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My friend told me that every equation possible with modern mathematical notation boils down to only $+, -,\times,\div$

What that means is that you can take any function and if you dive deep enough into it it will only have those operators

For example, if we have factorial, n!, it can be expressed as $n\times{(n-1)}\times{(n-2)},\,etc.$ Is this true with every equation in mathematics? Will everything when boiled down enough just be composed of those four operations?

EDIT: Things that can be expressed with these but an infinite number of them do not count (Ex, infinite sums) Because we can write these as: $n+n+n+n...$

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    Integrals? Derivatives? Exponentials (you can write this one with these operators, but need infinitely many of them)? Logarithm? – Clement C. May 21 '15 at 21:25
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    @ClementC.: Those both in turn depend on limits. You can define the rest of it in terms of the four basic arithmetic operations, but I don't see how you define the limit in terms of those. – Brian Tung May 21 '15 at 21:26
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    If you can perform computations in the symmetric group with only those operations, I will be mighty impressed :) More generally, most finite/combinatorial mathematics. – pjs36 May 21 '15 at 21:32
  • What about $\sqrt x$? How do you define it using those operations? – Kitegi May 21 '15 at 21:41
  • There is a formula to solve it, the fact that it takes multiple iterations doesn't make a difference. –  May 21 '15 at 21:43

2 Answers2

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No. First of all, functions don't even need to associate numbers at all, but we'll ignore that for now.

"Functions" are a lot more general than people tend to think. Functions need to take any member from its domain and associate it to exactly one element of its codomain - that's it. They do not even need to have a sort of "rule" that allows you to tell me what a given element of the domain maps to, I can just give you a set of ordered pairs. $\{(1,1),(2,7),(9,1)\}$ is a function.

The relation whose domain is the integer interval $[10,99]$ defined by switching the first and second digit is a function.

The relation whose domain is the reals that maps odd numbers to $1$, even numbers to $2$, and non-integers to $42$ is a function.

The relation whose domain is the naturals defined by having a six-sided die tossed on the n'th throw, and mapping n to the result of the die toss is a function.

The relation whose domain is the natural numbers which maps the n'th day of my life to my favorite number on that particular day is a function.

Surely it would be absurd to think that all of these are expressible with those operations.

Jonathan Hebert
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2

No, put simply. I cite piecewise equations, recursive functions, infinite sums, and trigonometric functions, amongst others. The list goes on and on and on.

Of course, you can intuitively define some functions with said "base" calculations; you (or your friend) cited $n!$ as an example, as $n(n-1)(n-2)\ldots(1)$ (which is better defined using the $\Gamma$ function or at least using $\prod$). I challenge your friend to try and express the equality $\pi=\pi$ using only the "base" calculations and numbers.

I guess the most obvious counter-example might be any $x^n$ for $n\notin\mathbb{R}$; how might you define $\sqrt2$ or $10^{4.326}$ with "base" calculations? It really is an absurd invalid claim, when you think about it. Though it does make intuitive sense, seeing how rigorous mathematicians can be, when all it requires is some of the basics. (Like trying to prove that $\sqrt2$ is irrational using some obscure calculus field, or something. Hehe.)

Specific examples

$$\int_{-3}^{3}x^2+x^3-x^{\lfloor x\rceil}\text{d}x$$ $$f(x)=\sin(x)$$ $$\lceil x\rceil$$ $$\lim_{n\to\infty}\left(1+\frac1n\right)^n$$

Edit

This is a bit of a necropost, but is done for completness's sake. I read in the comments that you allowed for iterations of some sense. This is a different matter entirely.

There exist Taylor series for a good deal of functions ($\sin,\cos,\exp,\sqrt{n},\pi,e$, i.e.), so this may seem like a more plausible claim. Moreover, if by iteration, you also mean recursion, even more is possible.

However, I am not sure of the extent of sole recursion etc.; I would think that an arbitrary integral still would not have a general method using recursion and base methods.

Conor O'Brien
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