Why dy/dx is not a ratio?

[Buffu]

I read people saying that dy/dx is not a ratio because it is a limit or standard part of a ratio. $${dy\over dx} = \lim_{h \to 0} {f(x +h) - f(x) \over h}, \ \ \ {dy\over dx} = st\left( {f(x +h) - f(x) \over h}\right)$$

what I get is ${f(x +h) - f(x) \over h}$ is a ratio but putting a limit or st makes it something else than a ratio.

So my question is $\lim_{x \to 0} \frac 4 5$ a ratio ? If it is then how would you put an formal argument against dy/dx as a ratio ?

 

[PeroK]

I read people saying that dy/dx is not a ratio because it is a limit or standard part of a ratio. $${dy\over dx} = \lim_{h \to 0} {f(x +h) - f(x) \over h}, \ \ \ {dy\over dx} = st\left( {f(x +h) - f(x) \over h}\right)$$

what I get is ${f(x +h) - f(x) \over h}$ is a ratio but putting a limit or st makes it something else than a ratio.

So my question is $\lim_{x \to 0} \frac 4 5$ a ratio ? If it is then how would you formal put an argument against dy/dx as a ratio ?

I was going to answer, but first define "ratio".

 

[Buffu]

I was going to answer, but first define "ratio".

I never thought about this and honestly I don't know the definition. Let me try anyways,

We say two numbers $p, q$ are in ratio if we can find a third number $r$ such that $p = rq$.

 

[PeroK]

I never thought about this and honestly I don't know the definition. Let me try anyways,

We say two numbers $p, q$ are in ratio if we can find a third number $r$ such that $p = rq$.

Unless $q = 0$, numbers are always in a ratio. I was expecting you to say something like a ratio is one number divided by another. In particular:

$\frac{f(x_0+h) -f(x_0)}{h}$ is a ratio of two numbers.

But, as $dx$ and $dy$ are not numbers, $\frac{dy}{dx}$ is not a ratio of two numbers.

That's important because you may have proved that for any (non-zero) numbers, you have: $\frac{a}{b} \frac{b}{c} = \frac{a}{c}$.

But, it doesn't immediately follow that $\frac{dy}{dx} \frac{dx}{dz} = \frac{dy}{dz}$.

Instead, you have to prove this (as separate and additional to your proof for numbers) using the definition of these things as limits.

From a pure maths point of view, making things look like already familiar things doesn't imbue them with the same properties; those properties must be proved in themselves.

 

[Buffu]

But, as dx and dy are not numbers, \frac{dy}{dx} is not a ratio of two numbers.

Why ? Let's say ${dy \over dx} = {f(x+ h) - f(x)\over h}$ then it is a ratio as ${f(x+ h) - f(x)\over h}$ is a ratio. But ${dy \over dx} = \color{red}{\lim_{h \to 0}} {f(x+ h) - f(x)\over h}$ then is not a ratio. So it must be something with the limit right ?

So if putting a limit make things something other than ratio then $\lim {4\over 10}$ is not a ratio. How will we justify $\lim {4\over 10}$ as a ratio ?

 

[PeroK]

Why ? Let's say ${dy \over dx} = {f(x+ h) - f(x)\over h}$ then it is a ratio as ${f(x+ h) - f(x)\over h}$ is a ratio. But ${dy \over dx} = \color{red}{\lim_{x \to h}} {f(x+ h) - f(x)\over h}$ then is not a ratio. So it must be something with the limit right ?

So if putting a limit make things something other than ratio then $\lim {4\over 10}$ is not a ratio. How will we justify $\lim {4\over 10}$ as a ratio ?

This is such a confusion of ideas, it's hard to answer. $\frac{4}{10}$ is no more a ratio than any other number. It's $0.4$. It's just a plain old number. Like all numbers, it can be written as a ratio in an infinite number of ways.

You can't make something a number by wishing it to be a number. How would you define, for example, $dx$ as a number? What number is it for a given function? Let's say the simple function $f(x) = x^2$. What is $dx$ at $x=1$?

 

[PeroK]

Since this isn't homework, let me explain the problem. $dx$ is an "infinitesimal" change in $x$ and likewise $dy$. Now, infinitesimals are not numbers. Why? Well, a number is either $0$ or it isn't. And, if you take an infinitesimal as so small that it's 0, then all you have is $dx = dy = 0$ in all cases (for all functions at all points). And that's no good.

But, if $dx$ is not $0$, then it's just some finite number. But, then, it's just some finite change in $x$. To keep things simple, let's assume that $dx = 1$. If we take the function $f(x) = x^2$ we have:

$f(x+1) - f(x) = (x+1)^2 - x^2 = 2x + 1$

So, for this function, we have $dy = 2x + 1$ and $\frac{dy}{dx} = 2x + 1$.

And that's no good either. That's just an approximation of the derivative. If we took $dx$ smaller, we would get a better approximation. But, we can never get $\frac{dy}{dx} = 2x$.

In fact, if $dx$ is a number we get $\frac{dy}{dx} = 2x + (dx)^2$.

And that's the problem: an infinitesimal like $dx$ can't be $0$ and it can't be non-zero, thus it can't be a number.

In fact, in a formal development of derivatives, you would tend to use $f'(x)$ for the derivative. And, once you proved the properties of the derivative and the integral, especially in applied maths, you tend to adopt the $\frac{dy}{dx}$ notation, because it can be easier to work with in things like the chain rule and the fact that you can cancel infinitesimals as though they were numbers is of considerable appeal.

But, never at any stage do $dx$ and $dy$ become real numbers.

 

[Buffu]

@PeroK Ok I get your point that dy and dx are not numbers and so dy/dx is not a ratio but let's forget that we denote dy/dx for a derivative. Let's say we write the first principle for derivative everytime we differentiate, like instead of ${d(x^2)\over dx} = 2x$ we write $\lim_{x \to 0} {(x+h)^2 - x^2\over h} = 2x$. Now there is no dy or dx to worry about.

We also did not change anything other than we replaced the shorthand for derivate with its definition. Now it is still not a ratio of any sort. The $\lim$ part is what disabling it to be a ratio.

We apply the same argument to the things like $\lim 4/3$ or whatever number you like. By that logic we say this is also not a ratio, but then there is nothing like a ratio because we can just replace 4/3 with anything and argument still holds the same. This conclusion can be easily contradicted with the fact that $4 = 2 \times 2$ thus $4:2 = 2:1$ is a ratio.

So what went wrong in my thinking ?

Sorry I think I did not have phrased the question properly, the real problem that is bugging me is that we say $\lim_{\delta x \to 0} {f(x + \delta x) - f(x) \over \delta x}$ is not ratio but we say $\lim 4/3$ as a ratio.

 

[PeroK]

@PeroK Ok I get your point that dy and dx are not numbers and so dy/dx is not a ratio but let's forget that we denote dy/dx for a derivative. Let's say we write the first principle for derivative everytime we differentiate, like instead of ${d(x^2)\over dx} = 2x$ we write $\lim_{x \to 0} {(x+h)^2 - x^2\over h} = 2x$. Now there is no dy or dx to worry about.

We also did not change anything other than we replaced the shorthand for derivate with its definition. Now it is still not a ratio of any sort. The $\lim$ part is what disabling it to be a ratio.

We apply the same argument to the things like $\lim 4/3$ or whatever number you like. By that logic we say this is also not a ratio, but then there is nothing like a ratio because we can just replace 4/3 with anything and argument still holds the same. This conclusion can be easily contradicted with the fact that $4 = 2 \times 2$ thus $4:2 = 2:1$ is a ratio.

So what went wrong in my thinking ?

Sorry I think I did not have phrased the question properly, the real problem that is bugging me is that we say $\lim_{\delta x \to 0} {f(x + \delta x) - f(x) \over \delta x}$ is not ratio but we say $\lim 4/3$ as a ratio.

You need to put aside the idea that there are special numbers called "ratios". The limit is the limit of a numerical function (it's actually a function of $h$ with $x$ constant while you take the limit). So:

$f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$

There is nothing special about numbers that are ratios. This is just a functional limit and should be understood as such.

 

[Buffu]

You need to put aside the idea that there are special numbers called "ratios". The limit is the limit of a numerical function (it's actually a function of $h$ with $x$ constant while you take the limit). So:

$f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$

There is nothing special about numbers that are ratios. This is just a functional limit and should be understood as such.

Sorry I can relate this to my previous post. Can you dumb this down a bit ? please.

 

[PeroK]

Sorry I can relate this to my previous post. Can you dumb this down a bit ? please.

Perhaps you are not ready for real analysis. It's not the easiest thing to pick up and grasp. Are you a maths student?

 

[Buffu]

Perhaps you are not ready for real analysis. It's not the easiest thing to pick up and grasp. Are you a maths student?

Yes.

 

[PeroK]

Yes.

Have you done limits yet?

 

[Buffu]

@PeroK
Yes (I once answered a question about limits) . I know multivariable calculus that is required to do electrodynamics.
I have also asked some questions about series, so I have done some real analysis at least.
But I have never learned that a dy/dx is not a ratio, so I was a bit confused. Sorry if I troubled you much.

 

[JonnyG]

You will need to learn about differential forms to understand what dx and dy is. I suggest either Spivak's Calculus on Manifolds or Munkres' Analysis on Manifolds.

 

[member 587159]

$\frac{dy}{dx}$ should not be interpretated as a ratio, but simply as notation for the derivative. The notation is useful because in some cases it corresponds with our intuition when we manipulate fractions. For example, without mentioning the conditions when this is true:

$\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$.

If $f: (a,b) \rightarrow \mathbb{R}$, then $\frac{df}{dx}: (a,b) \rightarrow \mathbb{R}: x \mapsto \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$

if these limits exist for all $x \in (a,b)$ (otherwise we must change the domain)

 

[Buffu]

$\frac{dy}{dx}$ should not be interpretated as a ratio, but simply as notation for the derivative.

If $f: (a,b) \rightarrow \mathbb{R}$, then $\frac{df}{dx}: (a,b) \rightarrow \mathbb{R}: x \mapsto \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$

if these limits exist for all $x \in (a,b)$ (otherwise we must change the domain)

Yes I know, I have the same definition in my high school book.

 

[member 587159]

Yes I know, I have the same definition in my high school book.

Then what is your exact problem?

 

[Buffu]

Then what is your exact problem?

Did you read post #8 ?

But I think I lack knowledge to understand this now.

 

[Drakkith]

You need to put aside the idea that there are special numbers called "ratios". The limit is the limit of a numerical function (it's actually a function of $h$ with $x$ constant while you take the limit). So:

$f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$

There is nothing special about numbers that are ratios. This is just a functional limit and should be understood as such.

PeroK, if I understand correctly, you're saying that limits only apply to functions?

 

[member 587159]

Did you read post #8 ?

But I think I lack knowledge to understand this now.

The limiting process where you write $\lim\limits_{x \to a} f(x)$ is not a quotient. It is exactly what is described in the $\epsilon - \delta$ - definition. The output however, if it exists, might be a real number that can be expressed as a fraction if it is not irrational. Therefore, $\lim\limits_{x \to a} 4/3$ is not a fraction. It means, intuitively, that we consider the constant function $f(x) = 4/3$ and look how it behaves around the point $4/3$ (read: what it converges to) This limit is obviously equal to $4/3$, which is a fraction.

I'm not sure whether I interpreted your question correctly though

 

[Svein]

Another way to look at it: $\frac{d}{dx}$ is a linear operator, mapping a subset of the continuous functions of one variable into another subset (see https://en.wikipedia.org/wiki/Differential_operator).

In the same way $\int dx$ is a linear operator, mapping a superset of the continuous functions of one variable into another superset. The dx specifies what measure we integrate against, in this case the interval-length measure (Riemann integration).

 

[PeroK]

PeroK, if I understand correctly, you're saying that limits only apply to functions?

There are two main types of limits: the limit of a sequence (whole number variable); and, the limit of a function (real variable). The derivative is the latter with $h$ as the variable, which tends to 0.

 

[pwsnafu]

There are two main types of limits: the limit of a sequence (whole number variable); and, the limit of a function (real variable). The derivative is the latter with $h$ as the variable, which tends to 0.

Clarification: a "sequence" is a function whose domain is the natural numbers. PeroK is using "function" in place of the phrase "a function whose domain and co-domain are the real numbers".

 

[Buffu]

The limiting process where you write $\lim\limits_{x \to a} f(x)$ is not a quotient. It is exactly what is described in the $\epsilon - \delta$ - definition. The output however, if it exists, might be a real number that can be expressed as a fraction if it is not irrational. Therefore, $\lim\limits_{x \to a} 4/3$ is not a fraction. It means, intuitively, that we consider the constant function $f(x) = 4/3$ and look how it behaves around the point $4/3$ (read: what it converges to) This limit is obviously equal to $4/3$, which is a fraction.

I'm not sure whether I interpreted your question correctly though

Yes you interpreted it correctly. Thanks for answering me.

 

[nuuskur]

It isn't a ratio. It's just notation. Likewise, we could have defined it as $[\mbox{d}y ; \mbox{d}x]$ and read it as "the derivative (what ever that is) of $y$ with respect to the variable $x$. There are some rules involved and it turns out the notation chosen is optimal, because we can describe the chain rule (derivative of a product of functions) with familiar notation. I.e if $y = y(x)$ and $x = x(t)$, then
$$\frac{\mbox{d}y}{\mbox{d}t} = \frac{\mbox{d}y}{\mbox{d}x}\cdot\frac{\mbox{d}x}{\mbox{d}t}$$
and it turns out we can cancel like terms just like we do when dealing with fractions. (or multiply the "fraction" with $1$ in a suitable manner.)

 

[Biker]

Since this isn't homework, let me explain the problem. $dx$ is an "infinitesimal" change in $x$ and likewise $dy$. Now, infinitesimals are not numbers. Why? Well, a number is either $0$ or it isn't. And, if you take an infinitesimal as so small that it's 0, then all you have is $dx = dy = 0$ in all cases (for all functions at all points). And that's no good.

Sorry for bumping the thread, But I have a couple of arguments more like question to ask.

When I first tried to calculate the slope of functions in physics in our first year, I used the method of infinitesimal and later on when I took calculus, we were introduced to limits..etc. The thing is that they still use the notation dy and dx as if they still have meaning and in my opinion they do. I read about the history of calculus to find more about it. A lot of famous mathematician considered them as this and in physics it gives you the feeling that you know what you are doing

But using the word ratio has consequences, One has to be careful about how he interpret it in different kind of situations.

For a starter in calculus like me, It makes every bit of intuitive sense for all operation they give us like separation of variables, area under the curve, summation of infinitely small intervals...etc and most theorems are just natural ( Some will say they first derived these theorems and then got that optimal notation)

and as far as I know some people justified this in their books like non-standard calculus, Calculus made easy(not sure if that is the book I think I read but it was something similar)..etc. So as long as it works and it is intuitive why did they just neglect it and move to the standard calculus in which they wave dx and dy as if they were nothing but a trick to get the right answer. They just leave the student asking about these stuff and others just accept the way. Thousands of articles and sites on the internet about this topic because in most textbooks they do that without giving justification. If two methods results in the same answer why not use them both?

 

[Buffu]

Sorry for bumping the thread, But I have a couple of arguments more like question to ask.

When I first tried to calculate the slope of functions in physics in our first year, I used the method of infinitesimal and later on when I took calculus, we were introduced to limits..etc. The thing is that they still use the notation dy and dx as if they still have meaning and in my opinion they do. I read about the history of calculus to find more about it. A lot of famous mathematician considered them as this and in physics it gives you the feeling that you know what you are doing

But using the word ratio has consequences, One has to be careful about how he interpret it in different kind of situations.

For a starter in calculus like me, It makes every bit of intuitive sense for all operation they give us like separation of variables, area under the curve, summation of infinitely small intervals...etc and most theorems are just natural ( Some will say they first derived these theorems and then got that optimal notation)

and as far as I know some people justified this in their books like non-standard calculus, Calculus made easy(not sure if that is the book I think I read but it was something similar)..etc. So as long as it works and it is intuitive why did they just neglect it and move to the standard calculus in which they wave dx and dy as if they were nothing but a trick to get the right answer. They just leave the student asking about these stuff and others just accept the way. Thousands of articles and sites on the internet about this topic because in most textbooks they do that without giving justification. If two methods results in the same answer why not use them both?

Maybe because it might require twice the time for students to understand both the methods. But I am in for teaching of non-standard analysis.

 

[YashOP]

@PeroK Ok I get your point that dy and dx are not numbers and so dy/dx is not a ratio but let's forget that we denote dy/dx for a derivative. Let's say we write the first principle for derivative everytime we differentiate, like instead of ${d(x^2)\over dx} = 2x$ we write $\lim_{x \to 0} {(x+h)^2 - x^2\over h} = 2x$. Now there is no dy or dx to worry about.

We also did not change anything other than we replaced the shorthand for derivate with its definition. Now it is still not a ratio of any sort. The $\lim$ part is what disabling it to be a ratio.

We apply the same argument to the things like $\lim 4/3$ or whatever number you like. By that logic we say this is also not a ratio, but then there is nothing like a ratio because we can just replace 4/3 with anything and argument still holds the same. This conclusion can be easily contradicted with the fact that $4 = 2 \times 2$ thus $4:2 = 2:1$ is a ratio.

So what went wrong in my thinking ?

Sorry I think I did not have phrased the question properly, the real problem that is bugging me is that we say $\lim_{\delta x \to 0} {f(x + \delta x) - f(x) \over \delta x}$ is not ratio but we say $\lim 4/3$ as a ratio.

When you are saying lim4/3 to what are you appliying the limit In the derivative case you are applying the limit to x that delta x tends to 0 but in this case you can't apply the limit because 4/3 is a constant number which never changes no matter how many limits you apply