# Is $\dfrac{dx}{dt}$ a fraction or not?

I am new to calculus and during my mathematics class my sir defined $\dfrac{dx}{dt}$ as $$dx/dt=\lim_{t\to t_1}\dfrac{f(t)-f(t_1)}{t-t_1}$$ and my sir made a clear statement that

$\dfrac{dx}{dt}$ is not a fraction it only behaves like a fraction!

(it means $\dfrac{dx}{dt}$ is just a notation to represent that big limit!) and he made a statement that

$dx$ or $dt$ does not have any meaning it is just $\dfrac{d}{dt}(x)$ which has meaning but we treat it as $\dfrac{dx}{dt}$.

but at same time my physics sir, to derive velocity he stated

let the particle be at position $x$ at time $t$ and after an infinitesimal change in position and time it reaches $x+dx$ at time $t+dt$. Now velocity is $\dfrac{displacement}{time}$ , so we will get $v=\dfrac{dx}{dt}$

and this expression purely tells that $\dfrac{dx}{dt}$ is a fraction!

• Also seen in the comics: smbc-comics.com/index.php?db=comics&id=2675#comic – dmckee --- ex-moderator kitten May 13 '16 at 19:34
• Related question on MathSE : math.stackexchange.com/q/21199 – Mo_ May 13 '16 at 19:47
• In my humble opinion, $dx/dt$ is not a fraction. You can treat it like a fraction in some algebraic manipualations, but that only works in some cases and it's better to just understand what $dx/dt$ is from the calculus point of view. – DanielSank May 13 '16 at 20:18
• If you know what differential forms are, that's your license to treat $dx$ and $dt$ like separate variables. – Robin Ekman May 13 '16 at 20:21
• I think saying "$dx$ or $dt$ has no meaning" is one of the reasons why many fresh students in physics and engineer have so much trouble when dealing with (physical) problems involving integrals. They are not able to identify integrations elements. – Diracology May 13 '16 at 23:27

In physics there are no infinitesimals, so dx is always treated as a "small but finite interval" during discussions, and only when the actual calculation is being done do we switch to mathematical mode, and "take the limit."

During the 17th and 18th centuries, mathematicians and physicists both did this all the time. As they say in sports "no harm, no foul!" However, as analysis moved beyond its origins in physics, definitions and methods were tightened up, and proofs became more rigorous, including the definition of limits, and the methods required tend to overwhelm the physical content if included in a physics class.

So to recapitulate: mathematical rigor is always required in a mathematics, but physics is ultimately an experimental subject, and math may be the language of physics, but it is just a tool.

• Peter, imagine you have a computer the size of the universe computing the entire time allotted to our civilisation and using the fastest growing functions in our logical arsenal including the superbusy beaver. Now imagine a number that is bigger than the biggest number your computer can express. Is there such a number "in physics"? Since it is apparently undetectable by any imaginable experiment, it is not. Therefore numbers that are "ordinary" mathematically speaking can also have the distinction of not being "in physics", just like infinitesimals... – Mikhail Katz May 29 '16 at 12:54
• For this reason I find it difficult to determine what precise meaning to attribute to your comment about infinitesimals. – Mikhail Katz May 29 '16 at 12:54
• One can make infinitesimals precise with surreal numbers. As an applied physicist I deal with large and small numbers, but always within a realm that corresponds to experimental results. – Peter Diehr May 29 '16 at 13:45
• Doing calculations within a range is precisely what the hyperreal framework allows you to do. In fact we just published a paper in a mathematical physics journal in connection with infinitesimal oscillations of the pendulum that explains this in more detail; see this paper. – Mikhail Katz May 29 '16 at 13:57
• I don't see in what sense surreal numbers would be experimentally more detectable than the hyperreal ones or for that matter than the hypothetical integer I described above in terms of a fairly large computer. – Mikhail Katz May 29 '16 at 14:00

What an interesting question! It depends. In modern calculus, $\frac{df}{dt}$ is just a symbol for the derivative

$$\lim_{h \to 0} \frac{f(t+h)-f(t)}{h}$$

As a matter of fact, mathematicians prefer different notations for the derivative of a function $f$, as $D f$ or $f'$.

But the above definition of derivative became rigorous only when the concept of limit became rigorous, and this happened only with Weierstrass around 1870 and with his (in)famous "epislon-delta" notation. But we all know that calculus was already used (and maybe invented), in a primitive form, by Newton and Leibniz in the 17th century!

Newton and Leibniz thought of derivatives in different ways: for Newton, it represented a fluxion, a rate of flux or change. For Leibniz, it was the ratio of infinitesimal (really, really small) differences, a differential quotient. In fact, it was Leibniz who first introduced the "quotient" notation $\frac{df}{dt}$!

For example, Leibniz argued that $\frac{d(x^2)}{dx}=2x$ because

$$\frac{(x+dx)^2-x^2}{dx} = \frac{2 x dx + dx^2}{dx} = 2 x + dx$$

and, since $dx$ is infinitesimal, we can ignore it: Q.E.D. ... or not?

The point is that a reasoning such as this is not rigorous enough and can lead to every kind of inconsistencies ($dx$ cannot be exactly $0$, otherwise the quotient would not exist...). So, after a while, mathematicians said goodbye to those treacherous "infinitesimals" and adopted once and for all the more rigorous notation of Weierstrass.

In physics, we are way more laid-back kind of guys: sometimes mathematical rigor just isn't our thing (I remember one of my professors saying this once: "This theorem is false...but we are going to use it anyway!"). We secretly stuck to Leibniz's notation, and we like to use it still today. And do you know why? Because it works. Yes: treating the "differential quotient" as an actual quotient works.

For example, let's say that I have to calculate the derivative with respect to $t$ of

$$f(x(t))$$

How can I do it? Well, easy:

$$\frac{d f}{dt} = \frac{df}{dx} \frac{dx}{dt}$$

Or maybe I want to solve the differential equation

$$\frac{d y}{d x} = \frac{x^2}{y}$$

Piece of cake:

$$y \ dy = x^2 dx \to \int y \ dy = \int x^2 dx \to \frac{y^2}{2} = \frac{x^3}{3} + c \to y=\pm \sqrt{\frac 2 3 (x^3+c)}$$

And so on. In fact, there is even a theory in which this kind of notation is made rigorous: it is called non-standard analysis.

So, if you're a mathematician, then try to avoid using $\frac{df}{dt}$. But if you are a physicist, then...go on, and have no fear!

I can't name any physical system that might be modeled in the domain of rational numbers so I would have to say that $\frac{dy}{dx}$ is not a fraction.
• But $\mathbb{Q}$ is dense in $\mathbb{R}$... In all numerical computations, only rational numbers are used; they get sufficiently close to possibly non-rational solutions. – anderstood May 13 '16 at 20:22
• The question asks about the intuition of viewing the derivative as the ratio of two numbers, not about the rationality of the derivative. Note also that a "fraction" $a/b$ need not be rational if $a$ or $b$ is irrational. "Fraction" is a more elementary way of naming a ratio of numbers, based on the context of the question. – Szmagpie May 13 '16 at 23:34