Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

Does this set form a vector space?

OpenStudy (anonymous):

I think the answer is No. But not sure

OpenStudy (anonymous):

maybe bcz if you have 3*p(6) = 6 i.e. not in set

OpenStudy (rational):

Exactly ! one contradiction is sufficient to end the proof

OpenStudy (anonymous):

and also p(6) + p(6) = 4 i.e. not in set either. So it contradicts both additive and multiplication property

OpenStudy (rational):

also, if u want to check another property that fails : if p(t) and q(t) are in vector space, then the polynomial defined by p(t) + q(t) doesnt exist in this vector space because : p(6) + q(6) = 2+2 = 4 \(\ne\) 2

OpenStudy (anonymous):

thats what I said above, haha thank you

OpenStudy (rational):

Ah ok, the replies are getting jumbled at my side

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!