Can you proof

if $$y\propto x$$ at z constant

and $$y\propto z$$ at x constant

then

$$y\propto xz$$

why not

$$y^2\propto xz$$

thank you

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if $$y^2 \propto xz$$

then you would get

$$y \propto \sqrt{xz}$$

so keeping fx. x constant you have

$$y \propto \sqrt{z}$$

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

assume:

$$y \propto y$$ for constant z

and

$$y \propto z$$ for constant x

this must meen that we can write

$$y(x,z) = f(z) x$$ for some function f and
$$y(x,z) = g(x) z$$ for some function g

then

$$g(x) x = f(x) x$$ so for x different from zero you have

$$g(x) = f(x)$$

that is

$$y(x,z) = f(z) x$$
$$y(x,z) = f(x) z$$

so

$$y(x,1) = f(1) x$$
$$y(x,1) = f(x) 1$$

from which you get

$$f(x)= f(1) x$$, inserting this you have

$$y(x,z) = f(1) z x$$

which is to say

$$y(x,z) \propto z x$$

maybe the proof is flawed did it pretty sloppy.

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why not

$$y^2\propto xz$$

Because then you'd have $y\propto \sqrt{x}$ for a fixed z. However, you can have $y \propto x f(x)$ where $f(z)$ is just-about-any function of z.

thanks

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.

Hootenanny
Staff Emeritus
Gold Member
Yes the constant of proportionality can take any value, positive, negative, real, complex.

ok the constant of Hooke's_law

$$F=-KX$$

k positive, negative, real, complex. or not

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Hootenanny
Staff Emeritus
Gold Member
ok the constant of Hooke's_law

$$F=-KX$$

k positive, negative, real, complex. or not
Real and positive.

if $$y\propto x$$ at z constant

and $$y\propto z$$ at x constant

then

$$y\propto xz$$

why not

$$y^2\propto xz$$

thank you

Would that not lead to:-

$$y\propto y^2$$

Irregardless of the first two statements.

Or have I grossly missed the point