Printable View
Well,.. if a irrational number like Sqroot(8) can have a rationalising
factor like Sqroot(2) what is the rationalising factor for pi ??
sqr(8) = 2*Sqr(2)
Sqr(8)*sqr(2) = 2*Sqr(2)*Sqr(2) = 4{rational}
---
My question.
pi * x = rational number
what is x ? is it possible ?
2*Sqr(2)*Sqr(2)=4
Haven't a clue about your question though.
Sorry, that was a genuine mistake !
I was thinking of putting 2 * 2 but
ate the half of it :( .
---
I actually want to know if there is a posibility that
x exists.
Well the only solution to this should be:
x=1/Pi
or a lot other expressions that contain Pi or somehow the definition of Pi (eg. 2/Pi,1/(2Pi) ...)
I guess, you can also proof that the rationalising factor has to contain Pi in some way. But my theoretical math lessons are far gone...
X=A/(atn(1)*B)
:D
Actually I did not want such solutions...
For the same reason I am not taking
x = 1/Sqr(8) or x = 1/Sqr(2) as a Required solution for
Sqr(8) * x = rational
...
Arctan(1)=45 degrees = Pi/4, so we're back at Pi
what are you intending to do then? pi is unique not like squareroots
are you saying x>1 v (x<0 ^ x>-1)
Pi, e, and certain other real numbers are called transcendental numbers. In some branch of mathematics, there is a theorem which proves that such numbers belong in a class by themselves.
They cannot be expressed using a finite number of simple arithmetic operations without using transcendental numbers (roots and powers allowed). They cannot be the roots of a polynomial with rational coefficients.
On the basis of the above, I believe that your search for a so called rationalizing factor is doomed.
Oddly bit of related information.
e^i*Pi = -1
Where i is SquareRoot(-1), and e is 2.71828...
BTW: I am pretty sure that numbers which can be roots of a polynomial are called algebraic numbers. The algebraic numbers are a bit stranger than you might expect. For 2nd, 3rd, & 4th order polynomials, there are direct formulae or algoithms which result in determing the roots. Except for special cases, the roots of higher order polynomials cannot be determined by any direct formulae or algorithms. You can use numerical methods to get accurate approximations, but there is no formula or expression which precisely describes the roots using a finite number of arithmetic operations. There is nothing remotely similar to the formula for the roots of a quadratic.
IMHO, that's not at all odd.. ;)Quote:
Originally posted by Guv
Oddly bit of related information.
e^i*Pi = -1
Where i is SquareRoot(-1), and e is 2.71828...
It's just the exponential form of the complex number -1+i*0 (or)
1*[Cos(pi)+iSin(pi)] (or) 1*[Cos(180o)+i Sin(180o)]
...
r*[Cos(theta)+iSin(theta)] can be expressed as r*e(i*theta)
Where r =Sqr(a^2+b^2) is the modulus of complex number a+ib
and theta is the principal argument of the complex number such that
r*cos(theta) = a and r*sin(theta)=b
....
Yes.. I have to agree with this. :pQuote:
On the basis of the above, I believe that your search for a so called rationalizing factor is doomed.
What that means is..
I can similarly show that... e(2*pi*i)= 1
Here is another
e(pi/2)*i = i