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Correct my simple error (property of scalar multiplication)

  1. Jan 18, 2014 #1
    say i have some vector ##\vec{v}## multiplied by a scalar k.

    the norm of ##\vec{v}## would be just ##||\vec{v}||## and the norm of ##k\vec{v}## is claimed to be ##|k|||\vec{v}||## ie, ##||k\vec{v}||=|k|||\vec{v}||## ie the sign of the constant is irrelevant.

    when i work it out..

    $$||k\vec{v}||=\sqrt{(kv_{1})^2+(kv_{2})^2+(kv_{3})^2+...}$$
    $$=\sqrt{k^2(v_{1})^2+k^2(v_{2})^2+k^2(v_{3})^2+...}=\sqrt{(k)^2[(v_{1})^2+(v_{2})^2+(v_{3})^2+...]}$$

    from this point, yes i could say that since all k values are squared, the square root must be a positive value. but take k to be a negative number (-a) and say we don't actually compute the square of k and we just leave it in ##(-a)^2## form. we can do the following

    (continued from previous)

    =$$\sqrt{(-a)^2}\sqrt{(v_{1})^2+(v_{2})^2+(v_{3})^2+...}$$

    in which case, the ##\sqrt{}## and the ##(-a)^2## can cancel out which would then return our original -k value which is negative. (where -k=-a)

    i'm pretty sure there's an error some where in my line of reasoning though, could someone please find it?
     
    Last edited by a moderator: Jan 18, 2014
  2. jcsd
  3. Jan 18, 2014 #2

    Mark44

    Staff: Mentor

    I don't know if you're aware of it, but there were four copies of this post where it looked like you were trying different things in LaTeX. I deleted all but this one, which was the best of the lot.
    No.
    These operations don't just cancel out. ##\sqrt{x^2} = |x|##.

    Think about it. Although ##\sqrt{4^2} = 4##, it's NOT true that ##\sqrt{(-4)^2} = -4##.
     
  4. Jan 18, 2014 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The norm of a vector is defined to be non-negative. So the norm of the vector -3 i is 3 and not -3.

    The real operation √x is defined for non-negative x values. √x is the non-negative quantity which has the square equal to x:(√x)2=x. √x ≥0.

    If a = -2, for example, and x=(-2)2, √x=√4=2.

    √(x2)=|x| (the square root does not have memory :smile:)

    ehild
     
  5. Jan 18, 2014 #4
    well, if this is expressed in the following way:

    $$\sqrt{x^2}=x^\frac{2}{2}$$

    are you guys saying that if i have any fractional exponent $$\frac{x}{y}$$

    where $$x\in even numbers , y\in\Re$$

    that i can't simply simplify the exponent if y is an even number? if x is an even number then the quantity must be positive?

    because i have done homework problems in the past where i have a quantity such as this

    $$(-x)^\frac{2}{6}$$ and i reduced this to $$(-x)^\frac{1}{3}$$, solved the problem and have gotten them correct. how is this any different from my original inquiry??
     
  6. Jan 18, 2014 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You can not express it that way.

    ehild
     
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