Відео

just use phi:C->V with phi(a+bi)=a+(b/pi)*j

You are write, problem with j² and other higher powers, I will fix it in my next video

It's true for real numbers √(-2)²= ±2

💕

Thanks

watch second video of e^j video

I also tried using the method p | (m, floor(m/2) menus 1), but it may not work for all cases. I'm currently working on an alternative method to find all prime

Yes, it's not perfect, but it works well. In my next video, I will create a formula for primes using a different method.

Must divide set S9 , all elements

Then a language barrier? Correct way is to say the whole set is not divisible by 9.

yes , 9 must divide all elements of S9 as R(S9,9) must be 0

Glad you find it interesting!

you can try your self , Ej =0 , Oj=j , E is even O is odd and j^n=0 for n>1

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We can write 0/0 = -1 as 0= 0×-1 0/0 is not 1.. It's 0k in singularity numbers

It means 1/0 is rational?

​@@log_menus_1 you demolished & annihilated that oversmart coward 😂

You can contact me on my Facebook page, check out UA-cam bio

Are u talking about primes formula or virtual numbers?

Today I learnt that X/log(X) gives the number of primes below X. Crazy!!!!

Any numbers? 😂😂😂 Can it be 2?

If any number then this will not satisfy the equation but I think only singularity numbers satisfy the equation

Multiply by 2

We can also find the nth prime. For example, if π(m) = 100, this represents the 100th prime. By using the π formula, we can determine the value of m and then proceed with the calculation.

@ That's not at all what it means to have a general formula!! If you want to compute P(100000), you need to first know the smallest m such that π(m)=100000, and that m is the output of P. Looking through the π function for such a value is not an "efficient technique" for calculting P, you are offloading the work of finding primes onto your π function.

We can find m using pi function

@@log_menus_1 Looking through the π function for such a value is not an "efficient technique" for calculting P, you are offloading the work of finding primes onto your π function.

We can find all the primes using formula.. I will make it better in future videos

Watch virtual numbers video to understand

@@log_menus_1 I watched that video after posting my comment. I found the video hard to understand, since you use some properties of the Real and Complex numbers but ignore others, and it is not clear which properties you are keeping, which ones you are not, and which new properties you are adding. (Those things are clear for a few properties but not for all.) Is there a reference that gives the axioms for the Virtual Numbers?

I understand how the use of familiar properties from real and complex numbers, along with new concepts like Virtual Numbers, can create some confusion. The goal is to establish a new framework that extends beyond the traditional number systems. I'll be sure to clarify the axioms of Virtual Numbers in future videos and provide more concrete examples.

No 👎

@@log_menus_1 At 4:50, that's a division by 0.

1/0 is singularity unit which is k

Infinity is not a number; it's a concept that represents something extremely large that we cannot write or count. If infinity were a number, we could add 1 to it, but that would contradict the idea of infinity. Additionally, there is no such thing as positive or negative infinity-it's simply a concept that doesn't have a sign. Remember, infinity is just a concept, not a number.

I will create a comprehensive video on the existence of God using mathematics in future videos. So stay tuned and subscribe to the channel. This video was just a short preview, and I didn’t explain much.

If I had asked you to solve x² = -1 before the discovery of complex numbers, would you still ask me to specify the domain? My question is straightforward: find x if x / x = -1.

@@log_menus_1 If you do not specify the domain, then symbol "/" has no meaning and the problem is nonsense. And yes, to pose problem x^2=-1 you have to either specify the domain, or to explicitly request to invent new domain in which symbols -1 and ^2 have meaning. Again: your formulation is dishonest and intends to confuse those who do not have serious mathematical education.

​@@log_menus_1😂😂😂😂 "discovery of complex numbers"...

The question of whether mathematics is discovered or invented is indeed a long-standing debate. In the case of complex numbers, I believe they were discovered, not invented. While the formalization of complex numbers as we know them today involved human ingenuity, the concept itself already existed, much like negative or irrational numbers. These mathematical ideas were always part of the structure of the universe; we simply needed to recognize and understand them. So, complex numbers weren’t invented-they were uncovered.

@@xgx899 The question is simple: find x if x / x = -1. You can use any domain or even invent a new one, as long as it satisfies the equation, just like how we extended the domain from real numbers to complex numbers.

Well... 0 = j^2 = j*j = j*ln-1 = ln((-1)^j) = ln(1 + j*ln(-1)) = ln(1 + j*j) = ln(1+ 0) = ln(1) = 0 ...does follow. Also, ln(j^2) would be ln(0), which was not defined anyway, so... You're kind of misusing the notation like that.

If my method is wrong, feel free to try your own method. Find .

@log_menus_1 What method??? As if it wasn't enough absurd with the imaginary numbers now we get the singularity numbers!

I understand that the concepts of Virtual Numbers and Singularity Numbers might seem unconventional at first, especially when compared to more familiar ideas like imaginary numbers. The goal here is to explore new ways of thinking about numbers and challenge traditional frameworks.

@log_menus_1 Traditional framework is just fine and it's the base for everything. Imaginary, singularity, etc numbers belong to marvel universe and serve other purposes.

I couldn't write these equations in PowerPoint, so I used ChatGPT. However, ChatGPT doesn't work on its own-it follows the instructions I provide on what to do and how to do it.

It seems you haven’t watched my video on Virtual Numbers. To clarify, j is the unit of Virtual Numbers, and ζ(1) refers to the sum of reciprocals of natural numbers: ζ(1) = 1 + 1/2 + 1/3 + 1/4 + ... This series is typically labeled as divergent, but I call it an irrational series, as its sum is unknown and not finite. Another example is the Gregory-Leibniz series for π: π = 3.14159265359... π= 3 + .14159265359... π - 3 = 1/10..0 + 4/10..0 + 1/10..0 + ... This series diverges but still converges to the value of π, showing how infinite processes can yield meaningful results. The k in the video is the Singularity Unit, and terms like divergent, infinity, and undefined simply reflect our current lack of understanding or methods to solve such concepts.

@log_menus_1 from what I understand you coined the term virtual number yourself (I couldn't find it anywhere on the internet). Could you rigourously define it? The harmonic sum IS divergent to infinity no matter how you look at it. And it IS NOT zeta of 1. The sum IS known. From what i searched the Gregory Leibniz series is just arctangent Taylor expansion evaluated at 1. What you wrote is just a sequence of decimal expansions of subsequent approximations of pi. It converges to pi. You can define every real number as limit of sequence of rationals.

The term 'sum/sequence divergent to infinity' has a strict definition, so don't yap about 'current lack of understanding'

The concepts of Virtual Numbers and Singularity Numbers are not just imaginative ideas-they’re part of an effort to push the boundaries of traditional mathematical thinking. While these concepts may seem unconventional, they are based on logical frameworks that challenge existing number systems. As for the channel, we focus on providing explanations and theoretical foundations. If you're open to it, I'd be happy to explain the reasoning behind these ideas in more detail!

AI is simply a tool, much like a calculator, used to assist in tasks like script writing and text-to-speech. As English is not my native language, I use AI to help ensure my message is clear and understandable to a wider audience. The purpose is to make sure everyone can follow along with the ideas I’m presenting. If my concepts are difficult to grasp, it might be due to a lack of familiarity with advanced mathematical ideas, but I encourage you not to misguide others. Ultimately, viewers will decide whether they find the videos valuable-if they do, they will watch; if not, they are free to move on. My goal is to share these concepts with those who are interested.

j = iπ π = j/i

@log_menus_1 exactly what i've said, thank you for your videos, could you make more about this new number's family ? i really love your channel

Wait for another video on (-1)! 😊

In my future videos

No... 0k = 0/0 which is not 1 .. K = 1/0 Multiply by 0 on both sides 0k = (1/0)*0/1 0k = 0/0 So singularity numbers={...,-2k,-k, 0k,1k,2k,...}

Is this is true 0=1/k

??

If I had asked you to solve the equation x² = -1 before the discovery of complex numbers, you would have said the question is invalid because no number satisfies that equation. Similarly, my question is: find x if x/x = -1. This is analogous to asking for x when x² = -1 before complex numbers were introduced. The point is, my question is valid-you just don’t have an answer for it in the complex numbers framework you're using.

@@log_menus_1 My man, complex numbers are a model that works mathematically and is consistent with basic arithmetic and geometric rules. You just made up a random ass number, said it works, and then broke math. You can always introduce new definitions in mathematics, but you MUST make sure that they are consistent or at least put some sort of boundaries. When using definitions that have some errors, results, at best, need some huge asterisks. However this is complete nonsense. Also are you an AI? Is this a social experiment?

I didn’t break math at all; virtual numbers are consistent and have their own framework where all operations are done, similar to how complex numbers and real numbers each have their own mathematical system. In my introduction to virtual numbers, I showed how they work when we extend the domain of functions from real to virtual. For example, 0! = 1 makes sense if we consider n! = n * (n-1) * (n-2) * ... * 1. When we put n = 0, we get 0! = 0 * (-1) * (-2) * ... * (-n). This results in 0! = 0. Would you say I’m wrong here as well? Also, negative numbers are undefined in the for factorial, but when we extend the domain of integers to complex numbers, we get different results. For instance, (-1/2)! = √π. The results with virtual numbers may seem counterintuitive, but they are correct due to the extension of the real domain to the virtual numbers system. I hope this makes sense.

Nothing is undefined

Undefined means it can have more than one definition. Depends on how you want to define it.

1/0 is also undefined but we use it often as infinity instead of writing 1/0

AI is simply a tool, much like a calculator, used to assist in tasks like script writing and text-to-speech. As English is not my native language, I use AI to help ensure my message is clear and understandable to a wider audience. The purpose is to make sure everyone can follow along with the ideas I’m presenting. If my concepts are difficult to grasp, it might be due to a lack of familiarity with advanced mathematical ideas, but I encourage you not to misguide others. Ultimately, viewers will decide whether they find the videos valuable-if they do, they will watch; if not, they are free to move on. My goal is to share these concepts with those who are interested.

j is actually i*pi, if you do j+j, or 2*j, you'll get 2*i*pi

Actually it follows the basic arithmetic as well j + j = 2j = 2ln(-1) = ln(-1)²= ln(1) = 0 Here 2j = 0 not 2 or j individually 0

Also j+j = 0 is true j = -j Ln(-1) = - ln(-1) Ln(-1) = ln(-1) j= j Actually -j = j So it follows basic arithmetic

2j = 0 so it means 2i*pi = 0?

@@log_menus_1 Ok, that's interesting. So you propose 2j=0 doesn't mean you can divide both sides by 2? Then it wouldn't follow that 2ipi=0, because in this case, you CAN divide by 2, giving ipi=0, which isn't true. This means that you might need to try redefining what it means to "multiply", so that the definition allows you to deal with both normal numbers like pi or 2 or i, and your new number j, and even k. For example, if 2j=0 but j isn't 0, what does the 2 mean? Without the 2, j=ln(-1), but with it, 2j=0. Can you find some rule to figure out what 3j is, then what 4j is, etc. without any contradictions arising? Are you able to do the same with K? How about numbers like j^k? Right now, your theory is interesting, but it isn't standing on many foundations. People are pointing out "there's contradictions" or "there's inconsistencies", because the rules for what you can and can't do with j and k are unclear. If you can figure out what these rules are, and prove they're perfectly consistent, then you'll have a theory that some experts might start considering, and something maybe no one has done before. If you fail, maybe it wasn't possible, but it would be one of the greatest learning experiences, and you'd have a much deeper understanding of why you can't divide by 0 than many other people who just read textbooks. Learning from actually researching and proving things yourself is incredibly valuable and something not many people can do. And if you're curious about how you're even supposed to build these rules for j and k, you might want to look into areas of math called "foundations of mathematics", "mathematical logic" or "proof theory". These areas deal with axiomatic systems, or those rules you place, and how to prove they're reliable rules.

Watch my other video title e^j = 1+j

No, all numbers. Every number is complex. Complex numbers can be expressed in polar or standard form, so we’ll go with standard: a+bi 6=6+0i has a real and imaginary part and is therefore complex. You must think that a and/or b can’t be zero, but even in that case, x/x is always 1 Plus, that wasn’t even the question

But question was x/x = -1

I know always x/x = 1 , My question was find x if x/x = -1

@@log_menus_1 you’re actually stupid

thanks

It's correct that Ln(-1) = πi Multiply both sides by 2 .. 2Ln(-1) = 2πi 0 = 2πi Also j² = 0.if you watch my video on e^j = 1+j u will understand