- #1

ryan8642

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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

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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)

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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!