is π rational? Understanding π = j / i

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  • เผยแพร่เมื่อ 26 ธ.ค. 2024

ความคิดเห็น • 34

  • @dominiqueubersfeld2282
    @dominiqueubersfeld2282 วันที่ผ่านมา +4

    This video deserves a nomination to the Terrence Howard Prize

  • @massimoesposito7014
    @massimoesposito7014 วันที่ผ่านมา +4

    if j=i*Pi but in this same algebra j^2=0 if you mix i and j you obtain a contraddiction in which(i*Pi)^2= -Pi^2=0, the Virtual numbers are a different algebra with different axioms but you fell i a confusion if you mix the two in a same reasonment.

    • @log_menus_1
      @log_menus_1  22 ชั่วโมงที่ผ่านมา +1

      I don't want to use both , I want to use logarithmic functions using j instead of i, i= e^j/2 in virtual system

    • @massimoesposito7014
      @massimoesposito7014 22 ชั่วโมงที่ผ่านมา +2

      The concept that ln(-1)=i*Pi prevails on Virtual numbers because all our universe works on Complex numbers, starting from Quantum Mechanic, but exploring new axioms and classify new algebras is an interesting and fascinating intellectual activivty, that tomorrow could reveal working in some real phenomena.😀

  • @DeltaInternetFaaaan
    @DeltaInternetFaaaan วันที่ผ่านมา +3

    j = ln(-1)
    -j = -ln(-1)
    -j = ln(-1)
    -j = j divide both sides by j
    -1 = 1
    But give this man a chance, maybe he can fix or mathematicaly explain all this mistakes🙏🙏🙏🙏
    People also used to hate complex numbers

  • @frog-d9w
    @frog-d9w 15 วันที่ผ่านมา +13

    Duh, j=3 and i=1. Didn't you know that pi is just equal to 3? (this is a joke)

    • @log_menus_1
      @log_menus_1  15 วันที่ผ่านมา +2

      3.14.../1

    • @aipers1846
      @aipers1846 15 วันที่ผ่านมา +9

      pi=e=3 ez

  • @AriosJentu
    @AriosJentu 3 วันที่ผ่านมา +3

    e^j + 1 = 0
    e^j = 1 + j
    Subtract both equations
    j = -2
    Which is impossible
    You should create also new axioms for consistency. I think there’s no working axioms like “(a + bj) + (c + dj) = (a+c) + (b+d)j” and “(a + bj) * (c + dj) = ac + (bc + ad)j”.
    So even the “e^j + 1 = 0” is wrong, or just j^2 is not 0.

    • @log_menus_1
      @log_menus_1  3 วันที่ผ่านมา +1

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

  • @luisalfredovidaldecarvalho876
    @luisalfredovidaldecarvalho876 16 ชั่วโมงที่ผ่านมา +1

    This guy is crazy!!!

  • @Daniel-x6b6w
    @Daniel-x6b6w 16 วันที่ผ่านมา +6

    I think there might be an issue with this proof. if "j = ln(-1)", then we can perform a few action to disprove that "pi = j/i". I may be wrong, however.
    If you assume that "ln(-1) = i•pi", then "2ln(-1) = 2i•pi". This can be written as "ln((-1)²) = 2i•pi". We know that "(-1)² = 1", so "ln(1) = 2i•pi". This then leads to "0 = 2i•pi", which is false. It would be fair to assume that "ln(-1) is not equal to i•pi" and therefore "pi is not equal to j/i".
    I hope this helps you understand the possible mistake in assuming that "pi = j/i" and that "ln(-1) = i•pi".

    • @log_menus_1
      @log_menus_1  16 วันที่ผ่านมา +3

      You know that I derived this equation from Euler's formula, which makes us question whether Euler's formula itself might be flawed. All I did was replace ln(-1) with j. Here’s the process:
      From Euler's formula:
      e^(pi*i) = -1
      Taking the natural logarithm:
      pi*i = ln(-1)
      Instead of squaring, I multiplied both sides by 2:
      2pi*i = 2ln(-1)
      According to this interpretation:
      2pi*i = 0
      This result seems contradictory, but it directly follows from the steps.

    • @log_menus_1
      @log_menus_1  16 วันที่ผ่านมา +1

      I believe there might be a flaw in the way math handles certain concepts. In the future, I plan to explore and address this. Personally, I don't think complex numbers are valid; to me, they seem like part of a mathematical scam.

    • @awesome7006
      @awesome7006 15 วันที่ผ่านมา +1

      Ln(-1) does not equal 0; Ln(-1)=ln|1|+ ipi= 0+ ipi= ipi.
      Thus 2Ln(-1)= 2(ipi)= 2ipi. However, ln(-1) is a multi values function and has an infinite number of solutions. In order to evaluate the logarithmic function, one must perform a branch cut to evade ambiguity and differing interpretations. Thus ln(-1)=ipi, 3ipi, 5ipi, -ipi, -3ipi, -5ipi. Often times we take the principle branch cut of the ln function which is why Ln(-1)= ipi.
      The same concept arrives with the square root function as (1)^0.5= +/-1. Because there are two possible values for (1)^0.5, we refer to the principle branch the +1 value.
      Logarithmic functions are the inverse exponential functions and since the complex exponential function repeats the same values an infinite amount of cycles, the argument of e grows to infinity and negative infinity referred to as complex infinity around a singularity. Defining a branch cut is critical for proper evaluation.

    • @minerscale
      @minerscale 15 วันที่ผ่านมา +4

      ​@@log_menus_1 I think you're just missing the concept of a multi-valued function.
      There are infinitely many solutions to the equation e^z = -1.
      Namely t = 2i*pi*n where n is an integer.
      This implies that ln(-1) is infinitely valued, that is, ln(-1) represents all the solutions to the equation e^z = -1 at once.
      Some mathematicians get around the fact that ln(-1) is multi-valued by introducing the concept of a "principal branch" where you choose a single value by some convention. In this case the "principal value" of ln(-1) is i*pi.
      But if we only consider the principal branch when performing certain operations we can end up with various paradoxes. So we must be mindful of what our equations actually represent.
      As for the claim that pi = j/i implies pi is rational, it is worth noting that a rational number is defined by p/q where p and q are integers and q is not zero.
      Since neither i or j is rational, this statement says nothing about the rationality or irrationality of pi.

    • @log_menus_1
      @log_menus_1  10 วันที่ผ่านมา +1

      2ln(-1) = 0 which means pi=0

  • @applimu7992
    @applimu7992 15 วันที่ผ่านมา +5

    How does e^j = -1 and e^j = 1 + j both work?

    • @log_menus_1
      @log_menus_1  14 วันที่ผ่านมา

      In the video, I replaced ln(-1) with j, which makes the equation mathematically consistent. I also believe that j² and higher powers of j are zero. Why? Because 1 × 1 = -1 / -1. That’s why e^j = 1 + j seems valid.
      If we avoid using j and instead write the equation as ln(-1) = πi, we encounter the same paradox you mentioned. For example, multiplying both sides by 2 gives 2ln(-1) = 0, which leads to 2πi = 0 on the other side. This shows the inconsistency in traditional interpretations.

    • @ElectroBanaana
      @ElectroBanaana 5 วันที่ผ่านมา +1

      @@log_menus_1 this shows the inconsistency in YOUR interpretations, you assume that with virtual numbers one can still use the property a*ln(b) = ln(b^a). WHICH also explains why the result of your previous video is mistaken, because you have again used this property there : e^j = 1 + j you said in the previous video, and now we have e^j + 1 = 0 => e^j = -1, replacing it in the first equation we get : -1 = 1+ j => j = -2 ... you have to chose between one of these 😂And your "Why? Because 1 × 1 = -1 / -1. That’s why e^j = 1 + j seems valid." doesn't mean Anything 🤣

  • @gamespotlive3673
    @gamespotlive3673 12 ชั่วโมงที่ผ่านมา

    interesting thought exercise but needs some clearing.

  • @gamespotlive3673
    @gamespotlive3673 12 ชั่วโมงที่ผ่านมา

    this is huge

  • @malakmeleek3847
    @malakmeleek3847 วันที่ผ่านมา

    But pi in the euler's identity is in radians unit, an other way to say 180 degrees, not the number pi itself (3.14159265358979...). We're talking about angles not constants, correct me if i'm wrong

    • @log_menus_1
      @log_menus_1  วันที่ผ่านมา

      You're right that in Euler's identity, π represents 180 degrees. However, in the virtual number system, the rotation is 180 degrees from the foundation of virtual numbers, while in the complex number system, the rotation is 90 degrees. Also, in the virtual number system, j is the logarithmic unit, and e^(j/2) equals i, the imaginary unit in the complex number system.

  • @NoName-vz4wu
    @NoName-vz4wu วันที่ผ่านมา

    Isn't 2j equal to 0 as well?
    This would not make a number system unless in base 2

  • @ferlywahyu342
    @ferlywahyu342 5 วันที่ผ่านมา +1

    Irational a/b only if a and b is integer lol 😂

    • @log_menus_1
      @log_menus_1  4 วันที่ผ่านมา +1

      It means 1/0 is rational?

    • @sudhamishra6596
      @sudhamishra6596 4 วันที่ผ่านมา

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

  • @hwaansswaanh3511
    @hwaansswaanh3511 6 วันที่ผ่านมา

    j=ln(-1)=iπ
    iπ/i=π

    • @log_menus_1
      @log_menus_1  5 วันที่ผ่านมา +1

      j = iπ
      π = j/i

    • @hwaansswaanh3511
      @hwaansswaanh3511 5 วันที่ผ่านมา +1

      @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

  • @chrisking14469
    @chrisking14469 2 วันที่ผ่านมา +1

    given that j is just the imaginary unit with extra steps:
    j = ln(-1) = ln(i^2) = 2ln(i)
    all the properties of imaginary numbers should correspond to properties of j. if you're serious about finding applications of j, you should stick to writing j in its complex form to avoid you and the viewers' confusion.

    • @log_menus_1
      @log_menus_1  วันที่ผ่านมา +1

      I propose using virtual numbers as a replacement for complex numbers.