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Vectors and axioms

  1. Aug 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Let V be the set of all ordered pairs of real numbers. Suppose we define addition and scalar
    multiplication of elements of V in an unusual way so that when

    u=(x1, y1), v=(x2, y2) and k∈ℝ

    u+v= (x1⋅x2, y1+y2) and

    k⋅u=(x1/k, y1/k)

    Show detailed calculations of one case where V
    i) satisfies one of the addition axioms (1 – 5)
    ii) fails to uphold one of the addition axioms (1 – 5)
    iii) satisfies one of the scalar multiplication axioms (6 – 10)
    iv) fails to uphold one of the scalar multiplication axioms (6 – 10)


    2. Relevant equations

    3. The attempt at a solution

    i)u+v=v+u: Axiom 2
    u+v= (x1⋅x2, y1+y2)
    v+u= (y1⋅y2, x1+x2)
    ∴u+v≠v+u

    ii) u+(-u)= (-u)+u=(0,0): axiom 4
    (x1, y1)+(-x1,-y1)= (x1⋅-x1, y1+-y1)
    = (-x1, 0)

    iii) 1u=u: axiom 10
    ku= (x1/k, y1/k) if k=1
    1u= (x1/1, y1/1)= (x1,y1)=u

    iv)??

    I don't know if I am even on the right track with the three that I have calculated. I feel confident in the third but the others, not so much. I also have struggled to demonstrate the fourth.
    Any help would be greatly appreciated.
     
  2. jcsd
  3. Aug 26, 2015 #2
    Ok, I think number three is correct, and I am not an expert, but I don't think that vectors break the commutative property of addition (i.e. u+v is the same as v+u).
     
  4. Aug 27, 2015 #3

    HallsofIvy

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    (i) asked you to give a case in which one of the axioms was satisfied. In any case, with u= (x1, y1), v= (x2, y2) (although you don't say that) swapping u and v swaps the "1" and "2" NOT "x" and "y".

    x1 times -x1 is NOT, in general, equal to -x1!

     
  5. Aug 27, 2015 #4

    pasmith

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    Here would have been an excellent place to tell us what you are given as axioms 1 to 10.

    The operation on the first component is real multiplication, which is associative and commutative.
    The operation on the second component is real addition, which is associative and commutative.

    Therefore vector addition is associative and commutative.

    The "zero" vector is (1,0): (1,0) + (x,y) = (1x, y + 0) = (x,y) = (x,y) + (1,0) for every (x,y).

    But can you always find a unique (a,b) such that (x,y) + (a,b) = (ax, b + y) = (1,0)?

    (When you can, is it actually equal to (-1)(x,y)? If it isn't, that suggests a problem with scalar multiplication...)

    Correct.

    But note that we can't multiply any vector by [itex]0 \in \mathbb{R}[/itex]. That again suggests a problem with scalar multiplication.
     
  6. Aug 27, 2015 #5
    If these are the kind of axioms he is talking about, then numbers 2 and 4 would be impossible to do, correct?
     
  7. Aug 27, 2015 #6
    Thank you for your help. I didn't think my calculations looked right algebraically but I rushed and wrote x1 time -x1 = -x1. I will look further into this equation.
    It makes sense to not swaps the "x" and "y" now that you say it. Thank you again, I appreciate the time you have taken to help.
     
  8. Aug 27, 2015 #7
    I have looked into it and you are correct, they don't break the commutative property of addition. Thank you for your advice.
     
  9. Aug 27, 2015 #8
    Thank you for your input, I very much appreciate it.
     
  10. Aug 27, 2015 #9
    I think I may have made a mistake and posted my responses within your post. I am still trying to work out this site. I apologise.
     
  11. Aug 27, 2015 #10
    I would say that number 2 and 4 are impossible. All those axioms would with any vectors/scalars. I don't know how your teacher expects you to do this. It's like asking you to name a living thing that doesn't require some form of energy source.
     
  12. Aug 27, 2015 #11

    SammyS

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    Axioms 2 and 4 are true. In fact pasmith showed what the zero vector is.
     
  13. Aug 28, 2015 #12
    Sorry for the confusion. Problems 2 and 4 (see post number 1) are impossible, because all 10 axioms are always true.
     
  14. Aug 28, 2015 #13

    pasmith

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    "Axiom" here means "defining property", not "statement accepted as true without proof". A set V is a vector space over a field F if and only if the vector addition and scalar multiplication operations, whatever they are defined to be, satisfy all of those 10 axioms.

    Here it is indeed the case that at least one of the first 5 axioms doesn't hold for vector addition as defined in the OP, and it is indeed the case that at least one of the second 5 axioms doesn't hold for scalar multiplication as defined in the OP.
     
  15. Aug 28, 2015 #14

    pasmith

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    The trouble with this sort of question is that we are so used to the standard vector operations on [itex]\mathbb{R}^2[/itex] that we just assume that [itex]\mathbf{0} = (0,0)[/itex] and [itex]-(x,y) = (-x,-y)[/itex]. But if the operations are defined differently then we have to go back to the axioms and ask: What do [itex]\mathbf{0}[/itex] and [itex]-\mathbf{v}[/itex] actually mean? And the answers are "additive identity" and "additive inverse" respectively.

    Axiom 5 says that every vector [itex]\mathbf{v} \in V[/itex] has an additive inverse, ie. a [itex]\mathbf{u} \in V[/itex] such that [itex]\mathbf{v} + \mathbf{u} = \mathbf{0}[/itex]. Thus here Axiom 5 holds if and only if for every [itex](x,y) \in\mathbb{R}^2[/itex] there exists an [itex](a,b) \in \mathbb{R}^2[/itex] such that [itex](x,y) + (a,b) = (1,0)[/itex]. Is that the case?

    Giving further hints comes far to close to doing the entire question for you, so you're on your own now.
     
  16. Aug 29, 2015 #15
    Some of these problems can be made easier to plug in a few arbitrary values, then solve for the general case if it is not apparent. Although, after a while you won't need to this. You can follow the axioms after understanding what they are.

    Let u=(1,1) and v=(2,2) for starters. Then preform the field axioms.

    (a+b) U=?

    Where a and b are scalers.
     
  17. Sep 2, 2015 #16

    micromass

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    What do the field axioms have to do with this?
     
  18. Sep 2, 2015 #17
    meant to put vector space axioms, was thinking of something else.
     
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