I once heard someone say that analysis is $99 \%$ algebra. He was, of course, referring to the amount of algebraic manipulations in the exercises from any calculus course.

I know that in topology, combinatorics or, oddly, (abstract) algebra, some interesting things can be said without writing down a single equation. However I found that I don't know any such examples in analysis.

Hence I ask for your help. Anything resembling analysis is welcome, but I'd prefer if the example was comprehensible to someone who has taken only a year or two of real analysis.

  • $\begingroup$ I wiki-hammered and retagged since you are asking for a list of example. I am not entirely sure about using the (algebra-precalculus) tag for this question, though I can see why it may be appropriate, so I am leaving that on the question. $\endgroup$ Apr 25, 2013 at 12:08
  • 4
    $\begingroup$ Anything you can say with an equation you can say without one by just expressing as an English sentence, so your question needs some refining. Also, perhaps an example of the kind of statement you have found in topology/combinatorics/algebra might help us get the idea. $\endgroup$ Apr 25, 2013 at 12:32
  • $\begingroup$ "interesting things can be said without writing down a single equation" Well it is certainly easy to state results without equations... but do you really mean that interesting things can be proven without using equations? $\endgroup$
    – rschwieb
    Apr 25, 2013 at 13:37
  • $\begingroup$ @GerryMyerson, I hoped it would be clear that it's not notation I'm talking about. There are such things as, say, combinatorial proofs. I want proofs by showing a natural equivalence, by constructing a counterexample or simply by only using abstractions of algebraic manipulation. An example could be the proof of bounds of chromatic number of unit distance graph - a proof by picture. $\endgroup$ Apr 25, 2013 at 18:45
  • 1
    $\begingroup$ I don't know what you mean when you say the Weierstrass function is a statement, or that it has reasoning. In what sense can a function be said to be a statement? In what sense can a function be said to have reasoning? $\endgroup$ Apr 27, 2013 at 4:46

2 Answers 2


Calculus is all about the interplay between differentiation and integration. Looking at the definition of the derivative, you'll see that it only really uses two properties of the real line: its linear structure (so that we can form the difference $f(x + h) - f(x)$ and its topological structure (so that we can take a limit). So in this sense, calculus is exactly half algebra.

Of course, that doesn't explain why it feels like mostly algebra when you're actually taking a calculus class. When you get right down to it what you use on a day-to-day basis in a calculus class is the relationship between differentiation and integration, and this relationship can be abstracted as follows. Begin with the ring $A$ of all smooth functions on $C^\infty(\mathbb{R})$. We have two linear maps $d$ (differentiation) and $I$ (integration) on $A$ which satisfy the following properties:

  • $d(1) = 0$
  • $d(f \cdot g) = df \cdot g + f \cdot dg$
  • $d(f \circ g) = dg \cdot df \circ g$
  • $d \circ I = id$
  • $I \circ d = id + C$

Taking these five properties as axioms, you can differentiate and integrate basically any function that you encounter in your first year of calculus. For instance, the second and third give you the quotient rule and the second and fifth give you integration by parts. In fact, with a bit of extra effort one can embed these axioms into an algebraic package which completely characterizes $C^\infty(\mathbb{R})$ and hence most of calculus.

A final remark. Notice that in your first year of calculus you don't actually encounter all that many functions. Here is a nearly exhaustive list:

  • Polynomials
  • Exponentials
  • Products, quotients, compositions, and inverses of the above

You may object that I forgot trigonometric functions, but thanks to de Moivre's equation $e^{ix} = cos(x) + i sin(x)$ these are really just exponentials. Polynomials are certainly algebraic objects, and exponential functions can be characterized as the only group homomorphisms from the additive group of the real line to the multiplicative group of the ray $(0,\infty)$. So it is not unreasonable to expect to do a lot of algebra given that you're only working with algebraic objects.

  • $\begingroup$ I guess none of this answers your actual question since I didn't give an example of a statement in analysis which doesn't involve writing down any equations. There are of course many such statements - such as "every differentiable function is continuous" - but my answer explains why it is unlikely for the proofs to be free of equations (or at least inequalities). Algebra really is built into the structure of calculus. $\endgroup$ Apr 25, 2013 at 13:39
  • $\begingroup$ $@$Paul: "In fact, with a bit of extra effort one can embed these axioms into an algebraic package which completely characterizes $C^{\infty}(\mathbb{R})$ and hence most of calculus." That sounds interesting: could you elaborate? $\endgroup$ Apr 27, 2013 at 5:06

I still don't entirely understand what OP wants, but let's try the Hahn Banach Separation Theorem:

Let $K$ be the real or complex numbers. Let $V$ be a topological vector space over $K$. If $A$ and $B$ are convex, non-empty disjoint subsets of $V$, then

  1. If $A$ is open, then there is a continuous linear map $\lambda:V\to K$ and a real number $t$ such that ${\rm Re}(\lambda(a))\lt t\le{\rm Re}(\lambda(b))$ for all $a$ in $A$ and all $b$ in $B$.

  2. If $V$ is locally convex, $A$ is compact, and $B$ is closed, then there exists a continuous linear map $\lambda:V\to K$ and real numbers $s$ and $t$ such that ${\rm Re}(\lambda(a))\lt t\lt s\lt{\rm Re}(\lambda(b))$ for all $a$ in $A$ and all $b$ in $B$.

As given on Wikipedia. I think I can safely assert that it's not trivial.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.