Little Help? (1 Viewer)

[Giant Lobster]

For a polynomial P, its true that if P is over rational numbers then any irrational roots occur in conjugate pairs

How do u prove this?

 

[abdooooo!!!]

LOL. this one... i tried to prove it in my polynomial assessment then i realised the question just says state why... ahahaha.

what are you man... its not in the course the proving of it.

just use google man... because its a well known theorem.

 

[Giant Lobster]

man! stop replying to my threads without the proof!

 

[abdooooo!!!]

Originally posted by Giant Lobster
man! stop replying to my threads without the proof!

ahahaha... nah man... i just felt sorry for you since no one else is gonna respond to your lonely thread.

 

[J0n]

Let P(x) be any polynomial with rational coefficients.
If a + b*sqrt(c) is a root of the equation then another root is a - b*sqrt(c)

Proof
Assume a + b*sqrt(c) is one root. Must show a - b*sqrt(c) is also a root.
If b = 0 then of course the other root is the conjugate (same thing), so b<>0
Let T(x) = (x - (a + b*sqrt(c)))*(x - (a + b*sqrt(c))) = (x-a)² - b²
Then T(x) is a quadratic polynomial with rational coefficients.
Now there is a polynomial Q(x) and R(x) such that:
P(x) = T(x)*Q(x) + R(x)
Where R(x) will be either a degree 1 or 0 .'. We can represent R(x) as Cx + D where D and E are rational
Since P(x) and T(x) have rational coefficients, so does Q(x) and R(x)
So:
P(x) = T(x)*Q(x) + Dx + E where D and E are rational
Now let x = a + b*sqrt(c)
D(a + b*sqrt(c)) + E = 0 .'. D = E = 0
Now we have:
P(x) = T(x)*Q(x)
Now let x = a - b*sqrt(c)
P(x) = 0*Q(x) = 0
.'. a - b*sqrt(c) is also a root

 

[Grey Council]

bullshit your in class of '04.

 

[Giant Lobster]

D(a + b*sqrt(c)) + E = 0 .'. D = E = 0

edit:: sorry i understand now

wow! what an elegant proof, amazing im up against u this year
what school do u terrorise with your uber math?