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.
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".
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.
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.
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.
@@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.
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.
@@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 🤣
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.
You are write, problem with j² and other higher powers, I will fix it in my next video
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".
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.
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.
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.
@@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.
2ln(-1) = 0 which means pi=0
Duh, j=3 and i=1. Didn't you know that pi is just equal to 3? (this is a joke)
3.14.../1
pi=e=3 ez
How does e^j = -1 and e^j = 1 + j both work?
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.
@@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 🤣
Irational a/b only if a and b is integer lol 😂
It means 1/0 is rational?
@@log_menus_1 you demolished & annihilated that oversmart coward 😂
j=ln(-1)=iπ
iπ/i=π
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