My Math Forum 4 times pi squared

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 June 2nd, 2016, 08:40 AM #1 Banned Camp   Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 Math Focus: horses,cash me outside how bow dah, trash doves 4 times pi squared is this number transcendental? Thanks from manus
 June 2nd, 2016, 08:51 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Yes, of course. If it were not it would satisfy a polynomial equation, say nth degree, with integer coefficients. But since any integer power of 4 is an integer that would then give a polynomial equation, of degree 2n, with integer coefficients, satisfied by $\displaystyle \pi$,which is impossible because $\displaystyle \pi$ is transcendental. Thanks from 123qwerty and manus
June 2nd, 2016, 09:00 AM   #3
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Math Focus: horses,cash me outside how bow dah, trash doves
ok sweet so if I then divide it by three, still transcendental?
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 June 2nd, 2016, 04:12 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Seriously? Since you simply asked whether a specific number was "transcendental", I assumed you knew what "transcendental number" meant! Are you now saying that you do not? A number, a, is "transcendental" if and only there exist a polynomial equation with integer coefficients (equivalently "rational coefficients") having a as a root. If a3 were NOT transcendental, then there would exist a Polynomial, of degree n, having a/3 as a root. Then that same polynomial, multiplied by 3 to the nth power, would have x as a root showing that a is not transcendental. Thanks from manus
June 4th, 2016, 10:56 AM   #5
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Quote:
 Originally Posted by Country Boy Seriously? Since you simply asked whether a specific number was "transcendental", I assumed you knew what "transcendental number" meant! Are you now saying that you do not? A number, a, is "transcendental" if and only there exist a polynomial equation with integer coefficients (equivalently "rational coefficients") having a as a root.
This should have been "if and only if there does not exist...", of course.

Quote:
 If /a3 were NOT transcendental, then there would exist a Polynomial, of degree n, having a/3 as a root. Then that same polynomial, multiplied by 3 to the nth power, would have x as a root showing that a is not transcendental.

 July 31st, 2016, 04:13 AM #6 Banned Camp   Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 Math Focus: horses,cash me outside how bow dah, trash doves yea its ok don't stress I was taking the piss over the gamma(2/3) thing.

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