# Direct proportion

This seems to be a simple question. But I'm rather dubious at a particular step.
The question is: Prove
$$\frac{1}{x}-\frac{1}{y}$$ is directly proportional to $$\frac{1}{x}$$

This how I went about it:
$$\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}$$
$$\frac{y-x}{xy}=[\frac{y-x}{y}]\frac{1}{x}$$

Can we say the quantity on the right hand side is proportional to $$\frac{1}{x}$$?

Well, what does proportional mean?

I'll give a more practical definition: If the quantity on the right hand side increases or decreases then the quantity on the left hand side also increases or decreases.There is a constant which can be any real number which relates the two quantites.

Now, in context to this question: Can you call the term in the parenthesis a 'constant' since both 'x' and 'y' are variables?

So when you say

$$a \propto b$$

you mean that there's some $k$ independent of $a$ and $b$ such that

$$a = k b$$?

or to be more precise, $a \propto b$ iff $\frac{da}{db}$ is constant.

Then

$$\frac{1}{x} - \frac{1}{y} \propto \frac{1}{x}$$

is indeed false.

For example, just let $u = \frac{1}{x}$, and $v = \frac{1}{y}$. Then your question is to prove

$$u - v \propto u$$

which is silly unless $v \propto u$ (easy to prove), and presumably $y$ is independent of $x$ in your question~

I imagine that your teacher isn't using that definition for proportionality, because otherwise he/she is being very silly~

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I know this a silly sum. But I think more precise definition of proportionality would be the geometrical one:

$$y \propto x$$

$$y = m x$$
m is the slope. With variable y on y-axis and variable x on x-axis.

Yeah, that's the same as the one I gave in a different form

$$y \propto x \Longleftrightarrow \frac{dy}{dx} \ \mbox{is constant}$$

dextercioby
Homework Helper
Nope,in the way it's formulated,it definitely doesn't match our understanding of direct proportionality.

Daniel.

Of course the x and y above are different from the question.
The result of this question seems obivious by your second method and you can get a straight line on a graph considering :
$$\frac{1}{x} - \frac{1}{y}$$
on y-axis &
$$\frac{1}{x}$$
on x-axis.

Actually, my definition doesn't work! Ignore it! That's what I get for staying up all night writing lab reports.

Your geometrical one is fine though. They aren't the same, I'm just silly when I haven't slept in a day

dextercioby
Homework Helper
Nope,direct proportionality means

$$a \propto b \leftrightarrow a=k b,k\in R^{*}_{+}$$

Daniel.

What's $$R^{*}_{+}$$?

'k' can be any real number, right?

dextercioby
Homework Helper
The real positive semiaxis.

$$\mathbb{R}_{+}^{*}=:\{x\in \mathbb{R}| x>0 \}$$

Daniel.

But the spring constant 'k' in the restoring force of an oscillating body is given by:
F = -kx , x-displacement

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dextercioby
Homework Helper
That's not a correct formula.F=-kx.It's not a direct proportionality.As "x" increases,the force decreases & viceversa.

Daniel.

BobG
Homework Helper
Reshma said:
This seems to be a simple question. But I'm rather dubious at a particular step.
The question is: Prove
$$\frac{1}{x}-\frac{1}{y}$$ is directly proportional to $$\frac{1}{x}$$

This how I went about it:
$$\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}$$
$$\frac{y-x}{xy}=[\frac{y-x}{y}]\frac{1}{x}$$

Can we say the quantity on the right hand side is proportional to $$\frac{1}{x}$$?
No, because $$\frac{y-x}{y}$$ is not a constant. A change in x changes both terms.

The solution

I have found the solution to this question. The problem was the crucial 'missing' component. Now, don't blame me, the source of this question is a very bad text. Let me demonstrate how 'excruciatingly' simple the solution is.

The missing component:

$$y \propto x$$

Now the proof:

$$y \propto x$$
Hence;
$$\frac{1}{y} \propto\frac{1}{x}$$
So;
$$\frac{1}{y} \ = k \frac{1}{x}$$
k= constant

Consider,
$$\frac{1}{x}-\frac{1}{y}$$

$$\frac{1}{x}-\frac{1}{y} = \frac{1}{x} - k \frac{1}{x}$$

Hence;
$$\frac{1}{x}-\frac{1}{y} = (1-k)\frac{1}{x}$$

Finally,
$$(\frac{1}{x}-\frac{1}{y})\propto\frac{1}{x}$$

Voila.... !

dextercioby
Homework Helper
You mean "voilà"... Yeah,hopefully you'll be more careful in the future.You had us all puzzled...

Daniel.

HallsofIvy