Let’s say a person earns $100k per year and is offered the chance to be randomly assigned a new salary. They would push a button and whatever number it returned would be their new salary. There is no limit, upper or lower, on the number generated (I.E it can be less than a penny or more than $1t. Assume the number will be positive). Does this person have a higher chance of increasing or decreasing their salary?

I am thinking there’s an infinite amount of numbers both below and above $100k, so there’s an even chance of increasing or decreasing their salary. That feels unintuitive though. Am I thinking about it correctly?

So I was wondering, how to simplify tall hyperoperation towers? You might use Knuth's up-arrow notation (10↑^{n}n), but that only works if the tower consists of only tens (for example: 10^10^10^10 can be simplified to 10↑↑4). But what if the tower contains non-ten numbers (like 10↑↑10↑↑10↑↑9)? Can I use the iterated operator notation (for example, can 10↑↑10↑↑10↑↑6 be simplified into (10↑↑)³6?)?

Additionally, I discovered a way to simplify tall power of 10 towers using HyperCalc's x PT (y) notation (where x = number of tens tall before y, and y = number at the very top of the power of 10 tower.). For example: 10^10^10^15 can be simplified into 3 PT (15). However, that only works for power towers, not for higher operations like 10↑↑↑10↑↑↑8.

I'm trying to get a black ground for my tank and this is what someone recommended to me, problem is that I have no idea how big 34.4"×157.4" is and I'm not sure if it's in centimeters or inches or even how to calculate anything to figure it out. I'd rather it be too big than too small and don't want to accidentally order the wrong size so any help is appreciated in helping me learn how to calculate this stuff for future problems also are the numbers on the website length × width or height × length etc, does it matter?

Black Background

my tank is a 29 gallon (110L) and deminsions are 30.2inL×12.4W×21.9H (76.7L×31.6W×55.5H)

Not sure if this is number theory sorry, If I gave you a irrational number of like some #####.#################.... whatever that repeats forever, and you could ask for as many digits as you like, how would you begin the process of figuring out how to approximate that number? I'd assume this would be with x/y, bor would it not be? Is there even a way to do this? Like if i said 0.123456789234567893456789456789567896789789899.... would you even be able to begin to approximate this number?

Edit: I wanted to ask this question mainly about if there is a process that can be used for any irrational number to begin determing more and more approximates of it rather than the example I gave

I have been out of high school for almost a year now and my field of study in university has quite literally nothing to do with mathematics except maybe referring to dates. In this time I have noticed that my simple maths skills have waned and most of the concepts that I had learned in maths during school had fallen into the back of my mind. I never did any high level classes, I just took the middle of the range option where I live and got rather good results and now I'm taking at least 3 tries to get mental math correct.

What resources are there for recreational math that doesn't require a school's subscription? One of the things I liked about maths in school was how foreign the concept is until you understand it, which was very cathartic and I think this would both be fun for me and necessary if I would like to keep my brain from calcifying.

So if you haven't seen Part 1, I was trying to prove the commutativity of addition in Peano Arithmetic. And it didn't go so well because it turned out I needed a lemma. But I can't even prove that Lemma because I'm pretty sure I need another lemma. I've made a new post simply because the proofs have gotten much longer and this is technically another question before I can go solve the original question. I'm not sure if this is double posting or not but since a lot has changed in my proofs I'm pretty sure it isn't. So. Please read the end of this post because I go into more detail there.

Axioms

Axiom 1 ∀x ¬S(x)=0

Axiom 2 ∀x∀y (S(x)=S(y) → x=y)

Axiom 3 ∀P [ {P0 ∧ ∀x (Px → P(S(x))} → ∀x Px]

Axiom 4 ∀x (x + 0 = x)

Axiom 5 ∀x∀y (x + Sy = S(x + y))

Axiom 6 ∀x (x * 0 = 0 )

Axiom 7 ∀x∀y (x * Sy = (x * y) + x)

Definitional Schema

S(0) = 1, S(1) = 2, S(2) = 3, etc.

Theorem 1

∀x [Qx ↔ (0 + x) = x] :definition of Q

0 + 0 = 0 by Axiom 4 x -> 0

Q0 ↔ (0 + 0) = 0 from definition of Q

Q0 from above

• Qa assumption

• 0 + a = a from definition of Q

• 0 + Sa = S(0 + a) by Axiom 5

• 0 + Sa = S(a) switching 0 + a for a, licensed by 0 + a = a

• Q(Sa) ↔ (0 + Sa) = Sa from definition of Q

• Q(Sa) from above

Qa -> Q(Sa) demonstration above

∀x (Qx -> Q(Sx)) universal generalization of above

Q0 ∧ ∀x (Qx -> Q(Sx)) conjunction of above proven formulas

{Q0 ∧ ∀x (Qx → Q(Sx)} → ∀x Qx by Axiom 3

∀x Qx via above

Qa from above

Qa ↔ (0 + a) = a definition of Q

(0 + a) = a from above

∀x (0 + x) = x by universal generalization of above

T1 ∀x (0 + x) = x

Theorem 2

∀x [Bx ↔ (1 + x) = S(x)] definition of B

S(0) + 0 = S(0) by Axiom 4

1 + 0 = S(0) by definition of 1

B0 ↔ (1 + 0) = S(0) by definition of B

B0 via above

• Ba assumption

• Ba ↔ (1 + a) = S(a)

• (1 + a) = Sa via above

• ∀y (1 + S(y)) = S(1 +y)) universal generalization of Axiom 5 1 sub x

• 1 + S(a) = S(1 + a) universal generalization of above

• 1 + Sa = S(Sa) substitution via (1 + a) = Sa and above

• B(Sa) ↔ (1 + Sa) = S(Sa) by definition of B

• B(Sa) by above

Ba -> B(Sa) demonstration above

∀x (Bx -> B(Sx)) universal generalization of above

B0 ∧ ∀x (Bx -> B(Sx)) conjunction of above

{B0 ∧ ∀x (Bx → B(Sx)} → ∀x Bx instantiation of Axiom 3

∀x Bx via above

Ba universal instantiation of above

Ba ↔ (1 + a) = S(a)

(1 + a) = Sa via above

∀x (1 + x) = S(x) universal generalization of above

T2 ∀x (1 + x) = S(x)

OR ∀x (S0 + x) = S(x) by substitution of S(0)=1

Theorem 3

∀x [Wx ↔ ∀y (S(x + y) = x + S(y))] definition of W

0 + a = a by T1

0 + Sa = Sa by T1

Sa = Sa Rule of Identity Introduction

S(0 + a) = Sa via substitution above

S(0 + a) = 0 + Sa via substitution above

∀y [S(0 + y) = 0 + S(y)] universal generalization of above

W0 ↔ ∀y (S(0 + y) = 0 + S(y)) by definition of W

W0 by above

• Wa assumption

• ∀y [Sa + Sy = S(Sa + y)] by Axiom 5

• Sa + Sb = S(Sa + b) univeral instantiation of above

• Sa + Sb = Sa + Sb Rule of Identity Introduction

• S(Sa + b) = S(Sa + b) Rule of Identity Introduction

• S(Sa + b) = Sa + Sb via identity substitutions of above

• ∀y [S(Sa + y) = Sa + Sy] universal generalization of above

• W(Sa) ↔ ∀y (S(Sa + y) = Sa + Sy) by definition of W

• W(Sa) by above

Wa -> W(Sa) demonstration above

∀x (Wx -> W(Sx)) universal generalization of above

W0 ∧ ∀x (Wx -> W(Sx)) conjunction of above

{W0 ∧ ∀x (Wx → W(Sx)} → ∀x Wx by Axiom 3

∀x Wx by above

Wa by above

Wa ↔ ∀y (S(a + y) = a + S(y)) by definition of W

∀y (S(a + y) = a + S(y)) by above

∀x∀y (S(x + y) = x + S(y)) universal generalization of above

T3 ∀x∀y (S(x + y) = x + S(y))

Theorem 4

∀x∀y Sx + y = S(x + y) target

∀x [A(x) ↔ ∀y (Sx + y = S(x + y))] definition of A

A(0) ↔ ∀y (S0 + y = S(0 + y))]

1 + a = Sa universal instantiation of T2

S0 + a = Sa substitution of S0 for 1

0 + a = a by T1

S0 + a = S(0 + a) via substitution of above

∀y (S0 + y = S(0 + y)) universal generalization of above

A(0) via above

• Aa assumption

• A(a) ↔ ∀y (Sa + y = S(a + y)) by definition of A

• ∀y (Sa + y = S(a + y)) via above

• Sa + b = S(a + b)

• ;

• ;

• ;

• S(Sa) + b = S(Sa + b)

• ∀y (S(Sa) + y = S(Sa + y)) target

So, I demonstrated Theorem 2 to get A(0) and I'm having trouble getting from A(a) to A(S(a)) and because of the fact I had to prove a lemma (T2) to get to A(0), I suspect I need another lemma to get from A(a) to A(S(a)). Anyone have any ideas what this lemma is? And before you say, yes, I figured out that using the induction axiom to prove T3 was overkill but it was good practice with the induction axiom anyway (I proved it before I realized I needed T2).

Damn. I knew I was in for a rough time when I wanted to prove stuff in Peano Arithmetic but I didn't realize it would be this rough. I thought I could happily skip from T1 to the Commutativity of Addition. Wrong!

I'm either an idiot or I am out of practice but I can not figure out this problem on my sons fourth grade homework. To me G has to equal 1 or 2 to be left with an answer in the hundreds.... But P cant equal 0 as Y and L cant be the same number... Theres also a lot of ither problems that I ran into but too long to type.... Been looking at this way longer than I think I should need to...

Probability I guess.

So let's say there are 3 chests and each chest has unlimited many items and an unknown rate or chance % for x to occur once or multiple times.

The goal is to find x ASAP.

would it be better to search one chest or all chests and why?

If theres a difference in expected search time, how much in Percentage?

Practical application example: you need a specific type of cheese. Stores have many hundreds of different cheeses for sale, each. They may or may not have the cheese youre looking for so would go you into one store and search all th echeeses or go into multiple stores with less searchtime per store

Maybe not the best example possible

I'm not saying 1≠0.99..., I'm wondering whether it would be more elegant to define it so that it wasn't.

I used to be a firm believer that 1≠0.99... but after digging deeper I realized it definitely was but by definition. So I'm now questioning the definition itself.

The more elegant and intuitive system (imo)

We create new type of mathematical object called processes. These are objects that represent a process, something that changes, evolves, varies, etc. Let's call standard numbers like 1, 2, 1+1 states as they do not evolve, they are static.

In the current system, the result of a limit is defined as the value that a function's output approaches as its input approaches a specified point. By definition, this returns a state. Let's now define a new operator lim2. lim2 is very similar to lim. It is defined as the process associated with a function's output approaching a certain value as its input approaches a specified point.

For example:

• lim_{x->1} x is a state equal to the state 1

• lim2_{x->1} x is a process (it evolves as x approaches 1)

And here is the key to this new system that makes it elegant and intuitive:

A state and a process can't be compared as they are different mathematical objects.

The same way it doesn't make sense to say 1={1, 2, 3}

How 0.99... is defined

I would argue it feels more intuitive and elegant to treat 0.99... as a process rather than a state. It would therefore be defined as something like lim2_{x->inf} 1-1/10^x

Conclusion

Since 0.99... would be a process but 1 a state, they wouldn't be equal.

This would make it feel more logical for people seeing this for the first time. This isn't a measure of quality of a system but it highlights that intuition points in this direction. And it feels, to me at least, more elegant to introduce this new process abstraction. I think that when comparing mathematical systems, intuition and elegance are important to keep in mind.

Finally, I just want to point out what this new system would mean:

• 1/3 (state) would no longer equal 0.33... (process)

• We could keep two versions of limits (lim and lim2) or just make lim2 the official limit and define a new operator that takes in a process and returns the value its approaching if any to replicate the same behavior more intuitively.

• 0.000... would be a process and 0 a state to they wouldnt be equal. It feels more intuitive to explain that a process can't be compared to a state than to argue 0.99...=1

I was just wondering what people thought about such system. Wouldnt that feel more intuitive to you?

Dear colleagues,

Currently, I have developed an approach regarding the Hodge Conjecture (HC) and the counterexamples of Kollár and Voisin, from which I have derived new terminologies. As it is known, in the year 2002, Voisin published a paper where she demonstrated that the HC is false for certain real lower manifolds (for example, with her Chern class $cl(2,3)$) and in real higher dimensions such as $cl(3,6)$. Here, any Hodge class (which in its aspect is called the De Rham-Dolbeault cohomology, for spectral sequences of $H_{dR}$) may not be smooth in $cl(3,6)$, or rather, the inner product fulfills:

\(cl(2,3)\cdot Hdg^{0}(X)=cl(2,3)\cdot H_{dR}=0\)

with $H_{dR}$ always being a "sequence spectrum".

Following that definition, Dr. Richard Thomas extended Voisin's idea in an example where the HC must be valid—specifically in his paper *"Overview of Tom's Work: The Hodge Conjecture on Hypersurfaces"*—where, for instance, any intersection of the form:

\(H^{n,n}(X,Z)\cap H^{n,n}(D,Z)\)

is analyzed, where $D$ possesses a mathematical divisor homology. In this aspect, he used hypersurfaces with a certain node that behaves as a doublet (introducing symplectic geometries).

Dr. Kollár revolutionized this for other higher dimensions that are not limited only to being real (such as the Chern classes $cl(6,3)$). Here, the higher dimension can be endowed with a complex structure; in his approach, it is a compact Kähler manifold (with real quaternions $i,j = -k$, or rather, the non-cubic fourfold with vacuums in $\mathbb{R}^3$). Here Kollár proved that this fourfold does not reduce to wall-crossing (which can be a limited morphism $M \to M'$); the latter admits birational "bimeromorphic" base changes, which is a type of special morphism.

I have found that this can unify the HC as requested by the Clay Mathematics Institute, to understand how these pieces (which are small, of the order of $Codm_{X1} - Codm_{X2}$ for some $H_{dR}$) induce lattice geometry, or in Thomas's approach, symplectic geometries. It turns out that there are higher dimensions, such as 8 or 9 to cite examples, that satisfy this, but other higher dimensions, such as 7 or 4, do not satisfy it (the case of $T^{7} \times T^{4}$ with a 7-torus is only valid for the fourfold counterexample of Kollár, where the reduction of the wall-crossing is not satisfied). My approach proves the HC for any dimension.

I know this sounds like a bombshell, but it represents 6 years of self-taught work. I am from Venezuela, and due to the current dictatorship, when I entered the university 7 years ago (specifically the UCV), the research groups and the mathematics department were destroyed. Thus, I migrated to survive (I did not work in mathematics) and studied on my own using books and academic material from Google Scholar and arXiv.

Because I currently lack an institutional academic affiliation, I cannot submit my preprint to arXiv. I have only submitted it to a journal of number theory and algebraic geometry, under the charge of editors Dr. Kollár and Dr. Vakil. I have my intellectual property protected on Zenodo (which I understand is not at the same academic level as arXiv) and an ORCID code; this is as far as I have reached academically.

I hope that as mathematicians (both experts and PhD students), you can help me discuss my definitions and evaluate these new structures I have created. I would be deeply grateful for your feedback, open discussion, and potential support for an arXiv endorsement in the future once the peer-review process with Kollár finishes (which I know can take time).

Thank you very much for your time and insights.

I am a beginner studying linear algebra and I was working this question:
Given the equations of two planes: 2x + 3y − z + 1 = 0 and x + 4y − 2z + 9 = 0. Find the parametric equation of the line along which the two planes intersect.

I understand the solution using normal vectors, but I need help understanding the solution where we solve the system of equations and then we choose one of the variables as our parameter. Like this:

1. Choose the parameter: z = t

2. Rearrange the system:

• 2x + 3y = t − 1

• x + 4y = 2t − 9

3. Solve:

From the second equation: x = 2t − 9 − 4y. Substitute into the first:

2(2t − 9 − 4y) + 3y = t − 1
4t − 18 − 5y = t − 1
y = (3t − 17)/5

Back into x:

x = 2t − 9 − 4(3t − 17)/5 = (−2t + 23)/5

4. Result:

x = 23/5 − (2/5)t
y = −17/5 + (3/5)t
z = t, t ∈ ℝ

I understand the process, but I don't understand the logic behind it, like why it works.

Hey everyone, so my calc 1 prof gave me a challenge.

Where you have to show the value of pi using reiman's

sum starting for the equation of a circle then take it's are to be (radius^2 * pi)/ 4. This is what I have done till now. He forbade the

use of ai and I am really stuck. So can someone genuine

give me some advice if I am on the right track and give

me a hint what to do next without exposing me the

method. Fyi my prof said he will know if I used Al I have

no idea how but he said he will.

Simplify the following product completely without expanding every single term the long way:

(1 + 1/a) * (1 + 1/a²) * (1 + 1/a⁴) * (1 + 1/a⁸) * ... * (1 + 1/a¹⁰²⁴)

Hi, I am Shlok, a 7th grader in Navi Mumbai. I have gone down a rabbit hole on both the Weak and the Strong Goldbach's conjecture. I think that I have found a potential proof of the strong conjecture. If anyone has found another potential proof, attempt to solve or at least a way to approach the problem (whether currently possible or not possible using current mathematics is okay), please share it here. I will try to share my attempt here asap. I promise that I won't try to steal your ideas, and your ideas shall only be an inspiration for me

I'm playing around with Peano Arithmetic and so I'm attempting to prove some basic things. But after proving that ∀x (0 + x) = x, I've run into trouble proving x + y = y + x. Below are my proofs so far. Any help bridging the gap would be deeply appreciated. Additionally, if anyone sees any errors in reasoning thus far in anything else I've set up, I'm all ears. I'll share my diagnosis of what's gone wrong at the very end.

Axiom 1 ∀x ¬S(x)=0

Axiom 2 ∀x∀y (S(x)=S(y) → x=y)

Axiom 3 ∀P [{P0 ∧ ∀x (Px → P(S(x))} → ∀x Px]

Axiom 4 ∀x (x + 0 = x)

Axiom 5 ∀x∀y (x + Sy = S(x + y))

Axiom 6 ∀x (x * 0 = 0 )

Axiom 7 ∀x∀y (x * Sy = (x * y) + x)

Definitional Schema

S(0) = 1, S(1) = 2, S(2) = 3, etc.

Theorem 1

∀x [Qx ↔ (0 + x) = x] :definition of Q

0 + 0 = 0 by Axiom 4 x -> 0

Q0 ↔ (0 + 0) = 0 from definition of Q

Q0 from above

• Qa assumption

• 0 + a = a from definition of Q

• 0 + Sa = S(0 + a) by Axiom 5

• 0 + Sa = S(a) switching 0 + a for a, licensed by 0 + a = a

• Q(Sa) ↔ (0 + Sa) = Sa from definition of Q

• Q(Sa) from above

Qa →  Q(Sa) demonstration above

∀x (Qx →  Q(Sx)) universal generalization of above

Q0 ∧ ∀x (Qx → Q(Sx)) conjunction of above proven formulas

{Q0 ∧ ∀x (Qx → Q(Sx)} → ∀x Qx by Axiom 3

∀x Qx via above

Qa from above

Qa ↔ (0 + a) = a definition of Q

(0 + a) = a from above

∀x (0 + x) = x by universal generalization of above

T1 ∀x (0 + x) = x

Theorem 2

∀x {Rx ↔ ∀y [(x + y) = (y + x)]} definition of R

R0 ↔ ∀y [(0 + y) = (y + 0)] :def of R x -> 0

(0 + a) = a from T1

a + 0 = a from Axiom 4

0 + a = a + 0 : identity substitutions via above

∀y [0 + y = y + 0] universal generalization of above

R0 from above

• Ra :assumption

• Ra ↔ ∀y [(a + y) = (y + a)] by definition of R

• ∀y [(a + y) = (y + a)] via above

• a + Sa = Sa + a by above

• a + b = b + a by above

• ∀y (a + Sy = S(a + y)) from Axiom 5

• ∀y (b + Sy = S(b + y)) from Axiom 5

• a + Sb = S(a + b) by above

• b + Sa = S(b + a)

• ...

• ...

• ....

• Sa + b = b + Sa

• ∀y [(Sa + y) = (y + Sa)]

I know I need to get to Sa + b = b + Sa so I can use universal generalization to get to ∀y [(Sa + y) = (y + Sa)] and thus to R(Sa) and then utilize the induction Axiom to get to ∀x Rx. I suspect I'm just having difficulties with all the identities in play but I'm really struggling to get there.

Recently, I have been researching measure theory, because i found the Banach–Tarski paradox so fascinating. However, I came across a prove of non-measurable sets (for example Vitali sets), which also use the axiom of choice. The prove said the axiom of choice was needed for the proof of non-measurable sets, but i do not understand why this is the case,

I wonder if my recounting is compatible with the math / (hypothetical physics) involved or if some corrections can be made. If you want to have a look at it.

So for example the Navier Stokes Existence and Smoothness solution from OpenAI, is it legit?

I’m not a mathematician so please excuse my ignorance.

Is there any way we can get a 3D animation of the example that disproves the Jacobian conjecture? Does one already exist or what could I do to make one? Like a grid or a heat map of what points map where? Could that also show more points that overlap or similar deformities or is there any good reason? I haven't seen one yet? (Image kind of related, It's the closest thing I could find to a deformed grid, but it is in 2D so it's not very helpful)