Proving Direct Proportion of \frac{1}{x}-\frac{1}{y} with \frac{1}{x}

[Reshma]

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 $$\frac{1}{x}$$?

 

[Data]

Well, what does proportional mean?

 

[Reshma]

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?

 

[Data]

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~

 

[Reshma]

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.

 

[Data]

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]

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

Daniel.

 

[Reshma]

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.

 

[Data]

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]

Nope,direct proportionality means

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

Daniel.

 

[Reshma]

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

'k' can be any real number, right?

 

[dextercioby]

The real positive semiaxis.

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

Daniel.

 

[Reshma]

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

 

[dextercioby]

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

Daniel.

 

[BobG]

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.

 

[Reshma]

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]

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

Daniel.

 

[HallsofIvy]

Or as teenagers nowadays like to say "Viola"! I figure about 10% of them know it's a joke, the other 90% really think that's the correct pronunciation of "voilà".

 

[Reshma]

Well, I don't have a French PC.

 

[thedy]

What is direct proportionality?
In ordinary life I understand it.For example if I buy more tickets i will more pay.But in physical formula I don t understand it.For example in formula of momentum we have p=m*v.But why is there " * "?Why not "+"?That means p=m+v?I know that is stupid question,but I m down.
From what is this formula,p=m*v derived?
Thanks