happens when I take the root of 2?
>you can't do that
Yes I literally and obviously can and just did, so what now? The number is irrational, it's a real number. How do you deny e and pi and the other transcendental numbers? I really don't get this the finitists have been wrong for hundreds of years and there are proofs this is actually just denying reality.

I don't undestand, the continuum of a line seems counterintuitive to you?

3/3 = 1
1/3 = [math]0.\bar{3}_{\frac{1}{3}}[/math]
1/3 × 3 = [math]0.\bar{9}_{\frac{3}{3}}[/math]
[math]0.\bar{9}_{\frac{3}{3}} = 0.\bar{9}_{1}[/math]
[math]0.\bar{9}+_{1} = 1[/math]
[math]0.\bar{9}_{\frac{n}{n}} < 0.\bar{9}_{\frac{n+1}{n+1}}[/math]
[math]0.\bar{9} \stackrel{=}{\neq} 1[/math]

Sci is never gonna fix it's latex

[math]\frac{1}{3} = 0.\bar{3}_{\frac{1}{3}}[/math]

Dont bring manmade laws of nature into math. Theres also nothing explicitly wrong with infinity, just its implements in math thus far. It needs retuning. In any case, infinity is not a number. Infinity even has a problem of breaking time, for example, something normal numbers cant do. An infinite sum should be completely unsolveable because the work never ends, and any slice of the most recent work accomplished is instantly outdated and wrong by continued partial sums and the eternal future history of partial sums. An infinitely long significand of unique numbers not falling in a pattern can only be proven true in its infinite lack of pattern by doing the work, which if it were truly infinite this work would never end but if it were finite to some insane degree, the work would end. In either case, if the work has not yet ended or cannot end, neither can predict when or if the work will end. Both are equally valid and neither can be invalid. If i told you I was immortal and you asked me to prove it, what could I do? We'd just have to wait a sufficient amount of time before enough has passed to convince yourself I hadn't aged. Same problem here. We can only wait for an answer to if numbers like pi or e have an end to them, but ironically numbers like these present a concept of infinity that is ignored when [math]\infty[/math] is used in place of limits - the concept of time. Infinite sums ignore the infinite time requirement of infinity, and in doing so, ignores the value of infinity outright allowing any little shitter to substitute an answer that fits the bill regardless of proveable authenticity.

>In any case, infinity is not a number.
Wrong.

>wrong

I don't have any problems with e, √2 or π.
You can have all this numbers in a set and it will still be countable.
All numbers that you can define are countable.

What makes the real numbers so much "bigger" than the set of natural number is a huge amount or "numbers" that you can't even define. And that is what I have a problem with.

>What makes the real numbers so much "bigger" than the set of natural number is a huge amount or "numbers" that you can't even define. And that is what I have a problem with.
there's a model where all reals are definable (note that the notion definability is always something that must be external from a model)

No. But do we need all those undefinable numbers to have the properties of a continuum line? The rational numbers already make a line "dense". Computable numbers already make so that most numbers that you would excpect to be in the line like √2 are already there. And if you consider all possible definable numbers, you get even more points in the line, while still being countable. "Between any two points that you can define there are infinitely many more than all definable numbers in the line, so many more numbers that you can't even make a bijection between them" seems counterintuitive for me.

What model?
I wasn't aware of this. If there is a way to define this numbers, I won't have a problem with them anymore. Could you please tell me where I can find about such model user?

Are you going to answer my question?

>it means infinity is not a number
How so?

>it means infinity is not a number and is no more valid a concept to math than the word "many".
I don't follow your logic, if infinity + 1 = infinity implies infinity is not a number then does 0 * 1 =0 imply that 0 isn't a number?

You can’t even define number and yet you think infinity is not a number. When will you accept that freshman calculus students are not smart enough to argue against the important of the reals?

>Infinite sums ignore the infinite time requirement of infinity
What do you mean?

...

>something which is true for the finite cases must be true for infinite cases
>no, infinity doesn't follow the rules of finity (by definition)
>b-but the finite cases...

Yes, the partial sums are less than, but we're not talking about partial sums, now are we?
>what is a limit

Are you retarded or do you not understand that for every countable finite n partial sum, of which there are infinitely many, they all evaluate inequal.

Mathlets literally understand infinity so poorly that they cannot keep a consistent concept for it. In one moment its treated by normal number rules, in another moment it obeys arbitrary irrational rules instead.

Garbage game. Not interested in it. If you dont want to use infinity correctly then we're just as well off not using it at all.

>In one moment its treated by normal number rules, in another moment it obeys arbitrary irrational rules instead.
What do you mean?

It is never to be treated by normal number rules; noone claimed that.
It is true that each of the infinitely many partial sums, the relationship is strictly equal, but again, this is not about partial sums. The face that the sequence of partial sums has a limit which is outside the sequence is not a contradiction of anything other than an evidently poor intuition of limits.

It must be very hard to live your daily life. You have my condolences user. Perhaps you will get lucky and a drunk driver will hit you while you walk down the street.

>But do we need all those undefinable numbers to have the properties of a continuum line?
Yes, we do. Doesn't your intuition say that every Cauchy sequence on the line should converge somewhere on the line?

You actually cannot take the square root of two without some concept of infinity. It requires an infinite process in order to compute.

Why aren't you allowed to take the square root of the negative integers, they're not so far from the natural numbers? Because it produces an imaginary number? The fact of the matter is that both operations take you out of your original set of discourse.

>You actually cannot take the square root of two without some concept of infinity.
What do you mean?

Saved mathematics. RIP, Vladimir Voevodsky!

[math] \sqrt{2} [/math] can be considered/defined as the element [math] x [/math] of the field [math] \mathbb{Q}[x]/ \langle x^2-2 \rangle [/math] or of the field [math] \mathbb{Z}[x]/ \langle x^2-2 \rangle [/math]
You don't need a concept of "infinity" (notion of distance, limits, etc.) to define it, like how you have to with π. It's an algebraic number.

out of curiosity, what does "snow patrol" refer to? can't find anything on urban dictionary

Try taking the square root of 43 and writing it down in decimal form without a calculator. You can't because the operation is basically an approximation that continues infinitely until some limit is reached. You can only reach that limit if the concept of infinity exists. This is also where the finitist problem with irrationals and other infinitely long numbers comes from; you can't actually compute them, or do any operations with them, without some concept of infinity.

Consider doing sums with irrationals. We typically denote irrationals either with specific operators (root 2) or with symbols (e or pi), or with infinitely long decimals. You can't actually ever write out infinitely long decimals, and if you try to add them together that requires and infinite number of computations. If you sum together e and pi, you can't actually simplify it to a single irrational number unless you convert to infinitely long decimals, and you run into the same problem.

Contrast this with natural numbers. We're agreed upon the specifications for these numbers; we use the arabic numerals and we can generate arbitrarily large numbers using this form. If you sum together any two natural numbers, you can produce a single number in a finite number of steps, although it might take a lot of time. There is no computational problem for us.

It's a valid position to take, although it's limiting to a certain degree.

do tell, what's Q[x], if infinity is not allowed?

...

The x that satisfies those criteria does not exists in the fields you've given, so the set would be null in both cases. You have to extend your field to algebraic numbers in order to actually have the number exists.

Furthermore, even after this extension you still require an infinite number of steps to actually simply expressions involving this number.

>out of curiosity, what does "snow patrol" refer to?
Presumably the band.

>Try taking the square root of 43 and writing it down in decimal form without a calculator.
Does that make it require the "concept of infinity"?

Does writing 1.000... mean that the decimal representation of 1 requires the "concept of infinity"?

Actually, the continuum is not really "bigger". There is just no bijection between the natural numbers and the continuum.
And you can in fact not resolve this dilemma by switching to computable reals - you won't find a computable bijection between the naturals and the computable reals.
And this, however, can't be resolved by saying: "Ok, but there is a classical set-theoretical bijection." because then you look at the computable standpoint from a higher standpoint, i.e. you go to a stronger set theory and act as if there were only computable reals.

>The x that satisfies those criteria does not exists in the fields you've given
What do you mean?

The decimal representation you gave would require the concept of infinity, because it would require an infinite process in order to write down that number.

However, the number 1 does not require this process.

>However, the number 1 does not require this process.
Neither does the square root of 43.

Please be bait. Don't tell me Veeky Forums can't even read basic notation.

I'm currently reading the notation you've used as "element of the rationals/integers which satisfies this equations."

Based on what operations you allow, you can construct pi with a finite number of steps, just like you can with root 2. For instance, if sin(x) and arcsin(x) are legitimate operations, you can simply write sin(x/2)=1 and solve for pi with two steps.

However, the sin function, like the square root function, requires an infinite number of steps to actually compute.

How about these (below) ancient caveman symbols with new "trips dubs" accoutrements?

jesus christ you're so dumb
just say "ok I don't know sorry" instead of this bullshit

Nice quads.

Ok. Let's do a test. I'm going to let you pick two natural numbers, and I'll add them together and produce a single natural number.

Afterwards, I'm going to produce two irrational numbers, and I want you to write down that number in a single representation.

The rules are that both of us have to use the symbols associated with our respective numerical fields. For natural numbers, I can only use the symbols {0,1,2,3,4,5,6,7,8,9}. I cannot put 0's at the beginning of numbers. I can combine numbers in any order I like.

For the irrationals, you are only allowed to use pi, e, the square root symbol, or any other common operator, and you can also use any decimal representation. You are allowed to use the same digits as above, except you are also allowed to put in a decimal point between your numbers.

I'll wait for your two natural numbers.

As far as only cardinality is concert, it's the set in OP's pic, which is the set of all natural numbers.
There's no mention of "infinity" in it, whatever "infinity" means.

Huh? What do you mean it does not exist?
Also, by x I meant the coset x+.

OP's pic is the axiom of infinity, genius

That's just its name. All it says is that there is an inductive set (the natural numbers {1,2,...}).
Yes, you can't biject it with a finite set.
In any case, when we talk about infinity we have to be precise about what we are talking.
I believe that the guy I responded to was saying that you can't define sqrt(2) without a limiting process because "It requires an infinite process in order to compute" which is simply incorrect.

That's not what I claimed, I know that √2 isn't a transcendental number, I simply claimed that just with the acceptance of the naturals we can get the reals by taking the √2 which can not be expressed in N Z or Q

what are you even trying to say? if you accept N, you accept Q[sqrt(2)]

√2 is not in Q it's in R, I think I'm misunderstanding what you're saying.

>with the acceptance of the naturals we can get the reals by taking the √2
How does "taking" the square root of 2 "gets" you the real?
>√2 which can not be expressed in N Z or Q
Yes it can, I showed you how: That's also how you can construct i, or any algebraic number for that matter.

Two things to define a number. You can count it, and you can do math with it. You can't do proper division and multiplication on 0 (even though you can do addition and subtraction so w/e) but you can count no things to acknowledge there is zero of whatever you're looking for. By this abbreviation, infinity falls away from even 0. You can't count an infinite amount of things, and you can't perform logical arithmetic on it. It is simply, pimply, dimply not a number. Again, replace "infinity" with "many" and see how far thay gets you in trying to derive finite results for something like infinite sums. Its stupid shit.

When mathematicians use "infinity" i imagine a bunch of shirtless rednecks in lousiana trying to figure how many beers they'll need to buy to hoot and holler all night. "A lot" does not suffice as an accurate result, and it's pure shibrained laziness to concatenate infinite work and assume any finite number result.

Its treated as a normal number in limits, as you can iterate over n finitely to get a finite result but iterating over n infinitely logically does not return a finite result at any time because there is infinite work to be done. That a result is claimed from infinite limits anyway directly implies mathematicians treat infinity like a finite number because retardation.

It really is pick abd choose flip and flop how mathematicians attempt to justify the existence of infinity in maths. It's just fuckin gay and lame.

>When mathematicians use "infinity" i imagine a bunch of shirtless rednecks in lousiana trying to figure how many beers they'll need to buy to hoot and holler all night. "A lot" does not suffice as an accurate result, and it's pure shibrained laziness to concatenate infinite work and assume any finite number result.
You are obviously are an engineering shitter who doesn't even know what [math] \lim\limits_{n \to \infty} a_n = a [/math] means.

>You are obviously are an engineering shitter
This, fucking engineers are so annoying.

that's because you just don't care I guess, you were already shown how to construct Q[sqrt(2)]: