# is $dx$ greater than $\frac{dx}{2}$?

I wanted to ask if $dx$ is greater than $\frac{dx}{2}$? i will make conclusions i am sure they are wrong : a) if yes then why in integration we do not use smaller than $dx$ like its half ? b) if you said they are equal then does it mean $1 = \frac{1}{2}$? some may say you can't divide by $dx$ but we do it a lot in solving differential equations ? so who is greater? might seem low question because i am not math major like you

• $dx$ is not a number, so I don't know what you mean by "greater than". This is the source of your confusion, I believe. – Chris Gerig May 21 '15 at 23:27
• is the operation of < and > not defined ? but we learnt in physics dx is an interval and we can compare intervals ?? – Mohamed Osama May 21 '15 at 23:28
• In practice in physics, it doesn't really matter which infinitesimal $dx$ actually is, the point is that for any given $dx$ you have some property. That is, you never work with particular infinitesimals, instead just with arbitrary ones. For instance when you write $dy=f'(x) dx$, that is really understood as an equation which holds for all infinitesimal $dx$. – Ian May 21 '15 at 23:29
• but we can compare functions at a certain number or limited domain ? – Mohamed Osama May 21 '15 at 23:30
• Even if $dy = f'(x)\,dx$, somehow considered numbers, it could be that $dx$ is negative, and then $dx/2$ is greater than $dx$. – GEdgar May 21 '15 at 23:35

To fully answer this question would require a many-volume narrative of the history of mathematics/physics since Newton and Leibniz! :)

But/and I would say that the question is eminently reasonable, rarely addressed directly in textbooks, and, indeed, subtle to answer "correctly".

As a ridiculously short sketch of what humans know about this, to the best of my own knowledge (and I am interested in such things for some years now):

Newton and Leibniz did argue/think genuinely in terms of "infinitesimals", and, yes, in that context, $dx/2$ is half as large as $dx$. (Yes, $dx$ is itself problemmatical in modern terms... though not at all impossibly so, in various ways, as "differential form", or as Skolem-Robinson-Nelson "infinitesimal").

Yes, tangential to foundational issues, differential equations can be solved by treating the various $d(whatever)$ as things existing in their own rights, without explaining what they are. That is, a heuristic succeeds in producing outcomes that are checkable.

The last 150 years of didactic tradition has been in a different direction, for somewhat artifactual reasons. That is, the popular style of calculus makes an exaggerated show of disparaging "infinitesimals" (despite Skolem-Robinson-Nelson's complete legitimization of them!), and of disparaging the symbol-manipulations that ... jeez! resolved zillions of questions over at least two centuries!

In short, the question is profoundly reasonable... but/and the accumulation of some centuries' artifacts about accepted mathematics does, indeed, seriously confuse anyone's understanding of ... for example... eminently reasonably heuristics in physics texts...

The operational answer is: try to think not in terms of "rules", but that the mathematics is mostly, and, certainly, initially, exactly a narrative, a description, of things. Then we hope that our subsequent manipulations of this description give us further information.

That is, no, we cannot deduce by pure logic what the minimum legal parking distance away from a fire hydrant might be. But we can easily understand that there is some reasonable distance.

... sorry, yes, a seemingly vague answer, but, so far as I know, after some experience, maybe to the point.

• it is very vague specially for a non math major :) but the answer i get is you can't treat them as numbers hence you can't compare is it right or read your answer again? – Mohamed Osama May 22 '15 at 0:09
• @paulgarrett There's a reason that mathematicians since Cauchy have spent time emphasizing that $dx$ and $dy$ are not infinitesimals -- because the infinitesimals that Newton and Liebniz and even Euler used were not consistent mathematical objects. It was only with Robinson being very careful was it discovered that in fact one could make infinitesimals rigorous, but that's a relatively new part of mathematical history and as most students won't ever need to learn nonstandard analysis, it's still worth emphasizing that in standard analysis $dx$ is NOT an infinitesimal. – user137731 May 22 '15 at 0:13
• Dear Mohamed, I think the point is two-fold: your curiosity is completely reasonable, but/and the ultimate issues are subtle. That is, we can drive a car without understanding the thermodynamics of an internal combustion engine... we can use the internet without knowing how it works, or even how a transistor works. Yes, it is honorable to be curious about those things... but it would be foolish to refuse to use the internet until one knew how transistors, ... worked. The point is that the question is substantial, fully "weighted"... and in fact cannot be trivially answered. :) – paul garrett May 22 '15 at 0:15
• @paulgarrett that's way i said dear that i just use them in physics and engineering without understand them :) majoring math is kinda far from me but the question just poped but it seems it is far beyond my level thanks alot for your answer. – Mohamed Osama May 22 '15 at 0:17
• @user57404 If $a+dx = a$ for all $a\in \Bbb R$, then $dx = 0 + dx = 0$. So while I've never studied nonstandard analysis, I doubt that's one of the properties of infinitesimals. I also think you mean $adx \ne 0$ for $a\ne 0$ and $|dx| \lt \Bbb R^+$, not $dx \lt \Bbb R$. Again, those are just my guesses, but if they're not the case then this system loses most of the useful properties of the reals -- to the point where I have a hard time believing they form a consistent arithmetic. – user137731 May 22 '15 at 16:45

In indefinite integral it is just a symbol. But it exist because indefinite and definite integral by Riman has very cool relationship as Newton-Leibniz theorem said.

In definite integral by Riman's definition it's small subarea in function domain that has infitine small diametr and take part in infinite summation. So in fact split your domain in which you're interesting as you wish for such dx-s... BUT: that parts shouldn't overlap, and max(length(dx)) in limit should go to measure_of_length(point)=0....

Theorems which use integration are based only on such definition. So split as you want.

Do define definitte classical integral you need to know classic definition of the limit. And define: function, function domain, how to measure area in function domain, what is a length, what is an operation of summation

Beside Riman's integral defintion, it is also exist Lebega integral which is used in theory of probability. As Software Engineer I know only that two integrals.

• Riemann integration does not sum infinitesimal lengths. It's defined as the limit of the lower and upper Riemann sums as the norm of the partition goes to $0$. – user137731 May 22 '15 at 0:06
• 1. That you said about lower|upper Riemann summ I know as Darbu lower|upper summ. And It is not a definition. – bruziuz May 22 '15 at 0:16
• 2. By russian mathematician as Feihtengoltz (1970) it is defined such as I mentioned. – bruziuz May 22 '15 at 0:17
• But I'm a a software engineer, not a professional mathematician, so I wouldn't argue) – bruziuz May 22 '15 at 0:25
• I guess you're right that Darboux was the one who use upper and lower sums -- Riemann just chose any point in each interval of the partition in his sums. Even so, Riemann integrals are defined as the limit of Riemann sums as the norm of the partition goes to $0$ -- NOT via infinitesimals. Mathematicians (with a few exceptions) haven't really used infinitesimals since Bolzano gave the world the $\epsilon$-$\delta$ definition of limits in 1817 -- nearly a decade before Bernhard Riemann was even born. – user137731 May 22 '15 at 0:53