- #1
ryan8642
- 24
- 0
u and v are contained in V
Lets say the scalar multiplication is defined as:
ex.
ku=k^2 u or ku = (0,ku2) u=(u1,u2)
does this mean that this is also the same for different scalar m?
mu=m^2 u or mu = (0,mu2) u=(u1,u2)
and does this mean the same for any vector v
kv=k^2 v or kv = (0,kv2) v=(v1,v2)
Is this correct?
Axioms 7,8,9 contain the 2 different scalars as well as vectors. it really confuses me.
Can someone please put me on the right track :s
_____________
so u guys know and I am not confusing you guys i showed 2 examples there to help show my problem.
Ex 1.
Lets say the scalar multiplication is defined as:
ku = (0,ku2) u=(u1,u2)
does this then mean that this is also the same for different scalar m?
mu = (0,mu2) u=(u1,u2)
and also this for any vector v
kv = (0,kv2) v=(v1,v2)
_____________________
addition u+v=(u1+v1, u2+v2)
ex.. axiom 8 (to help explain my problem)
using what is described above.
(k+m)u = ku + mu
(k+m)(u1,u2)=k(u1,u2) + m(u1,u2)
(0,(k+m)u2)=(0,ku2) + (0,mu2)
(0,(k+m)u2)=(0,ku2+mu2)
(0,(k+m)u2)=(0,(k+m)u2)
LS=RS therefore axiom 8 holds for the set.
now using just ku=(0,ku2)
(k+m)u = ku + mu
(k+m)(u1,u2) = k(u1,u2) + m(u1,u2)
((k+m)u1, (k+m)u2) = (0,ku2) + (mu1,mu2)
((k+m)u1, (k+m)u2) = (0+mu1, ku2+mu2)
(ku1+mu1,ku2+mu2) = (mu1, ku2 +mu2)
LS ≠ RS so axiom 8 doesn't hold for the set.
hopefully that helps explain my problem...
which way is correct?? please help!
Lets say the scalar multiplication is defined as:
ex.
ku=k^2 u or ku = (0,ku2) u=(u1,u2)
does this mean that this is also the same for different scalar m?
mu=m^2 u or mu = (0,mu2) u=(u1,u2)
and does this mean the same for any vector v
kv=k^2 v or kv = (0,kv2) v=(v1,v2)
Is this correct?
Axioms 7,8,9 contain the 2 different scalars as well as vectors. it really confuses me.
Can someone please put me on the right track :s
_____________
so u guys know and I am not confusing you guys i showed 2 examples there to help show my problem.
Ex 1.
Lets say the scalar multiplication is defined as:
ku = (0,ku2) u=(u1,u2)
does this then mean that this is also the same for different scalar m?
mu = (0,mu2) u=(u1,u2)
and also this for any vector v
kv = (0,kv2) v=(v1,v2)
_____________________
addition u+v=(u1+v1, u2+v2)
ex.. axiom 8 (to help explain my problem)
using what is described above.
(k+m)u = ku + mu
(k+m)(u1,u2)=k(u1,u2) + m(u1,u2)
(0,(k+m)u2)=(0,ku2) + (0,mu2)
(0,(k+m)u2)=(0,ku2+mu2)
(0,(k+m)u2)=(0,(k+m)u2)
LS=RS therefore axiom 8 holds for the set.
now using just ku=(0,ku2)
(k+m)u = ku + mu
(k+m)(u1,u2) = k(u1,u2) + m(u1,u2)
((k+m)u1, (k+m)u2) = (0,ku2) + (mu1,mu2)
((k+m)u1, (k+m)u2) = (0+mu1, ku2+mu2)
(ku1+mu1,ku2+mu2) = (mu1, ku2 +mu2)
LS ≠ RS so axiom 8 doesn't hold for the set.
hopefully that helps explain my problem...
which way is correct?? please help!