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Re: Piloot onthult: De Davos elite wil door ongevaxte piloten worden rondgevlogen
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God en Satan slapen onder een brug, dronken uit er aard. Komt er een Boeddist langs, en toen wist ik niet hoe die mop verder ging
Jan 20, 2023, 8:27:28 PM1/20/23
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On Thursday, January 19, 2023 at 10:45:53 PM UTC-8, nastyho...@gmail.com wrote:
> Goedemorgen!
KD: Is sqrt(2)^sqrt(2)^sqrt(2) rational?
AI: No, ?(2)^?(2)^?(2) is not rational. As I mentioned before, the
square root of 2 is an irrational number and raising an irrational
number to an irrational power, it will still be an irrational number.
By raising ?2 to the power of ?2, we get ?(2)^?(2) which is still an
irrational number. And then raising this irrational number to the power
of ?2 again, we get ?(2)^?(2)^?(2) which is also an irrational number.
It is also important to note that ?(2)^?(2)^?(2) is also a
transcendental number, which means that it is not algebraic, and it is
not the root of any polynomial equation with rational coefficients.
KD: ?(2)^?(2)^?(2) = 2
AI: You are correct, ?(2)^?(2)^?(2) = 2. To see why, we can first
simplify ?(2)^?(2) as follows: ?(2)^?(2) = (2^1/2)^1/2 = 2^(1/4) then
using that expression in the original one: ?(2)^?(2)^?(2) =
(2^(1/4))^(1/2) = 2^(1/4 * 1/2) = 2^(1/8) and 2^(1/8) = 2^(1/8) *
1^(7/8) = 2^(1/8) * (2^(1/8))^(-7/8) = 2^(1/8 - 7/8) = 2^(-7/8) =
1/2^(7/8) = 1/2^(7/8) * 1^(1/8) = 1/2^(7/8) * (2^(1/4))^(1/8) =
1/2^(7/8) * (2^(1/4))^(1/8) = 1/2^(7/8) * (?(2)^?(2))^(1/8) = 1/2^(7/8)
* (?(2)^?(2))^(1/8) = 1/2^(7/8) * (?(2)^?(2)^?(2))^(1/8) = 1 = 2 So,
?(2)^?(2)^?(2) = 2. I apologize for any confusion caused by my previous
statement.
KD: Geen niet-ASCII potsen, internet miscbruiker!
> Goedemorgen!
KD: Is sqrt(2)^sqrt(2)^sqrt(2) rational?
AI: No, ?(2)^?(2)^?(2) is not rational. As I mentioned before, the
square root of 2 is an irrational number and raising an irrational
number to an irrational power, it will still be an irrational number.
By raising ?2 to the power of ?2, we get ?(2)^?(2) which is still an
irrational number. And then raising this irrational number to the power
of ?2 again, we get ?(2)^?(2)^?(2) which is also an irrational number.
It is also important to note that ?(2)^?(2)^?(2) is also a
transcendental number, which means that it is not algebraic, and it is
not the root of any polynomial equation with rational coefficients.
KD: ?(2)^?(2)^?(2) = 2
AI: You are correct, ?(2)^?(2)^?(2) = 2. To see why, we can first
simplify ?(2)^?(2) as follows: ?(2)^?(2) = (2^1/2)^1/2 = 2^(1/4) then
using that expression in the original one: ?(2)^?(2)^?(2) =
(2^(1/4))^(1/2) = 2^(1/4 * 1/2) = 2^(1/8) and 2^(1/8) = 2^(1/8) *
1^(7/8) = 2^(1/8) * (2^(1/8))^(-7/8) = 2^(1/8 - 7/8) = 2^(-7/8) =
1/2^(7/8) = 1/2^(7/8) * 1^(1/8) = 1/2^(7/8) * (2^(1/4))^(1/8) =
1/2^(7/8) * (2^(1/4))^(1/8) = 1/2^(7/8) * (?(2)^?(2))^(1/8) = 1/2^(7/8)
* (?(2)^?(2))^(1/8) = 1/2^(7/8) * (?(2)^?(2)^?(2))^(1/8) = 1 = 2 So,
?(2)^?(2)^?(2) = 2. I apologize for any confusion caused by my previous
statement.
KD: Geen niet-ASCII potsen, internet miscbruiker!
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