This is simply incorrect. Natural log is not an inverse function when it comes to complex numbers. Furthermore, a^(xy) = (a^x)^y does not always hold in complex numbers (and it doesn't hold here) These are the most obvious mistakes, with a couple more subtle ones (like canceling out 2nπ which is a mistake that involves switching the order of quantifiers) 1^z = 2 does not have solutions. It doesn't have real, it doesn't have complex solutions. 1^z=1 for all complex z
I don't know who's right but at least one of you is wrong (this is because you both have "opinions" that are contradictory to each other (I used speech marks here to highlight the fact that these are not really opinions, as that would imply some sort of subjectivity to the discussion, which is obviously not the case)).
@mffbataineh ah, this is an interesting point, but it still falls to a mistake. Here you suggest that if x satisfies 2^(1/x) = 1 that then it must satisfy 2=1^x However, here you are assuming that [2^(1/x)]^x = 2^(x/x) = 2 However, this is still the same mistake you made in the video, where you assume that b^(xy) = (b^x)^y It is true that there exists complex z different than 0 such that 2^z=1 However, this does not imply 1^(1/z) = 2
@@mffbataineh I will also demonstrate how you switched the order of quantifiers in your solution. Your solution is x=ln2/(2niπ) for all integers n≠0. This would mean that x=ln2/(2iπ) is one of the solutions. Then, assuming b^xy=(b^x)^y, we have 2=1^x=[e^(4iπ)]^(ln2/2iπ) = = e^(ln2*4iπ/2iπ) = e^2ln2 = 4 So one of your solutions for x can result in 1^x=4=2 In fact, any of your solutions for x can result in any power of 2 if we assume (b^x)^y=b^xy So, according to your own logic, 1=2=4=8=16=... Most honest mathematicians would conclude that now we have arrived at a contradiction and the "solution" must be false. You have made (several) mistakes in your proof
@@methatis3013 This is a very intricate and well thought out argument. I am looking forward to seeing the opponent's response. Edit: I have myself concluded that your argument is analytically robust and it follows that Dr. Bataineh must either admit his error and resign or simply ignore the undeniable logic presented to him. Either way I would say you are the victor.
This is simply incorrect. Natural log is not an inverse function when it comes to complex numbers. Furthermore,
a^(xy) = (a^x)^y does not always hold in complex numbers (and it doesn't hold here)
These are the most obvious mistakes, with a couple more subtle ones (like canceling out 2nπ which is a mistake that involves switching the order of quantifiers)
1^z = 2 does not have solutions. It doesn't have real, it doesn't have complex solutions. 1^z=1 for all complex z
Try 1 = 2^(1/x) and see what you get.
I don't know who's right but at least one of you is wrong (this is because you both have "opinions" that are contradictory to each other (I used speech marks here to highlight the fact that these are not really opinions, as that would imply some sort of subjectivity to the discussion, which is obviously not the case)).
@mffbataineh ah, this is an interesting point, but it still falls to a mistake.
Here you suggest that if x satisfies
2^(1/x) = 1
that then it must satisfy
2=1^x
However, here you are assuming that
[2^(1/x)]^x = 2^(x/x) = 2
However, this is still the same mistake you made in the video, where you assume that
b^(xy) = (b^x)^y
It is true that there exists complex z different than 0 such that 2^z=1
However, this does not imply
1^(1/z) = 2
@@mffbataineh I will also demonstrate how you switched the order of quantifiers in your solution.
Your solution is
x=ln2/(2niπ) for all integers n≠0. This would mean that
x=ln2/(2iπ) is one of the solutions.
Then, assuming b^xy=(b^x)^y, we have
2=1^x=[e^(4iπ)]^(ln2/2iπ) =
= e^(ln2*4iπ/2iπ) = e^2ln2 = 4
So one of your solutions for x can result in 1^x=4=2
In fact, any of your solutions for x can result in any power of 2 if we assume (b^x)^y=b^xy
So, according to your own logic, 1=2=4=8=16=...
Most honest mathematicians would conclude that now we have arrived at a contradiction and the "solution" must be false. You have made (several) mistakes in your proof
@@methatis3013 This is a very intricate and well thought out argument. I am looking forward to seeing the opponent's response.
Edit: I have myself concluded that your argument is analytically robust and it follows that Dr. Bataineh must either admit his error and resign or simply ignore the undeniable logic presented to him. Either way I would say you are the victor.
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No solutions.
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Wow
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