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

Can you proof

  1. Apr 14, 2008 #1
    if [tex]y\propto x[/tex] at z constant

    and [tex]y\propto z[/tex] at x constant


    [tex]y\propto xz[/tex]

    why not

    [tex]y^2\propto xz[/tex]

    thank you
    Last edited: Apr 14, 2008
  2. jcsd
  3. Apr 14, 2008 #2
    if [tex] y^2 \propto xz [/tex]

    then you would get

    [tex] y \propto \sqrt{xz} [/tex]

    so keeping fx. x constant you have

    [tex] y \propto \sqrt{z} [/tex]

    which is wrong. Maybe a proof could go like this:


    [tex] y \propto y [/tex] for constant z


    [tex] y \propto z [/tex] for constant x

    this must meen that we can write

    [tex] y(x,z) = f(z) x[/tex] for some function f and
    [tex] y(x,z) = g(x) z[/tex] for some function g


    [tex] g(x) x = f(x) x [/tex] so for x different from zero you have

    [tex] g(x) = f(x) [/tex]

    that is

    [tex] y(x,z) = f(z) x[/tex]
    [tex] y(x,z) = f(x) z[/tex]


    [tex] y(x,1) = f(1) x[/tex]
    [tex] y(x,1) = f(x) 1[/tex]

    from which you get

    [tex] f(x)= f(1) x[/tex], inserting this you have

    [tex] y(x,z) = f(1) z x[/tex]

    which is to say

    [tex] y(x,z) \propto z x[/tex]

    maybe the proof is flawed did it pretty sloppy.
    Last edited: Apr 14, 2008
  4. Apr 14, 2008 #3
    Because then you'd have [itex]y\propto \sqrt{x}[/itex] for a fixed z. However, you can have [itex]y \propto x f(x)[/itex] where [itex]f(z)[/itex] is just-about-any function of z.
  5. Apr 15, 2008 #4

    but a have another

    Definition of directly proportional - can k be negative?

    In almost all textbooks, "directly proportional" is defined by saying
    that a is directly proportional to b if and only if a = kb for some
    constant k. That's perfectly sensible, but taking the definition
    literally, it would seem to imply that any k will do, even negatives.

    However, in every example that I have seen to illustrate the concept,
    the term "directly proportional" is always applied to the relationship
    between two positive quantities or two negative quantities--never
    between a positive quantity and a negative quantity.
  6. Apr 15, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes the constant of proportionality can take any value, positive, negative, real, complex.
  7. Apr 16, 2008 #6
    ok the constant of Hooke's_law


    k positive, negative, real, complex. or not
    Last edited: Apr 16, 2008
  8. Apr 16, 2008 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Real and positive.
  9. Apr 16, 2008 #8
    Would that not lead to:-

    [tex]y\propto y^2[/tex]

    Irregardless of the first two statements.

    Or have I grossly missed the point :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Can you proof
  1. How can you prove? (Replies: 9)

  2. Can you find 0! (Replies: 4)