Can you proof

  • Thread starter santa
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18
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Main Question or Discussion Point

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

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

then

[tex]y\propto xz[/tex]



why not

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


thank you
 
Last edited:

Answers and Replies

246
1
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:

assume:

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

and

[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

then

[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]

so

[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:
quadraphonics
why not

[tex]y^2\propto xz[/tex]
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.
 
18
0
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
Science Advisor
Gold Member
9,598
6
Yes the constant of proportionality can take any value, positive, negative, real, complex.
 
18
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ok the constant of Hooke's_law

[tex]F=-KX[/tex]

k positive, negative, real, complex. or not
 
Last edited:
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
ok the constant of Hooke's_law

[tex]F=-KX[/tex]

k positive, negative, real, complex. or not
Real and positive.
 
84
0
if [tex]y\propto x[/tex] at z constant

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

then

[tex]y\propto xz[/tex]



why not

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


thank you
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:
 

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