Is the axiom for addition in the set of polynomials of degree >=3 true?

stunner5000pt
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Determine if this is a vector space with the indicated operations

the set of V of all polynominals of degree >=3, togehter iwth 0, operations of P (P the set of polynomials)

now all the scalar multiplication axioms hold.
the text however says that the axion
\mbox{For u,v} \in V, \mbox{then} \ u+v \in V does not hold

well ok take two polynomials
u(x) = a_{3} x^3 + ... + a_{n} x^n
v(x) = b_{3} x^3 + ... + b_{k} x^k
where both n,k>= 3, then suppose k< n
u(x) + v(x) = (a_{3} + b_{3}) x^3 + ... + (a_{k} + b_{k}) x^k + ... + a_{n} x^n
which is certainly a polynomial or degree >= 3 isn't it?
It also applies for n<k and n = k
is the textbook wrong?
 
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What is the degree of the following poly

x^4+1

now, do you see your error?
 
ok let me correct that then
n,k >= 3
u(x) = a_{0} + a_{1} x + ... + a_{n} x^n
v(x) = b_{0} + b_{1} x + ... + b_{k} x^k
then for n< k
u + v = (a_{0} + b_{0}) + ... + (a_{k} + b_{k}) x^k + ... + a_{n} x^n

stil lseems to be of degree three to me
however if k=n and an= -bn then the polynomial is no more degree 3
is this corret?
 
Why don't you just find a counter example? two polys of degree 3 or greaterwhose sum isn't? A single counter examplem suffices.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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