39i hope David McNaughton is well known.
http://dlmcn.com/sin10-irr.html
see dat.
its not that i cant give the proof , but th e article he has given is really good to look at
14 Answers
39
62awesome work Celestine...
That made it a piece of cake :)
I was thinking of 100's of different combinations ;)
but i guess we still need to prove that the roots of the cubic equation formed here will be irrational..
341that's where the rational roots theorem comes in handy.
The equation is 8t^3 -6t^2+1 = 0 as in the article linked by b555.
To make it easier consider the reciprocal equation
t^3 -6t^2+8 = 0
By RRT, the only rational roots possible are \pm1, \pm 2, \pm 4, \pm 8 all of which are easy to verify as not being solutions
62prophet.. you also missed out celestine's post
expansion of sin 3x gives that equation
there is no need to refer to the post by bhargav :)
341i did see celestine's post. but the equation is already there, so i thot i would take it from there
9yes sir i left it as wrk for soumik
i was sure he wud learn more by doing that himself ( cos ive seen some of his gud posts )
we have
8x3 - 6x + 1 = 0
take for contradiction x=p/q
with p , q comprime
we find that cases
p even ,q odd p,q both odd dont satisfy
then in the case p odd , q even we can find that q =2m where m is necessarily odd
again we'll see its odd - odd+odd = 0 which is impossible
9wow the rational root theorem is a surprisingly beautiful result
anyone trying to prove it ?
In algebra, the rational root theorem (or 'rational root test') states a constraint on rational solutions (or roots) of the polynomial equation
P(x) = anXn + ..... + a0
with integer coefficients.
If a0 and an are nonzero, then each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., gcd(p,q) = 1) satisfies
p is an integer factor of the constant term a0, and
q is an integer factor of the leading coefficient an.
Thus, a list of possible rational roots of the equation can be derived using the formula .
62Celestine why dont you start a new thread on it?!
I guess that could be a bit useful :)
btw I think we had discussed this one earlier
