Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Scalar multiplication axiom, quick question

  1. Sep 24, 2012 #1
    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
     
  2. jcsd
  3. Sep 24, 2012 #2
    so u guys know and im 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 doesnt hold for the set.

    hopefully that helps explain my problem...

    which way is correct?? please help!!!
     
  4. Sep 26, 2012 #3

    Stephen Tashi

    User Avatar
    Science Advisor

    It also confuses anyone who doesn't know what axioms 7,8 and 9 are, which I assume is almost everybody. Why don't you give a complete statement of the exercise that your are trying to work?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Scalar multiplication axiom, quick question
  1. Quick Question (Replies: 4)

Loading...