OpenStudy (darkprince14):

Abstract algebra.

2 years ago
OpenStudy (darkprince14):

2 years ago
OpenStudy (amistre64):

can you define a field?

2 years ago
OpenStudy (darkprince14):

a field is commutative division ring

2 years ago
OpenStudy (amistre64):

ive got no clear thought on this. my class work never got past groups. so i lack any practical experience in the matter.

2 years ago
OpenStudy (misty1212):

HI!!

2 years ago
OpenStudy (misty1212):

i have a suggestion try it with a simple polynomial of degree say 2 and see why it works

2 years ago
OpenStudy (misty1212):

if you want a bigger hint i can provide one, but i think it will pretty much give it away

2 years ago
OpenStudy (amistre64):

already considered that, too much trouble :)

2 years ago
OpenStudy (amistre64):

half a degree is all im willing to process at the moment

2 years ago
OpenStudy (darkprince14):

Is it okay if i divide f(a)by a^n?

2 years ago
OpenStudy (misty1212):

first off the notation is a bit odd, because they use \(a_i\) for the coefficients, and \(a\) for the zero

2 years ago
OpenStudy (misty1212):

so lets use \(b\) and you see that \[a_0+a_1b+a_2b^2+...+a_nb^n=0\] the see what you get when you multiply this by \(b^{-n}\) which i guess is another way of saying "divide by \(b^n\)"

2 years ago
OpenStudy (misty1212):

half a degree is better than none! it is what i have after all...

2 years ago
OpenStudy (darkprince14):

thanks:)

2 years ago
OpenStudy (misty1212):

\[\color\magenta\heartsuit\]

2 years ago
OpenStudy (anonymous):

consider \(g(x)=x^{-n}f(x^{-1})\) hence \(g(a)=0\implies a^{-n}f(a^{-1})=0\) and using the fact \(a\ne0\) we conclude \(f(a^{-1})=0\). so now just work backwards

2 years ago
OpenStudy (anonymous):

err \(g(x)=x^n f(x^{-1})\) and so \(a^n f(a^{-1})=0\)

2 years ago
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