Hey. How do I prove that each subfield of the field of complex numbers contains every rational number? Thanks. Please Register or Log in to view the hidden image!
Ummm,... Don't ask me,... heh! I do love Algebra, though... My GOD! * looks around Please Register or Log in to view the hidden image!* This website thingy is like a qweerer, spacier (as in outer, although it is roomy) version of GeekCulture! *GASP* Oh my,...it's all so shiny and new to me...What if I can't handle it Please Register or Log in to view the hidden image! !? Have you all gone MAD? Well!...I sure have,...and a long, long time age...teeh! I'm so glad I get to spread my good little escence here...OOP! Didn't mean that in a naughty way, no no.
I wish I can do that homework problem for Mallory but I don't remember squat about linear algebra anymore. That class is just proof after proof, you just have to struggle through it like I did. Please Register or Log in to view the hidden image!
My math teacher My teacher of Math, has been a teacher from 7th grade to college she almost had her Doctorit in Biomedical Engineering and has 2 Masters in Endomorphic Biology and General Science her email is <href a=evansjo@mukwonago.k12.wi.us>Jolene</a> just tell her you know Ben Tomasik
Maybe its my lack of english maths by i really dont understand your question How do you define a subfield in C ? If you consider it as a 2-d vector space, your propertie is wrong// plz, would be glad to have some light upon thatPlease Register or Log in to view the hidden image!
This is late, i know, but... OK. well your subfield must contain 1, and 0, otherwise it is not a subfield. also, it must be closed under addition, so it contains 1+1=2. by induction it contains every positive integer. the subfield must be closed under additive inverse, so every negative integer is also is in the subfield. the subfield must also be closed under multiplicative inversion, so it contains every number of the form 1/n, where n is an integer which is not the additive identity (0). finally, the subfield is closed under field multiplication, so it contains all numbers of the form m/n, where m is any integer, and n is any nonzero integer. thus any subfield of the complexes contains the rationals. note that i didn t use any properties that are specific to the complexes in this proof. just field axioms. it might appear then that every field must contain the rationals. but i assumed a few things in my proof. fact is, every field of infinite order and characteristic 0 must contain the rationals. finite fields obviously do not contain the rationals. A field is by definition a set with two binary operations, call them +,*. the set must form an abelian group with +, and a group (usually also abelian) under the set with the additive identity removed, with the * operation. a group is a set where every element has an inverse. also the * operation must be distributive over the + operation. a subfield is a subset of a field which is closed under the field operations.
