Question

et of all triples of real numbers with standard vector addition but with scalar multiplication defined by
k(x,y,z)=(k^2,k^y,k^z)
axiom 8 doesn't hold but I don't see how
(axiom 8 is (k+m)u=ku+km) which seems like it holds to me

Answer 1

it is distributivity. Take u in R^3, m,k in R
(k+m)u=((k+m)^2,(k+m)^y,(k+m)^z)
although to prove it is a ring (or vector space) you would have to use the binomial expansion, it is only enough to give a counter example to disprove the statement.
we see that the first coordinate must always be something^2
(k+m)^2=k^2+2km+m^2
but we see that we can arbitrary choose k and m so that they are not powers of two (like take 3 and 5). Thus the distributivity fails.

Answer 2

sorry, squares, not powers of 2

Answer 3

well I see it like this:

if we solve the LHS: (k+m)^2u1, (k+m)^2u2,(k+m)u^3
RHS: k^2u1, k^2u2,k^2u3 + m^2u1,m^2u2,m^2u3
--> k^2u1+m^2u1,m^2u2+k^2u2,k^2u3+m^2u3
--> u1(k^2+m^2)

ANDDDDD... isee where I went wrong... you cant do (k^2+m^2)=(k+m)^2.

Thank you kind stranger!